First Edition, 2009
ISBN 978 93 80168 42 5
© All rights reserved.
Published by: Global Media 1819, Bhagirath Palace, Chandni Chowk, Delhi-110 006 Email:
[email protected]
Table of Contents 1. Temperature Measurement 2. Derived Theory 3. Macroscopic Properties 4. Change of State 5. Heat Capacity 6. Heat and Temperature 7. The Reversibility 8. Physical Significance 9. Changing State
Temperature Measurement
1
1 Temperature Measurement Scale of Temperature In order to indicate the temperature of a body as a number, a ‘scale of temperature’ is adopted and the temperature of the body is estimated against this scale. The instrument on which such a scale is used, is called a ‘thermometer.’ The temperature scale is completed in the following steps: (i) First of all, some measurable physical property of a substance is chosen which changes with temperature : Such properties are (a) the length of a liquid column in a glass capillary, (b) the volume of a fixed mass of gas at constant pressure, (c) the pressure of a fixed mass of gas at constant volume, (d) the electrical resistance of a metallic wire, (e) the e.m.f. of a thermocouple, (f) the colour of a lamp filament, etc. The chosen substance and its property are called respectively the ‘thermometric substance’ and the ‘thermometric property’.
Thermal Physics
2
Suppose we have chosen a thermometric substance whose thermometric property is represented by X. Suppose the thermometer using this substance is in thermal equilibrium with a system. We arbitrarily define that the magnitude of this property varies linearly with the temperature T of the system. That is X = aT,
...(i)
where a is some constant. This means that equal changes in temperature correspond to equal changes in X. (ii) Now, an easily reproducible state of some standard system is chosen. The temperature corresponding to this state is called a ‘fixed-point’: This point is chosen to be the ‘triple point’ of water, i.e. the temperature and pressure at which ice, liquid-water and water-vapour coexist in equilibrium. (The water vapour-pressure at the triple point is 4.58 mm of mercury). (iii) The chosen fixed point is given an arbitrary numerical value for the temperature. The temperature at the triplepoint of water is arbitrarily given a value of 273.16 K (kelvin). The kelvin is a unit temperature-interval. (iv) Finally, the kelvins are numbered by using the chosen thermometric property X. The temperature is kept at the triple point (by dipping the thermometer in a triple-point cell), and the magnitude of the property X at this point (273.16 K) is measured. Let it be Xtr. Then, from equation (i), we have Xtr = α (273.16).
...(ii)
Then the thermometer is placed in contact of the system whose temperature T is to be measured, and the magnitude of X at this unknown temperature is measured. Let it be XT. Then, again from equation (i), we have XT = aT.
...(iii)
Temperature Measurement
3
Dividing equation (iii) by equation (ii), we have XT T = Xtr 273.16
∴
⎛ X ⎞ T = ⎜ 273.16 T ⎟ K. Xtr ⎠ ⎝
Thus T can be calculated from the two measurements Xtr and XT. This is the temperature defined on the scale using the thermometric property X. To emphasise this, we write Tx for T in the above equation : ⎛ X Tx = ⎜ 273.16 T X tr ⎝
⎞ ⎟K ⎠
...(iv)
Let us now define the temperature T on the scales given by actual thermometers using different thermometric properties:
Temperature on Liquid-in-glass Scale In a liquid-in-glass thermometer, the thermometric property X is the length of a liquid column, l, (in a capillary tube) which increases with temperature. If ltr be the length of the column at the triple-point, and lT the length at an unknown temperature, then the unknown temperature measured on the length scale is defined by [from equation (iv)] ⎛ l T1 = ⎜ 273.16 T l tr ⎝
⎞ ⎟ K. ⎠
Temperature on Constant-volume Gas Scale : At constant volume, the pressure p of a gas varies with temperature; and this is the thermometric property used in a constant-volume gas thermometer. If ptr be the pressure at the triple-point, and pT at an unknown temperature, then the unknown temperature measured on the gas pressure scale is defined by
Thermal Physics
4 ⎛ p ⎞ Tp = ⎜ 273.16 T ⎟ K . ptr ⎠ ⎝
Temperature on Constant-pressure Gas Scale : At constant pressure, the volume V of a gas varies with temperature and is the thermometric property used in a constant-pressure gas thermometer. As above, the temperature on the gas volume scale is defined by ⎛ V Tv = ⎜ 273.16 T V tr ⎝
⎞ ⎟ K. ⎠
Temperature on Resistance Scale : In a platinum resistance thermometer, the thermometric property is the electrical resistance R of a platinum wire which varies with temperature. If Rtr be the resistance at the triple point and RT at an unknown temperature, then the unknown temperature measured on the resistance scale is defined by ⎛ R TR = ⎜ 273.16 T R ⎝ tr
⎞ ⎟ K. ⎠
Similar expressions hold for other thermometric substances and thermometric properties. Need of a Standard Scale of Temperature : The different properties of matter utilised in different thermometers give different scales of temperature. Now, the difficulty is that if we measure the temperature of a body with different kinds of thermometers, we find differences among their readings. (All thermometers agree only at the fixed point 273.16 K). Even when different thermometers of the same kind are used, such as constant-volume gas thermometers filled with different gases, we obtain different temperature readings for the body. Hence, to obtain a completely definite scale of temperature, we select one particular kind of thermometer as the ‘standard’ and calibrate all the other thermometers against it.
Temperature Measurement
5
The Standard Scale (Ideal Gas Temperature Scale) : Out of the various types of thermometers, the smallest difference in readings is found among the constant-volume gas thermometers filled with different, so-called permanent gases. The difference is only of the order of 0-2 - 0.3%. It is noticeable that if the amount of the gas filled in the thermometers is reduced i.e. the pressure of the gas is lowered; the difference in readings among different thermometers further decreases. This can be seen from the following experiment: Suppose an amount of a gas, say O2, is filled in a constantvolume gas thermometer such that the pressure of the gas ptr is 1000 mm of Hg when the bulb of the thermometer is in a triple-point cell. Now the bulb is placed in condensing steam and the pressure of the gas (pT) at steam temperature is measured. The temperature of the steam is calculated by the formula ⎛ P Tp = ⎜ 273.16 T Ptr ⎝
⎞ ⎟ K, ⎠
where pT is the gas pressure. Now the quantity of the gas in the thermometer is reduced in steps (so that ptr becomes smaller and smaller) and each time the temperature of steam Tp is determined.
Thermal Physics
6
A graph is then plotted between Tp and ptr, which is practically a straight line. The line is produced back to intersect the axis where ptr = 0. The whole experiment is then repeated by filling the thermometer with other gases like air, N2, He, H2 turn by turn. These graphs show that the different gas thermometers do give different temperatures of the steam, but the difference decreases as the gas pressure is lowered. As pressure ptr→ 0, ‘all’ gas thermometers give the same value of 373.15 K for the steam temperature. The temperature measured by a constant-volume gas thermometer, and extrapolated to a value corresponding to ptr→ 0, is known as ‘ideal gas temperature’ and the corresponding scale is known as ‘ideal or perfect gas temperature scale’. It is defined by the relation lim PT ⎞ ⎛ T = ⎜ 273.16 ⎟ K. p tr →0 Ptr ⎠ ⎝
The standard thermometer is therefore chosen to be a constant-volume gas thermometer, using a temperature scale defined by the above relation. The International Bureau of Weights and Measures has accepted a ‘constant-volume hydrogen thermometer’ as the standard thermometer when the hydrogen has been filled at a pressure of 1 meter of mercury at 0°C. For measurement of low temperatures, a constant-volume helium thermometer has been recommended, which can be used upto 1 K. Thus, 1 K is the lowest ideal gas temperature that can be measured by a gas thermometer. (The temperatures below 1 K remain as yet undefined). The choice for H2 among all the gases is due to the fact that it is (except helium) nearest to an ideal (perfect) gas, and can be obtained in the pure state anywhere in the world. Near 1 K only helium can be used because all gases liquify. Absolute Scale of Temperature : An absolute scale of temperature is one which is independent of the properties of
Temperature Measurement
7
any particular substance. No scale given by any thermometer is absolute as it depends upon the properties of the thermometric substance. Even the scale defined above is not absolute, because although it is independent of the properties of any one particular gas, it still depends upon the properties of gases in general, that is, on the properties of an ideal gas. Lord Kelvin, from the consideration of the efficiency of an ideal heat engine, defined a scale which is absolutely independent of the properties of the working substance in the engine. This scale is called the ‘Kelvin’s absolute thermodynamical scale of temperature’. It is defined such that if an ideal engine absorbs a quantity of heat Q1 at temperature θ1 , and rejects a quantity of heat Q2 at θ2 , then
θ1 Q1 = . θ 2 Q2 It is, however, remarkable that in the temperature-range in which a gas thermometer can be used, the ideal gas scale and the Kelvin’s absolute scale are exactly identical in all respects. Hence, the realization of an ideal gas scale is in fact the realization of Kelvin’s absolute scale. This is why the temperatures on the ideal gas scale are expressed in K (kelvin). Relation of Celsius and Fahrenheit Scales with the Kelvin’s Scale : The Kelvin’s absolute scale has an absolute zero of 0 K and the Kelvin temperature of triple-point of water (Ttr) is 273.16 K. By experiment, the Kelvin temperature of the ice point (Tice) comes to be 273.15 K and that of the steam point (Tsteam) 373.15 K. The temperature scales in common use are, however, the Celsius scale and the Fahrenheit scale. Both of these scales have been defined in terms of the Kelvin’s scale. The Celsius temperature scale employs a degree of the same magnitude as that of the Kelvin’s absolute scale; but its
Thermal Physics
8
zero has been set at 273.15 K. Thus, if the temperature of a body on the Celsius scale is t (°C) and that on the Kelvin’s scale is T (K), then t = T- 273.15.
Thus, the triple point of water on the Celsius scale is ttr = Ttr- 273.15 = 273.16 - 273.15 = 0.01°C. Similarly, the temperatures of the ice-point and steampoint are tice = Tice - 273.15 = 273.15 - 273.15 = 0.00°C and
tsteam = Tsteam - 273.15 = 373.15 - 273.15 = 100.00°C.
A temperature on the Fahrenheit scale (tF) is related to the corresponding temperature on the Celsius scale (tC) as tF = 32° +
9 tc . 5
Clearly, the ice-point (0.00 °C), the triple-point of water (0.01°C) and the steam-point (100.00 °C) are 32.0 °F, 32.018 °F and 212.0 °F respectively on the Fahrenheit scale.
Derived Theory
9
2 Derived Theory Dulong and Petit Law : In 1819, Dulong and Petit established empirically that “the atomic specific heat of all solids is nearly 6 cal/(gm-atom-K) and is independent of temperature.” This is Dulong and Petit Law. Derivation from Classical (Kinetic) Theory : The atoms in a metallic solid are arranged in a crystalline array, held in position by strong electromagnetic forces exerted on one another. Thus they are not free to wander about (as they are in a gas). They can simply vibrate about their fixed positions under the electromagnetic interaction forces. Thus they have kinetic energy of vibration and potential energy of interaction. We know from the law of equi-partition of energy that the average kinetic energy per atom in each degree of freedom is ½kT. Now, in simple oscillatory motion, the average kinetic energy is equal to the average potential energy. Therefore, the average potential energy per atom in each degree of freedom will also be ½ kT. Hence the total energy per atom in each degree of freedom will be ½ kT+ ½kT = kT†. As the atom has
Thermal Physics
10
3 degrees of freedom in oscillatory motion, its total energy is 3 kT. Now, 1 gm-atom of a solid has N atoms in it, where N is Avogadro’s number. Therefore, the total energy of 1 gm-atom of the solid is given by U = N × 3kT. But N × k = R, the universal gas constant. ∴
U = 3RT.
The atomic specific heat Cv of a solid at constant volume is the energy required to raise the temperature of 1 gm-atom of the solid by 1°C (or 1K). Thus Cv =
dU . dT
Putting the value of U from above, we get CV =
d (3RT) = 3R. dT
We know that R
2 cal/(gm-atom-K).
∴
CV
6 cal/(gm-atom-K).
This is Dulong and Petit Law.
Variation of Specific Heat with Temperature Applicability of the Law: According to the Dulong and Petit law, the specific heat of all solids must be about 6 cal/ (gm-atom-K) and it should not vary with temperature. Actually, this is not true. The observed relationship between the atomic specific heat of solids and temperature. The non-metallic solids such as boron, carbon and silicon have specific heats appreciably different from 6 at ordinary temperatures. At very high temperatures, however, the specific heats of these solids approach the value given by Dulong and Petit law.
Derived Theory
11
The specific heats of metallic solids, however, lie near 6 cal (gm-atom-K) at ordinary temperatures and change little when the temperature is raised. Thus, metallic solids obey the Dulong and Petit law at ordinary as well as at high temperatures. But, the law completely fails at very low temperatures. The atomic heat at low temperatures is found to decrease with fall in temperature. Below a certain temperature (which is different for different solids), the atomic heat decreases rapidly with decreasing temperature. It tends to zero at absolute zero for all solids. The non-agreement of the result of the classical theory of specific heats with experiment led to the belief that classical mechanics breaks down in this case. Einstein’s Quantum Theory of Specific Heat : Einstein, in 1907, applied Planck’s quantum theory to the specific heats of solids. According to this theory, an atomic oscillator of frequency ν can only have energy values hν , 2 hν , 3hν ... where h is Planck’s constant, and not in between. On this basis, the average energy of an oscillator per degree of freedom is hν e hν / kT − 1
( not kT ) .
Einstein assumed that all atoms of a given solid vibrate independently of each other with the same frequency in each of the three degrees of freedom. Therefore, the total energy of a gm-atom of the solid consisting of N atoms is U = 3N
hν e
hν / kT
−1
.
Therefore, the specific heat is given by
CV
⎛ hν ⎞ e hν / kT ⎜ 2 ⎟ dU ⎝ kT ⎠ = 3 N hν = 2 hν / kT dT ( e − 1)
Thermal Physics
12 2
e hν / kT ⎛ hν ⎞ . = 3N k ⎜ ⎟ 2 ⎝ kT ⎠ ( e hν / kT − 1)
But Nk = R (the universal gas constant). 2
e hν / kT ⎛ hν ⎞ . CV = 3 R ⎜ ⎟ 2 ⎝ kT ⎠ ( e hν / kT − 1)
∴
Let us put
hν = ΘE , where ΘE is known as Einstein k
characteristic temperature. Then, we have ⎛Θ ⎞ CV = 3 R ⎜ E ⎟ ⎝ T ⎠
2
(e
e ΘE / T ΘE / T
− 1)
2
.
…(i)
This is Einsteins’s specific heat formula. (i) At high temperatures (T >> ΘE ), we have ΘE << 1 so that T Θ e ΘE /T ≈ 1 + E . The eq. (i) then takes the form T ΘE 2 1+ ⎛ ΘE ⎞ T ≈ 3R. CV ≈ 3R ⎜ ⎟ ⎝ T ⎠ ⎛ ΘE ⎞ 2 ⎜⎝ ⎟⎠ T
…(ii)
This is in agreement with Dulong and Petit law which agrees with experiment at high temperatures. ΘE >> 1, so that T >> 1. Now, the Einstein’s formula (i) gives
(ii) At low temperatures (T << ΘE ), we have e ΘE /T
2
⎛Θ ⎞ CV ≈ 3 R ⎜ E ⎟ e −ΘE / T . ⎝ T ⎠
...(iii)
This shows that as T → 0, the specific heat CV also approaches zero, a conclusion in general agreement with experimental results.
Derived Theory
13
Limitations : Although Einstein’s theory gives qualitative account of the experimental specific heat curves, there is no quantitative agreement throughout. The theory fails at very low temperatures where Cv is found to be nearly proportional to T3 rather than as given by eq. (iii). This disagreement is due to the neglect of the mutual forces exerted by the atoms upon one another. Taking this into consideration, Debye in 1912 developed a specific heat formula which gives excellent agreement with experiment over the whole temperature-range.
Debye’s Theory of Specific Heat of Solids Debye, presented a theory of the specific heat of solids which was a modification of the Einstein’s quantum theory. He considered solid as a continuous elastic body; in which the vibrations of the atoms generate stationary waves. The internal energy of the solid resides in these elastic stationary waves. Let us consider a unit volume of a continuous solid in which elastic waves are set up. It can be shown that the number of possible waves of any kind with frequencies between ν and ν + d ν is 4π 2 ν dν ν3 where ν is the wave-speed. There are two kinds of elastic waves that can occur in a solid; longitudinal waves with speed vl and transverse waves with speed vt moreover, there are two perpendicular directions of polarisation for a transverse wave. Hence the total number of possible elastic stationary waves with frequencies between ν and ν + dν is ⎛ 1 2 ⎞ 4π ⎜ 3 + 3 ⎟ ν 2 dν. vt ⎠ ⎝ vl
If V is the volume of a gm-atom of a particular solid, the number of waves with frequencies between ν and ν + dν in it will be
Thermal Physics
14 ⎛ 1 2 ⎞ 4πV ⎜ 3 + 3 ⎟ ν 2 dν. vt ⎠ ⎝ vl
Hence the total number of waves comprising all possible frequencies is ⎛ 1 2 ⎞ Vm 4πV ⎜ 3 + 3 ⎟ ∫ ν 2 dν. vt ⎠ 0 ⎝ vl
The upper limit of integration cannot be infinite, as this would make the number of waves and hence the internal energy of the solid infinite. Debye assumed that in a solid no frequency can go beyond a definite upper limit vm, which is characteristic of the substance. This limit is so chosen that the total number of waves is equal to the total number of degrees of freedom of the solid i.e. equal to 3N, where N is Avogadro’s number. Hence vm can be determined by putting the expression (i) equal to 3N. That is ⎛ 1 2 ⎞ V 2 3N = 4πV ⎜ v 3 + v 3 ⎟ ∫0 ν dν t ⎠ ⎝ l m
=
4π ⎛ 1 2 ⎞ V ⎜ 3 + 3 ⎟ ν 3m , 3 ⎝ vl vt ⎠
and hence ν m3 =
9N ⎛ 1 2 ⎞ 4πV ⎜ 3 + 3 ⎟ vt ⎠ ⎝ vl
.
…(i)
Thus the value of vm for a given solid can be calculated from a knowledge of its elastic constants. Let us now calculate the energy of solid. According to Planck’s theory, the average energy E per wave is given by hν E = e hν / kT − 1 ,
Derived Theory
15
where k is Boltzmann’s constant and T is Kelvin temperature. Therefore, the total internal energy of a gm-atom of the solid is U = E × total number of waves as given by (i). ⎛ 1 2 ⎞ Vm h ν3 d ν π + 4 V U = ⎜⎝ v 3 v 3 ⎟⎠ ∫0 e hν/ kT − 1 l t
Thus
⎛ 1 2 ⎞ Substituting the value of 4πV ⎜ 3 + 3 ⎟ from eq. (ii), we vt ⎠ ⎝ vl have 9N Vm hν 3 dν U = v 3 ∫0 e hν / kT − 1 . m
For convenience let us put dv =
kT dx. Then h
hν kT = x, so that ν = x and kT h
3
where xm =
hν m kT
3 xm x dx 9N ⎛ kT ⎞ ; U = 3 ⎜ ⎟ kT ∫0 x vm ⎝ h ⎠ e −1
3
⎛ k ⎞ 4 xm x 3 dx . ⎟ T ∫0 x e −1 ⎝ hν m ⎠
= 9 Nk ⎜
Let us now define a characteristic Debye temperature Θ D , as ΘD =
h νm hν Θ , so that xm = m = D . k kT T
In view of eq. (ii), ΘD can be independently calculated for a given solid. Now, the energy expression becomes U = 9R
T4 ΘD3
∫
( Θ )D / T
0
x 3 dx ex − 1
[Q R = Nk ]
Thermal Physics
16
Therefore, the atomic specific heat of the solid, CV, is given by ⎛ ∂U ⎞ CV = ⎜ ⎟ ⎝ ∂T ⎠v ⎡ ⎛ T ⎞ 3 (Θ)D / T x 3 dx ⎛ Θ ⎞ ⎤ 1 −⎜ D⎟ ⎥ x ∫ ⎟ 0 e − 1 ⎝ T ⎠ eΘ D / T − 1 ⎦⎥ ⎣⎢ ⎝ Θ D ⎠
= 9R ⎢ 4 ⎜
This is Debye’s specific heat formula. ΘD ⎛ hν ⎞ At high temperatures both x ⎜ = are very small ⎟ and T ⎝ kT ⎠
so that e x ≈ 1 + x , and e Θ gives
D
/T
≈ 1 + Θ D / T . Then Debye’s formula (iii)
⎡ ⎛ ⎞ CV ≈ 9 R ⎢ 4 ⎜ T ⎟ ⎢⎣ ⎝ Θ D ⎠
3
∫
( Θ )D / T
0
⎛Θ x 2 dx − ⎜ D ⎝ T
⎞ 1 ⎤ ⎥ ⎟ ⎠ Θ D / T ⎥⎦
⎡ ⎛ T ⎞ 3 ( Θ D / T )3 ⎤ ≈ 9R ⎢ 4 ⎜ − 1⎥ ⎟ 3 ⎢⎣ ⎝ Θ D ⎠ ⎥⎦
≈ 3R,
which is the Dulong-Petit value in agreement with experiment at high temperatures. At very low temperatures ΘD / T → ∞ , so that the second term in the brackets of eq. (iii) becomes negligible, and we can replace the upper limit of integration in the first term by ∞. Thus, we have CV But the integral
∫
∞
0
⎡ ⎛ T ⎞ 3 ∞ x 3 dx ⎤ ⎥. ≈ 9R ⎢ 4 ⎜ ⎟ ∫ x ⎢⎣ ⎝ Θ D ⎠ 0 e − 1 ⎥⎦
x 3 dx π 4 = . e x − 1 15
Derived Theory
17 ⎡ ⎛ T ⎞3 π4 ⎤ ⎥ ⎟ ⎢⎣ ⎝ Θ D ⎠ 15 ⎥⎦
CV ≈ 9 R ⎢ 4 ⎜
∴
≈
12 π4 R 3 T . 5 ΘD 3
Since ΘD is a constant for the solid, we obtain CV ∝ T3. Thus the specific heat at extremely low temperatures varies as the cube of the Kelvin temperature, a conclusion in excellent agreement with experiment. This is known as “Debye’s T3 law.” At intermediate temperatures the Debye’s formula has been evaluated numerically. The results have been found in excellent agreement with experiment for many substances.
PROBLEMS 1. Calculate Einstein temperature for a case for which = 2.49 × 1012 sec–1. Given : k = 1. 38 × 10–23 joule/K and ν h = 6.63 × 10–34 joule-sec. Solution: The Einstein temperature is given by ΘE
= =
hν k
( 6.63 × 10
−34
joule - sec )( 2.49 × 1012 sec −1 )
1.38 × 10 −23 joule/K
= 119.6 K. 2. Calculate the Einstein’s frequency for a case for which Einstein temperature ΘE = 236K. Given : k = 1.38 × 10–23 joule/ K and h = 6.63 × 10–34 joule-sec. Ans: 4.9 × 1012 sec–1.
Thermal Physics
18
3. Calculate Einstein’s frequency for copper. Given: ΘE = 230 K, h = 6.63 × 10–34 Js and k = 1.38 × 10–23 J/ K. Show that if T >> 230 K, then the classical result CV = 3R will hold good. Solution: By the definition of Einstein temperature ΘE = hν / k , the Einstein’s frequency is
ν=
−23 ΘE × k 230K × ( 1.38 × 10 J / K ) = = 4.79 × 1012 s−1 . h 6.63 × 10 −34 Js
If T >> 230 K ( ΘE ) , then
ΘE << 1 and T
ΘE e ΘE /T = 1 + T .
Substituting this in Einstein’s specific heat formula, we get Θ CV = 3 R ⎛⎜ E ⎞⎟ ⎝ T ⎠
or
CV
2
(e
e ΘE / T ΘE / T
− 1)
2
ΘE 2 1+ ⎛ ΘE ⎞ T ≈ 3 R. ≈ 3R ⎜ ⎟ 2 ⎝ T ⎠ ⎛ ΘE ⎞ ⎜ ⎟ ⎝ T ⎠
Macroscopic Properties
19
3 Macroscopic Properties System and its Surroundings : Any portion of matter which is considered as separated from its surroundings is called a ‘system’. All those things which are outside that system and influence its behaviour are known as the ‘surroundings’ of the system. For example, let a gas be filled in a cylinder fitted with a piston and heated by a burner. Here the ‘gas is the system’, while the ‘piston and burner are the surroundings’. The behaviour of the system can be described in terms of two types of properties ; the macroscopic and the microscopic. These are the properties which (i) describe the gross characteristics of the system, (ii) can be measured directly in the laboratory and (iii) can usually be directly experienced by our sense of perception. They are not concerned with the structure of the system. Thus, in the above example, the chemical composition (if any), pressure, volume, temperature, internal energy, entropy, etc. are the macroscopic properties of the gas (system).
20
Thermal Physics
Microscopic Properties : These are the properties which (i) describe the internal structure (atoms and molecules) of the system, (ii) cannot be measured directly in the laboratory, and (iii) cannot be directly experienced by our sense of perception. In the above example, the masses, velocities, energies, momenta, etc. of the molecules of the gas are the microscopic properties of the gas. Relation between Macroscopic and Microscopic Properties : The macroscopic and microscopic properties are simply the different ways of describing the same system. Hence, they must be related to each other. For example, the pressure of a gas is related to the average rate of change of momentum due to all the molecular collisions made on a unit area. Higher the rate of change of momentum, higher the pressure. Here, the pressure is a macroscopic property, whereas the rate of change of momentum due to molecular collisions is a microscopic property. Similarly, the temperature of a gas, macroscopic property, is related to the average kinetic energy of translation of its molecules, which is a microscopic property. Pressure, Volume and Temperature—Macroscopic or Microscopic : These properties have no meaning for an individual molecule, but only for a large collection of molecules. They can be directly experienced by our sense of perception and can be measured in the laboratory. Hence, they are intrinsically macroscopic concepts.
The Temperature Temperature : We can distinguish hot bodies from cold bodies by our sense of touch. For example, by touch, we can roughly arrange bodies in the order of their hotness, deciding that A is hotter than B, B is hotter than C, and so on. This is known as our ‘temperature’ sense and it is said that A is at a higher temperature than B, B is at a higher temperature than C, and so on. Thus, the temperature of a body is a measure of the hotness or coldness of the body.
Macroscopic Properties
21
Thermal Equilibrium A system is said to be in thermal equilibrium if, any two of the independent properties describing the state of the system remain constant as long as the surrounding conditions remain unchanged. For example, if a gas enclosed in a vessel has a pressure p and a volume V at the temperature of the surroundings, the values of p and V will remain constant as long as the temperature of the surroundings remains unchanged. The gas is said to be in thermal equilibrium with the surroundings. The values of pressure and volume are connected by a functional relationship f ( p , V ) = 0.
This is called the ‘equation of constraint’. For a gas obeying Boyle’s law, pV= constant, it would read pV- constant = 0. Let us consider two systems A and B isolated from the surroundings, but in thermal contact with each other. Their independent properties undergo a change until an equilibrium state of the combined system is reached when these properties assume constant values. The systems are now in thermal equilibrium with each other. Thus, thermal equilibrium is the state attained by two (or more) systems in thermal contact with one another and characterised by constant values of the properties of the systems. Zeroth Law of Thermodynamics : This fundamental law states that if two systems A and B are separately in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. Suppose there are three systems A, B and C. Systems A and B are isolated from each other but are in thermal contact with C. Experiments show that both A and B individually attain thermal equilibrium with C. If now A and B are put in thermal
Thermal Physics
22
contact of each other, no further change takes place. That is, A and B are found to be in thermal equilibrium with each other. A simple illustration of the law is provided when A and B are two gases enclosed in vessels and C is a mercury thermometer. The zeroth law says that if there is no change in the length of mercury thread when the thermometer C is placed in thermal contact of A nor when it is placed in thermal contact of B, then there will be no change if A and B are brought in thermal contact with each other.
Zeroth Law The zeroth law can be used to define temperature. It leads to that all systems in thermal equilibrium with one another have a common property having same value for all of them. This property is ‘temperature’. Thus, the temperature of a system is the property which determines whether or not the system is in thermal equilibrium with other systems. That is, two systems in thermal equilibrium with each other are at the same temperature. Conversely, if two systems have different temperatures, they cannot be in thermal equilibrium with each other.
PROBLEMS 1. Three systems (gases) 1,2, and 3 have coordinates (p1, V1); (p2 , V2) and (p3, V3). When 1 and 3 are in thermal equilibrium, the equation p1V1-nbp1 -p3V3 = 0 holds. When 2 and 3 are in thermal equilibrium, the equation holds, where n, b, b’ are constants.
p2V2 -p3V3 +
nb ' p3V3 =0 V2
(a) What equation expresses thermal equilibrium between 1 and 2 ? (b) What are three functions which are equal to one another at thermal equilibrium and each is equal to the empiric temperature T.
Macroscopic Properties
23
Solution : (a) The systems 1 and 3 are in thermal equilibrium: P1V1 - nbp1 − p1V3 = 0 P1 (V1 − nb) = p3V3 .
or
...(i)
The systems 2 and 3 are in thermal equilibrium : P2V2 − p3V3 +
or
nb ' p3 v3 =0 V2
⎛ nb ' ⎞ P2V2 − P3V3 ⎜ 1 − ⎟=0 V2 ⎠ ⎝
p2V2 = P3V3 . nb ' 1− V2
or
(ii)
According to zeroth law, 1 and 2 must also be in thermal equilibrium. To obtain the required equation, we must eliminate p3 and V3 from eq. (i) and (ii), when we get p1 (V1 − nb ) =
p2V2 nb ' 1− V2
...(iii)
(b) Looking at eq. (i), (ii) and (iii), the three functions equal to one another at thermal equilibrium are p1 (V1 − nb ) ,
p2V2 and p3V3 nb ' 1− V2
Since the systems are gases, each of these is equal to the empiric temperature T (by gas equation pV= RT).
4 Change of State Clausius-Clapeyron Equation : Clapeyron in 1834, and Clausius in 1850, deduced an important equation which describes conditions governing changes of state, such as melting of solids and boiling of liquids. This is known as, the “ClausiusClapeyron equation” or the “first latent heat equation.” Let us consider two isothermals ABCD and EFGH at close temperatures T and T- d T respectively for unit mass of a substance below its critical point. Along the sections AB and EF the substance is in the liquid state, along the sections BC and FG the liquid and vapour states coexist in equilibrium, while along CD and GH only the unsaturated vapour state exists. At B and F the substance is entirely liquid, while at C and G it is entirely just saturated vapour. Let p and p - d p be the saturated vapour pressures of the liquid at temperatures T and T— d T respectively. Let V1 and V2 be the specific volumes (volume per unit mass) of the substance at B and C respectively. Let us draw two adiabatics through B and C .They intersect the lower isothermal at points M and N (very close to F and
Thermal Physics
26
G) respectively. The figure BCNM represents the indicator diagram when the substance is enclosed in the cylinder of an ideal heat engine and is taken through a reversible Carnot cycle along BCNMB. The efficiency e of the engine is defined as e=
W (work done by the substance during the cycle . Q1 (heat taken in at temperature T)
...(i)
Now, during a Carnot cycle, the work done is given by the area BCNMB . As the adiabatics are very small, they can be taken straight and parallel. Thus, the figure BCNMB is a parallelogram, and so W = area BCNMB = BC × perpendicular distance between BC and MN = (V2 - V1) × d p. During the process BC, the unit mass of the substance is changed completely from the liquid state at B to the vapour state at C at constant temperature T. The heat Q1 taken in by the substance during the process BC is given by
Change of State
27 Q1 = L
where L is the latent heat of vaporisation at temperature T. Substituting these values of W and Q1 in eq. (i), we get e=
(V - V1 ) dp W = 2 Q1 L
In this expression, the numerator (V2 - V1) d p is in work unit, and so the latent heat L should also be put in work unit. If L is given in heat unit it must be converted into work unit by multiplying it by J. Now, by the definition of thermodynamic scale of temperature, the efficiency of a Carnot engine working between temperatures T and T- d T is given by e = 1-
T - dT dT = T T
Equating the last two expressions for e, we get (V2 - V1 ) dp dT = L T
or Infinitesimally,
dp L = dT T (V2 - V1 ) dp L = dT T (V2 - V1 )
This is Clausius-Clapeyron equation. In this equation, V1 and V2 are the specific volumes (volume/mass) of the substance in the liquid state and the vapour state respectively. If L is given is cal/gram or in kcal/kg, then dp JL = dT T (V2 - V1 )
The above equation also holds to the change from the solid to the liquid state ; but L will then represent the latent heat
Thermal Physics
28
of fusion, V1 the volume occupied by the unit mass of the substance in the solid state and V2 that in the liquid state. Effect of Pressure on Boiling Point of Liquids : If the part BC of the isothermal ABCD at temperature T represents the change of state of the substance from liquid to vapour, then T is the boiling point of the substance under the pressure p . Now , the specific volume V2 of a substance in the vapour state is much larger than the specific volume V1 in the liquid state. Therefore, in case of change of state from liquid to vapour state, the quantity (V2-V1) is always positive. Since L and T are necessarily positive, the Clausius-Clapeyron equation shows that dp/dT is positive. This means that the boiling point of every liquid rises with increase in pressure . Effect of Pressure on Melting Point of Solids : If the part BC of the isothermal ABCD represents the change from solid to liquid state, then T is the melting point of the substance under the pressure p .For substances like wax and sulphur, which expand on melting, the quantity (V2 - V1) is positive and so dp/dT is positive. The means that the melting point of waxtype substances rises with increase in pressure. But for a substance like ice, which contracts on melting, the quantity (V2 - V1) is negative. Therefore, dp/dT is negative which means that the melting point of ice-type substances is lowered with increase in pressure. Conditions in which Latent Heat become Zero : At the critical point the distinction between liquid and vapour phases disappears (V1 = V2). Hence according to Clausius-Clapeyron equation : L V2 − V1 = =0 T (dp / dT ) Hence either L = 0 or dp/dT= ∞ . Generally the specific volume of a liquid is much smaller than that of its vapour. Assuming the vapour to obey the perfect gas equation, prove that p = constant e-(L/RT)
Change of State
29
Solution : From the Clausius-Clapeyron equation, we have L dT . dp = (V2 - V1 ) T Ignoring volume V1 of liquid in comparison to volume V2 of vapour, we get L dT dp = ., V T where V2 has been replaced by V. From gas equation pV = RT, we may write dp =
or
pL dT L dT = RT/p T R T2
dp L dT = P R T2
Integrating : we get log e p =
or
L R
æ 1÷ ö çç- ÷ + constant çè T ÷ ø
p = constant e − L / RT
Clausius’ Latent Heat Equation Let us consider two isothermals ABCD and EFGH at close temperatures T and T- d T respectively for unit mass of a substance below its critical point. Along the sections AB and EF the substance is in the liquid state, along sections BC and FG the liquid and vapour states coexist in equilibrium, while along CD and GH only the unsaturated vapour state exists. The dotted curve represents the boundary which separates the different states. Thus, along BF the substance is entirely in the liquid state, and along CG it is entirely in the just saturated vapour state. Let p and p— d p be the saturated vapour pressures of the liquid at temperatures T and T- d T respectively. Let V1 and V2 be the volumes of the unit mass of the substance at B and C respectively.
Thermal Physics
30
Let the substance be taken round the cycle FBCGF. (This is not a true Carnot cycle because BF and CG are not a diabatics). Let s1 and s2 be its specific heats in the liquid and in the saturated vapour state respectively. If L be the latent heat at temperature T , and dL/dT be the rate of variation of latent heat with temperature, then the latent heat at temperature (T- d T) will be æ dL ÷ ö ççL÷ çè dT dT ÷ ø
As the substance is taken from F to B, its temperature rises by d T and so it takes in a quantity of heat s1 d T. Along BC, the substance changes from liquid to vapour state at constant temperature T and takes in a quantity of heat L. In passing from C to G, the temperature of the substance falls by d T and so it gives out a quantity of heat s2 d T. Finally, in passing along GF, it changes from vapour to liquid state at constant æ dL ö÷ dT ÷. temperature (T- d T), and gives out a quantity of heat çççèLdT ø÷
Thus , the net quantity of heat taken in during the cycle is given by
Change of State
31 æ dL ö÷ dQ = (s1 dT+ L) - çççs2 dT + L dT ÷ è ø dT ÷ édL ù = ê + s1 - s2 údT êëdT ú û
Now, the external work W done by the substance during the cycle is equal to the area of the cycle i.e. W = area BCGF. Since the two isothermals are very close together, the figure BCGF is practically a rectangle of area BC × d p . Therefore W = BC × d P = (V2 - V1) d p.
…(ii)
The substance, after the end of the cycle, returns to its initial state. Therefore , its internal energy remains unchanged. Hence, by the first law of thermodynamics , the net heat d Q taken in is equivalent to the external work W i.e. d Q = W.
Substituting the values of d Q and W from eq. (i) and (ii), we get édL ù ê + s1 - s2 údT = (V2 - V1 ) dP êëdT ú û 1 V ( 2 - V1 )
or
édL ù dP ê + s1 - s2 ú= êëdT ú û dT
dp L But dT = T (V - V ) by Clausius-Clapeyron equation. 2 1 \
or
édL ù 1 L ê + s1 - s2 ú = (V2 - V1 ) êëdT ûú T (V2 - V1 ) dL L + s1 - s2 = dT T
Thermal Physics
32 or
dL L = s2 - s1 dT T
This is known as “equation of Clausius” or “second latent heat equation.”
Specific Heat of Saturated Water Vapour From the Clausius equation, the specific heat of saturated vapour is given by dL L s2 = s1 + dT T The specific heat of (liquid) water s1 = 1 cal/(gm-K), and at the normal boiling point of water i.e. at T= 100 + 273 = 373 K, the latent heat L = 540 cal/gm. get
Putting the values of s1 ,L and T in the above equation, we dL 540 dT 373 dL =1+ - 1.45 dT dL = - 0.45 + dT
s2 = 1 +
We know that the latent heat of water diminishes with rise in temperature i.e. dL/dT is negative. Thus, the specific heat s2 of saturated water vapour at 100 °C is negative. Explanation of the Negative Specific Heat of Saturated Water Vapour : The specific heat of a saturated vapour is the quantity of heat required to raise the temperature of 1 gm of vapour through 1 °C (or 1 K) while at the same time the pressure and volume are varied such that the whole mass remains saturated. Thus the negative specific heat of a saturated vapour means that in order to raise the temperature of the vapour, while keeping it saturated , a certain quantity of heat is to be withdrawn from the vapour. That this is possible can
Change of State
33
be explained with the help of Fig. in which ABCD and EFGH represent two isothermals of a substance at temperatures T and T- d T respectively. At the point C the substance is in the saturated vapour state at temperature T. At the point G it is in the saturated vapour state at temperature T- d T. The curve (dotted) which passes through these points is called the ‘curve of saturation.’ Now, suppose 1 gm of vapour is to be taken from the saturated state G (at temperature T- d T) to the saturated state C (at temperature T) along the saturated curve GC. Thus it would remain saturated throughout the process. The figure shows that the volume V 2 of the saturated vapour at temperature T is less than the volume V2’ at temperature (T- d T). Therefore, to take the vapour from G to C , it must be compressed, and so heat will be generated. In the case of saturated water vapour this heat is so much that it will raise the temperature of the vapour above T. Hence some heat will have to be withdrawn from the vapour so that the temperature does not rise above T. This means that the specific heat of a saturated water vapour will be negative. Practical Application of Negative Specific Heat : We have seen that if a saturated water vapour is compressed adiabatically, it does not remain saturated and becomes superheated. Similarly, if the saturated water vapour is expanded adiabatically, it becomes super-saturated. If nuclei are present, the vapour condenses and forms a fog. The effect is used in observing the tracks of charged particles passing through a Wilson cloud chamber.
Triple Point of a Substance A substance is found to exist in three states solid, liquid and gas. For each substance there is a set of temperatures and pressures at which any two of the three states of the substance may coexist in equilibrium.
34
Thermal Physics
Let us consider an enclosure filled with a substance, partly in the liquid state and partly in the state of saturated vapour. The saturated vapour pressure is found to be a function of the temperature and increases with increase in temperature. If we plot a graph between saturated vapour pressure and the corresponding temperature, we obtain a curve OA as shown in Fig. (a). It is called the “curve of vaporisation.” It represents the boundary between the liquid and the vapour states, the substance being liquid at all pressures above the curve for each temperature, while below it there exists only vapour.
Similarly, liquid and solid states can coexist in equilibrium at temperatures at which the solid melts under the given pressures. This pressure-temperature relationship can be represented by a curve OB or OB’ which is called the “curve of fusion”. It represents the variation of melting point with pressure. The curve OB which slopes to the left is for ice-type substances (melting point lowered with increase in pressure). The curve OB’ which slopes to the right is for wax-type substances (melting point raised with increase in pressure). On the right of the curve the substance is entirely liquid and on the left entirely solid. A solid and its vapour may also coexist in equilibrium. The temperatures and the corresponding pressures at which the solid and the vapour states coexist are represented by a curve OC . This curve is called the “curve of sublimation”. Above this curve the substance is entirely solid and below it entirely vapour.
Change of State
35
When all the three curves are plotted on the same diagram, they are found to intersect in a common point O . This point is called the “triple point”. The triple point for a substance may be defined as the point at which the temperature and pressure are such that the solid , liquid and vapour states of the substance may coexist in equilibrium. For water the triple point has coordinates 0.0075°C temperature and 4.58 mm (or 0.006 atmosphere) pressure. The curves OA , OB and OC, in case of water, are called the steam line, ice line and hoar-frost line respectively.
There is only one triple point for a substance : If the temperature or the pressure is even very slightly changed from its value at the triple point, one of the three states will disappear. For example, it the temperature is raised the ice will melt and the state of the system (water and saturated vapour) will be represented by a point on the curve OA.
Thermal Physics
36
To prove that there is only one triple point, suppose that the three curves intersect at three points O1 , O2 and O3 as shown in Fig. Let us consider the state of the substance within the triangle O1 O2 O3 . As it is below the curve of sublimation, the substance must be entirely vapour ; as it is above the curve of vaporisation, the substance must be entirely liquid ; and as it is to the left of the curve of fusion, the substance must be entirely solid. But these are three contradictory conclusions. Hence the triangle O1 O2 O3 cannot exist, and the three curves must intersect in a single point.
PROBLEMS 1. Find the change in the melting point of ice (or freezing point of water) at 0°C for an increase of pressure by 1 atmosphere. Latent heat of ice at 0 °C = 80 cal/gram . At melting point the volumes of 1 gram of water and ice are respectively 1.000 cm3 and 1.091 cm3; 1 atmosphere = 1.013 × 106 dyne/cm2 and J = 4.18 × 107 erg/calorie) Solution : The change in the melting point (dT) of a substance at absolute temperature T with change in pressure (dp) is given by Clapeyron equation dp L = dT T (V2 - V1 ) where V2 and V1 are the specific volumes (volume/mass) of the substance in the liquid and solid states respectively, and L is the latent heat of fusion in work unit. If L is given in heat unit, then we shall write dp JL = dT T (V2 - V1 )
So that
dT =
T (V2 - V1 ) dp JL
Here T = 0 + 273 = 273 K, V2 = 1.000 cm3/g, V1 = 1.091 cm3/g, dp = 1 atmosphere = 1.013 × 106 dyne/cm2 and L = 80 cal/g.
Change of State \ dT =
37
273 K × (-0.091 cm 3 /g) × (1.013 × 10 6 dyne/cm 2 ) ( 4.18 ´ 10 7 erg / cal) × (80 cal/g)
= - 0.0075 K (erg = dyne x cm) = - 0.0075 °C. The negative sign means that the melting point of ice, or the freezing point of water, is lowered by increase in pressure. 2. Calculate the change in the melting point of ice per atmosphere increase in pressure. The latent heat of ice is 80 kcal / kg and the ratio of the densities of ice and water is 10/ 11. Given : 1 atmosphere = 105 N/m2 and J = 4.2 joule/cal. Solution : By Clausius-Clapeyron’s equation, we have dp JL = dT T (V2 - V1 )
where L is in heat unit. From this, we have dT =
T (V2 - V1 ) dp JL
...(i)
Here T = 0 + 273 = 273 K, dp = 1 atmos. = 105 N/m2, L = 80 kcal / kg and J = 4.2 joule/cal = 4.2 × 103 joule/kcal. Now sp. volume of water at 0 o C V2 = V1 sp. volume of ice at 0 o C
=
density of ice at 0 o C 10 = o density of water at 0 C 11
But V2 = 1.000 cm3/g = 1.000 × 10-3 m3/kg (this value 11 should be remembered), so V1 = × (1.000 × 10-3) = 1.100 10 × l0-3 m3 kg.
Thermal Physics
38 Putting these values in eq. (i), we get dT =
273 K × (-0.100 × 10- 3 m3 /kg) × 10 5 N/m2 ( 4.2 ´ 10 3 J/kcal) × (80 kcal/kg)
= - 0.0081 K = - 0.0081 ºC The melting point will be lowered by 0.0081 °C . 3. Calculate the lowering of melting point of ice subjected to a pressure-increase of 50 atmospheres. The density of ice at 0 °C is 0.917 gram/cm3 and its latent heat is 334 joule/gram. Take 1 atmosphere = 1.013 x 106 dyne/cm2. Solution : Ice melts into water at 0 °C (T=273 K) under 1 atmosphere pressure. Under a pressure-change dp , the change in melting point, dT, is given by dp L = dT T (V2 − V1 ) where V2 and V1 are the specific volumes (reciprocal of densities) of water and ice respectively and L is latent heat in work unit. From this, we have dT =
T (V2 − V1 ) dp L
Now, T = 273 K , V 2 = 1.000 cm 3/g (it should be remembered), V1 = 1/0.917 = 1.091 cm3/g, dp = 50 atmos = 50 x (1.013 x 106) = 5.065 x 107 dyne/cm2 and L = 334 joule/gram = 334 x 107 erg/gram. ∴ dT =
273 K × (1.000 − 1.091)cm3/g × (5.065 × 107 dyne/cm2 ) 334 × 107 crg/g
273 × ( − 0.091) × (5.065 × 107 ) K 334 × 107 = − 0.38 K = − 0.38 °C. =
4. Calculate the change in the melting point of napthalene for one atmosphere rise in pressure, given that the melting point is 80 °C, latent heat of fusion is 4563 cal/mole and
Change of State
39
increase in volume on fusion is 18.7 cm3/mole. One calorie = 4.18 x 107 erg. Solution : In usual notations, the change in the melting point of napthalene is given by dT =
T (V2 − V1 ) dp L
Here T = 80 + 273 = 353 K, V2 - V1 = + 18.7 cm3/mole, dp = 1 atmos = 1.013 x 106 dyne/cm2 and L = 4563 cal/mole = 4563 x 4.18 x 107 erg/mole. .∴ dT = =
353 K x 18.7 cm 3/mole x (1.013 x 10 6 dyne/cm 2 ) 4563 x 4.18 x 107 erg/mole
353 x 18.7 x 1.013 x 10 6 ) K 4563 x 4.18 x 10 7
[Q erg = dyne x cm]
= + 0.035 K = + 0.035 °C.
dT is positive. It means that the melting point of napthalene rises with increase in pressure. 5. Calculate the change in the melting point of wax subjected to a pressure of 50 atmospheres from the following data : Melting point = 64 °C, volume of solid wax at 64 °C = 1.161, volume of liquid wax at 64 °C = 1.166 , density of solid wax at 64 °C = 0.96 g/cm3, latent heat of fusion = 97 cal/g, J = 4.18 x 107 erg/cal 1 atmos = 1.013 x 106 dyne/cm2. Solution : Wax melts at 64 °C (T=337K) under 1 atmosphere pressure. According to Clausius-Clapeyron latent heat equation, the change in the melting point of wax (dT) with increase in pressure (dp) is given by dT =
T (V2 − V1 ) dp L
...(i)
where V2 and V 1 are volume/mass of liquid and solid wax respectively and L is the latent heat of fusion in terms of work unit.
Thermal Physics
40
Since volume is inversely proportional to density, we have density of liquid wax volume of solid wax = density of solid wax volume of liquid wax ∴ density of liquid wax =
1.161 x 0.96 (g/cm3) 1.166
= 0.956 g/cm3. Thus, here we have T= 64 + 273 = 337 K, V2 = 1/0.956 = 1.046 cm3/g, V1 = 1/ 0.96 = 1.042 cm3/g, dp = 50 -1 = 49 atmos = 49 x 1.013 x 106 dyne/cm2 and L = 97 cal/g = 97 x 4.18 x 107 erg/g. Substituting these values in eq. (i), we get dT =
=
337 K x (1.046 - 1.042) cm 3/g x (49 x 1.013 x 10 6 dyne/cm 2 ) 97 x 4.18 x 107 erg/g 337 x 0.004 x 49 x 1.013 x 10 6 K 97 x 4.18 x 107
[Q erg = dyne x cm]
= 0.0165 K = 0.0165 °C.
6. Calculate the pressure required to make water freeze (or ice melt) at -1 °C. Given : density of ice = 0.917 g/cm3, latent heat of ice = 80 cal/g and J = 4.18 x107 erg/cal. Solution : Water freezes at 0 °C at a pressure of 1 atmosphere. Let dp be the change in pressure required to make the ice freeze at -1 °C. By Clausius-Clapeyron latent heat equation, we have dp =
L dT T (V2 − V1 )
where V2 and V1 are specific volumes (volume/mass or reciprocal of density) of water and ice respectively and L is
Change of State
41
latent heat in work unit (as erg/gm). If L is in heat unit (as cal/gm), then we must multiply it by J, that is dp =
JL dT T (V2 − V1 )
Here L = 80 cal/g, T= 0 + 273 = 273 K, dT = -1oC = - 1 K, V2= 1.000 cm3 /g , V1=1/0.917 = 1.091 cm3/g so that V2- V1 = -0.091 cm3/g . ∴ dp =
(4.18 × 107 erg/cal)(80 cal/g)(-1 K) 273 K (-0.091 cm3/g)
= 134.6 × 106 erg/cm3 =134.6 × 106 dyne/cm3 .
Now , 1 atmosphere pressure = 1.013 × 106 dyne/cm2 . ∴ dp =
134.6 × 10 6 = 133 atmosphere. 1.013 × 10 6
Thus the pressure must be increased by 133 atmosphere to lower the freezing point of water (or melting point of ice) by 1 °C. This means that water will freeze at - 1 °C under a pressure of 134 atmosphere. 7. Calculate the specific gravity of solid sulphur from the following data: Melting point of sulphur = 115 °C ; latent heat of fusion of sulphur = 9.3 cal/g, volume of 1 g of liquid sulphur = 0.513 cc, rate of change of melting point with pressure = 0.025 °C per atmosphere. (Atmospheric pressure = 106 dyne/cm2 and J = 4.18 x107 erg/cal) Solution : The Clausius-Clapeyron equation for the melting of a solid substance is dp JL = dT T (V2 − V1 ) .
Thermal Physics
42
where T is melting point (in Kelvin) of the substance, L is latent heat of fusion in heat unit (cal/gm) and V2 and V1 are specific volumes (volume/mass) in liquid and solid state of the substance respectively. From this, the rate of change of melting point with pressure is dT T(V2 − V1 ) = dp JL V2 − V1 =
or
JL dT T dp .
Here L = 9.3 cal/g, T = 115 + 273 = 388 K and =
dT 0.025 o C = dp 1atmos
0.025 K . 10 dyne/cm2 6
∴ V2 − V1 =
(4.18 × 107 erg/cal)(9.3 cal/g) 0.025 K × 6 388 K 10 dyne/cm 2
= 0.025 cm3/g. [∴ erg = dyne x cm] Thus, specific volume of solid sulphur is V1 = V2 – 0.025 = 0.513 – 0.025 = 0.488 cm3/g. Hence sp. gr. of solid sulphur =
1 1 = = 2.05 g/cm 3 . V1 0.488
7. Calculate the change in the boiling point of water when the pressure of steam on its surface is increased from 1.0 atmosphere to 3.1 atmospheres. The latent heat of water at 100 °C is 537 cal/g and the volume of 1 g of steam is, 1676 cm3. Take J = 4.2 × 107 erg/cal and 1 atmosphere = 1.0 × 106 dyne/cm2. Solution : Water boils at 100 °C (T= 100 + 273 = 373 K) under one atmosphere pressure. Let dT be the change in boiling point when pressure is increased by dp. From ClausiusClapeyron equation, we have
Change of State
43 dp L = dT T (V2 − V1 )
where V1 and V2 are specific volumes (volume/mass) of water and steam respectively and L is latent heat in work unit. If L is given in heat unit (cal/gm), then we shall write dp JL = dT T (V2 − V1 )
dT =
or
T (V2 − V1 )dp JL
Here T = 373 K, V1 = 1 cm3/g (it should be remembered), V2 = 1676 cm3/g, dp= 3.1 - 1.0 = 2.1 atmos = 2.1 x 106 dyne/ cm2, J = 4.2 x 107 erg/cal and L = 537 cal/g . ∴ dT =
373 K x 1675 cm 3/g x (2.1 x 10 6 dyne/cm 2 ) (4.2 x 107 erg/cal) x (537 cal/g)
= 58 K = 58 °C. The boiling point of water will be raised by 58 °C. 8. Calculate the change in the boiling point of water due to an increase of pressure of 1 cm of mercury. Datas as in the last problem. Solution : Refer to last problem. dT =
T (V2 − V1 )dp JL
Here dp = 1 cm of Hg = 1 x 13.6 x 981 dyne/cm2. ∴ dT =
373 K x 1675 cm 3/g x (1 x 13.6 x 981 dyne/cm 2 ) (4.2 x 107 erg/cal) x (537 cal/g)
= 0.37 K = 0.37 °C. The boiling point of water will be raised by 0.37 °C .
Thermal Physics
44
8. One gram of water vapour at 100 °C and one atmospheric pressure occupies a volume of 1640 cm3. Calculate the vapour pressure of water at 99 °C. Latent heat of vaporisation = 536 cal/g and density of water at 100 °C 1 g/cm3. Solution : The vapour pressure of water at 100 °C (T = 100 + 273 = 373 K) is 1 atmosphere. Let dp be the change in the vapour pressure when the temperature decreases from 100 °C to 99 °C, that is, dT= -1°C = - 1 K . Now, the rate of change of vapour pressure with temperature is given by the Clausius-Clapeyron equation dp JL = dT T (V2 − V1 ) where V2 and V1 are the specific volumes in the vapour state and in the liquid state respectively, and L is the latent heat of vaporisation in heat unit (cal/g). Here L = 536 cal/g, V2 = 1640 cm3/g and V1 = 1 cm3/g. J may be taken 4.18 x 107erg/cal . Substituting these values in the above equation, we get dp = =
=
(4.18 × 107 erg/cal) × (536 cal/g) × (-1 K) 373 K × (1640 -1) cm3/g 4.18 × 107 × 536 erg 373 × 1639 cm3 _
3.66 x 104 dyne/cm2.
[Q erg = dyne x cm]
Now, 1 atmosphere pressure is the pressure of 76 cm mercury column, that is, 76 x 13.6 x 981 = 1.01 x 106 dyne/cm2. Q dp =
3.66 × 10 4 = −0.0362 atmos. 1.01 × 10 6
The minus sign shows that the vapour pressure will decrease by 0.0362 atmos. Hence the vapour pressure of water at 99°C will be 1- 0.362 = 0.964 atmos. This is equivalent to 733 mm of mercury.
Change of State
45
9. Calculate the change in the vapour pressure of water as the boiling point changes from 100 °C to 102 °C. The specific volume of steam is 1671 cm2/g and the latent heat of steam is 540 cal/g. J = 4.18 × 107 erg/cal and 1 atmosphere = 1.01 × 106 dyne/cm2. Solution : The vapour pressure of water at its normal boiling point 100 °C (T=373 K) is 1 atmosphere. Let dp be the change in vapour pressure as the boiling point changes from 100 °C to 102 °C, that is, when dT= 2 °C = 2 K. From Clausius-Clapeyron equation, we have JL dT dp = T (V2 − V1 ) where V2 and V1 are specific volumes in the vapour state and liquid state respectively. Q dp =
(4.18 x 107 erg/cal) × (540 cal/g)(2 K) (373 K) × (1671 -1) cm 3/g
= 7.25 × 10 4 dyne/cm 2 =
7.25 × 10 4 = 0.0718 atmos. 1.01 × 106
The vapour pressure will increase by 0.0718 atmosphere or by 0.0718 760 = 54.6 mm of mercury. 10. Calculate under what pressure water would boil at 120 °C. one gm of steam occupies a volume of 1677 cm3. Latent heat of steam = 540 cal/g, J = 4.2 × 107 erg/cal, 1 atmosphere pressure = 1.0 × 106 dyne/cm2 . Solution : Water boils at 100 °C (T = 373 K) at 1 atmosphere pressure. Let dp be the change in pressure required to make water boil at 120 °C, that is, for dT= 20 °C = 20 K. From Clausius-Clapeyron equation, we have dp =
JL dT T (V2 − V1 )
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46
where V2 (specific volume of steam) is 1677 cm3/g and V1 (specific volume of water) is 1 cm3/g . Substituting the values, we get dp =
(4.2 x 107 erg/cal) × (540 cal/g)(20 K) (373 K) × (1677 -1) cm 3/g
4.2 × 107 × 540 × 20 erg 373 × 1676 cm3 6 2 = 0.725 × 10 dyne/cm =
=
0.725 × 106 = 0.725 atmosphere. 1.0 × 106
Thus, in order to make water boil at 120 °C, the pressure should be increased by 0.725 atmos, that is, it should be increased to 1.725 atmosphere . 11. Calculate the specific volume of the vapour of carbon tetrachloride at the boiling point from the following data. Boiling point = 77 °C at 1 atmosphere, latent heat = 46 cal/ g, density of liquid = 1.6 g/cm3, dp/dT = 23 mm of mercury/ °C, J = 4.18 × 107 erg/cal. Solution : The Clausius-Clapeyron equation for the boiling point of a liquid is dp JL = . dT T (V2 − V1 )
where T is boiling point (in Kelvin) of the liquid, L is latent heat of vaporisation in heat unit (cal/gm) and V2 and V1 are specific volumes of vapour and liquid respectively. From this, we have V2 − V1 =
JL . T (dp / dT )
Here L = 46 cal/g, T = 77 + 273 = 350 K and
dp 23 mm of Hg = dT °C
Change of State
47 2.3 = cm × 13.6 g/cm3 × 980 dyne/g K 3.065 × 104 dyne/cm2 = K
=
∴ V2 − V1 =
(4.18 x 107 erg/cal) x (46 cal/g) (3.065 x 10 4 dyne/cm 2 ) 350 K x K
= 179.2 cm3/g.
Now, the density of liquid is given 1.6 g/cm3. Therefore, the specific volume (volume/mass) is 1 V1 = = 0.6 cm 3/g . 1.6 g/cm 3 V2 = 179.2 + V1 = 179.8 cm3/g .
∴
12. Calculate the latent heat of steam at 100 °C under atmospheric pressure. Given : specific volume at saturated steam and water = 1677 cm3/g and 1 cm3/g respectively, dp/dT = 27.1 mm of Hg per °C, J = 4.18 x 107 erg/cal. Solution : Refer to the last problem. L=
T (V2 − V1 ) dp . J dT
Here T = 100 + 273 = 373 K, (V2- V1) = (1677-1) = 1676 cm3/g, dp 27.1mm of mercury = dT °C
=
2.71 cm × 13.6 g/cm 3 × 980 dyne/g K
3.61 x 10 4 dyne/cm 2 . K 3 373 K x 1676 cm /g 3.61 x 10 4 dyne/cm 2 ∴L= × 4.18 x 107 erg/cal K
=
= 540 cal/g.
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13. A Carnot’s engine having 100 g water-steam as working substance has, at the beginning of the stroke ; a volume 104 cm3 and pressure 788 mm (B.P. = 101 °C). After a complete isothermal change from water into steam, the volume is 167, 404 cm3 and the pressure is then lowered adiabatically to 733.7 mm (BP =99 °C). If the engine is working between 99 °C and 101 °C, calculate the latent heat of steam. Solution : Volume of 100 g of water in the beginning is 104 cm3. 104 ∴ specific volume, V1 = = 1.04 cm 3/g . 100 After the isothermal change, the volume of 100 g of steam is 167,404 cm3. ∴ specific volume, V2 =
167,404 = 1674.04 cm 3/g . 100
∴ change in specific volume is
(V2- V1) = 1674.04 -1.04 = 1673 cm3/g. Now, change in pressure, dp = 788 - 733.7 = 54.3 mm = 5.43 cm = 5.43 x 13.6 x 980 = 7.24 x 104 dyne/cm2, change in boiling point, dT= (273 + 101) - (273 + 99) = 2 K, 99 + 101 + 273 = 373 K. 2 According to Clapeyron equation
and
mean temperature, T =
dp JL = dT T (V2 − V1 ) ∴L =
T(V2 − V1 ) dp J dT
373 K × 1673 cm3/g 7.24 × 104 dyne/cm2 × 4.18 × 107 erg/cal 2K = 540 cal/g. =
Change of State
49
13. Calculate the specific heat of saturated vapour (steam) at 100 °C if the latent heat L varies with temperature T according to the equation L=800 - 0.705 T, and the specific heat of water at 100 °C is 1.0 cal/(g-°C) Solution : By the equation of Clausius, the specific heat s2 of a saturated vapour is given by s2 = s1 +
dL L − dT T
where s1 is the specific heat in the liquid phase. Here L = 800 - 0.705 T so that Also, at
dL = 0.705. dT
T= 100 + 273 = 373 K, we have L = 800 - 0.705 x 373 = 537.
Putting the values of dL/dT, L/T and s1 in eq. (i), we get 537 373 = 1.0 − 0.705 − 1.44 = −1.145 cal/(g − °C).
s2 = 1.0 − 0.705 −
Thus the specific heat of saturated steam is negative. 14. The latent heat of Water diminishes by 0.64 cal/g for each degree centigrade rise in temperature in the neighbourhood of 100 °C and the latent heat of water vapour at 100 °C is 540 cal/g. Find the specific heat of saturated steam at 100 °C. The specific heat of water at 100 °C is 1.01 cal/g- °C . Solution : By the equation of Clausius, the specific heat s2 of a saturated vapour is given by s2 = s1 +
dL L − . dT T
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50
where s1 is the specific heat in the liquid state and dL/dT is the rate of change of latent heat with temperature. Here dL/dT = - 0.64 cal/g-°C , T = 100 + 273 = 373 K, L = 540 cal/g and s1=1.01cal/g-°C. ∴
s2 = 1.01 + (-0.64)-
540 373
= 1.01- 0.64 - 1.447 = - 1.077 cal/(g-°C). 15. Calculate the specific heat of saturated steam at 100 °C. Given : L at 90°C = 545.25 cal/g, at 100°C = 539.30 cal/ g and at 110°C = 533.17 cal/g. The specific heat of water at 100 °C is 1.013 cal/(g-°C). Solution : By the equation of Clausius, the sp. heat s2 of saturated steam is given by s2 = s1 +
dL L − dT T
where s1 is the sp. heat of water and L is the latent heat of steam at temperature T. Here s1 = 1.013 cal/(g-K), T= 100 + 273 =373 K and
L (at 100 °C) = 539.30 cal/g ,
dL 533.17 − 545.25 12.08 = = = - 0.604 dT 110 − 90 20
cal (g-K) ∴ s2 = 1.013 − 0.604 −
539.30 373
= 1.013 - 0.604 - 1.446 = -1.037 cal/(g-K). 16. The specific heat of a certain liquid is given by s1 = 0.53 + 0.0004 t and its latent heat of vaporisation is given by L = 94 - 0.06 t - 0.0006t2,
Change of State
51
where t is temperature in °C. Calculate the specific heat of saturated vapour at 50 °C. Solution : By the equation of Clausius, the specific heat of saturated vapour is given by s2 = s1 +
dL L − dT T
...(i)
From the given equations, at 50 °C, we have s1 = 0.53 + (0.0004 x 50) = 0.55 cal/(g-K), L = 94 -(0.06 x 50) - (0.0006 x 50 x 50) = 89.5 cal/g, T = 50 + 273 = 323 K and
dL dL = = - 0.06 - 0.0006 (2 t)[dT= dt, Q 1 °C = 1 K] dT dt
= - 0.06 - 0.0006 x 2 x 50 = - 0.12 cal/(g-K). Substituting all these values in eq. (i), we get s2 = 0.55 − 0.12 −
89.5 323
= 0.55 - 012 - 0.277 = 0.153 cal/(g-K). The specific heat of the given vapour is thus positive . 17. The vapour pressure p (in mm of Hg) of solid ammonia is given by loge p = 23.03 - (3754/T) while that of liquid ammonia is given by loge p = 19.49 - (3063/T) where T is in Kelvin. Calculate the triple point of ammonia. Solution : At the triple point the solid, liquid and vapour phases co-exist in equilibrium, i.e. vapour pressure will be same for solid and liquid states. Thus log e p = 23.03 −
3754 3063 = 19.49 − T T
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52 23.03 − 19.49 =
or
T=
3754 − 3063 T 691 = 195.2 K. 3.54
Substituting this value of T in the given equation, we get 3754 195.2 = 23.03 − 19.23 = 3.8
log e p = 23.03 −
p = 45 mm of Hg.
∴
18. The coordinates of the triple point of water are t = 0.0075°C and p = 0.006 atmos. Calculate the slope of the ice line in atmos/°C. Solution : At triple point, all the three states of matter are in equilibrium. The slope of the ice line is dp JL = dT T (V2 − V1 )
where L is latent heat of ice, T is melting point and V2 and V1 are the specific volumes of water and ice respectively. Putting the known values : dp (4.18 × 107 erg/cal) × (80 cal/gram) = dT 273K × ( − 0.091 cm2/gram)
= - 13.46 × 107 dyne/(cm2-K) =−
13.46 × 107 = − 132.8 atmos/K , 1.013 × 10 6
because 1 atmos = 1.013 × 106 dyne/cm2. clearly, the ice line is almost vertical.
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53
5 Heat Capacity Kilo-calorie: 1 Kilo-calorie is the quantity of heat required to raise the temperature of 1 kilogram of water through 1°C (from 14.5 to 15.5°C). Calorie: 1 calorie is the quantity of heat required to raise the temperature of 1 gm of water through 1°C (from 14.5 to 15.5°C). Evidently, 1 kilo-calorie = 1000 calories. Materials differ from one another in the quantity of heat required to produce a given rise in temperature in a given mass. If a quantity of heat Δ Q raises the temperature of a body through Δ T, then the ratio of Δ Q to Δ T is called the ‘heat capacity’ C of the body. C=
ΔQ . ΔT
If Δ T= 1°, then C = Δ Q . Thus, the heat capacity of a body is numerically equal to the quantity of heat required to raise its temperature by 1°.
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54
Specific Heat : The heat capacity per unit mass of a body is called the ‘specific heat’ of the material of the body. If we denote it by c, then c=
heat capacity ΔQ / Δt ΔQ = = . mass m m ΔT
...(i)
If m = 1, Δ T = 1, then c = Δ Q. Thus, the specific heat of a material is numerically equal to the quantity of heat required to raise the temperature of unit mass of that material through 1°. It is expressed in cal-gm-1–°C-1 or kilocal-kg -1–°C-1 . From the definitions of cal and kilo-cal it is clear that 1 calgm –°C-1 = 1 kilocal-kg-1–°C-1 . -1
Strictly speaking, the specific heat of a material depends on the location of the temperature-interval in the temperature scale. The eq. (i) defines the average specific heat of a material over the temperature-interval Δ T. The true specific heat of a material at any temperature is defined from eq. (i) by considering an infinitesimal temperature-rise dT, so that C=
or
dQ m dT
dQ = mc dT.
Thus the heat required to raise the temperature of a material of mass m and specific heat c from Ti to Tf is Tf
Q = m ∫ c dT , Ti
..(ii)
where c is a function of the temperature. At any ordinary temperature and over small temperature-intervals, c can be taken constant.
Molar Heat Capacity or Gram-molecular Sp. Heat The amount of heat required to raise the temperature of 1 gm-molecule (or 1 mole) of a substance through 1° is called
Heat Capacity
55
the ‘molar heat capacity’ of the substance. It is equal to the product of the specific heat and the molecular weight of the substance. Specific heat and Molar heat capacity depend also on the conditions under which the heat is added to the substance. Hence we must specify the conditions also, such as specific heat at constant pressure (cp), sp. heat at constant volume (cV), molar heat capacity at constant pressure (Cp), molar heat capacity at constant volume (Cv), and so on. For solids and liquids the difference between cp and cv or the difference between Cp and CV is negligible, but certainly not so for gases. Suppose the specific heat c of a substance is found to vary with temperature in a parabolic fashion, that is c = A + BT2 where A and B are constants and T is the Celsius temperature. (i) Calculate the quantity of required to raise m gm of the substance from a temperature T1 to T2 (ii) Compare the mean specific heat of the substance in the temperature-range T=0 to T=T with the true specific heat at T/2 The true specific heat of a substance is given by c = A + BT2.
...(i)
The heat required to raise the temperature of a mass m of the substance from temperature T1 to T2 is given by T2
T2
T1
T1
O = m∫ c dT = m∫
( A − BT ) dT 2
T2
⎡ T3 ⎤ B ⎡ ⎤ = m ⎢ AT + B ⎥ = m ⎢ A ( T2 − T1 ) + ( T23 − T13 ) ⎥ 3 ⎦ T1 3 ⎣ ⎦ ⎣ B ⎡ = m ⎢A + 3 ⎣
(T
2 2
⎤ + T12 ) ⎥ ( T2 − T1 ) . ⎦
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56
(i) The mean specific heat in temperature-range T= 0 to T= T is given by cmean =
1 T 1 T cdt = ∫ ( A + BT 2 ) dt ∫ 0 T T 0
=
T3 ⎤ BT 2 1⎡ AT + B ⎥ = A + . ⎢ T⎣ 3 ⎦0 3
T
The true specific heat at T/2 is obtained by putting T= T/2 in eq. (i). Thus BT 2 . cT/2 = A+B(T/2)2 = A+ 4 ∴
⎛ BT 2 ⎞ ⎛ BT 2 ⎞ BT 2 . cmean − cT / 2 = ⎜ A + ⎟−⎜A+ ⎟= 3 ⎠ ⎝ 4 ⎠ 12 ⎝
Thus, the mean specific heat exceeds that at T/2 by BT2/12.
Specific Heat of Water and Climate The specific heat of water is about five times higher than that of clay or sand. Therefore, if same quantity of heat be given to (or taken from) the same mass of water and earth, the water will have a much smaller rise (or fall) in temperature than the earth. This has a very important effect on the climate of islands and coastal areas which have a sea or ocean near them. During a hot day the temperature of the sea rises more slowly than that of the land, and likewise falls more slowly when heat is radiated during the night. Therefore, during the day the cooler sea keeps the land near it cool, while during the night it tends to keep the land warm. This transfer of heat takes place through the movement of air. Thus, the land near the sea undergoes a smaller day and night temperature-variation than the land elsewhere. Similarly, the land near the sea remains cooler during the summer and warmer during the winter than the land far from the sea.
Heat and Temperature
57
6 Heat and Temperature The temperature of a body is a measure of its degree of hotness or coldness. A body appearing hotter to our sense of touch than the other is said to be at a higher temperature than the other. When two bodies A and B, the body A being at a higher temperature, are placed in contact, then after some time they both acquire the same temperature which is somewhere between the two initial temperatures. This means that something from A has been transferred to B. This ‘something’ is called heat. Thus, ‘heat is that which is transferred from one body to the other, because of a temperature difference between them’. In fact heat is a form of energy which is transferred from one body to the other because of a temperature difference between them. There is a distinction between the temperature of a body and the heat that it contains. The heat that a body contains depends upon its mass as well as upon its temperature. The sparks from a blacksmith’s hammer are white hot (i.e. at a very high temperature) but they do not burn the hand since their
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58
mass is very small, and therefore they contain little heat. On the other hand, a jug of hot water, though at a much lower temperature than the spark, causes a severe burn because it contains more heat. Heat and temperature can be compared with water and water-level, if two vessels filled with water upto different heights are connected, water flows from the higher to the lower level irrespective of the quantities of water in the two vessels. Similarly, heat flows from a body at higher temperature to a body at lower temperature, irrespective of the quantities of heat in the two bodies. Hot and Cold Bodies Placed in Contact : When a hot body and a cold body are put in mutual contact, heat flows from the hot to the cold body until they attain a common temperature. This means that the temperature of the hot body falls and that of the cold body rises. These temperature-changes are not necessarily equal because the masses (and also the specific heats) of the two bodies may be different. Suppose 10 gm of hot water at 50°C is added to 20 gm of cold water at 20°C and acquire a common temperature T °C. Thus heat lost by hot water = heat gained by cold water or ∴
10 × 1 × (50 - T) = 20 × 1 × (T - 20) T = 30°C.
Thus, the temperature of hot water falls by 20°C and that of cold water rises by 10°C. We cannot say that the temperature has passed from hot to cold water. It is the ‘heat’ that has passed.
Equivalence of Heat and Work Heat, a form of Energy: Formerly it was believed that heat was a fluid called ‘caloric’ which was filled in the spaces between
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59
the molecules of matter, and a body at higher temperature contained more caloric than that at a lower temperature. This theory, called ‘caloric theory of heat’, could, however, not explain many phenomena concerning heat and hence it was rejected. Today, it is an established fact that heat is a form of energy rather than a substance. We know that in a situation involving friction, heat is always produced. When the brakes of a moving car are applied, the brake-shoes and drums become hot. When air is pumped into a cycle tyre, the pump becomes hot. The match-stick on being rubbed becomes hot to burn. In all these situations, and in many others, mechanical energy (work) disappears and heat is produced. Rumford’s Experiments : The first evidence of relationship between work and heat came from Count Rumford’s experiments. He noticed, while supervising the boring of cannons, that large amount of heat was generated during the boring process, and that the generation of heat continued even when the boring tool became so dull that it was no longer cutting the metal. If heat were a substance (caloric) it would not be possible to evolve it indefinitely from the metal. Rumford, therefore, hinted that the mechanical work done by the borer against the friction of the metal of cannons created the heat. Davy’s Experiments : Davy liquefied two pieces of ice by rubbing them together. It was well known that a considerable amount of heat must be supplied to ice to melt it. Since there was no external source of heat, it was considered that the mechanical work done in rubbing the ice pieces against each other created the necessary heat. Joule’s Experiments : Joule performed accurate quantitative experiments to establish relation between work and heat. He tried to find the amount of heat produced by a certain amount of work done. He rotated a brass paddle-wheel in water contained in a copper calorimeter, with the expenditure of a
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60
measured amount of mechanical work, and measured the rise in temperature of the water. He found that the amount of heat (thermal energy) produced was always proportional to the amount of work done. Joule performed may more experiments to measure the heat produced by measured amounts of mechanical work. In one experiment, he stirred mercury contained in an iron vessel by an iron paddle and measured the heat produced. In another experiment he rubbed two iron rings against each other under mercury. He also measured the heat produced when electrical current was flown in a metallic wire. All these experiments gave practically the same result. Thus, Joule established that a given amount of mechanical work done in any way always produces the same amount of heat; and that the disappearance of that very amount of heat always gives the same amount of mechanical work. This establishes the fact that ‘heat is a form of energy. Mechanical Equivalent of Heat : Whenever mechanical work is transformed into heat, or heat into mechanical work, there is a constant ratio between the work and the amount of heat. This ratio is called “mechanical equivalent of heat” and is denoted by J . Thus, if W be the amount of work done and Q the amount of heat produced, we have W =J Q
or
W = JQ.
If Q = 1 unit then J = W. Therefore, J is numerically equal to the mechanical work required to produce one unit of heat. The value of J is 4.186 joule/kilo-calorie (or 4.186 joule/ calorie or 4.186 x 107 erg/calorie). This means that 4.186 joule of mechanical work will produce 1 kilo-calorie of heat. But 1 kilo-calorie of heat is defined as the heat which raises the temperature of 1 kg of water from 14.5 to 15.5 °C. Hence, we
Heat and Temperature
61
may say that 4.186 joule of mechanical work, when transformed into heat, will increase the temperature of 1 kg of water from 14.5 to 15.5 °C. It is clear from the definition of J that it is not a physical quantity. It is only a conversion factor used to convert energy from heat units (kilo-calorie or calorie) to mechanical units (joule or erg) and vice-versa. Thermodynamic System : A thermodynamic system is one which may interact with its surroundings in at least two distinct ways and one of these is necessarily a transfer of heat in or out of the system. The other (or others) may be some other means of transfer of energy, say by performance of mechanical work by or on the system, or through electromagnetic interaction such as magnetisation. A gas contained in a cylinder, a vapour in contact with its liquid, a stretched wire, a piece of magnetic material are examples of thermodynamic system. Thermodynamic Equilibrium : When there are no unbalanced forces between the system and its surroundings, the system is said to be in ‘mechanical equilibrium’. If the system has no tendency to undergo a change in internal structure and also has no tendency to transfer matter from one part of the system to another, it is said to be in ‘chemical equilibrium.’ Again, if all parts of the system are at the same temperature which is equal to the temperature of the surroundings, the system is said to be in ‘thermal equilibrium’. When the system is under all the three types of equilibrium, it is said to be in ‘thermodynamic equilibrium’. In this condition any change in the state of the system or of the surroundings would not occur. If, however, the conditions for any one of the three types of equilibrium are not satisfied, the system is said to be in a ‘non-equilibrium state’. When this is the case, the phenomena like acceleration, eddies, chemical reaction, heat-transfer with surroundings, etc. would take place.
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State Variables : The state of a thermodynamic system in equilibrium can be completely specified in terms of certain measurable (macroscopic) quantities. In the case of a gas, for example, these quantities are pressure p, volume V and temperature T. They are functions of the state of the system alone and return to the same values whenever the system returns to the same equilibrium state. They are known as ‘state functions’ or ‘state variables’ or ‘thermodynamic coordinates’ of the system. Equation of State : The equation of state of a thermodynamic system in equilibrium is a functional relationship among the state variables of that system. The general form of the equation of state of a gas, for example, may be written as f (p,V,T) = 0. It can be solved with respect to any one of the variables. Its three equivalent forms are p = p(V,T) V = V(p,T) T = T(p,V). This implies that any two of the three variables p , V and T are enough to specify the state of a gas and to fix the value of the third variable. If, however, the system is in a non-equilibrium state then this state cannot be described in terms of thermodynamic coordinates. Dependence of Temperature on Pressure and Volume of a System : If the pressure and volume of a system in thermodynamic equilibrium are given, then its temperature is also fixed and is determined by the equation of state. If, however, the system is in a non-equilibrium state, then no equation of state exists and the temperature cannot be uniquely determined.
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63
Temperature of an Isolated System : When a system is not influenced in any way by its surroundings, it is said to be ‘isolated’. The temperature of an isolated system is conserved provided no internal changes like acceleration, eddies, chemical reaction, heat-transfers between different parts are taking place in the system. Relation between Isothermal Bulk Modulus, Volume Expansion and Pressure Expansion : Let p, V and T be the state variables of a system in thermodynamic equilibrium. The general equation of state is f (p,V,T) = 0. We can solve it for p and V, to get p = p (V,T) and
V = V(p,T).
From partial differentiation, we get
and
⎛ ∂p ⎞ ⎛ ∂p ⎞ dp = ⎜ ⎟ dV + ⎜ ⎟ dT ⎝ ∂V ⎠T ⎝ ∂T ⎠V
...(i)
⎛ ∂V ⎞ ⎛ ∂V ⎞ dV = ⎜ ⎟ dp + ⎜ ⎟ dT . ⎝ ∂T ⎠ p ⎝ ∂p ⎠T
...(ii)
If the pressure is increased on the system at constant temperature (dT = 0), we get from eq. (i) ⎛ ∂p ⎞ dp = ⎜ ⎟ dV ⎝ ∂V ⎠T
If we consider the relative change in volume, we can define the isothermal bulk modulus of elasticity as ET = −
dp ⎛ ∂p ⎞ = −V ⎜ ⎟ dV /V ⎝ ∂V ⎠T
...(iii)
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64
If the system is heated at constant pressure (dp = 0), we get from eq. (ii) ⎛ ∂V ⎞ dV = ⎜ ⎟ dT . ⎝ ∂T ⎠ p
...(iv)
The coefficient of volume expansion can, therefore, be defined by
α=
dV / V 1 ⎛ ∂V ⎞ = ⎜ ⎟ . dT V ⎝ ∂T ⎠ p
...(iv)
If the system is heated at constant volume (dV= 0), we get from eq. (i) ⎛ ∂p ⎞ dp = ⎜ ⎟ dT . ⎝ ∂t ⎠V
The coefficient of pressure expansion can, therefore, be defined by
β=
dp / p 1 ⎛ ∂p ⎞ = ⎜ ⎟ dT p ⎝ ∂T ⎠V .
...(v)
Now, by the reciprocity theorem for the variables p, V and T, we have ⎛ ∂p ⎞ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎟ = −1 ⎟ ⎜ ⎜ ⎟ ⎜ ⎝ ∂V ⎠T ⎝ ∂T ⎠ p ⎝ ∂p ⎠V
Making substitutions from eq. (iii), (iv) and (v), we get ⎛ 1 ⎞ ⎛ ET ⎞ ⎟ = −1 ⎜ − ⎟ (α V ) ⎜ ⎝ V ⎠ ⎝ βp⎠
or
ET α
β
=p.
This is the required relation. Change in Pressure for a Change in Temperature at Constant Volume
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65
By the reciprocity theorem for the variables p, V and T of a gas, we have ⎛ ∂p ⎞ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎟ = −1 ⎟ ⎜ ⎜ ⎟ ⎜ ⎝ ∂V ⎠T ⎝ ∂T ⎠ p ⎝ ∂p ⎠V
or
⎛ ∂p ⎞ ⎛ ∂p ⎞ ⎛ ∂V ⎞ ⎟ . ⎜ ⎟ = −⎜ ⎟ ⎜ ⎝ ∂T ⎠V ⎝ ∂V ⎠T ⎝ ∂T ⎠ p
⎛ ∂p ⎞ 1 ⎛ ∂V ⎞ Now ET = – V ⎜ ∂V ⎟ and α = ⎜⎝ ∂T ⎟⎠ ⎝ ⎠T p V
∴
⎛ ∂p ⎞ ⎜ ⎟ = ET α . ⎝ ∂T ⎠V
The change in pressure dp corresponding to a change in temperature dT at constant volume is therefore given by dp = ET α dT = (104N/m2) (0.0037/K) (0.1 K) = 3.7 N/m2.
Equation of State of a Stretched Wire Let us consider a stretched wire, and let its state depend on its length L, temperature T and tension F. Let us write F = f (L,T). On differentiation: ⎛ ∂F ⎞ ⎛ ∂F ⎞ dF = = ⎜ ⎟ dL + ⎜ ⎟ dT. ⎝ ∂L ⎠T ⎝ ∂T ⎠L
If the wire undergoes an infinitesimal change from one state of equilibrium to another, then the differentials dF, dL and dT are perfect differentials. Of the three state variables F, L and T, only two are independent. Choosing F and T as independent variables, the
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66
above equation must be true for all sets of values dF and dT. Then, if dF = 0 and dT ≠ 0, we have ⎛ ∂F ⎞ ⎛ ∂F ⎞ ⎜ ⎟ dT = − ⎜ ⎟ dL ⎝ ∂T ⎠ L ⎝ ∂L ⎠T ⎛ ∂F ⎞ ⎛ ∂F ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ . ⎝ ∂T ⎠ L ⎝ ∂L ⎠T ⎝ ∂T ⎠ F
or
Now, L π r2
...(i)
1 ⎛ ∂L ⎞ ⎜ ⎟ = λ (coefficient of linear expansion) and L ⎝ ∂T ⎠ F
⎛ ∂F ⎞ ⎜ ⎟ = Y (isothermal Young’s modulus), where π r 2 is ⎝ ∂L ⎠T
the area of cross-section of the wire. Making these substitutions in eq. (i), we get ⎛ ∂F ⎞ 2 ⎜ ⎟ = −π r Yλ . ⎝ ∂L ⎠L
External Work and Internal Work When a system undergoes a displacement under the action of a force, ‘work’ is said to be done, its magnitude being equal to the product of the force and the component of the displacement parallel to the force. If the system ‘as a whole’ exerts a force on its surroundings and a displacement takes place, then the work done by the system is called ‘external work’. For example, when a gas contained in a cylinder expands and pushes out the piston, external work is done by the gas on the piston. When the work is done by one part of a system on another part of the same system, then it is called ‘internal work’ . For example, the molecules of an actual gas attract one another. Therefore, when a gas expands, work is done against the mutual attraction between its molecules. This is internal work.
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Internal work has no place in thermodynamics. Only the external work which involves an interaction between a system and its surroundings is important in thermodynamics, and is a macroscopic concept. If W is the (external) work done by a system against its surroundings, then - W will stand for the work done on the system by its surroundings. If in a process the volume of a system increases, then the work is done by the system. If volume decreases, then the work is done on the system.
Performance of External Work Versus A system which can interact with its surroundings can pass from one state to another by two different processes : by doing work and by transfer of heat. The water, its container and the paddle constitute the system, and the weight constitutes the surroundings. There is no temperature difference between the system and the surroundings. When the weight is let fall a certain distance, the paddle rotates and the water is churned. Thus the state of the system changes as a result of the work done by the surroundings (falling weight) on the system. “We may therefore define work as the energy which is transferred between a system and its surroundings when no temperature difference exists between them”. The system has some water in a container, which is in contact with a Bunsen flame which constitutes the surroundings. The temperature of the flame is higher than that of the water, so that heat flows from the flame into the water. Certainly no work has been done. Thus, in this case, the state of the system (water + container) changes because heat is transferred from the surroundings (flame) into the system. “We may therefore define heat as the energy which is transferred between a system and its surroundings only by virtue of a temperature-difference between them”.
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Naturally, a decision as to whether a particular change of state is due to the work done or due to the transfer of heat cannot be taken until we decide, which is the system and what are the surroundings’. This may be clarified by an example. A falling weight which turns a generator G, which in turn sends an electric current through a resistor R immersed in a water container. Suppose the water plus the resistor plus the generator is chosen as the system, and the weight as the surroundings. Now, the temperature of the surroundings does not differ from that of the composite system. Hence during the falling of weight the surroundings do work on the system, there being no transfer of heat between the two. Again, let us choose the water along as the system, and the resistor plus the generator plus the weight as the surroundings. Now there is a temperature-difference between the surroundings (resistor) and the system (water). Hence heat will flow from the resistor into the system during the falling of the weight, there being no work done by the surroundings on the system Thus, in this case, there is a transfer of heat between the system and the surroundings.
Heat given without Temperature-rise If a substance in a condensed state, say ice, is put in contact of a flame, the heat from the flame will flow into the ice by virtue of a temperature-difference between them. But it would not cause the temperature of ice to rise. Instead, the condensed state begins to change into the liquid state (ice melts) at constant temperature. Thus, the addition of heat without temperaturerise is possible, and this does not contradict the concept of heat as energy transferred by virtue of a temperature-difference. This also does not violate energy conservation because the given heat is stored as potential energy of the molecules of the substance.
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Heat, a Form of Stored Energy The heat added to the melting ice is used in displacing the molecules from their positions and making their order irregular. That is, it is stored as the internal potential energy of the molecules. Thus, heat can also be considered as a form of potential energy and this again does not contradict the concept of heat as energy transferred by virtue of a temperaturedifference. Changing Temperature without Heat-transfer : In all dynamic processes involving friction (or resistance), heat is always produced causing temperature-rise, although no heat is transferred. For example, when water is churned by a rotating paddle wheel or current is sent through a wire, the temperature of water (or of wire) rises.
Dependence of Work Done on the Path Let a gas be contained in a cylinder with insulating walls and conducting base, and fitted with a movable piston which carries weights upon it. Let a heat-reservoir of adjustable temperature T be available on which the cylinder may be placed, when desired. Let the gas be the system, and the piston (with weights) or the heat-reservoir be the surroundings. Work can be done on the system (or by the system) by increasing (or decreasing) weights on the piston. Similarly, heat can be made to flow into the system (or out of the system) by placing the cylinder on the heat-reservoir and adjusting the temperature of the reservoir to be higher (or lower) than that of the system. Suppose initially the system (gas) is in equilibrium with its surroundings, having a pressure pi, and volume Vi. It is then made to interact with the surroundings with some ‘quasistatic’ process in which it expands and reaches a final equilibrium state with pressure pf and volume Vf. Let us compute the external work done by the gas during this process.
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Let p be the instantaneous pressure of the gas while expanding against the piston. The instantaneous force on the piston is pA, where A is the face-area of the piston. The work done by the gas in displacing out the piston through an infinitesimal distance ds is dW = p A ds (work= force × distance) = p dV, where dV (= A ds) is the infinitesimal increase in the volume of the gas. The total work done by the gas in expanding from the initial volume Vi to the final volume Vf would be
Vi
W = ∫ dW = ∫ p dV . Vi
The value of this integral can be obtained graphically by plotting the p-V curve of the gas. The nature of the curve depends upon the conditions under which the gas has expanded. It is clear that the integral
∫
Vf
Vi
p dV is the area under
the p-V curve between the initial and final states. This area is,
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therefore, the measure of the work done by the gas. Now, there are many processes (paths) by which the system (gas) can be taken from an initial state i (with coordinates pt, Vi, Ti) to a final state f (with coordinates pf, Vf, Tf). Let us represent these two states by points i and f on a p-V diagram Now, one path from i to f is the path iaf. The pressure is kept constant from i to a and then the volume is kept constant from a to f. In this case, the work done is equal to the area under the line ia. Another path is ibf, in which case the work done is the area under the line bf which is much smaller than the area under the line ia. The other possible paths are the continuous curve from i to f, and the series of short zig-zag paths from i to f. The areas under these paths are different from each other and also different from the first two paths i af and ibf. Thus, we see that the work done by or on a system depends not only on the initial and final states but also on the intermediate states, i.e. on the path of the process.
Dependence of Heat-transfer on Path : Similarly, the amount of heat flowed ( ∫ dQ ) into the system (or out of the system), when it passes from state i (at temp. Ti) to state f (at temp. Tf) also depends on how the system is heated. For example, we can heat the gas at constant pressure pi until the temperature rises to Tf , and then lower the pressure to pf, keeping the temperature constant at Tf . Alternatively, we can first lower the pressure to pf , and then heat it to the temperature Tf, keeping the pressure constant at pf. Each process would give a different value of
∫ dQ . Thus, we see that the heat flowed
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into (or out of) a system depends not only on the initial and final states but also on the intermediate states, i.e. on the path of the process. Both work and heat do depend on the path of the process i.e. neither one can be conserved alone.
Cyclic Process When a system in a given state goes through a series of different processes which bring it back to its initial state, the system is said to have undergone a ‘cycle’.
Suppose a gas goes from state i to state f along path iaf and returns to state i along path fbi. During the first process (iaf) the gas expands. The work is done by the gas and is equal to the area between the curve iaf and the V-axis. During the second process (fbi) the gas is compressed. Now the work is done on the gas and is equal to the area between the curve fbi and the V- axis. The net work done by the gas in the cycle is equal to the difference between the two areas, that is, equal to the area enclosed by the cycle. When the p-V cycle is traced clockwise, the net work is done by the system and is said to be positive. When the cycle is traced anti-clockwise, the net work is done on the system, and is said to be negative.
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Let W1 and W2 represent the work in cycle 1 and 2 respectively in W1 is positive (cycle 1 is clockwise) and W2 is negative (cycle 2 is anti-clockwise). Since area of cycle 1 is smaller, W1 < W2. Hence the net work is negative. A similar consideration the net work is positive, while the net work is zero.
Internal Energy of a System A system may seem to have no apparent mechanical energy but may still be capable of doing work. It is therefore said to possess ‘internal energy’. For example, a mixture of hydrogen and oxygen does not possess any external kinetic or potential energy but still it can do work on explosion. This is due to the internal energy of the system. The internal energy is associated with microscopic mechanical energy. For example, the internal energy of an actual gas is the sum of the kinetic and potential energies of its molecules. The molecules of an ideal gas do not possess potential energy (because they do not attract one another), and hence, the internal energy of an ideal gas is only the kinetic energy of its molecules. This is, however, not an operational definition, as microscopic mechanical energy cannot be measured directly.
Measurement of Internal Energy Let us formulate a macroscopic definition of internal energy which provides a means of measuring it. Let a thermodynamic system interact with its surroundings, and pass from an initial equilibrium state i to a final equilibrium state f through a certain process (path). Let Q be the heat absorbed by the system and W the work done by the system during this process. The quantity (Q - W) can be computed. It is experimentally found that if the system be carried from the state i to f through
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74
different paths, the quantity (Q - W) is found to be the same, although Q and W individually are different for different paths. Thus, when a thermodynamic system passes from state i to state f, the quantity (Q - W), depends only upon the initial and final (equilibrium) states and not at all on the path taken by the system between these states. This quantity is defined as the ‘change in the internal energy of the system. ’Thus, if Ui and Uf be the internal energies of the system in the initial and the final states respectively, we have Uf-Ui = Δ U = Q-W.
…(i)
We conclude that there is a function U of the thermodynamic coordinates of the system whose final value minus initial value is definite and equals Q — W in the process, by whatever path the system went from the initial to the final state. This function is called ‘internal energy function .’ It is a point function. Eq. (i) can be used to measure the change in the internal energy of the system. If some arbitrary value is assumed for the internal energy in some standard reference state, its value in any other state can be computed from the above equation. In practice, however, only the change in internal energy is important. The internal energy of a system in a given state is a function of that state only and does not at all depend upon the way by which that state has been acquired. Hence, we cannot tell whether the internal energy of the system has been acquired by heat-transfer or by performance of mechanical work or by both. In other words, it is impossible to divide the internal energy into a thermal part and a mechanical part.
First Law of Thermodynamics The first law of thermodynamics is simply the principle of conservation of energy applied to a thermodynamic system. In one form it is stated as follows :
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“Whenever other forms of energy are converted into heat or vice-versa, there is a fixed ratio between the quantities of energy and of heat thus converted “. Mathematically, the change in the internal energy of a system in passing from an initial state i to a final state f is Uf-Ui = Δ U = Q-W, where Q is the amount of heat absorbed by the system and W, the external work done by the system. This equation is taken as mathematical form of the first law of thermodynamics. If the system undergoes only an infinitesimal change in state absorbing an infinitesimal amount of heat dQ and doing an infinitesimal amount of work dW, then the infinitesimal change in internal energy dU would be given by dU = dQ-dW or
dQ = dU + dW.
This is the differential form of the first law of thermodynamics. (Q, U and W must all be expressed in the same unit). Significance : The mathematical formulation of the first law contains three related ideas : (i) heat is a form of energy in transit, (ii) energy is conserved in thermodynamic system, (iii) every thermodynamic system in a equilibrium state possesses internal energy, which is a function of the state only. Thus, the first law establishes an exact relation between heat and other forms of energy. For example, it tells us that a certain quantity of heat will produce definite amount of work, and vice-versa. It follows directly from this law that it is impossible to derive any work without expenditure of an equivalent amount of energy in some other form.
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Cycle of Processes When a closed system is taken from an initial to a final state by one or more processes, and then back to the initial state by some other one or more processes, the net change in internal energy of the system is zero ( Δ U = 0). This is because the internal energy of the system depends only on the state. Hence, by the first law of thermodynamics, Δ U = Q - W , we have 0 = Q-W or
Q = W.
In the differential form we may write
∫ dQ
=
∫ dW .
Thus, for a closed system undergoing a cycle of processes, the cyclic integral of heat is equal to the cyclic integral of work. This is also taken as a statement of the first law of thermodynamics. Isobaric Process : A process taking place at constant pressure is called an ‘isobaric’ process. For example, the boiling of water to steam or the freezing of water to ice taking place at a constant pressure (and also at a constant temperature) are isobaric processes. Boiling of Water : Let us consider the boiling of water at 100 °C and atmospheric pressure. Let m be the mass of water, Vl its volume in the liquid state , Vv in the vapour state and L the latent heat of vaporisation. The heat absorbed by the mass m during the change of state is Q = mL. The work done by the mass in expanding from volume Vl to volume Vv against the constant external pressure p is Vv
Vv
Vl
Vl
W = ∫ p dV = p ∫ dV = p (Vv − Vl ) .
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If Δ U be the change in the internal energy, then from the first law of thermodynamics, we have Δ U = Q-W
or
Δ U = mL-p (Vv-Vl).
This is the expression for the change in the internal energy of the system. We know that at 1 atmospheric pressure, 1.000 gm of water having a volume 1.000 cm3 becomes 1671 cm3 of steam when boiled. The heat of vaporisation of water at 1 atmosphere is 539 cal/gm. Thus, the heat spent is Q = mL = 1.000 × 539 = 539 cal. The external work done against the atmosphere is W = p(Vv - Vl) = (1.013 × 106) × (1671 - 1.000) = 169.2 × l07 erg. We know that 1 calorie = 4.18 × 107 erg. W =
169.2 × 10 7 = 40.5 cal. 4.18 × 107
Substituting these values in the above expression, we get Δ U = 539 cal - 40.5 cal = 498.5 cal,
which is positive. Thus, the internal energy of the system increases by 498.5 cal during the process. Of the 539 cal of heat needed to boil 1 gm of water, 40.5 cal are used in doing external work and 498.5 cal are added to the internal energy of the system. This energy is in fact the internal work done in overcoming the strong attractions between the H2O molecules in the liquid state. Freezing of Water : When a given quantity of water freezes into ice at 0 °C under atmospheric pressure, it gives out a
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definite quantity of (latent) heat. At the same time, it does work against the atmospheric pressure. We know that ice occupies a greater volume than an equal mass of water (i.e. water expands on freezing). At 1 atmospheric pressure, 1.000 gm of water, having a volume of 1.000 cm becomes 1.091 cm3 of ice on freezing. The heat of fusion of water at 1 atmosphere is 80.0 cal/gm. Thus, the heat given out is Q = - mL = - 1 x 80 = - 80 cal and the external work done is W = p(Vice-Vwater) = (1.013xl06) × (1.091-1.000) = 0.0922 × 106 erg =
0.0922×10 6 = + 0.0022 cal. 4.18 ×l07
∴ by the first law of thermodynamics, the change in internal energy is
ΔU = Q – W
= - 80-0.0022 = - 80.0022cal, which is negative. Thus, the internal energy of the system decreases by an amount greater than the heat extracted from it. Isochoric Process : A process taking place at constant volume ( Δ V = 0) is called an ‘isochoric’ process. In such a process, the work done on or by the system is zero (W = 0). Hence, by the first law of thermodynamics, Δ U=Q- W, we have Δ U = Q.
Thus, in an isochoric process, the heat added to (or taken from) the system becomes entirely the increase (or decrease) in the internal energy of the system.
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Adiabatic Process : When a system passes from an initial state i to a final state f through a process such that no heat flows into or out of the system, then the process is called an ‘adiabatic process’. Such a process can occur when the system is perfectly insulated from the surroundings, or when the process is very rapid so that there is little time for the heat to flow into or out of the system. Let a system pass from an initial state i to a final state f by absorbing heat Q from outside and by doing external work W . If Ui and Uf be the internal energies of the system in the initial and the final states respectively then, from the first law of thermodynamics, the change in energy is given by Δ U = Uf-Ui = Q-W.
...(i)
If the process be adiabatic then Q = 0 (because no heat enters or leaves the system in adiabatic process). Then the change in internal energy is Δ U = -W,
...(ii)
which is negative. Thus, if an amount of work is done by the system in an adiabatic process (i.e. W is positive), the internal energy of the system decreases by that amount. That is, the system does adiabatic work at the cost of its own internal energy. If, on the other hand, the work is done on the system in the adiabatic process, the energy of the system increases by that amount. Let us take a few examples. When a gas is suddenly compressed (adiabatic process), the work done on the gas is added to its internal energy so that its temperature rises. This is why the bicycle pump becomes hot when the air in it is compressed into the cycle tube. Similarly, when a gas is suddenly expanded, the work done by it against the surroundings is drawn from its internal energy which therefore decreases and the temperature of the gas falls. When a motor-car tyre bursts,
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the sudden expansion of its air into the atmosphere is adiabatic and the tyre is cooled. The eq. (ii) shows that during an adiabatic process the work done is exactly equal to the change in the internal energy of the system. But we know that the change in the internal energy of a system depends only on the initial and the final states of the system, and not at all on the path connecting them. Hence, in the adiabatic process, the work done will also depend only on the initial and the final states i.e. it will be same for all adiabatic paths connecting the two states. Free Expansion : If a system, say a gas, expands in such a way that no heat enters or leaves the system (adiabatic process) and also no work is done by or on the system, then the expansion is called the ‘free expansion’. Let us consider an asbestos-covered vessel with rigid walls and divided into two parts, one containing a gas and the other evacuated. When the partition is suddenly broken, the gas rushes into the vacuum and expands freely. Uf and Uf be the initial and final internal energies of the gas, we have, by the first law of thermodynamics Uf - Ui = Q - W. Since the vessel is heat-insulated and the process is sudden, no heat enters or saves the system i.e. Q = 0. Again, since the walls of the vessel are rigid and the gas expands into the vacuum, no external work is done i.e. W=0. Hence, from the last expression, we have Uf - Ui = 0 or
Ui = Uf.
Thus, the initial and the final internal energies are equal in free expansion.
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81 PROBLEMS
1. A mass of 5 kg falling vertically through a height of 50 meter rotates a paddle wheel which in turn churns 0.5 kg of water initially at 15°C. What is the final temperature of water ? sp. heat of water is 1.0 k- cal/(kg- °C) and g = 9.8N/kg. Solution : The falling mass loses potential energy which is used in rotating the wheel and is finally converted into heat. Now, the work done in rotating the wheel is W= mgh = 5 × 9.8×50 = 2.45×103 joule. The heat equivalent of this work is Q=
W 2.45 × 10 3 = cal J 4.18
( J=4.18 joule/cal )
= 586 cal = 0.586 k-cal. This heat is taken by the water. Let Δ T be the rise in temperature. Then Q = mass of water × specific heat × rise in temperature or ∴
0.586 = 0.5 × 1.0 × Δ T ΔT =
0.586 = 1.17°C. 0.5 × 1.0
2. The height of the Niagara falls is 50 meter. Calculate the difference between the temperature of water at the top and at the bottom of the fall, if J = 4.2 Joule/calorie Solution. Suppose m kg of water fall in one second. The potential energy lost in one second is W = mgh = m × 9.8 × 50 joule. This lost energy is converted into heat. If Q be the heat produced, then
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82 Q=
W (m × 9.8 × 50) joule = J 4.2 joule/calorie
= 117 m cal = 0.117 m k-cal. If this heat causes a temp-rise Δ T in water, then Q = mass × sp. heat of water x temp-rise or
0.117 m k-cal = m kg× 1.0 k-cal/(kg-°C) x Δ T ΔT =
0.117 m k-cal m kg×l.0k-cal/(kg-°C)
= 0.117°C 3. A 2.0-g bullet moving with a velocity of 200 meter/sec is brought to a sudden stoppage by an obstacle. The total heat produced goes to the bullet. Calculate the rise in temperature of the bullet. Sp. heat for the bullet may be taken d0.03 cal/ (g-°C). (J = 4.2 × 107 erg/cal) Solution. The kinetic energy in the moving bullet is W =
1 1 mv2 = ×2.0g × (200,00cm)2 2 2
= 4.0× 108 erg. By sudden stoppage of the bullet, the whole of the energy is converted into heat. The heat equivalent of this energy is Q=
W 4.0 × 108 = =9.52 cal. J 4.2 ×107
Δ T be the rise in temperature of the bullet due to this heat, then
Q = m x c × ΔT or ∴
9.52 = 2.0 × 0.03 × Δ T ΔT =
9.52 = 159°C 2.0 × 0.03
(c is sp. heat)
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4. A lead bullet at 100 °C strikes a steel plate and melts. What was its minimum speed ? Sp. heat of lead = 0.03 cal/ (g-°C), latent heat = 5 cal/g and melting point = 327 °C. The heat produced is shared equally between the plate and the bullet. (J= 4.2 × 107 erg/cal) Solution. Let m gm be the mass of the bullet and v cm/sec the velocity with which it strikes the plate. Its kinetic energy 1 mv2 erg. is 2 By sudden stoppage of the bullet, the whole of the energy is converted into heat. According to relation W = JQ , the heat generated is 1 mv 2 W 2 Q= = calories. J J
This heat is shared equally between the plate and the bullet. Hence the heat shared by the bullet is 1 mv 2 calories. 4 J
This heat is used up in raising the temperature of the bullet to its melting point and then in melting the bullet. Now, the heat required to raise the temperature of the bullet, from 100 °C to 327 °C (melting point) is mc Δ T = m × 0.03 x (327 -100) cal and the heat required to melt it is mL = m × 5 cal. Therefore, the total heat required to melt the bullet is Q = m × 0.03 × (327 – 100) + m × 5 = m (0.03 × 227 + 5) cal = 11.81 m cal. Clearly, this must be equal to the heat shared by the bullet. Thus, 1 mv2 11.81 m = 4 J
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84 or
v2 = 4J× 11.81 = 4×(4.2× 107)× 11.81 = 19.84×108
∴
v = √(19.84× 108) = 4.4 × 104 cm/sec.
5. A piece of ice of mass 50 kg is pushed with a velocity of 5 meter per sec along a horizontal surface. As a result of friction between the piece and the surface, the piece stops after travelling 25 meter. How much ice has been melted? (J = 4.2 joule/cal, latent heat of ice is 80 cal/g.) Solution : The initial kinetic energy in the piece is W=
1 1 mv2 = × 50 kg × (5 meter/sec)2 = 625 joule. 2 2
The energy becomes zero when the piece stops. This lost energy has been converted into heat, which is Q=
W 625 = =149 cal J 4.2
Let m gm of ice melt due to this heat. Then Q = mL or ∴
149 = m x 80 m=
149 = 1.86 g. 80
Here we have assumed that all the heat produced goes to melt the ice and that the ice melts after it is brought to rest. 6. An athlete consumes 4000 kilo-calories per day through his diet. Compute his power in watt. Solution : W = JQ, where J = 4.18 joule/cal. The work equivalent to 4000 kilo-calories (= 4000 × 1000 cal) per day is W = 4.18 × (4000 × 1000) joule per day
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85
4.18×(4000×l000) = 193.5 joule per sec. 24 x 60 x 60
Now, the power is the rate of doing work, and 1 watt = 1 joule per sec. ∴ power = 193.5 watt.
7. A metallic ball falls from a height of 10 meter on the ground and rebounds to a height of 0.50 meter. Has its internal energy changed ? Compute the rise in temperature of the ball. Take the specific heat of the ball as 0.12 cal-g-1-°C-1, g = 980 dyne/gram and J = 4.2 x 107 erg/cal. Solution. Let m gram be the mass of the ball. The energy released when it falls by 10 meter (=1000 cm) is mgh = m × 980 × 1000 erg. A part of this is used to rebound the ball to a height of 0.50 meter (=50 cm) and the rest is added to the internal energy of the ball. Energy used to raise the ball = m × 980 × 50 erg. ∴ energy added to the ball
= (m × 980 × 1000) - (m × 980 × 50) = m × 980 x 950 erg =
m × 980 x 950 cal. 4.2 ×l07
Let it be Q . Then if Δ T be the temperature-rise due to this, we have Q = m × c × ΔT or ∴
m × 980 x 950 = m×0.12×T 4.2 ×l07
Δ T=
980×950 = 0.18°C 4.2 ×l07 × 0.12
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8. 0.004 kg air is heated at constant volume from 2°C to 6°C. Find the change in its internal energy. The sp. heat of air at constant volume is 0.172 cal/(g-°C) and J = 4.18 joule/cal. Solution : By the first law of thermodynamics, the change in the internal energy of a system is given by Δ U = Q-W,
where Q is the heat absorbed by the system and W is the external work done by the system. Here the air is heated at ‘constant’ volume. Hence W = 0. Δ U = Q.
∴
The heat absorbed by the air is Q = mass x sp. heat × temp-rise = 0.004 kg ×0.172 kcal/(kg-°C) x 4°C = 2.752 x 10-3 kcal = 2.752 cal ∴ Δ U = Q= 2.752 cal
= 2.752 × 4.18 = 11.5 J. 9. 50 cal per gram heat is added to a system and the process takes place at constant volume. What will be the change in the specific internal energy in joule ? J = 4.18 joule/cal. Solution : At constant volume, the change in internal energy equals heat added: Δ U = Q.
Here Q = 50 cal/g. Therefore, change in internal energy per gram (i.e. specific internal energy) is Δ U = 50 cal/g
= 50 × 4.18 = 209 joule/gram.
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10. A system absorbs 1000 cal of heat and does 1675 joule of external work. The internal energy of the system increases by 2505 joule. Compute the value of J . Solution : By the first law of thermodynamics, we have Δ U = Q-W,
where Q is the heat absorbed and W the external work done by the system. Here Δ U = + 2505 joule, Q = + 1000 cal = + 1000 J joule, and W = + 1675 joule. Thus 2505 = 1000 J-1675 . ∴
J=
2505 + 1675 4180 = = 4.18 joule/cal 1000 1000
11. At atmospheric pressure 1.0 g of water, having volume 10 cm3, becomes 1671 cm3 of steam when boiled. Calculate the external work and the change in the internal energy. The heat of vaporisation of water is 539 cal/g. (J = 4.18 × 107 erg/cal and 1 atmospheric pressure = 1.013 × 106 dyne/cm2) Solution : The external work against the atmospheric pressure p is W = p dV = (1.013 × 106 dyne/cm2) (1671 cm3 - 1.0 cm3) = 169.2 × 107 erg. The heat taken by the water to become steam is Q = mL = 1.0 × 539 = 539 cal. In work unit, it is Q = 539 × (4.18 × 107) = 2253 × 107 erg. By the first law of thermodynamics, the change in internal energy is
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88 Δ U = Q-W
= 2253 ×l07-169.2 ×l07 = 2083.8 ×l07 erg = 2083.8 joule. 12. When water is boiled under 2 atmosphere pressure, the heat of vaporisation is 2.20 × 106 joule/kg and the boiling point is 120°C. At this pressure, 1 kg of water has a volume of 103 m3 and 1 kg of steam a volume of 0.824 m3 . Compute the work done and the increase in internal energy when 1 kg of water is converted into steam at 120°C (1 atmosphere = 1.013 × 105 N/m2). Solution : The heat taken in boiling 1 kg of water to steam at 120°C is Q = mL = 1 kg × (2.20 × 106 J/kg) = 2.20 × 106 joule. During this process the volume increases from 0.001 m3 to 0.824m3 at 2 atmosphere pressure. The external work done against the atmosphere is W = pdV = (2 × 1.013 × 105 nt/m2) × (0.823 m3) = 0.167 × 106 joule. By the first law of thermodynamics, the increase in internal energy is Δ U = Q -W = 2.20 × 106- 0.167 × 106
= 2.033×106 joule. 13. When a system is taken from state i to state f along the path iaf, it is found that the heat Q absorbed by the system is 50 cal and the work W done by the system is equal to 20 cal. Along the path ibf ; Q = 36 cal.
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89
(i) What is W along the path ibf ? (ii) If W = - 13 cal for the curved return path fi, what is Q for this path ? (iii) Take Ui = 10 cal. What is Uf ? (iv) If Ub = 22 cal, what are Q for the processes bf and ib ?
Solution : According to the first law of thermodynamics, the change in internal energy of a system in going from the state i to f is Δ U = Uf-Ui = Q-W,
...(i)
where Q is heat absorbed by the system. For the path iaf; we have Q = + 50 cal, W = + 20cal. ∴
Δ U = Q- W = 50 - 20 = 30 cal.
Since internal energy U is a state function, Δ U between i and f is always 30 cal, whatever path be adopted,
(i) For the path ibf; Q = 36 cal (given) and Δ U = 30 cal. ∴ W = Q - Δ U=36-30 = 6cal.
(ii) For the return curved path fi ; W = - 13 cal. (given) and Δ U= - 30 cal. ∴
Q = Δ U+W= - 30 - 13 = -43 cal.
(iii) Ui = 10 cal (given). Now, Δ U = Uf-Ui = 30 cal. ∴
Uf = 30 + Ui = 30+10 = 40 cal.
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(iv) For the process bf, the volume remains constant so that work done is zero. Hence from eq. (i) Q = Δ U=Uf-Ub. Now, Uf= 40 cal (obtained above) and Ub = 22 cal (given). ∴
Q = 40 - 22 = 18 cal.
For the path ibf , Q = 36 cal (given), while for the path bf, Q = 18 cal (just determined). Hence for the path ib, we have Q = 36 -18 = 18 cal. 14.When a system is taken from state a to state b along the path acb, 80 joules of heat flow into the system, and the system does 30 joules of work,
(i) How much heat flows into the system along path adb , if the work done is 10 joule ? (ii) When the system is returned from b to a along the curved path, the work done on the system is 20 joule. Does the system absorb or liberate heat and how much ? (iii) If Ua = 0 and Ud = 40 joule, find the heat absorbed in the processes ad and db . Solution : According to the first law of thermodynamics, the change in internal energy of a system in going from the state a to the state b is Δ U = Ub-Ua = Q-W,
...(i)
where Q is heat absorbed by the system and W is work done by the system.
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91
For the path acb , Q = 80 joule and W= 30 joule. ∴
Δ U = Ub - Ua = 80 - 30 = 50 joule.
(i) For the path adb , W= 10 joule and Δ U=Uh-Ua = 50 joule. Therefore, from eq. (i), we get Q = Δ U + W = 50 + 10 = 60 joule. (ii) For the curved path ba , W = - 20 joule and Δ U = Ua - Ub = - 50 joule. Therefore, from eq. (i), we get Q = Δ U+W = -50-20 = - 70 joule. Minus sign means that heat is liberated from the system. (iii) For the path ad, the work would be same as for adb (because no work is done along db ) i.e. W= 10 joule and Δ U = Ud - Ua = 40 joule (given). Thus, from eq. (i), we get Q = Δ U + W= 40 + 10 = 50 joule. For the path db , W=0 (volume is constant) and Δ U=UbUd=(Ub-Ua) - (Ud - Ua) = 50 - 40 = 10 joule. Thus, from eq. (i), we get Q = AU+W= 10 joule. 15. A system is taken from a state i to a state f along the path iaf, where path ia is isobaric and path af is isochoric. The heat absorbed by the system is 36 cal and the work done by the system is 6 cal. (i) If Ui = 10 cal and Ua = 22 cal, find the heat absorbed along ia and af. (ii) Will the heat rejected for any return path fi be just 36 cal ? If not, on what factor will it depend ? Ans. (i) 18 cal, 18 cal; (ii) no ; it will depend on the actual path fi which has been adopted.
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16. A gas-filled cylinder fitted with a piston is immersed in ice (0 °C). The piston is rapidly pushed down to compress the gas which is therefore heated. It is then left for some time so that the gas is once again at 0 °C. The piston is now slowly raised up to the initial position. If 100 g of ice is melted during this cyclic process, compute the work done on the gas. Represent the cycle on a p-V diagram.
Solution : Let A be the Initial state of the gas characterised by thermodynamic coordinates (p1, V1). It is first compressed adiabatically to B whose coordinates are (p2 V2). It is then left to reach the state C when its volume remains constant at V2, the temperature falls to 0 °C, and pressure falls to p3(>p1)Finally, it is expanded isothermally to reach the initial state A, its pressure falling to p1 and volume increasing to V1. Let us write the first law of thermodynamics for the entire process ; Δ U = Q – W.
Since the gas returns to the initial state, its internal energy is restored to the initial value i.e. Δ U = 0. ∴
W = Q.
The heat Q is used to melt 100 gm of ice. ∴
Q = mL = 100 × 80 = 8000 cal.
∴ work done on the gas during this cycle
W = Q = 8000 cal.
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17. A given system is subjected to three different processes as shown in the table below. Supply the missing information. All the data are in joule. Calculate the unknown values with the help of the first law of thermodynamics. Which of these processes is adiabatic? Process
Q
W
Ut
Uf
Δ U = Uf – Ui
1
35
-15
…
-10
…
2
-15
…
…
60
-20
3
…
-20
80
…
20
Ans. (1) Ui = -60, Δ U =50,(2) W = 5, Ui= 80, (3) Q = 0,Uf = 100. Process 3 is adiabatic because for it Q = 0.
7 The Reversibility Quasi-static Process : A process during which a system is never more than infinitesimally far away from an equilibrium state is a ‘ quasi-static ‘process. For example, a quasi-static expansion of a gas may be obtained by an infinitesimal change in pressure, or a quasi-static flow of heat may be obtained by an in-finitesimal change in temperature. In fact, quasi-static process is an ideal process which can never be exactly obtained, but can only be approximated under some circumstances.
Reversible Process A reversible process is one which can be reversed in such a way that all changes taking place in the direct process are exactly repeated in the inverse order and opposite sense, and no changes are left in any of the bodies taking part in the process or in the surroundings . For example, if an amount of heat is supplied to a system and an amount of work is obtained from it in the direct process; the same amount of heat should be obtainable by doing the same amount of work on the system in the reverse process.
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Conditions of Reversibility : A process can be reversible only when it satisfies two conditions : (i) Dissipative forces such as friction , viscosity, inelasticity, electrical resistance, magnetic hysteresis , etc. must be completely absent: Suppose a gas is contained in a cylinder fitted with a piston and placed in contact with a constant-temperature source. The piston is loaded so that the pressure exerted by the piston on the gas exactly balances the pressure of the gas on the piston. If the load on the piston is now decreased, the gas will expand, doing external work in pushing up the piston and also in overcoming the friction between the piston and the walls of the cylinder. The heat necessary for this work is taken from the source. If now the load on the piston is increased, the gas will be compressed. The work used in pushing up the piston during the expansion is now recovered. On the contrary, more work has to be done against the friction. The expansion is therefore irreversible. Similarly, other dissipative effects like inelasticity, electrical resistance , etc. make the process irreversible. (ii) The process must be quasi-static : When the gas expands, an amount of work is done by the gas to give kinetic energy to the piston. This work cannot be recovered during the reverse process, but on the contrary, more work is to be done to give kinetic energy to the piston. Hence in order to make the expansion of the gas reversible; the pressure of the gas on the piston should be only infinitesimally different from the pressure exerted by the piston on the gas. Under this condition the expansion or compression will take place infinitely slowly so that no kinetic energy will be produced. These conditions are never realised in practice. Hence, a reversible process is only an ideal conception.
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97
Irreversible Process Any process which is not reversible exactly is an irreversible process. All practical processes such as free expansion, JouleThomson expansion, electrical heating of wire, diffusion of liquids or gases, etc. are irreversible. All natural processes such as conduction, radiation, radioactive decay, etc. are also irreversible. Thus , irreversibility is a rule. Isothermal Expansion of a Gas : Let us imagine a gas contained in a cylinder having perfectly insulating walls but a perfectly conducting base, and fitted with a frictionless piston. The cylinder is placed on a heat-reservoir maintained at a constant temperature, which is the same as the temperature of the gas. The piston is loaded, so that the pressure exerted by the piston on the gas exactly balances the pressure of the gas on the piston. Suppose the load on the piston is decreased by an infinitesimally small amount. The gas will expand, doing external work in pushing up the piston and its temperature will tend to fall. It will thus very slightly deviate from equilibrium, but an amount of heat equivalent to the work done will immediately flow from the heat-reservoir to the gas which will again be at the temperature of the reservoir and attain equilibrium. Thus, the infinitely-slow isothermal expansion of a gas in the absence of any friction is an example of a reversible process. The conditions described above are, however, ideal. In practice, a very slow isothermal expansion is “approximately” reversible. Adiabatic Compression of a Gas : Again, if the cylinder containing the gas is a perfect insulator (including its base) and the gas is compressed infinitesimally slowly, the compression is reversible. In practice, it can be made only approximately reversible. Diffusion of Gases : When two or more gases diffuse into one another, there is a change in chemical composition and the process is irreversible.
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Transfer of Heat from Hot to Cold Body : This process is irreversible, because heat cannot be transferred back from the cold to the hot body without leaving any change elsewhere. Joule Expansion (or Free Expansion) of a Perfect Gas: In this process, the (perfect) gas changes from a volume Vi to a larger volume Vf without any change in temperature. To revert the gas to its initial state, it would have to be compressed to Vi by some external device. The external work so done would be converted into heat. To ensure that the gas retains its initial temperature and no changes are left in the surroundings, the heat produced would have to be extracted from the gas and converted completely into work. Since this last step is impossible, the process is irreversible. Joule-Thomson Effect : The Joule-Thomson expansion of a gas is an irreversible process. The reason is same as for free expansion. (Any heat cannot be completely converted into work.) Transfer of Heat by Radiation : Heat coming from a hot body by radiation cannot be radiated back to the hot body without leaving any change elsewhere. Hence the process is irreversible. Electrical Heating of a Wire : The electrical energy dissipated as heat in the wire cannot be fully converted into electrical energy and so the process is irreversible. Very Slow Extension or Contraction of Spring : In this process, if carried extremely slowly, the spring passes through states of thermodynamic equilibrium, which may be traversed just as well in one direction as in the opposite direction. The process is therefore approximately reversible. Dissipation of Mechanical Energy to Heat through Friction: Suppose a body moves on a surface from an initial position, spends its mechanical energy to overcome friction between itself and the surface, and again returns to its initial position. The energy spent is dissipated as heat. Now, if the body be
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99
allowed to go round its path in the reverse direction, its energy spent previously cannot be recovered. On the contrary, the body will have to further spend its energy against the friction in the reverse path. Hence the process is irreversible. Complete Conversion of Work into Heat : We can convert a given quantity of work ‘completely’ into heat. For example, when we rub two stones together under water, the work done against friction is converted into heat which is communicated to the surrounding water. Since the state of the stones is the same at the end of the process as at the beginning, the net result of the process is merely the conversion of mechanical work into heat with 100 per cent efficiency (W = Q). This conversion can be continued indefinitely. The reverse process is, however, not possible. We cannot make a device by which a given amount of heat can be ‘completely’ converted into work. At first thought, the isothermal expansion of an ideal gas can be considered as a process in which heat is converted completely into work . In this case D U = 0 (since the temperature remains constant) and so the heat absorbed by the gas is equal to the work done by the gas during the expansion (Q = W). But here the state of the gas changes . Its volume increases and pressure decreases until atmospheric pressure is reached at which the process stops. Thus the conversion of heat into work cannot be continued indefinitely. Further, any device which converts heat completely into work would lead to a decrease in entropy of the universe, which is impossible. Heat Engine : Any “cyclic” device by which heat is converted into mechanical work is called a heat engine. There are three main parts in an engine: a hot body called ‘source’, a working substance, and a cold body called ‘sink’. The working substance takes in heat from the source, converts
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100
a part of it into useful work, and gives out the rest to the sink. This series of processes is called a ‘cycle ‘ since the working substance returns to its original state. This is shown in Fig. By repeating the same cycle over and over again, work can be continuously obtained.
Suppose the working substance takes in an amount of heat Q1 from the source, and gives out an amount Q2 to the sink. Suppose W is the amount of work obtained. The net amount of heat absorbed by the substance is Q1 – Q2, which has been actually converted into work. Applying the first law of thermodynamics to one complete cycle , we get Q1 –Q2 = w. Thermal Efficiency: The ‘thermal efficiency‘ e of an engine is defined as the ratio of the work obtained to the heat taken in from the source , that is , e=
W Q1 –Q2 = Q1 Q1
e = 1-
Q2 Q1
This equation indicates that the efficiency of the heat engine will be unity (100%) when Q2 = 0 (no heat is given out to the
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101
sink). This is, however, not possible in practice. This means that the engine cannot convert all the heat taken in from the source into work. We cannot define the efficiency as W/Q2 , because in that case we shall have e=
Q W Q1 – Q2 = = 1 = –1 , Q2 Q2 Q2
so that the condition for the ideal value of efficiency (i.e. for e = 1) would be Q1= 2Q2 which is absurd. Indicator Diagram : If the pressure of a system undergoing a process is plotted against its volume, the resulting curve is called the ‘indicator diagram’ of the system. This diagram enables us to represent the working of an engine and to calculate the useful work obtained from the engine. Let a given mass of a working substance, say a gas, be contained in a cylinder closed by a frictionless piston which is loaded so that the pressure exerted by the piston on the gas exactly balances the pressure of the gas on the piston when the cylinder is placed in contact with a heat source. Let p and V be the pressure and volume of the gas and A the area of cross-section of the piston. Suppose the load on the piston is decreased by an infinitely” small amount. The gas will expand very slowly, pushing up the piston through an infinitely small distance ds and reach a new equilibrium state. The external work done by the gas will be dW = force exerted on the piston × displacement = pressure × area of piston face x displacement = p × A × ds . The pressure p has been assumed constant during the small change. But A x ds = dV is the increase in volume.
Thermal Physics
102 \
dW = pdV.
This work is drawn from the hot source. This process may be carried further by decreasing the load on the piston in infinitely small steps. Clearly, the external work done for an expansion from an initial state (p1, V1) to a final state (p2 , V2) will be V2
W=
ò p dV V1
Let us choose the coordinate axes the pressure and volume axes. Let the points A and B (Fig.) represent the initial state (p1, V1) and the final state (p2 , V2) of the gas, and the curve AB represent the expansion of the gas.
Let us consider a point M . Let p be the pressure and V the volume at this point. Let (V+dV) be the volume at an infinitely close point N . Then MN represents expansion of the gas through a volume dV. The strip MNRS will be a rectangle of area p dV . But this is external work done by the gas which is thus represented by the area of the strip. The whole area ABCD between the curve AB and the volume-axis is clearly V2
ò p dV which represents the external work done by the gas in V1
expanding from V1 to V2 . Thus , the area bounded between
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103
the p- V curve and the volume axis gives the work done directly . This result is very useful in the study of heat engines. Reversible Engine: In an engine the working substance goes through a cycle of processes. It takes in heat from a hot body, converts a part of it into work and gives out the rest to a cold body, returning to its initial state. During this cycle the conditions of the hot and cold bodies and of the surroundings change. If this cycle can be traversed in the reverse order such that all the parts of the engine completely recover their original conditions and no changes are left in the surroundings, the cycle is a “reversible cycle”, and the engine is a “reversible engine”. Such an engine can be realised if (i) the working parts of the engine are free from friction, (ii) the pressure and temperature of the working substance never differ appreciably from its surroundings at any stage of the cycle, so that all the processes involved in the cycle are quasi-static. These conditions can never be realised in practice. Hence a reversible engine is an ideal conception. (Carnot has presented an imaginary picture of such an engine.)
Carnot’s Heat Engine Carnot developed a plan of an idealised heat engine, free from all the imperfectness of an actual engine. As shown in Fig., the Carnot’s engine consists of four components : (i) A cylinder with perfectly heat-insulating walls but perfectly conducting base, and closed with a tight-fitting perfectly insulating and frictionless piston. A fixed mass of a gas (working substance) is filled in the cylinder, and some weights are placed on the piston of the cylinder. (ii) A hot body of infinitely large heat capacity at a constant temperature T1 , serving as ‘source’.
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Thermal Physics
(iii) A cold body of infinitely large heat capacity at a constant temperature T2 , serving as ‘sink’. (iv) A perfectly heat-insulating stand. The cylinder may be placed on any of the three bodies (ii), (iii) and (iv) and may be moved from one to the other without doing any work.
The working substance is imagined to go through a cycle of four processes, known as the “Carnot’s cycle”. Suppose it is in an initial state represented by the point A on the p-V diagram (Fig.).
Process 1 : The cylinder is put on the source and the weights on the piston are removed in infinitely small steps. The
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105
working substance thus expands infinitely slowly, doing work in raising the piston. During this process the substance takes in heat from the source by conduction through the base. The expansion is isothermal. It is continued until the state represented by the point B is reached. The curve AB represents the isothermal expansion of the substance at the constant temperature T1 (of the source). Process 2 : The cylinder is now put on the heat-insulating stand and, by further removing the weights on the piston , the substance is further expanded. This expansion is adiabatic because now no heat can leave or enter the substance through the (insulating) cylinder. The substance does work in raising the piston and its temperature falls. The expansion is continued until the temperature falls to T2 which is the temperature of the sink. It is represented by the curve BC. Process 3: The cylinder is now put on the sink. Weights are now placed on the piston in infinitely small steps. The substance is thus compressed isothermally until the state represented by the point D is reached. The curve CD represents the isothermal compression at the constant temperature T2 (of the sink). During this process work is done on the substance by the piston and a quantity of heat is given out to the sink. Process 4 : The cylinder is once more put on the heatinsulating stand and, by further placing the weights on the piston, the substance is compressed adiabatically. A further amount of work is done on the substance and its temperature rises. The process is continued until the temperature rises once more to T1 and the state A is recovered. The curve DA represents the adiabatic compression. The external works done by the substance during the processes AB and BC are represented by the areas ABB’ A’ and BCC B’ respectively. Similarly, the external works done on the substance during the processes CD and DA are represented by
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the areas CDD’ C and DAA’ D’ respectively. The difference between these two parts of areas is area (ABB’ A’ + BCC’ B’) – area (CDD’ C’ + DAA’ D’) = area ABCD. Thus, the area ABCD represents the net external work done W by the substance in one complete cycle. The net amount of heat absorbed by the substance in the cycle is Q1 — Q2 where Q1 is the heat taken in from the source during the process AB, and Q2 the heat given out to the sink during the process CD. Since the initial and final states of the substance are the same; there is no net change in its internal energy. Hence , by the first law of thermodynamics, W = Q1–Q2 for the cycle. The result of the cycle is that heat has been converted into work by the system. Any required amount of work can be obtained by simply repeating the cycle. It is not necessary to start with the cycle at the point A . Since the cycle is reversible at each step, we can choose any point in the cycle as the starting point. The net external work obtained will always be the same. Efficiency: The efficiency, e, of an engine is defined as the ratio of net work done by the engine during one cycle to the heat taken in from the source in one cycle. Thus e=
Q W Q1 –Q2 = 1– 2 Q1 Q1 Q1
...(i)
The expression is independent of the nature of the working substance. Let us now calculate the efficiency taking 1 mole of an ideal gas as the working substance in terms of T1 and T2 . Let (pa ,Va), (pb ,Vb), (pc ,Vc) and (pd ,Vd) be the coordinates of the points A, B, C and D respectively. The heat Q1 taken in by the
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107
gas will be used in increasing the internal energy of the gas in expanding isothermally along AB and in the external work done by the gas. But since the gas is ideal, the internal energy is wholly kinetic depending on the temperature only, and so it remains unchanged during the isothermal expansion AB. Therefore, the heat Q1 will be equivalent only to the external work done by the gas in expanding from A to B at temperature T1 i.e. Q1 =
ò
Vb
Va
p dV =RT1
ò
V dV V = RT1 [log eV ]Vb = RT1 log e b ...(ii) a V Va
Vb
Va
Similarly, Q2 will be equivalent to the external work done on the gas during compression from C to D at temperature T2 i.e. Q2 = – ò
Vd
Vc
p dV =RT2
ò
Vc
Vd
V dV V = RT2 [loge V ]Vcd = RT1 log e c , ...(iii) V Vd
minus sign indicating that work is done on the gas. The work done by the gas during the adiabatic expansion from B to C is exactly equal to the work done on the gas during the adiabatic compression from D to A , so the net work during adiabatic processes is zero. From (ii) and (iii), we have Vb Va Q1 T = 1 = V Q2 T2 log e c Vd log e
...(iv)
The points B and C, and similarly the points D and A lie on the same adiabatic. Therefore, we have, from Poisson’s law
and
T1 Vbg –1 = T2 Vcg –1 ,
...(v)
T1 Vag –1 = T2 Vdg –1 .
...(vi)
Dividing eq. (v) by eq. (vi), we get
Thermal Physics
108 Vb V = c. Va Vd
Using this result in equation (iv), we get Q1 T = 1. Q2 T2
Hence the efficiency, given by eq. (i), is e=1–
Q1 T =1– 2. Q2 T1
Thus , the efficiency of Carnot’s reversible engine is independent of the working substance and depends only on the absolute temperatures of the sink and the source. For the engine to have 100 per cent efficiency (e = 1), T2 must be zero. Since we cannot obtain a sink at absolute zero, an engine with 100 per cent efficiency is a practical impossibility. Efficiency of Carnot’s Engine in Terms of Adiabatic Expansion Ratio : We can obtain an alternative expression for the efficiency. From eq. (v), we have g –1
Vö T2 æ ÷ = ççç b ÷ ÷ T1 çèVc ÷ ø
Now,
= p (adiabatic expansion ratio). g –1
∴
T1 æ 1ö ÷ = ççç ÷ ÷ . T2 çè p ÷ ø
Making this substitution in the above expression, we get g –1
æ1 ö ÷ e = 1 – ççç ÷ ÷ çè p ÷ ø
.
Carnot Engine is not a Practical Possibility : Carnot engine has two ideal features :
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109
(i) The source and the sink are bodies of infinitely large heat capacities. Therefore, the working substance takes in all the heat at a constant temperature (of the source) and gives up all the heat at another constant temperature (of the sink). Thus it utilises the full temperature difference available. (ii) Each process in Carnot’s cycle is completely reversible. This is because of three reasons: (a) The working substance is contained in a cylinder with perfectly insulating walls and fitted with a perfectly insulating frictionless piston. Thus the dissipative forces are absent, (b) The base of the cylinder is perfectly conducting . So when it is placed on the hot or the cold body, the working substance differs in temperature with the body only by an infinitesimal amount, (c) The gas is expanded and compressed infinitely slowly , so the substance differs in pressure with its surroundings only by an infinitesimal amount. These features cannot be actually realised. Hence it is not possible to obtain a Carnot engine in practice. The way to Increase Efficiency of Carnot Engine : The efficiency of a Carnot engine operating between temperatures T1 and T2 is given by e=–
T2 . T1
To increase e , we must decrease T2/T1 . This can be done either by decreasing T2 or by increasing T1 . Since T2 < T1 , a decrease in T2 will be more effective than an equal increase in T1 . The working substance takes in heat from the body at temperature T1 and, after doing work, gives out the balance to the body at temperature T2 . This continues until T1 drops to Tf and T2 rises to Tf . Thus the total heat taken in from the hot body is
Thermal Physics
110 Q1 = Cp(T1 – Tf)
...(i)
and that given out to the cold body is Q2 = Cp(Tf–T2).
...(ii)
The balance Q1 - Q2 is converted into work W. Thus W = Q1– Q2 = Cp(Tl -Tf) – Cp(Tf - T2) = Cp(T1 + T2 - 2Tf). This is the required expression. The heat Q1 is taken in at a mean temperature the heat Q2 is given out at a mean temperature engine is the maximum efficient.
T1 +T f
T2 +Tf 2
2
and
. Carnot
So for maximum work, we must have (T1 +Tf ) / 2 Q1 = . (T2 +Tf ) / 2 Q2
From eq. (i) and (ii), this becomes T1 –Tf Tf –T2
=
T1 +Tf T2 +Tf
.
On simplifying: T f = T1T2 .
Two Carnot engines are operating in series. The first engine absorbs a quantity of heat Q1 at a temperature T1 and after doing work W1, rejects the remaining heat Q2 at a lower temperature T2. The second engine absorbs the heat Q2 at temperature T2 (rejected by first) and after doing work W2, rejects the remaining heat Q3 at a still lower temperature T3. Compute the efficiency of the combination.
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111
The total work by the combination in one cycle is W1 + W2 = (Q1– Q2) + (Q2– Q3) = Ql–Q3 . The heat absorbed by the combination is Q1. Therefore, the efficiency of the combination is e=
=
Q Q1 – Q3 =1– 3. Q1 Q1
=
Q Q Q1 – Q3 Q – Q3 =1– 3.= 1 =1– 3. Q1 Q1 Q1 Q1
But ∴
total work done heat absorbed
Q3 T3 = . Q1 T1 e = 1−
T3 . T1
This is the same as the efficiency of a single engine operating between T1 and T3.
Carnot’s Ideal Refrigerator Any device for removing heat from a cold place and adding it to a hotter place is called a “refrigerator”. It is essentially a heat engine running backwards. In a heat engine the working substance takes in heat from a body at a higher temperature, converts a part of it into mechanical work, and gives out the rest to a body at a lower temperature. In refrigerator, a working substance takes in heat from a body at a lower temperature, has a net amount of work done on it by an external agent, and gives out a larger amount of heat to a hot body (Fig.). Thus it continually transfers heat from a cold to a hot body at the expense of mechanical energy supplied to it by an external agent. The working substance is called a “refrigerant”.
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112
The engine employing the Carnot cycle may be adopted as a refrigerator. Each step in the cycle is reversible, therefore it is possible to reverse the entire cycle. Let Q2 be the heat removed from the cold body at temperature T2 , W the net work done on the refrigerant, and Q1 the heat delivered to the hot body at temperature T1 . Then, we have Q1 = Q2+ W or
W = Q1– Q2 æQ1 ö ÷ ç . ÷ = Q2 ççQ – 1÷ ÷ è 2 ø
As in Carnot’s engine, if we use an ideal gas as a working substance, we can show that Q1 T1 = . Q2 T2 \
æT ö T – T2 ÷ ÷= Q2 1 W= Q2 çç 1 –1÷ . ÷ çèT2 T2 ø
This is the expression for the work that must be supplied to run the refrigerator.
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113
Coefficient Performance This purpose of the refrigerator is to remove as much heat (Q2) as possible from the cold body with the expenditure of as little work (W) as possible. Therefore, a measure of the performance of the refrigerator is expressed by the “coefficient of performance” K which is defined as the ratio of the heat taken in from the cold body to the work needed to run the refrigerator. That is Q Q2 1 K= 2= . = Q W Q1 – Q2 1 –1 Q2 Q1 T1 = But Q2 T2 ∴
K=
1 T1 –1 T2
=
T2 . T1 – T2
This is the expression for the coefficient of performance. A good refrigerator should have a high coefficient of performance, typically 5 or 6. Thicker and high-quality insulation tends to increase the coefficient of performance. Relation between the Efficiency e of a Carnot Engine and the Coefficient of Performance K of a Carnot Refrigerator : Let a Carnot engine and a Carnot refrigerator work between the same temperatures T1 and T2 . Then, we have e=
and NOW ∴
T1 – T2 T1
K=
K+ 1=
e=
This is the required relation.
T2 T1 – T2
T2 T1 1 + 1= = . T1 – T2 T1 – T2 e
1 K+1
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Unattainability of Absolute Zero : A refrigerator is a device for producing cold. An analysis of Carnot’s ideal refrigerator shows that its performance coefficient is K=
T2 . T1 –T2
Obviously, K becomes vanishingly small as the temperature T2 of the reservoir from which heat is taken approaches absolute zero. This means that as the reservoir becomes more and more cold, the refrigerator finds it more and more difficult to run. Therefore, it is impossible to take the reservoir down to T2 = 0 . Heat Pump : A heat pump transfers heat from some cold place to a hotter place at the expense of energy supplied externally by a motor. Thus it does not violate any law of thermodynamics. Opening the door of Refrigerator : A room cannot be cooled by having opened the door of the refrigerator placed in it. On the contrary, the room will warm up. The reason is as follows : The refrigerator removes heat from its interior and expels it into the surrounding air, thus warming the air. For doing this, additional energy is supplied to the refrigerator by an electric motor. The heat expelled into the air is the sum of the energy from the motor and that removed from the interior of the refrigerator. In other words, the refrigerator adds more heat into the room than it removes from its interior. On opening its door it will run continuously and hence add even more heat to the room than when its door is closed. Purpose of the Second Law of Thermodynamics : Limitations of the First Law: The first law of thermodynamics states the equivalence of mechanical work and heat, when one is completely converted into other (W = Q). Thus , it is the principle of conservation of energy applied to a thermodynamic system.
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115
If, however, we propose to extract a certain quantity of heat from a body and convert it completely into work, the first law would not be violated. But, in actual practice this is found to be impossible. If this were possible, we could drive ships across an ocean by extracting heat from the water of the ocean. Thus , the first law simply tells that if a process takes place, energy will remain conserved. It does not tell us whether the process is possible or not. Similarly, if a hot body and a cold body are brought in contact, the first law is not violated whether the heat flows from the hot to the cold body or vice versa. By experience we know that heat never flows from cold to hot body. The purpose of the second law is to incorporate such experimental facts into thermodynamics. There are two equivalent statements of the second law of thermodynamics.
Kelvin-Planck Statement In a heat engine, a working substance takes in heat from a hot body, converts a part of it into mechanical work, and gives out the rest to a cold body. No engine has ever been designed which can operate in a cycle by taking heat from a body and converting all of it into work ; some heat must always be given to a colder body. This experience led Kelvin and Planck to state the following: It is impossible to construct a device which, operating in a cycle, will take heat from a body and convert it’ completely‘ into work, without leaving any change anywhere.
Clausius Statement In a refrigerator, a working substance takes in heat from a cold body, has a net amount of work done on it by an external agent (electric supply), and gives out a larger amount of heat to a hot body. It thus transfers heat from a cold body to a hot body with the aid of external supply. No refrigerator has ever
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been designed which can run without supply of external energy. This experience led Clausius to state the following : “It is impossible to construct a device which , operating in a cycle, will transfer heat from a cold body to a hot body without expenditure of work by an external energy source . In other words, heat cannot flow spontaneously from a colder to a hotter body.” Equivalence of the two Statements : We can show that these two statements of the second law are equivalent. Let us suppose that there is a refrigerator R (Fig.) which transfers an amount of heat Q2 from a cold body to a hot body without having any supply of external energy. It is thus against the Clausius statement. Now, suppose an engine E working between the same hot and cold bodies takes in heat Q1 from the hot body, converts a part W (= Q1 - Q2) into work, and gives up the remaining heat Q2 to the cold body. The engine E alone does not violate the law. But if the refrigerator R and the engine E are combined together, they form a device that takes in heat Q1 - Q2 from the hot body and converts all into work without giving up any amount to the cold body. This is clearly against the Kelvin-Planck statement. Similarly, let us suppose that there is an engine E (Fig.) which takes in an amount of heat Q1 from a hot body and converts it completely into work W (= Q1), without giving any heat to the cold body. It is against the Kelvin-Planck statement. Now, suppose a refrigerator R working between the same hot and cold bodies takes in heat Q2 from the cold body, has work W(= Q1) done upon it by an external agent, and gives out heat Q1 + Q2 to the hot body. The refrigerator R alone does not violate the law. But both E and R together form a device which transfers an amount of heat Q2 from a cold body to a hot body with no external energy source. This is clearly against the Clausius statement.
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117
The second law of thermodynamics supplements the first law. The first law simply tells us that any device cannot deliver more energy than it receives. It does not speak regarding any limitation , or any condition necessary for the deliver of energy. The second law, however, does it . For example, heat taken in by a substance cannot be all delivered as work, or heat cannot flow spontaneously from a colder to a hotter body. These phenomena are not disallowed by the first law, but they are disallowed by the second law. Production of 8000 K Temperature by Sun-rays : It is not possible to produce a temperature of 8000 K by focussing sunrays . The reason is that the creation of a temperature of 8000 K by transferring heat from a colder body (sun at 6000 K) by means of a lens is a violation of the second law of thermodynamics. Driving a Ship by Extracting Heat from Ocean : It is an attractive idea to drive ship on the energy drawn from the internal energy of water. At the start of its cycle the engine of the ship will draw some heat Q1 from the water, convert a part of it into work, but where it would reject the rest ? By the second law, it must reject some heat into a colder reservoir, but none is available at hand. Theoretically, it is possible if we
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118
can arrange some conveyance to the cold upper atmosphere but practical difficulties would make it almost impossible. From Carnot’s cycle also we see that the efficiency is e = 1-
T2 T1
Thus , e = 0 if T2 = T1 i.e. without a temperature-difference the conversion of thermal energy into mechanical work is impossible. Heat Conduction is Irreversible : Suppose there are two bodies 1 and 2 at temperatures T1 and T2 where T1 > T2. When they are brought into contact, heat flows by conduction from 1 to 2 till they reach a common temperature. Heat cannot flow in the reverse direction, from 2 to 1, because heat-flow by itself from cold to hot body is not allowed by the second law of thermodynamics. Thus heat-conduction is an irreversible phenomenon. Atomic Power Plant: An atomic energy power plant does not violate any law of thermodynamics. Carnot’s Theorem : It states that no engine working between two given temperatures can be more efficient than a reversible engine working between the same two temperatures, and that all reversible engines working between the same two temperatures have the same efficiency, whatever the working substance. Let us consider two engines A and B working between a given pair of source and sink. Let A be irreversible and B reversible. Let the quantities of working substance used in the two engines be such that they perform equal quantities of work per cycle. Let the engine A take in heat Q1 from the source, perform work W, and give out heat Q1 - W to the sink. Its efficiency will be W/Q1’x. Similarly, let the engine B take in heat Q1’ from the source, perform work W, and give out heat Q1’ - W to the sink. Its efficiency will be W/Q1’ .
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119
Let us suppose that the irreversible engine A is more efficient than the reversible engine B i.e. W W > Q1 Q1 '
or
Q 1 < Q1 .
...(i)
Now, let us imagine the two engines to be so coupled that the engine A works directly and it drives the engine B reversibly as shown in Fig. B now acts as a refrigerator, taking in heat Q1’-W from the sink, having work IV done on it, and giving out heat Q1’ to the source. The couple in this way forms a selfacting device, since all the work needed to run the refrigerator B is supplied by the engine A. The net heat taken in from the sink is thus (Q1’ - W) - (Q1 - 2) = Q1’ - Q1 which is positive since Q1 < Q1’ from (i). This is also the net heat given to the source. Thus the couple is transferring in each cycle a quantity of heat Q1’ - Q1 from the cold sink to the hot source, without the aid of any external agent. But this is against the second law of thermodynamics, hence impossible. Hence our supposition that the irreversible engine A is more efficient than the reversible engine B is wrong. Thus the first part of Carnot’s theorem is proved.
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120
The second part of the theorem follows as a corollary. Let us consider two reversible engines A and B working between the same source and sink. If we suppose that A drives B backward, then A cannot be more efficient than B . Similarly, if we suppose that B drives A backward, then B cannot be more efficient than A . Hence the two engines are equally efficient. Thus, the efficiency of a reversible engine depends only on the temperatures of the source and the sink, and is completely independent of the nature and properties of the working substance. Imagine an engine I and a Carnot engine R operating between the same two reservoirs. Suppose that they absorb from the hotter reservoir different amounts of heat, do different amounts of work, but reject to the cooler reservoir equal amount of heat. Prove Carnot’s theorem, assuming the efficiency of I to be greater than that of R and coupling the two engines. Suppose the engine I takes in heat Q1 from the source, performs work W, and gives out heat Q2 to the sink. Its efficiency is e1 =
W' Q1' - Q 2 Q = =1- 2 Q1 ' Q1' Q1
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121
Suppose the Carnot (reversible) engine R takes in heat Q1 from the source, performs work W and gives out heat Q2 (same as I) to the sink. Its efficiency is eR =
W' Q1' - Q 2 Q = =1- 2 Q1 ' Q1' Q1
Let us assume that I is more efficient than R, that is, 1-
or
Q2 Q > 1- 2 Q1 Q1 '
Q1 > Q1’
...(i)
Now, let us imagine the two engines to be so coupled that I works directly and it drives R reversibly as shown in the Figure above. Now R acts as refrigerator taking in heat Q2 from the sink, having work W done on it, and giving out heat Q1’ to the source. The couple formed by I and R takes in net heat Q1 - Q1’ from the source, performs net work W- W’, and gives out no net heat to the sink. Now Q1 - Q1’ = (W + Q2)- (W’ + Q2) = W - W’ and is positive by eq. (i). This means that the couple converts the heat taken in from the source completely into work, without giving out any heat to the sink. This is against the KelvinPlanck statement of the second law. Hence our assumption that I is more efficient than R is wrong i.e. no engine can be more efficient than a reversible engine working between the same temperatures. This is Carnot’s theorem. Absolute Scale of Temperature : A temperature scale which is independent of the properties of any particular substance is called an absolute scale of temperature. No scale furnished by any thermometer is absolute as it depends upon the properties of the thermometric substance. According to Carnot’s theorem, the efficiency of a reversible engine is independent of the working substance and depends
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122
only on the two temperatures between which it is working. Taking this hint, Lord Kelvin defined a temperature scale which does not depend upon the properties of any particular substance. This is the “Kelvin’s absolute thermodynamic scale of temperature”. Let a reversible engine take in heat Q1 from a source at temperature t1 and give out heat Q2 to a sink at temperature t2, where the temperatures t1 and t2 have been measured on Q2 any scale. The efficiency of this engine is e = 1 - Q ' which 1 depends on t1 and t2 only. We may thus write e=1-
Q2 f (t1 ,t2 ) Q1
where f ’ is an unknown function. From this, we may also say that Q1/Q2 must be a function of t1 and t2 only. Thus Q1 = f' (t1 , t2 ) Q2
...(i)
where f ’ is some other unknown function. Similarly, for a reversible engine taking in heat Q2 at temperature t2 and giving out heat Q3 at temperature t3, we have Q2 = f' (t2 , t3 ) Q3
...(ii)
where the function f’ remains unchanged. The heat Q2 given out by the first engine is taken in by the second. Thus both engines, working together, form a third engine which takes in heat Q1 at temperature t1 and gives out heat Q3 at temperature t3, where Q1 = f' (t1 , t3 ) Q3
...(iii)
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123
Q1 Q1 / Q3 Since Q = Q / Q , we have from eq. (i), (ii) and (iii) 3 2 3 f'(t1 ,t3 ) f ' (t2 ,t3 )
f '(t1 ,t2 ) =
...(iv)
This equation does not contain t3 on the left hand side. It means that the function f ’ must be such that t3 cancels out in the right-hand side also. Hence we choose f ’ in the following form :
and
f '(t1 ,t3 ) =
f (t1 ) f (t3 )
f '(t2 ,t3 ) =
f (t2 ) f (t3 )
where f is some other function. If we substitute these values of f’(t1, t3) and f’ (t2 , t3) in the right hand-side of eq. (iv), we shall have f '(t1 ,t2 ) =
f (t1 ) . f (t2 )
...(v)
Now, from eq. (i) and (v), we get Q1 f (t1 ) = Q2 f (t2 )
We know that Q1 > Q2. Hence the function f (t1) > f (t2) when t1 > t2. It means that the function f (t) increases as the temperature rises. Hence it can be used to measure temperatures. Let us suppose that f (t) denotes a temperature q on a new scale. Then we may write. or
Q1 q1 = Q2 q2
Thermal Physics
124 q1 Q1 = q2 Q2
...(vi)
This equation defines the Kelvin’s absolute thermodynamic scale of temperature. That is, the ratio of any two temperatures measured on this absolute scale is equal to the ratio of the quantities of heat taken in and given out by a Carnot’s reversible engine working between these temperatures. It is independent of the properties of any particular substance. In order to complete the definition of the Kelvin’s absolute scale, we assign the arbitrary value of 273-16 K to the temperature of the triple point of water, i.e. qt
r
= 27316 K.
Now, for a Carnot’s engine working between a source at temperature 8 and a sink at temperature qt r , we have, from eq. (vi), q Q = qt r Qt r
or
q=
Q ´ qt r Qt r
or
q=
Q ´ 273.16 K. Qt r
Comparing this with the corresponding equation for the ideal (perfect) gas temperature T, which is T = æ ö Lim pT ÷ çç . ÷ 273 16 ÷ ç ÷ K we see that on the Kelvin’s absolute scale, p t r® 0 p t r ÷ çè ø Q acts as a thermometric property. Absolute Zero : We can find out the zero of the absolute scale. The last expression shows that the heat Q transferred isothermally between two given adiabatics decreases as the temperature q decreases. Conversely, the smaller the value of
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125
Q, the lower is the corresponding temperature q . The smallest possible value of Q is zero. The corresponding value of q (= 0 K) is called ‘absolute zero’. Thus , the absolute zero is the temperature at which a reversible isothermal process takes place ‘without any transfer of heat’. This means that at absolute zero the isothermal and adiabatic processes are identical. Comparison of Absolute Scale and Ideal Gas Scale : Let us consider a reversible engine with one mole of an ideal gas as the working substance and performing a Carnot cycle ABCD (Fig.). Let (pa , Va), (pb , Vb), (pc , Vc) and (pd , Vd) be the coordinates of the points A, B, C and D respectively. Let T1 and T2 be the temperatures of the source and the sink respectively on the ideal gas scale. Let Q1 be the heat taken in from the source during the isothermal expansion AB, and Q2 the heat given out to the sink during the isothermal compression CD.
Let us now calculate Q1 and Q2 in terms of T1 and T2 . Since the working substance is an ideal gas, its internal energy, depending on the temperature only, will remain unchanged during the isothermal expansion AB. Therefore, the heat Q1 absorbed by the gas will be equivalent only to the external work done by the gas in expanding from A to B at temperature T1 i.e. Q1 = ò VVba p dV = RT1 ò VVba
V dV V = RT1 [log e V ]Vb = RT1 log e b a V Va
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126
Similarly, Q2 will be equivalent to the work done on the gas during compression from C to D at temperature T2 i.e. Q2 = ò VVdc p dV = RT2 ò VVdc
V dV V = RT2 [log e V ]Vc = RT2 log e c d V Vd
minus sign indicating that the work is done on the gas. From the last two expressions, we get V log e b Q1 T1 Va = Q2 T2 log Vc e Vd
...(vii)
The points B and C, and similarly the points D and A, lie on the same adiabatic. Therefore, we have, from Poisson’s law T1Vbg - 1 = T2Vcg - 1
and
T1Vag - 1 = T2Vdg - 1
Dividing we get, Vb Vc = Va Vd
Using this result in eq. (vii), we have Q1 T1 = Q2 V2
Now, if q1 and q2 be the temperatures of the source and the sink measured on the Kelvin’s absolute scale, they would be defined by Q1 q1 = Q2 q2
Comparing the last two expressions , we get q1 T1 = q2 T2
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127
i.e. the ratio of temperatures on the absolute scale is the same as the ratio on the ideal gas scale. If q and T be the temperatures of a given body on the absolute and ideal gas scales respectively, and qt r and Ttr the respective temperatures of the triple point of water, we have T q = 1 qt r Tt r
Since qt r = Tt r = 273-16 K , we have q= T
Hence numerically q and T are also the same. Thus the two scales are identical in all respects.
Practical Realization of Absolute Scale The Kelvin’s absolute scale cannot be directly realised in practice. It is, however, exactly identical to an ideal gas scale. Hence the temperatures measured by a constant-volume (or constant pressure) gas thermometer filled with an ‘ideal’ gas would exactly be the same as the temperatures measured on an absolute scale. But, in practice, no gas is ideal, hence we cannot have an ideal gas thermometer. However, the constant-volume hydrogen thermometer is the closest approach to an ideal gas thermometer. Any reading taken on the hydrogen thermometer can be corrected to obtain the corresponding reading on the ideal gas thermometer. Thus, Kelvin’s absolute scale which coincides with the ideal gas scale, is realised in practice. No Negative Temperatures on the Absolute Scale : The efficiency of a Carnot’s engine is e = 1-
T2 T1 ,
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128
where T1 is the temperature of the source and T2 that of T2 q2 the sink. Since T = q , we may write 1 1 q e = 1- 2 q1 To obtain 100% efficiency (i.e. e= 1), q2 must be zero. That is, if a sink at absolute zero were available, all the heat taken in from the source would have been converted into work. Clearly, a negative temperature on the absolute scale would mean a temperature of the sink at which the efficiency of the engine is greater than unity. This would be a violation of the second law of thermodynamics. Hence, a negative temperature on the absolute scale is impossible .
Steam Engine A steam engine is one in which water is used as the working substance. It consists essentially of a boiler (source), a heatinsulating cylinder closed with a lubricated piston, a condenser (sink) and a feed-pump (Fig.). Its working differs from that of a Carnot engine in which the working substance remains confined in the cylinder throughout all the processes. In the steam engine the working substance takes in heat in the boiler, performs work in the cylinder, gives out heat in the condenser and is transferred back to the boiler by the feed-pump. Thus its cycle, called the “Rankine’s cycle” consists of six processes:
The Reversibility
129
1. The working substance (water) is heated in the boiler to its boiling temperature T1 at constant pressure p1 (the vapour pressure of water at T1). The process is represented by A A‘ in the p-V diagram (Fig. b). 2. The boiling water is converted into steam at the constant temperature T1 and constant pressure p1. The process is represented by A’B’ in (Fig. b). 3. The steam so formed is passed through gas-heated pipes (not shown) where it is superheated to a temperature T3 (say). It is then admitted into the cylinder through the valve V1. This process is represented by B’B in (Fig. b). 4. The valve V1 is automatically closed and the steam reached in the cylinder expands adiabatically, thus doing external work against the piston until its temperature and pressure fall to T2 and p2 respectively (say). During this process some of the steam condenses to form water droplets. This adiabatic expansion is represented by BC in (Fig. b). 5. The piston now returns, opening the valve V2 . The steam in the cylinder completely condenses into the condenser at constant temperature T2 and constant pressure p2 . The complete condensation is represented by CD in the p-V diagram. The latter stage of this process differs from the Carnot’s corresponding process in which the condensation stops at a point which lies on the adiabatic through A . 6. The water obtained by the condensation of steam is transferred from the condenser to the boiler by means of the feed-pump and the pressure is again increased from p2 to p1 . During this process the temperature changes slightly. This process also is different from the corresponding Carnot process. It is represented by DA in (Fig. b) . The cycle is thus complete.
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130
To obtain the net work done during the cycle, the work done on the feed-pump must be allowed for. If the volume of the condensed water returned in each cycle by the feed-pump from the condenser at pressure p2 to the boiler at pressure p1 is Vw , the amount of work involved is (p1 - p2) Vw . This is given by the shaded area ADp2 p1; the net work done by the working substance being given by the area ABCD . If reversibility is assumed for all the processes, the Rankine cycle represents the maximum possible efficiency for the steam engine. Efficiency : Let Q1 denote the total amount of heat taken in by the working substance during the three processes A ® A’, A’ ® B’ and B’ ® B. Since all these processes take place at constant pressure, we have Q1 = Hb - Ha where Hb and Ha are the total heat functions at the points B and A respectively. Similarly, the heat Q2 given out during the process C ® D is Qa = Hc - Hd . The efficiency is given by e=
Q1 - Q2 (H b - H a ) - (H c - H d ) = Q1 Hb - H a
...(i)
The values Hb, Hc and Hd are obtainable in the steam tables. The value of Ha is, however, not obtainable. Hence let us eliminate it. The equation defining the total heat function is H = U + pV. Differentiating: dH = dU + pdV+Vdp. But dU + p dV=dQ (by the first law of thermodynamics) \
dH = dQ+Vdp.
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131
For the adiabatic process DA , we have dQ = 0 . \
dH = Vdp.
Integrating between the limits D and A , we get H a - Hd =
ò
A
D
V dp .
Now, V is practically constant during the process D ® A, and is equal to Vw, the volume of the water in the condenser. A
\
Ha - Hd = Vw ò dp = Vw (P1 - P2 ).
\
H a = H d + Vw (P1 - P2 ).
D
Putting the value of Ha in eq. (i), we get e=
H b - Hc - Vw (P1 - P2 ) Hb - H d - Vw (P1 - P2 ) ,
where the term Vw (p1 - p2) is the ‘feed-pump term.’ Actual steam engines have an efficiency only about 60% of the efficiency so deduced because the actual processes are not strictly reversible. The adiabatic expansion of the steam is p generally not complete and the valves take time in their opening and closing. Hence in actual curve the angles at A, B , C and D are rounded (Fig.), instead of being sharp. Clearly, the actual curve has an area smaller than the Rankine’s curve.
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Thermal Physics
Internal Combustion Engine There are two general types of heat engines, the “steam engines” and the “internal combustion engines”. In both the engines, a working substance contained in a cylinder performs a cycle of processes in which it takes in heat from some source, converts a part of it into external work and gives out the rest to a sink. In the steam engine the heat is supplied to the working substance by burning the fuel outside the cylinder (in a furnace under the boiler). In the internal combustion engine, the heat is developed by combustion taking place inside the cylinder.
One type of the internal combustion engine is the Otto engine, realised by Otto in 1876, whose plan is shown in Fig. It consists of a cylinder fitted with a piston and provided with two valves, one inlet valve I and the other out-let valve O. These valves are automatically opened and closed at the right moments by a suitable mechanism. The working substance is air.
The Reversibility
133
The engine operates on a cycle, called ‘Otto cycle’, consisting of six processes. Out of these six processes, four involve the motion of the piston and are called ‘strokes’. Hence the cycle is a four-stroke cycle. 1. Suction Stroke : The inlet valve I is opened and the outlet valve O is closed. The piston moves down, and a mixture of 98% air and 2% petrol vapour is sucked into the cylinder at atmospheric pressure through the valve I (Fig. a). The process is indicated by EC in the indicator (p-V) diagram (Fig.). 2. Compression Stroke : Both the valves are closed. The piston moves up and the mixture is compressed adiabatically to about 15 of its original volume, its temperature rising to 600 °C (Fig. b). The process is represented by CD in the p-V diagram. 3. Combustion : Just when the compression stroke ends, an electric spark ignites the mixture. Hence a large amount of heat of combustion is produced. This creates a high pressure and raises the temperature of the air to 2000 °C at constant volume . The process is indicated by DA in the p-V diagram. (The piston remains stationary during the process). 4. Expansion Stroke : The valves still remain closed. The hot combustion products now expand adiabatically, so that the piston moves down (Fig. c). As a result, the pressure as well as the temperature of the combustion products drops. It is this stroke in which the engine does external work and hence it is also called as ‘working stroke’. It is represented by AB in the p-V diagram.
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Thermal Physics
5. Valve Exhaust: Just when the working stroke ends, the outlet Valve O is opened. The combustion products (still at a pressure and temperature higher than outside) escape through O until the pressure falls to the atmospheric pressure. In this process heat is given up. It is represented by BC in the p-V diagram. (The piston remains stationary during this process also). 6. Exhaust Stroke : The piston moves up so that the burnt gases escape out through the valve O (Fig. d). This is represented by CE in the p-V diagram. The next cycle then occurs with a fresh charge of air and petrol vapour. The first and last processes EC and CE cancel each other. Therefore, we can imagine as if a fixed mass of air say, 1 mole, is always in the cylinder and is going round the cycle CDABC, taking in heat at constant volume along DA and giving up heat at another constant volume along BC . Efficiency: Let Ta ,Tb , Tc and Td be the absolute temperatures corresponding to the points A , B , C and D respectively. Then if Cv is the molar specific heat of air at constant volume, the heat Q1 taken in during the process D ® A is
The Reversibility
135 Q1 = Cv (Ta - Td ),
and the heat Q2 given out during the process B ® C is Q2 = Cv(Tb-Tc). Hence the thermal efficiency is e = 1-
T - Tc Q2 = 1- b . Q1 Ta - Td
...(i)
Let V1 and V2 be the volumes of the mixture at C and D respectively. Since the points B and A , and similarly the points C and D, lie on the same adiabatic ( TV g- 1 = constant), we have
and
TbV1g - 1 = TaV2g - 1
...(ii)
TcV1g - 1 = TdV2g - 1
...(iii)
Subtracting, we get (Tb - Tc )V1g - 1 = (Ta - Td )V2g - 1 g- 1
or
ö (Tb - Tc ) æ V ÷ ÷ = ççç 2 ÷ (Ta - Td ) èV1 ÷ ø
g- 1
æ1 ÷ ö ÷ = çç ÷ çè r ÷ ø
where r = V1/V2 , and is known as the compression ratio’. Putting this value in eq. (i), we get g- 1
æ1 ö ÷ e = 1 - ççç ÷ ÷ èr ÷ ø
This expression shows that the efficiency increases with increasing compression ratio r . In the actual engine, r cannot be made greater than about 10, otherwise the rise in temperature during the adiabatic compression of the air-fuel mixture will be large enough to cause an explosion before the spark strikes. For a typical compression ratio of 8 and g = 1 -4 , the efficiency would be
Thermal Physics
136 0 .4
æ1 ö e = 1 - ççç ÷ ÷ = 0.56 = 56%. è8 ø÷
This is the theoretical efficiency predicted for an engine operating in the idealised Otto Cycle. In real engines the efficiency is much lower, probably less than half the ideal value because of effects such as acceleration, turbulence, friction, heat loss to the cylinder walls, and incomplete combustion of the air-fuel mixture. Otto Engine and Carnot Engine: The efficiency of Otto engine can also be expressed in terms of a ratio of temperatures. From eq. (ii) and (iii), we have g- 1
æV ÷ ö ÷ çç 1 ÷ çèV2 ÷ ø
=
Ta Td = Tb Tc g- 1
æ1 ÷ ö e = 1 - çç ÷ ÷ çèr ÷ ø
∴
1-
g- 1
æV ÷ ö = 1 - ççç 2 ÷ ÷ ÷ èV1 ø
Tb T = 1- c . Ta Td
During the Otto cycle, the lowest temperature is Tc and the highest temperature is Ta. The efficiency of a Carnot engine operating between these two extreme temperatures would be e(Carnot) = 1 -
Tc . Ta
which is greater than the efficiency of the Otto cycle. Comparison of Steam Engines and Petrol Engines : As compared to steam engine, the internal combustion (petrol) engine is (i) cleaner, (ii) lighter, (iii) more compact, (iv) has no separate boiler, (v) has easy fuel supply, and (vi) possesses a higher thermal efficiency. Therefore these engines are used in most forms of modern transport such as aeroplanes. The steam engines are, however, cheaper than petrol engines because the fuel (coal) of the steam engines is cheaper
The Reversibility
137
than the fuel (petrol) of the petrol engines. Hence they are used in railways in which a very large amount of work is to be obtained. Internal Combustion Engine : An internal combustion engine is one in which heat supplied to the working substance is produced by combustion taking place inside the cylinder. There are two types of internal combustion engines ; the Otto engine in which heat is supplied at constant volume , and the Diesel engine in which heat is supplied at constant pressure. Diesel Engine: The plan of this engine, proposed by Diesel in 1900, is shown in Fig. It consists of a cylinder closed by a piston and provided with three valves, air inlet valve I, oil inlet valve I’ and outlet valve O . These valves are automatically opened and closed at the right moments by a suitable mechanism. The working substance is air.
The engine operates on a four-stroke cycle, known as “Diesel cycle.” (i) Suction Stroke : The air inlet valve I is opened, the oil inlet valve I’ and the outlet valve O are closed. The
138
Thermal Physics piston moves down, and pure air is sucked into the cylinder at atmospheric pressure (Fig. a). This is indicated by EC in the p-V diagram (Fig.).
(ii) Compression Stroke : All the valves are closed. The piston moves up and compresses the air adiabatically to about 1 of its original volume, its temperature rising to 17 1000 °C. The is indicated by CD in the p-V diagram.
Just at the end of this stroke i.e. at the point D the oil inlet valve I’ is opened. (iii) Working Stroke : A heavy oil (fuel) is now injected into the cylinder through I’. As the temperature inside the cylinder is very high, this oil immediately burns. Its rate of supply is so adjusted that as the piston moves down, the burning oil supplies heat to the air at constant pressure. This portion of the working stroke is represented by DA in the p-V diagram. At A , when the temperature is about 2000 °C the supply of oil is cut off and for the rest of the working stroke the expansion is adiabatic and so the temperature falls. This portion of the working stroke is shown by AB in the p-V diagram.
The Reversibility
139
At B the outlet valve O is opened so that the pressure immediately falls to atmospheric pressure. The heat is given up to the outside during the process which is indicated by BC in the p-V diagram. (iv) Exhaust Stroke : The piston moves up so that the burnt gases escape out through the valve O (Fig.). This is represented by CE in the p-V diagram. The next cycle then occurs with a fresh supply of air. The processes EC and CE cancel each other. Therefore, we can imagine as if a fixed mass of air, say 1 mole, is always in the cylinder and is going round the cycle CD ABC, taking in heat at constant pressure along DA and giving up heat at constant volume along BC. Efficiency : Let Ta ,Tb ,Tc ,Td be the absolute temperatures at the points A, B, C, D respectively. Then if Cp be the molar specific heat of air at constant pressure, the heat Q1 taken in at constant pressure during the process D ® A is Q1 = Cp(Ta - Td). Similarly, if Cv be the molar specific heat of air at constant volume, the heat given out during the process B ® C is Q2 = Cv(Tb - Tc). Hence the thermal efficiency is e = 1-
where
Q2 Q1
= 1-
Cv (Tb - Tc ) cp (Ta - Td )
= 1-
ö 1æ çç (Tb - Tc ) ÷ ÷ , ÷ ç g èç(Ta - Td ) ÷ ø
γ = Cp/Cv.
...(i)
Thermal Physics
140
Let V1 , V2 and V3 be the volume of air at C, D and A respectively. Since the points B and A , and similarly the points C and D, lie on the same adiabatic, we have Tb V1g - 1 = TaV3g - 1
and
Tc V1g - 1 = TdV2g - 1
Subtracting, we get (Tb - Tc )V1g - 1 = Ta V3g - 1 - Td V2g - 1 g- 1
or
g- 1
æV ö æV ö ÷ ÷ Tb - Tc = Ta ççç 3 ÷ - Td ççç 2 ÷ ÷ ÷ èV1 ÷ ø èV1 ø÷ g- 1
æ1 ö ÷ = Ta ççç ÷ ÷ èçr ÷ ø e
g- 1
æ1 ö ÷ - Td ççç ÷ ÷ ÷ èçr ø c
where ρe (= V1/V3) is the ‘expansion ratio’ and ρc (= V1/V2 is the ‘compression ratio’. Substituting this value of Tb - Tc in (i), we get g- 1 é æ1 ö æ1 ö÷ ùú êT çç ÷ çç ÷ ÷ T êa ÷ d çr ÷ ú è c ø÷ ú 1 çèr e ø÷ e = 1- êê ú. Ta - Td gê ú ê ú êê ú ú ë û
Since the pressure along DA is constant ; we have Ta V3 V3 V1 r c = = ´ = Td V2 V1 V2 r e Ta = Td
ρc ρe
Putting this in eq. (ii), we get
...(ii)
The Reversibility
141 g- 1 é r æ1 ög - 1 æ1 ö÷ ù êT c çç ÷ çç ÷ ú ÷ T d ÷ ÷ ú êd ø èçr c ø÷ ú 1 ê r çèr ÷ e = 1- ê e e ú r gê ú Td c - Td ú ê re êê ú ú ë û
or
é 1 ög êæ ç ÷ ÷êç ÷ ø 1 êçèr e ÷ e = 1- ê gê 1ê êê r e ë
g æ1 ö÷ ù çç ÷ ú ÷ú çèr ÷ cø ú 1 ú ú. rc ú ú ú û
In Diesel engine, the air is simply compressed and there is no danger of explosion. Hence its compression ratio can be made much higher than that of an Otto engine. Thus, the efficiency of the Diesel engine can be made greater than that of an Otto engine. But, on account of the higher working pressures, it must be more robust in construction. However, it runs cheaper as it requires lesser fuel and runs on crude oil.
PROBLEMS 1. In a Carnot cycle the isothermal expansion of an ideal gas takes place at 400 K and the isothermal compression at 300 K. During then this case expansion 500 calories of heat energy are transferred to the gas. Let’s determine (i) the work performed by the gas during the isothermal expansion, (ii) the heat rejected from the gas during the isothermal compression, (iii) the work done on the gas during the isothermal compression. (J = 4 18 joule/cal). Solution : (i) The work performed by the gas during the isothermal expansion is equivalent to the heat Q1 absorbed by the gas. Here Q1 = 500 cal. \
work = 500 × 4.18 = 2090 joule.
Thermal Physics
142
(ii) Let Q2 be the heat rejected. We know that Q1 T1 = Q2 T2 Q2 = Q1 ´
or
T2 T1
= 500 cal ×
300 K = 375 cal. 400 K
(iii) The work done on the gas during the isothermal compression is equivalent to the heat Q2 rejected and is thus 375 × 4.18 = 1567.5 joule. 2. A reversible engine works between two temperatures whose difference is 110°. If it absorbs 746 joule of heat from the source and gives 546 joule to the sink, let’s calculate the temperatures of the source and the sink. Solution : For a reversible (Carnot) cycle, we have Q1 T1 = Q2 T2
where Q1 is the heat taken in from the source at temperature T1 and Q2 is the heat rejected to the sink at temperature T2 in one cycle. Here , Q1 = 746 joule and Q2 = 546 joule . 746 T1 = ∴ 546 T2 . Also, T1-T2= 110 (given). ∴
T1 746 = 546 T1 - 110
Solving, we get T1 = 410 K and
T2 = 300 K.
The Reversibility
143
3. An ideal (Carnot’s) heat engine takes 1000 calorie of heat at 627 ºC from a source and rejects a part at 27 °C to the sink. The work done by the engine per cycle, the heat rejected to the sink per cycle and the efficiency of the engine may be found out. Solution : For a Carnot cycle, we have Q1 T1 = Q2 T2
where Q1 is the heat taken in from the source at temperature T1 and Q2 is the heat rejected to the sink at temperature T2 in one cycle. Here , Q1 = 1000 cal, T1 = 627 + 273 = 900 K and T2 = 27 + 273 = 300K. ∴
1000 900 = Q2 300
or
Q2 =
1000´ 300 = 333.3 cal. 900
The balance of heat is converted into work W. W = Q1 - Q2 = 1000 - 333.3 = 666.7 cal = 666.7 × 4.18 = 2.79 × 103 joule. [Q 1 cal = 418 joule] The efficiency of the Carnot engine is e = 1-
= 1-
T2 T2
300 = 0.67= 67%. 900
4. A Carnot engine takes in 100 cal of heat from the source at temperature 400 K and gives up 80 cal to the sink. The temperature of the sink and Thermal efficiency of the engine is as follows.
Thermal Physics
144 Solution : For a Carnot engine, we have
Here Q{ = 100 cal, Q2 = 80 cal, T1 = 400 K, T2 = ? T2 = T1 ´
∴
Q2 Q1
= (400 K) ×
The efficiency is
e=1
Q2 Q1
=1-
80 cal = 320 K. 100 cal
æ ö ççor e = 1 - T2 ÷ ÷ ÷ çè T1 ÷ ø
80 = 0.2 = 20% 100
5. A reversible (Carnot) engine takes 200 k-cal of heat from a source at 527 ºC , converts a part of it into work and rejects the remaining to a sink at 27 °C. Its efficiency work done by the engine per cycle in joule in kilo-watt-hour in electron-volt, Given : J = 4.2 x 103 joule/kilo-calorie and leV = 1.60 × 10-l9J are calculated as follows. Solution : If T1 and T2 be the Kelvin temperatures of source and sink, the efficiency of a reversible engine is e =1-
T2 . T1
Here T1 = 527 + 273 = 800 K and T2 = 27 + 273 = 300 K. ∴
e=1-
300 5 = = 0.625 = 62.5%. 800 8
We know, that the efficiency of an engine is defined as e=
W , Q1
where W is the work done by t he engine per cycle and Q1 is the heat taken from the source per cycle.
The Reversibility
145
Here Q1 = 200 kilocalorie. Therefore, W= eQ1 = 0.625 × 200 = 125 kilocalorie = 125 × (4.2 × 103) [Q 1 kilocalorie = 4.2 × 103 J] = 5.25 × l05 joule. Now, 1 kilowatt-hour = 1000 x 3600 = 3.60 x 106 joule. \
W =
5.25 × 10 5 3.6 0 × 10 6 = 0.146 kilowatt-hour.
Again , 1 electron-volt = 1.6 × 10-19 joule. \
W=
5.25 × 10 5 = 3.28× 10 24 eV. 1.60 × 10 -19
6. An ideal engine operating between 227 °C and 27 °C develops 74600 watt. The efficiency of the engine, the heat taken per second from the hot reservoir and the heat rejected per second to the cold reservoir may be found out as follows. e =1-
T2 . T1
Solution : The efficiency of an ideal (Carnot) engine operating between temperatures of T1 and T2 Kelvin is given by Here T1 = 227+ 273 = 500 K and T2 = 27 + 273 = 300 K. \
e = 1-
300 = 0 .4 = 40% 500
Now, if Q1 be the heat taken from the hot reservoir (source) and Q2 the heat rejected to the cold reservoir (sink), then the efficiency of the engine is given by e-
Q1 - Q2 W = , Q1 Q1
...(i)
Thermal Physics
146
where W is the useful work produced by the engine. Here the rate of useful work produced (power) is 74600 watt. Thus W = 74600 joule/sec.[Q 1 watt = 1 joule/sec] Hence, by eq. (i), the heat taken from the source is Q1 =
W 74600 = = 1.865´ 10 5 joule/sec e 0 .4
=
1.865 × 10 5 = 4.46 × 104 cal/sec. 4.18
The heat rejected to the sink is Q2 = Q1 - W = 1.865 x 105 - 0.746 x 105 = 1.119 × 105 joule/sec =
1.119 × 10 5 = 2.68 × 104 cal/sec. 4.18
7. A reversible engine takes in heat from a reservoir of heat at 527°C and gives out heat to a sink at 127°C. Let’s find as to how many calories per second it must take from the reservoir in order to produce useful mechanical work at the rate of 750 watts. Solution : Let T1 and T2 be the kelvin temperatures of the reservoir and the sink respectively. If the working substance takes Q1 heat from the reservoir, performs work W and rejects the rest to the sink, then the efficiency of the engine is given by e=
W . Q1
Since the engine is reversible, we have e=1-
(127 + 273) T2 = 1= 0.5 T1 ( 527+ 273
The Reversibility
147
\ eq. (i) gives
Q1 =
W 750 watt = = 1500 joule/sec. e 0.5
Now 1 cal= 4.18 joule. ∴
Q1 =
1500 = 359 cal/sec. 4.18
8. A Carnot engine works between 100° C and 0°C. If the work done per cycle is 1200 kg-m2/s2, then the heat (in calories) taken in from the source may be calculated. (J = 4 18 joule/ cal) Hint: Q1 = W/e. A heat engine operating between temperatures 137 °C and - 43 °C receives 250 kcal of heat per minute. The inventor of the engine claims to develop 12 HP with this engine. Let’s find whether his claim is agreeable or not. 1 HP = 746 watt. Solution : The maximum work obtainable from the engine is the work obtainable from a ‘Carnot’ engine operating between the same temperatures 137 °C and -43°C. Thus, the maximum possible efficiency of the engine is e=1-
T2 T1
Here T1, = 137 + 273 = 410 K and T2 = - 43 + 273 = 230 K. ∴
e=1-
230 = 0.44 410
Now, if Q1 be the heat taken from the source and W the useful work produced by the engine, then we have e=
W . Q1
Here Q1 = 250 kcal/min . \ W = e Q1 = 0.44 × 250 = 110 kcal/min.
Thermal Physics
148
Now , 1 kcal = 4180 joule and 1 min = 60 sec. ∴
W=
(110 × 4180) joule 60 sec
= 7663 joule/sec = 7663 watt =
7663 = 10.3 H.P. 746
[Q 1 H.P. - 746 watt]
This is the maximum possible rate of work (power) that can be produced by the engine. Hence the claim of developing 12 HP is wrong . 9. For a Carnot engine using an ideal gas, the adiabatic expansion ratio is 5 and the value of y= 1.40. The efficiency of the engine may be calculated here. Solution : The efficiency of Carnot engine in terms of adiabatic expansion ratio p , is given by g- 1
æ1 ö ÷ e = 1- çç ÷ ÷ çèr ø÷ Here p = 5 and g =1.40.
0.40
∴ Let us put
æ1 ö e = 1 - ççç ÷ ÷ ÷ è5 ø
= 1 - (0.2)0.40
x = (0.2)0.40
so thatlog x = 0.40 log 0.2 = 0.40 × (-0.6990) = -0.2796 = 1 .7204. ∴
x = 0.52.
Making this substitution in eq. (i), we get e = 1 - 0.52 = 0.48 = 48%
...(i)
The Reversibility
149
10. The thermal efficiency of a reversible engine working between temperatures 0 °C and 100 °C may be calculated as follows. Solution : If Q1 be the heat taken in by the working substance from the source at Kelvin temperature T1, and Q2 be the heat given up to the sink at Kelvin temperature T2 , then the efficiency of any engine is given by Q e=1- 2 Q1 Q2 T2 For a Carnot (reversible) engine, Q = T 1 1 T2 e = 1∴ T1
Here , T2 = 0 + 273 = 273 K and T1 = 100 + 273 = 373 K. e=1-
∴
273 = 0.27 = 27% 373
11. The efficiency of a reversible heat engine working between the temperatures 72 ‘C and 187 °C may be calculated as follows. Solution : The efficiency of a “reversible” (Carnot) engine is maximum and given by e=1-
T2 T1
where T2 and T1 are absolute temperatures of the sink (lower temp.) and the source (higher temp.) respectively. Here, T2 = 72 + 273 = 345 K and T1 = 187 + 273 = 460 K. ∴
Q = 1-
345 = 0.25 = 25% 460
12. One of the most efficient engines ever developed operates between 2100 K and 700 K. Its actual efficiency is 40%.
Thermal Physics
150
Percentage of its maximum possible efficiency is this is as follows : Solution : The maximum possible efficiency of a heat engine operating between Kelvin temperatures T1 and T2 is given by e=1-
T2 T1
Here T1 = 2100 K and T2 = 700 K. ∴
e=1-
700 = 0.67 = 67% 2100
The actual efficiency is only 40%. ∴
percentage fraction =
=
actual efficiency × 100 maximum efficiency 0.40 ´ 100 = 60% 0.67
13. Let two Carnot engines working between the temperatures (a) 1000 K and 500 K, (b) x K and 1000 K are equally efficient compute x . Solution. For 1000 K and 500 K , we have e =1-
T2 500 = 1= 0.5 T1 1000
For x K and 1000 K, we have e =1\
1000 = 0.5 (given) x
x = 2000K.
14. A, B, C, D are four baths. An engine AC (working between A and C) has efficiency halfway between the efficiencies of AB and AD. Show that the absolute temperature of C is half-way of those of B and D. All engines are reversible.
The Reversibility
151
Solution : The efficiencies of AC , AB and AD are 1 and 1 -
TC TB , , TA TA
TD respectively. We are given that TA TC 1 ïì æ T ö ÷+ = ïí ççç1 - B ÷ ÷ TA 2 ïîï è TA ÷ ø T T + TD 1- C = 1- B 2TA TA TC T + TD = B TA 2TA T + TD TC B 2
1-
or or or
æ çç1 çè
öïüï TD ÷ ÷ ý ÷ TA ÷ øþïï
15. The temperature of the sink of a Carnot engine is 27 °C. If the efficiency of the engine is 40%, the temperature of the source may be found in following way. Solution : If T1 and T2 be the Kelvin temperatures of source and sink respectively, the efficiency of the Carnot engine is e =1-
T2 T1
Here e = 40% = 0.40, T2 = 27 + 273 = 300 K. ∴
T1 =
T2 300 = 1-e 1 - 0.40
= 500 K = 227 ºC. 16. The efficiency of a Carnot cycle is 1/6. On reducing the temperature of the sink by 60°C , the efficiency increases to 1/3. Let us find the initial and final temperatures between which the cycle is working. Solution : Let T1 and T2 be the initial Kelvin temperatures of the source and the sink respectively. Then , the efficiency is given by
Thermal Physics
152 e =1-
T2 1 = T1 6
...(i)
When T2 is decreased to T2 - 60 (1 ºC = 1 K in size), the new efficiency is e'=1-
T2 - 60 1 = T1 3
From eq. (i) and (ii), we have respectively T2 1 5 = 1- = T1 6 6 T2 - 60 1 2 = 1- = T1 3 3
and Dividing, we get
T2 5 = T2 - 60 4
Solving, we get T2 = 300 K = 27 °C. Putting the value of T2 (in Kelvin) in eq. (i), we get 1-
300 1 = T1 6
This gives T1= 360 K = 87 °C. The cycle is initially working between 87 °C and 27 °C. Finally, the temperature of the sink is reduced by 60 °C, so that the cycle works between 87 ºC and - 33 °C. 17. A reversible engine converts one-sixth of heat absorbed at the source into work. When the temperature of the sink is reduced by 82 °C, the efficiency is doubled. The temperatures of the source and the sink would be as follow.
The Reversibility
153
Solution : Suppose the engine absorbs heat Q1 from the source at Kelvin temperature T1 and rejects heat Q2 to the sink at Kelvin temperature T2 . Then the heat converted into work is Ql — Q2 . It is given that Q1 - Q2 1 = Q1 6
1-
For a reversible engine, 1 -
\
Q2 1 = Q1 6 Q2 T = 2 Q1 T1 T2 1 = T1 6
which is the efficiency of the engine. When T2 is reduced to T2 - 82 (a change of 82 °C is same as a change of 82 K), the efficiency is doubled, that is, it becomes 1/3. Thus T - 82 1 1- 2 = T1 3 On solving eq. (i) and (ii), we get T1 = 492 K = 219 ºC and
T2 = 410 K = 137 ºC.
18. A Carnot engine has an efficiency of 50% when its sink temperature is 17 °C. Let’s the change in its source temperature for making efficiency 60%. Solution : Let T1 be the initial temperature of the source. The temperature of the sink is T2 = 17 °C = 290 K and the efficiency e = 50% = 0.5 . Now e=1\
T2 290 =1= 0.5 T1 T1
T1 = 580 K .
Thermal Physics
154
Now, let the temperature of the source be raised to (580 + D T) K , when the efficiency becomes 60% i.e. 0.6 . Thus 1-
290 = 0.6 580 + D T D T = 145 K
\
The temperature of the source should be raised by 145 K or 145 °C. 19. A Carnot engine whose low-temperature reservoir is at 7 °C has an efficiency of 50%. It is desired to increase the efficiency to 70%. We may find as to by how many degree should the temperature of the high-temperature reservoir be raised . Solution : Let T1 be the initial absolute temperature of the high-temperature reservoir (source). The temperature of the low-temperature reservoir (sink) is T2 = 7 + 273 = 280 K, and the efficiency e is 50% = 0.5 . Now e=1\
T2 280 =1= 0.5 T1 T1
T1 = 560 K.
Now, let the temperature T1 of the source be raised to (560 + D T) K in order to increase the efficiency to 70% i.e. 0.7. Thus 1-
Solving, we get
280 = 0.7 560 + D T D T = 373 K.
The temperature of the source should be raised by 373 K or 373 °C. 20. Two Carnot engines A and B are operated in series. The first one, A , receives heat at 900 K, and rejects to a reservoir
The Reversibility
155
at temperature T K. The second engine , B , receives the heat rejected by the first engine and in turn rejects to a heat reservoir at 400 K. Let’s calculate the temperature T for the situation when (i) the work outputs of the two engines are equal , (ii) the efficiencies of the two engines are equal. Solution. (i) Let the engine A take in heat Q1 at temperature T1 and reject heat Q at temperature T ; and the engine B take in heat Q at temperature T and reject heat Q2 at temperature T2 . Then , the works done by them are given by WA = Q1 - Q and
WB = Q - Q2 .
But here WA = WB. ∴
or
Q1 - Q = Q - Q2 Q1 = Q2 =2 Q
...(i)
As both the engines are Carnot’s we have Q1 T Q T = 1 and 2 = 2 Q T Q T
or
Q1 + Q2 T2 + T2 = Q T
From eq. (i) and (ii), we get T1 = T2 = 2 T Here T1= 900 K and T2 = 400 K. ∴ or
900 + 400 = 2 T
T = 650 K.
(ii) If efficiencies are equal, then e=1-
T T =1- 2 T1 T
...(ii)
Thermal Physics
156 or
T T = 2 T1 T
or
T = (T1 × T2)½ = (900 × 400)½ = 600 K.
21. A Carnot’s engine working as a refrigerator between 260 K and 400 K extracts 600 calories of heat from the reservoir at lower temperature. The amount of heat delivered to the reservoir at the higher temperature and the work that must be supplied to operate the refrigerator are calculated as follows. Solution : For a Carnot’s refrigerator, we have Q1 T1 = Q2 T2
where Q2 is the heat taken in from a cold body at temperature T2 and Q1 is the heat delivered to the hot body at temperature T1. Here, Q2 = 600 cal, T1, = 400 K and T2 = 260 K . Thus Q1 400 = 600 260
or
Q=
600 ´ 400 = 923 cal. 260
Thus the excess of heat delivered to the reservoir at higher temperature is Q1 - Q2 = 923 - 600 = 323 cal. This comes from the work W done by the motor running the refrigerator. Thus W = 323 cal. Now, 1 cal = 4.18 joule . W = 323 x 4.18 = 1350 joule. 22. A refrigerator is to be maintained at —73 °C and the outside air is at 27 ºC. The minimum amount of work that
The Reversibility
157
must be supplied to remove 1000 joules of heat from inside the refrigerator may be calculated as follows. Solution : The refrigerant extracts heat from refrigerator’s inside space (which act as the cold body), has a net amount of work done on it by the electric motor fitted in the refrigerator, and delivers a larger amount of heat to the outside (room) air (which acts as the hot body). For computing minimum work to be supplied, we treat the given refrigerator as Carnot refrigerator. Suppose its refrigerant extracts heat Q2 from the cold body at absolute temperature T2 , has a net supply of work W and delivers heat Q1 to the hot body at absolute temperature T1 . Then Q1 = Q2 + W or
æQ ö ÷ W = Q1- Q2 = Q2 ççç 1 - 1÷ ÷ ÷ èQ2 ø
Q1 T1 But Q = T (for Carnot refrigerator). 2 2
Here Q2= 1000 joule , T2 = - 73 + 273 = 200 K and T1 = 27 + 273 = 300K. \
æ300 - 200 ö÷ ÷= 500 joule. W = 1000 joule çççè ø 200 ÷
23. A refrigerator is maintained at 0 °C and the outside temperature of the room is 27 °C. The minimum energy (work) to be supplied to freeze 1 kg of water already at 0 °C is calculated below. The coefficient of performance may also be checked heat of ice = 80 cal/gram and 1 cal = 418 joule. Solution : For computing the “minimum” energy to be supplied, the refrigerator must be regarded as “Carnot” refrigerator. The heat that must be removed from inside and discarded to the room in order to freeze one kilogram of water is
Thermal Physics
158
mL = 1000g × 80cal/g = 80 × 104cal. For Carnot refrigerator, the energy (work) that must be supplied to remove an amount of heat Q2 from its freezing compartment at temperature T2 K is given by W =Q2
T1 - T2 T2
where T1 K is the temperature of the outside room air. Here, Q2 = mL = 8.0 × 104 cal , T1 = 27 + 273 = 300 K and T2 = 0 + 273 = 273 K. \
W = (8.0 × l04cal)
300 - 273 273
= 7-9xl03cal = (7.9 x 103) × 4.18 = 33 x 103 joule. The coefficient of performance is given by K=
Q2 8 .0 ´ 10 4 cal = W 7 .9 ´ 10 3
10.
K=
T2 273 = T1 - T2 300 - 273
10.
Alternatively;
24. A refrigerator works between - 10 °C and + 27 °C, while the other between - 20 °C and + 17 °C, both removing 2000 joule heat from the freezer. The calculation shows us which one normally one chooses. Solution : For a (Carnot) refrigerator, the energy W that must be supplied to remove heat Q 2 from the freezer at temperature T2 K is
The Reversibility
159 K=
T2 273 = T1 - T2 300 - 273
10
where T1 is the temperature of the outer atmosphere. For the first refrigerator: T1 = 27 + 273 = 300 K, T2 = -10 + 273 = 263 K. \ W = (2000 joule) ×
300 - 263 = 281 joule . 263 263
For the second refrigerator : T1 = 17 + 273 = 290 K, T2 = - 20 + 273 = 253 K . \ W = (2000 joule) x
290 - 253 = 292 joule . 263
Since the first refrigerator runs at less energy input, hence it will be our choice. 25. A refrigerator is fitted with a motor of 200 watt. The temperature of its freezing compartment is - 3 °C and that of the room air is 27 °C. The maximum amount of heat which the refrigerator can remove from the freezing compartment in 10 minutes is calculated as follows. Solution : For an ideal (Carnot) refrigerator the work (energy) that must be supplied to remove an amount of heat Q2 from the freezing compartment at temperature T2 K is given by
W = Q2
T1 - T2 T2
where T2 K is the temperature of outside air. Therefore, the heat removed is given by
Thermal Physics
160 Q2 = W
T2 T1 - T2
Here T2 = - 3 + 273 = 270 K , T1 = 27 + 273 = 300 K and W = 200 watt = 200 joule/sec = 200 × 600 joule (in 10 minutes) = 1.2 ×l05 joule. Substituting these values in eq. (i), we get Q2 = (1.2 × l05 joule) ×
270 300 - 270
= 10.8 × 105 joule =
10.8 ´ 10 5 = 2.58 × l05 cal. 4.18
26. A refrigerator is driven by a 1000-watt electric motor, which is operating at an efficiency of 60%. If the refrigerator can be treated as a reversible heat engine operating between 0 °C and room temperature, which is 20 °C, the time required by it to freeze 100 kg of water, which is at 0 °C is calculated as below. Heat losses may be neglected. (Latent heat of fusion of ice is 80 cal/gm and j = 4.18 joule/cal.) Solution : The work (energy) that must be supplied to the refrigerator to remove an amount of heat Q2 from the freezing compartment is given by
W = Q2
T1 - T2 T2
where T1 and T2 are the Kelvin temperatures of the outer air and the freezing compartment respectively. Therefore, the heat removed is
The Reversibility
161 Q2 = W
T2 . T1 - T2
Now, the refrigerator is being driven by a motor of 1000 watt with 60% efficiency 60 = 600 watt. 100
∴ energy input W = 1000 x
Now,
T1 = 20 + 273 = 293 K and T2 = 0 + 273 = 273 K .
Therefore, the heat extracted from the freezing compartment is Q2 = 600 ×
=
273 = 8190 joule/sec 293-273
8190 = 1.96 × 103 joule/sec. 4.18
But the heat which is to be extracted from 100 kg of water to freeze it is mL = 100,000 × 80 = 8 × 106 cal. Hence the time required to extract 8 × 106 cal of heat is =
8 ´ 10 6 cal = 4082 sec. 1.96´ 10 3 cal / sec
27. The ice in a cold storage melts at the rate of 2 kg per hour when the external temperature is 20 °C. The power of the motor driving the refrigerator which may just prevent the ice from melting is found out in following way. (Latent heat of fusion of ice = 80 cal/gm, J = 4 2 joule/cal). Solution : The energy that must be supplied to the refrigerator to remove an amount of heat Q2 from the freezing compartment is given by W = Q2
T1 - T 2 T2
...(i)
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162
where T1 and T2 are the Kelvin temperatures of the outer air and the freezing compartment respectively. Now, the ice melts at the rate of 2 kg/hour, or 2000/ (60 x 60) = 5/9 gram/sec. Therefore, the heat that must be removed to prevent the ice from melting would be Q2 =
5 gram cal 400 cal × 80 = 9 sec gram 9 sec
Also, T1 = 20 + 273 = 293 K and T2 = 0 + 273 = 273 K . Substituting these values in eq. (i), we get W=
400 293 - 273 ´ 9 273
= 3.256 cal/sec = 3.256 × 4.2 = 13.7 joule/sec = 13.7 watt.
Physical Significance
163
8 Physical Significance Entropy If a substance takes in an amount of heat Q in a reversible process at a constant temperature T, then Q/T is called the “increase in entropy” of the substance. Similarly, if the substance gives up an amount of heat Q at constant temperature T, then Q/T is called the “decrease in entropy” of the substance. The change in entropy is denoted by Δ S. Thus Δs =
Q . T
If the temperature of the substance does not remain constant during the process, we may consider the heat to be taken in or given up in successive small elements dQ such that the temperature remains sensibly constant for each element. The change in entropy will then be Δs = ∫
dQ . T
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Like energy U and temperature T, the entropy S of a system is a physical property that can be measured in the laboratory. It provides an alternative statement of the second law of thermodynamics according to which only those processes are possible for a system in which the entropy of the system plus surroundings increases . Thus we can decide whether or not an event will occur spontaneously. An event would occur provided it will cause the entropy of the universe (system + surroundings) to increase. Events causing decrease in entropy of the universe are impossible. For example, heat can flow from higher to lower temperature, but the reverse is not possible. Free expansion of a gas is possible, free compression is not. Therefore, on account of the processes occurring in nature, the entropy of the universe is continuously increasing. Entropy does not obey conservation law (energy does). Change in Entropy in Reversible Cycle—Clausius Theorem : The Clausius theorem states that “in any reversible cycle the net change in entropy is zero,” that is,
∫
dQ =0 . T
Let us first prove it for a (reversible) Carnot cycle (Fig.). Let us start with working substance in the state A . During the isothermal expansion AB, the working substance takes in an amount of heat Q1 from the source at the constant temperature T1 of the source. Its entropy therefore increases by Q1/T1 . This is also the decrease in the entropy of the source. During the adiabatic expansion BC no heat is taken in or given up, so the entropy remains unchanged. During the isothermal compression CD , the working substance gives up an amount of heat Q2 to the sink at constant temperature T2. Its entropy therefore decreases by Q2/T2 . This is also the increase in the entropy of the sink. During the adiabatic compression DA there is again no change in entropy. Thus, for the whole cycle, the net increase in the entropy of the working substance is
Physical Significance
165
Δs =
Q1 Q2 − T1 T2 .
This is also the net decrease in the entropy of the sourcesink system. But, by the definition of Kelvin’s scale of temperature, Q1 Q2 − T1 T2 . Therefore Δs =
Q1 Q2 − =0 . T1 T2
Thus , the net change in the entropy of the working substance, and also of the surroundings, during a complete Carnot cycle is zero. If we regard the heat taken in by the substance as positive and heat given up as negative, then Q1 will be positive and Q2 will be negative. The above equation then becomes Δs =
Q1 Q2 + =0 . T1 T2
...(i)
This equation indicates that the sum of the quantities Q/ T is zero for a Carnot cycle. Let us now consider the general case of any reversible cycle, indicated by the curve A → B → A (Fig.). We may consider this cycle to be made up of a large number of
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elementary Carnot cycles abcd, efgh, ijkl, etc. Let us imagine that the substance, instead of tracing the smooth curve A → B → A , traces successively the cycles abcd, efgh, ijkl. .........In following these cycles the portions bh ,fl,jp, nt and so on , are traversed twice in the reverse order, and their effects are thus can celled out. The net effect of the whole process is that the substance goes along the closed zigzag path abcfi ... ghcda . Now, for each elementary Carnot cycle the above relation (i) holds and may be written in the form
δs =
δQ1 δQ2 + =0 . T1 T2
Taking the sum of such results for all the cycles we conclude that for the closed zigzag path abefi, ......ghcda , we have Δs = Σ
δQ =0 . T
If the adiabatics of the elementary Carnot cycles be infinitesimally close together, they will be connected with infinitesimally small isothermals. Then the zigzag path will coincide with the smooth curve A → B → A , and we may write Δs = ∫
dQ =0 . T
Physical Significance The symbol
∫
167 denotes the integration over the whole
cycle. Thus , in any reversible cycle, the net change in entropy is zero . This is ‘Clausius theorem.’ Change in Entropy in Irreversible Cycle : The efficiency of a Carnot reversible cycle working between absolute temperatures T1 and T2 is given by e =1−
Q2 Q1 ,
where Q1 is the amount of heat taken in at temperature T1 and Q2 that given up at temperature T2. Since the cycle is reversible, Q2 T2 we have Q = T . Thus 1 1 e =1−
Q2 T =1− 2 Q1 T1
This is the maximum possible efficiency of an engine working between temperatures T1 and T2. If the cycle of the engine be irreversible, the efficiency will be lowered. Thus, in this case e =1−
Q2 T <1− 2 Q1 T1
or
Q2 T2 > Q1 T1
or
Q2 Q1 > T2 T1
Now, during a complete irreversible cycle, the entropy of the source decreases by Q1/T1 , while that of the sink increases by Q2/T2. The net change in the entropy of the working substance in this case also is zero because when the cycle is completed, the working substance recovers its initial state. Thus the total
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increase in the entropy of the system (working substance) plus the surroundings (source and sink) is ΔS =
Q2 Q1 − T2 T1 ,
Q2 Q1 which is positive since T > T . Thus, in an irreversible cycle 2 1 the entropy of the system plus its surroundings always increases .
In general, we can write ΔS (universe) > 0,
where the equality sign holds for reversible process and the inequality sign for irreversible process. Here the word universe means system + surroundings. Motion Involve a Change : In a purely mechanical motion there is no exchange of heat and hence no change in entropy. The mixing of two gases is an irreversible process. Hence the entropy will increase i.e. the change in entropy will be positive. Entropy is a State Function : Let two points i and f (Fig.) represent any two states of a system, characterised by definite values of pressure, volume and temperature. Suppose the system can pass ifreversibly from the initial state i to the final state f, either by the path 1 or by the path 2. Now, if the system goes through the complete cycle i → f → i, the net change in entropy for the cycle as a whole is zero, that is
Physical Significance
169 dQ = 0 (Clausius theorem) T
∫
Let us put it as a sum of entropy-changes in two parts, i → 1 → f and f → 2 → i. Then we write
∫
i dQ dQ + 2∫ =0 . f T T
f
1 i
Now, if the direction of traverse of the part f—¥ 2—ti be reversed, the entropy-change will be reversed in sign, since the heat-change will now take place in the reverse direction. That is 2
∫
dQ = −2 T
i f
f
∫
i
dQ T
Substituting this in eq. (i), we get
∫
1 i
f
f dQ dQ = 2∫ . i T T
dQ is the change in entropy in passing T from state i to state f. Thus , the change in entropy between any two equilibrium states of a system is independent of the reversible path connecting these states. In other words, the entropy of a system is a definite function of its state and is independent of the way by which that state has been achieved. Thus entropy is a ‘state variable’.
The quantity
∫
f
i
Formulation of the Second Law of Thermodynamics : Let Si and Sf be the entropies of a system in an initial state i and a final state f (measured from some arbitrary zero). Then , the entropy-change will be ΔS = S f − Si = ∫
i
f
dQ . T
This relation applies only for reversible path. If the two states are infinitesimally close to each other, the above equation will be written as
Thermal Physics
170 dS =
or
dQ T
dQ = TdS.
This equation is the mathematical representation of the second law of thermodynamics. Entropy and the Second Law of Thermodynamics : The second law of thermodynamics is concerned with the direction in which any physical or chemical process involving energychange takes place. It has emerged out from experience, and cannot be proved theoretically. The consideration of entropy-changes in reversible and irreversible processes enables us to express the second law in terms of entropy in the following way : All natural (or irreversible) processes that proceed from one equilibrium state to another of a system takes place in a direction that causes the entropy of the system plus surroundings to increase in the limiting (ideal) case of a reversible process the entropy of the system plus surroundings remains constant , that is, ΔS (universe) ≥ 0. A process for which ΔS (universe) < 0 is impossible. The above entropy statement of the second law is consistent with both the Kelvin-Planck statement and the Clausius statement of the law. According to Kelvin’s statement of second law, we cannot have a ‘perfect’ heat engine in which the working substance would convert into work all the heat Q taken in from the source at temperature T. If this were so, the entropy of the source would decrease by Q/T, whereas that of the working substance would remain unchanged (because it returns to its initial state after completing the cycle). Thus, there would be a net decrease in the total entropy of the system plus surroundings, which is against the entropy statement.
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171
Similarly, according to Clausius statement of second law, we cannot have a ‘perfect’ refrigerator in which the working substance would transfer heat Q from a cold body at temperature T2 to a hot body at temperature T1; without taking energy from an external source. If this were so, the entropy of the cold body would decrease by Q/T2, that of the hot body would increase by Q/T2, and that of the working substance would remain unchanged because it undergoes a cycle. Thus, ⎛Q Q⎞ as T2< T1, there would be a net decrease of ⎜ T − T ⎟ ⎝ 2 1 ⎠
in the
total entropy of the system plus surroundings, which violates the entropy statement of the second law. Thus, all three statements of the second law of thermodynamics are equivalent.
Entropy Change of an ‘Isolated’ System According to the entropy-statement of the second law of thermodynamics, the total entropy of a thermodynamic system plus surroundings, that is, the entropy of the universe either remains constant (in reversible processes ) or increases (in irreversible processes): ΔS (universe) ≥ 0 .
Processes involving ΔS (universe) < 0 are impossible. If, however, the system is completely isolated from its surroundings {i.e. it can neither exchange heat nor work with its surroundings) then the entropy of the system alone is the entropy of the universe which should either remain constant or increase. Thus ΔS (isolated system) ≥ 0 .
Processes involving ΔS (isolated system) < 0 are impossible, that is, the entropy of an isolated system can never decrease.
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172
Cosmic Rays Falling on Earth Some processes in nature appear to involve a decrease in entropy and thus violate the second law of thermodynamics. For example, it is said that the cosmic rays falling on earth decrease the entropy of the earth. Actually, this is not a violation of the second law of thermodynamics. The reason is that the earth is not an isolated system. Its entropy can decrease but then somewhere else in the universe there will be an increase in entropy. In the given example, the entropy of the earth decreases but at the same time the entropy in some remote parts of the universe from where the cosmic rays come increases. Isentropic Processes: A reversible process during which the entropy of a system remains constant is called “isentropic process”. A reversible adiabatic process is an isentropic process. For such a process, we have dQ = 0 (no exchange of heat) so that or
dS =
dQ =0 T
S = constant.
Thus, the entropy of a system remains constant during a reversible adiabatic process in the system , and the process is isentropic. This is why the adiabatics are also known as “isentropics”. Isothermal and Isentropic Processes in T-S Diagram : For the heat transferred in a reversible process, we have the relation dQ = TdS. It therefore follows that the total heat transferred in a reversible process carrying a system from an initial state i to a final state f is given by Q =
∫
i
f
TdS .
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173
This integral can be interpreted graphically as the area under a curve on T-S diagram, in which T is plotted along the y-axis and S along the x-axis. An isothermal process (T constant) would be a horizontal line (parallel to S-axis) on a T-S diagram. Anisentropic (adiabatic) process (5 constant) would be a vertical line (parallel to T-axis) on a T-S diagram.
Carnot Cycle on T-S Diagram A Carnot cycle consists of two reversible isothermal processes and two reversible adiabatic processes (Fig. a). Hence it forms a rectangle on a T-S diagram, no matter what the working substance is (Fig. b).
During the isothermal expansion AB at constant temperature T1, the entropy of the working substance increases from S1 to S2. During the adiabatic expansion BC , the temperature falls to T2 , but the entropy remains constant. During the isothermal compression CD at constant temperature T2 , the entropy decreases from S2 to S1. Finally, during the adiabatic compression DA , the temperature rises to T1, the entropy remaining constant.
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174
Let Q1 be the amount of heat absorbed by the working substance during isothermal expansion AB and Q2 that rejected during isothermal compression CD . By the relation dQ = TdS, we have Q1 = T1 (S2 – S1) = AH × AB = area ABGH and
Q1 = T2 (S2 -S1) = DH × DC = area DCGH.
The difference Q1 – Q2 is converted into useful work and is thus the available energy per cycle . It is given by Q1– Q2 = area ABGH – area DCGH = area ABCD . Thus, the area of the cycle on T-S diagram represents available energy. Practical Utility : The T-S diagrams are very useful for the study of engines. On account of simplicity of figure the area of the cycle can be easily computed. Efficiency of Carnot Engine : The efficiency of the engine is defined as e=
=
area ABCD area ABGH
=
AB × BC AB × AH
=
or
Q1 − Q2 W = Q1 Q1
=
BC AH
T1 − T2 T1 e = 1−
T2 T1
This is the required expression.
Physical Significance
175
In the above given figure the entropy of a system is increasing at constant temperature . Hence the process is isothermal. In Fig. the temperature of a system is increasing, the entropy remaining constant. Hence the process is isentropic (or adiabatic).
If Q1 be the heat absorbed during the process in which entropy increases, and Q2 the heat given up during the process in which entropy decreases, then the efficiency is given by e=
Q1 − Q2 areaADC = , Q1 Q1
where Q1 – Q2 is the available energy which is equal to the area of the cycle on the T-S diagram. In Fig.(a) and (b) the areas of the cycles are equal. However, in each cycle, Q1 is given by the area ABED , which is greater for the cycle (a). Hence the efficiency of cycle (b) is larger.
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176
For the heat transferred in a reversible process, we have the relation dQ = TdS.
...(i)
If 1 mole of a gas is heated at constant volume (isochoric process) at temperature T, then dQ = CvdT.
...(ii)
Equations (i) and (ii) yield dT T = , dS Cv
...(iii)
where dT/dS is the slope of the isochoric curve on T-S diagram. If the gas is heated at constant pressure (isobaric process), then dQ = CpdT which gives
...(iv)
dT T = dS Cp
...(v)
Thus the slopes of isochoric and isobaric curves on T-S diagram are T/Cv and T/Cp respectively. Since Cv < Cp , the slope of the isochoric curve is greater. The ratio of the slopes is slope(isochoric ) slope(isobaric) =
T T
Cv Cp
=
Cp Cv
=ϒ
The Entropy of the Universe is increasing (Principle of Increase of Entropy) : Whenever the entropy of a system changes, the entropy of its surrounding bodies also changes. The sum of all these entropy-changes of all the bodies taking part in the process is called the ‘entropy-change of the universe’. According to the principle of increase of entropy, whenever any natural, that is, irreversible process takes place, the entropy of the universe increases .
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177
Let us consider the entropy-change of the universe when a system goes from an initial equilibrium state i to a final equilibrium state f through a natural (irreversible) process. There is an entropy-difference S f—S i between the two equilibrium states, but we cannot calculate it from the relation S f − Si = ∫
i
f
dQ , because this relation applies only to reversible T
paths. Hence, instead of the actual irreversible path , we choose some reversible path connecting the states i and f and calculate the entropy-change for that path. We are justified in doing so because entropy is the characteristic of the state only, no matter how the state has been achieved. Thus S f − Si = ∫
i
f
dQ T
(over any reversible path).
Irreversible processes are of various types. As an example, let us consider heat conduction. Heat Conduction (Equalisation of Temperature) : Let a hot body A at temperature T1 be placed in contact with a cold body B at temperature T2 in a box which is thermally insulated from the surroundings. An amount of heat Q (say) flows irreversibly from the hot to the cold body so that both reach a common intermediate temperature Tm (say). To calculate the entropy-change of the system, we have to think of some reversible process by which heat Q could be conducted from the hot (T1) to the cold (T2) body so that both reach the temperature Tm. This is possible if we have a heatreservoir whose temperature can always be changed and adjusted to any value. We adjust its temperature to T1 and put the hot body A in contact with it. Then we lower the reservoir temperature infinitesimally slowly to Tm , so that heat (Q) from the hot body (A) flows reversibly to the reservoir until the temperature of the body A falls to Tm. (The flow is reversible because at every stage the
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178
temperature of the reservoir is only infinitesimally lower than that of the hot body.) The loss in the entropy of the body A is ΔS1 = −
Q T1m ,
where T1m is the average of T1 and Tm . Now, we adjust the reservoir temperature to T2 and put the cold body B in contact with it. Then we raise the reservoir temperature infinitesimally slowly to Tm , so that heat (Q) from the reservoir flows reversibly to the cold body B until the temperature of the body B rises to Tm. The gain in the entropy of the body B is ΔS2 = +
Q T2m
where T2m is the average of T2 and Tm . Both bodies are now at the same temperature Tm . The change in entropy for the system is thus ΔS = ΔS1 + ΔS2
Q Q Q Q = – T +T = – T −T , 1m 2m 2m 1m
which is positive (since T1m > T2m). This is the increase in entropy of the system due to the reversible process selected. The same would be the increase in entropy of the system due to the actual irreversible heat conduction. Now, in the actual heat conduction the entropy-change of the surroundings is zero because the system is kept thermally insulated. Hence the entropy-change of the universe is ⎛ Q ⎛Q Q⎞ Q ⎞ ΔS (universe) = ⎜ T − T ⎟ = ⎜ T − T ⎟ , ⎝ 2m ⎝ 2 1m ⎠ 1 ⎠
Physical Significance
179
which is positive (since T1m > T2m). Thus the entropy of the universe increases in the actual irreversible conduction of heat, as required by the principle of increase of entropy. All natural processes, like conduction and radiation, are irreversible. These processes tend to equalize temperatures between various bodies. It, therefore, follows that the entropy of the universe is continually increasing. The Available Energy of the Universe is Decreasing : Heat energy is available for conversion into work only when it is let down from a higher to a lower temperature. For example, in a perfectly reversible Carnot engine, a quantity of heat Q1 is taken in from a hot body at temperature T1, a quantity Q2 is given up to a cold body at temperature T2 , the balance Q1– Q2 being converted into work and is called the ‘available energy’. Thus available energy = Q1 – Q2 ⎛ Q2 ⎞ = Q1 = ⎜ 1 − Q ⎟ ⎝ 1 ⎠ ⎛ T2 ⎞ = Q1 = ⎜ 1 − T ⎟ . ⎝ 1 ⎠
⎡ Q2 T2 ⎤ = ⎥ ⎢Q ⎣ Q1 T1 ⎦
Clearly, the lower the temperature T2 of the cold body, the greater will be the available energy. If T0 is the temperature of the coldest body available, then ⎛ T0 ⎞ maximum available energy = Q1 = ⎜ 1 − T ⎟ . ⎝ 1 ⎠
Let us now consider the (irreversible) conduction of heat Q from a hot body at temperature T1 to a cold body at temperature T2. Let T0 be the lowest available temperature. Before conduction, the heat Q was available in a body at a higher temperature T1. Hence the maximum amount of energy ⎛ T0 ⎞ available from this was Q = ⎜ 1 − T ⎟ . But, after conduction, the ⎝ 1 ⎠
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180
heat Q is available at the lower temperature T2. Hence the ⎛ T0 ⎞ maximum available energy has reduced to Q = ⎜ 1 − T ⎟ . Thus ⎝ 2 ⎠ the energy that has become unavailable for work ⎛ T0 ⎞ ⎛ T0 ⎞ = Q =⎜1− T ⎟ – Q =⎜1− T ⎟ ⎝ ⎝ 1 ⎠ 2 ⎠ ⎛ T0 T0 ⎞ = Q =⎜ T − T ⎟ ⎝ 2 1 ⎠ ⎛Q Q⎞ = T0 = ⎜ T − T ⎟ ⎝ 2 1 ⎠
= T0 ( ΔS ). ⎛ Q Q⎞ where ΔS ⎜ = T − T ⎟ is the increase in entropy of the universe ⎝ 2 1 ⎠ brought about by the (irreversible) conduction of heat. Hence, we conclude that the energy that becomes unavailable for work during an irreversible process is To times the increase in entropy of the universe due to that irreversible process .
Since irreversible processes are continually going on in nature, energy is continually becoming unavailable for work i.e. the available energy of the universe is continually decreasing. This is known as the ‘degradation of energy’.
Heat energy is available for work only when there is a difference of temperature, so that heat can flow. In A there is no “available energy”. Hence the entropy of A is maximum (As entropy increases, the available energy decreases). Change in Entropy during Free Expansion : Let μ moles
Physical Significance
181
of an ideal gas be contained in a vessel which is connected to another ‘evacuated’ vessel by means of a stop-cock, both being thermally insulated from the surroundings. As the stop-cock is suddenly opened, the gas rushes into the vacuum of the second vessel. This is the irreversible free expansion of the gas. As the system is thermally insulated, there is no exchange of heat with the surroundings (Q = 0). Also no work is done against the vacuum (W = 0). Hence, from first law ΔU = Q – W, we have ΔU = 0.
The internal energy of the gas remains unchanged. Because the gas is ideal, its temperature T remains constant. Suppose we wish to calculate the entropy-change of the system due to the irreversible free expansion from an initial volume Vi to a final volume Vf. We can do so by considering some reversible process between the same initial and final states of the system. Because in the free expansion of the gas the temperature remains constant, a convenient reversible process, which can be substituted for the irreversible free expansion, is the isothermal expansion of the gas from the initial volume Vi to the final volume Vf. (This we can do by bringing the gas in contact of a heat-source at temperature T which supplies heat to the gas). The entropy-change is ΔS = Sf – Si =
∫
Vf
Vi
dQ . T
...(i)
Now, by the first law of thermodynamics, we have dQ = dU + dW. But dU = 0 (as the gas is ideal, there is no change in internal energy in isothermal expansion). ∴
dQ = dW = pdV = μ RT
dV (∴ pV = μ RT) V
Thermal Physics
182 dQ dV = μR . T V
Substituting this value of or
ΔS = μR ∫
Vf
Vi
dQ in eq. (i), we have T
dV V = μ R [log e V ]Vif V ΔS = μR log e
Vf Vi
(reversible isothermal expansion) which is positive, since Vf > Vi. This is the increase in the entropy of the ‘system’ (ideal gas) undergoing reversible isothermal expansion. Note that this is the increase in the entropy of the system (gas) alone due to the reversible isothermal expansion. The entropy of the universe , however, remains unchanged during the process because there is an equal decrease in the entropy of the heat-source. The above will also be the increase in entropy of the gas in the actual irreversible free expansion from volume Vi to Vf. Now, in free expansion, the entropy-change of the surroundings is zero because there is no exchange of heat with the surroundings. Hence the increase in entropy of the universe due to the free expansion is ΔS(universe ) = μR log e
Vf Vi
(free expansion).
Thus we conclude that the entropy of the universe increases during free expansion (an irreversible process).
Reversible Isothermal Compression When 1 mole of an ideal gas undergoes reversible isothermal compression from volume Vj to Vf, the change in entropy of the gas is
Physical Significance
183 ΔS = R log e
Vf Vi
.
Since Vf< Vi (compression), ΔS is negative which means that the entropy of the gas decreases . There is also an equal increase in the entropy of the heatreservoir in whose contact the gas has been put to carry out the isothermal compression. Thus, the total entropy of the universe (gas plus surroundings) remains unchanged, a characteristic of the reversible process. Reversible Adiabatic Expansion : In adiabatic expansion (or compression) the gas does not exchange heat with the surroundings. Hence there is no change in entropy. Free Expansion: When 1 mole of an ideal gas undergoes free expansion from volume Vi to Vf, the change in the entropy of the universe is Vf ΔS (universe) = R log e V . i
Since Vf > Vi (expansion), ΔS is positive which means that the entropy of the universe increases, a characteristic of irreversible process. In free expansion, the increase in the entropy of the universe is same as that of the gas alone because in this case the entropy of the surroundings does not change. Entropy-change on Heating a Substance : Suppose a liquid of mass m and specific heat c is heated from T1 to T2. We can connect the initial and final states by a reversible process of heating, making use of an infinite number of heat-reservoirs with temperature ranging from T1 to T2 . If dQ be the amount of heat taken in by the liquid for an infinitesimally small temperature-rise dT, then we have dQ = mc dT.
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184
The corresponding entropy-change of the liquid is dS =
dQ dT = mc . T T
The entropy- change in the entire process is therefore ΔS (liquid) =
∫
T2
T1
T2
dS = mc ∫ T
1
dT T
T2 = mc [log e T]TT12 = mc loge T , 1
which is positive (since T2 > T1). Thus the entropy of the liquid increases . The same would be the increase in the actual irreversible heating. The above is the increase in the entropy of the system (liquid) alone due to the reversible heating. The entropy of the universe (system + surroundings) remains, however, unchanged because for every dQ absorbed by the liquid (system) at T, there is an equal amount of heat given up by the reservoir (surroundings) at T, so that there is an equal decrease in the entropy of the surroundings. In the actual irreversible heating the liquid is placed in contact with a single reservoir. In this case the decrease in entropy of the reservoir (surroundings) is less than the increase in entropy of the liquid (system). Thus in the actual irreversible process the entropy of the universe increases. Entropy-change on Cooling : When a substance is cooled from temperature T1 to T2 , its entropy decreases, because T2 ΔS = me loge T , 1
and T2 < T1 (cooling), so that ΔS is negative. The total entropy of the universe, however, can never decrease ; if always increases . Entropy-change in Melting of Ice : Suppose ice, of mass m
Physical Significance
185
and latent heat L, melts at a (constant) Kelvin temperature T. It takes an amount of heat Q = mL . We can connect the initial and final states by a reversible process of melting, by bringing ice in contact of a heat-reservoir at a temperature only infinitesimally higher than T. Since the temperature remains constant, the entropy-change in the entire process of melting is ΔS =
∫
dQ Q mL = = , T T T
which is positive. This is the increase in entropy of the system (ice) alone . There is an equal decrease in entropy of the surroundings (heat-reservoir), so that in reversible melting the entropy-change of the universe is zero. In the actual irreversible melting, however, the entropy of the universe increases. Entropy-change when Ice is converted into Steam : Let T1 be the Kelvin temperature at which ice melts into water, and T2 the Kelvin temperature at which water is boiled to steam. Then, the entropy-change when ice is converted into water is ΔS1 =
mL1 T1 ,
where L1 is the latent heat of fusion of ice. The entropy-change when water is heated from T1 to T2 is T2 ΔS2 = mc loge T , 1
where c is the specific heat of water. The entropy-change when water is converted into steam is ΔS3 =
mL2 T2 ,
where L2 is the latent heat of vaporisation of water. Thus, the total entropy-change is ΔS = ΔS1 + ΔS2 + ΔS3
Thermal Physics
186 =
mL1 T2 mL2 + mc log , + e T T1 T2 . 1
This is the entropy-change of the system (ice) alone. Entropy-change on Mixing Liquids at Different Temperatures: Let c be the specific heat of the liquid. On mixing equal masses m of the same liquid at temperatures T1 and T2(T1 > T2), heat flows from the hotter to the colder liquid. Let T be the equilibrium temperature of the mixture. By the principle of calorimetry, we have mc (T1 - T) = mc (T - T2)
∴
T1 + T2 2
T=
The hotter liquid gives up heat in cooling from T1 to T. Replacing the actual irreversible process by a reversible process involving an infinite number of heat-reservoirs with temperatures ranging from T1 to T, the entropy-change is given by ΔS1 =
∫
T
T1
dQ = mc T
∫
T
T1
dT T = mc loge T T1
.
Similarly, the entropy-change in the colder liquid which takes in heat in rising from T2 to T is given by ΔS2 =
∫
T
T2
dQ = mc T
∫
T
T2
dT T = mc loge T T2
.
The entropy-change of the system is ΔS =
ΔS1 + ΔS2
⎡ T T⎤ = mc ⎢log e T + log e T ⎥ ⎣ 1 2 ⎦ T2 = mc log e T T 1 2
[∴ logA + logB = logAB]
Physical Significance =
187 ⎛ T mc loge ⎜⎜ ⎝ T1T2
⎞ ⎟⎟ ⎠
2
T = 2mc loge T T 1 2
= 2mc loge
(T1 + T2 ) 2 T1T2
This is the required expression. We know that the arithmetic mean of two numbers is T +T always greater than their geometric mean. Thus 1 2 > T1T2 2 . Hence ΔS is necessarily positive which means that the entropy of the system increases in the irreversible process of mixing. Since the system is thermally insulated, there is no entropychange in the surroundings, and ΔS , as calculated above , represents the entropy-change of the universe. We conclude that the entropy of the universe increases .
Entropy-change in Irreversible Heat Transfer Let T1 > T2. Then heat will flow from the body m1 to the body m2 until both reach at an equilibrium temperature T. By the principle of calorimetry, we have m1c1(T1 - T) = m2c2(T - T2). ∴
T =
m1 c1T1 + m2 c2 T2 m1c1 + m2 c2 . (T1 > T > T2)
Replacing the actual irreversible process by two reversible processes involving infinite succession of reservoirs, we have ΔS1 =
m1c1
∫
T
T1
dT T
T = m1c1 loge T < 0 1
(cooling from T1 to T)
Thermal Physics
188 and
T
ΔS2 = m2c2 ∫T
2
T dT = m2c2loge T > 0 T 2
(heating from T2 to T). :. total change in the entropy of the system is ΔS = ΔS1 + ΔS2
⎛
⎞
= log e ⎜ T ⎟
m1 c1
⎝ T1 ⎠
⎛T⎞ + log e ⎜ ⎟ ⎝ T2 ⎠
m2 c2
⎡⎛ T ⎞m1c1 ⎛ T ⎞ m2 c2 ⎤ ⎟ ⎜ ⎟ ⎥ > 0. ⎢⎣⎝ T1 ⎠ ⎝ T2 ⎠ ⎥⎦
= log e ⎢⎜
The entropy of the system increases.
Entropy of a Perfect Gas Let us consider one mole of a perfect gas at pressure p, Kelvin temperature T and volume V. When an infinitesimal amount of heat dQ is added to it, the increase in entropy is dS =
dQ . T
If dU be the increase in the internal energy of the gas and dW the external work done, then by the first law of thermodynamics, we have dQ = dU + dW. If Cv be the molar specific heat of the gas at constant volume, dT the rise in temperature and dV the change in volume, then dU = CvdT dV . V
and
dW = pdV = RT
:.
dQ = Cv dT + RT
dV V
[∴ pV = RT]
Physical Significance and so
189
dS =
dQ dT dV = Cv +R T T V
...(i)
The entropy of the gas, measured from an arbitrary zero, is then given by S =
∫
dQ dT dV = Cv ∫ + R∫ . T T V
From gas eq. pV = RT, we have dT =
...(ii)
pdV + Vdp and from R
Mayer’s relation, R = Cp – Cv. Thus, from eq. (ii), S = Cv ∫
pdV + Vdp dV + (Cp − Cv )∫ RT V
Again, RT = pV, so that S = Cv ∫
or
pdV + Vdp dV + (Cp − Cv )∫ pV V
= Cv ∫
dp dV dV + Cv ∫ + (Cp − Cv )∫ V p V
= Cv ∫
dp dV + Cp ∫ p V
S = CV logep + Cp loge V + constant....(iii)
This is the required expression. If a gas passes from an initial state pi, Vi, Tf, to a final state Pf,Vf, Tf then, from eq. (iii), the change in entropy of the gas is Pf dp Vf dV ΔS = Cv ∫Pi p + Cp ∫Vi V
or
Pf Vf ΔS = Cv loge P + Cp loge V . i i
This is the entropy-change in terms of pressures, volumes and molar specific heats.
Thermal Physics
190
Entropy-change in a Perfect Gas : Proceeding as in last question, we reach the following equation (i): so that
ds =
Cv dT dQ dV +R = . T T V
If the gas passes from an initial state i having Pi, Vi, and T1, as its pressure, volume and temperature respectively, to a final state f with Pf,Vf, Tf as the corresponding quantities, then the change in entropy is ΔS =
∫
i
f
dS
= Cv ∫i
f
∫
f
i
dV ⎞ ⎛ Cv dT +R ⎜ ⎟ V ⎠ ⎝ T
f dV dT + R∫ i V T
Tf Vf ΔS = Cv loge T + R loge V . i i
or
...(i)
This is the expression for the entropy-change in terms of temperatures, volumes and specific heat at constant volume. piVi Ti
Now,
=
p f Vf Tf
Tf
p f Vf
or T = p V and R = Cp – Cv. i i i
∴
pf Vf ⎛ + log log ⎜ e e = C ΔS v pi Vi ⎝
or
ΔS = Cv log e
pf pi
+ Cp log e
Vf ⎞ ⎟ + (Cp − Cv )log e Vi ⎠
Vf Vi
...(ii)
This is the expression for the entropy- change in terms of pressures, volumes and specific heats. Again,
Vf Vi
=
p f Tf pi Ti
, and so eq. (i) can be written as
⎛ Tf Tf pi + + log log log C R ⎜ v e e e = ΔS ⎜ pf Ti Ti ⎝
⎞ ⎟ ⎟ ⎠
Physical Significance
191 Cv log e
Tf Ti
− R log e
pf pi
+ R log e
Tf
pf
Tf Ti Tf
= (Cp – R) log e T − R log e p + R log e T i i i Tf pf ΔS = Cp log e T − R log e p . i i
...(iii)
This is the expression for the entropy-change in terms of temperatures, pressures and specific heat at constant pressure. Entropy- change in a Van der waals’ Gas : The equation of state for one mole of a Van der waals’ gas is a ⎞ ⎛ ⎜ p + 2 ⎟ (V-b) = RT, V ⎠ ⎝
where a and b are Van der waals’ constants. When an infinitesimal amount of heat dQ is added to the gas at Kelvin temperature T, the increase in entropy is dS =
dQ . T
By the first law of thermodynamics, we have dQ = dU + dW. If Cv be the molar specific heat at constant volume, then the increase in internal energy is dU = CvdT and the work done including internal work against the molecular attraction, is a ⎞ RT ⎛ dV . dW = ⎜ p + 2 ⎟ dV = V ⎠ V −b ⎝ ∴
dS =
dT dV dU + dW +R = Cv T V −b T
Thermal Physics
192
The change in entropy when the gas passes from the state (Vi, Ti) to (Vf, Tf) is sf – si =
∫
i
f
dS = Cv ∫i
f
f dV dT + R∫ i V −b T
Tf
Vf − b
= Cv log e T + R log e V − b . i i Let a small amount of heat dQ be given to 1 gm of water at absolute temperature T. The change in entropy is dS =
dQ . T
If the temperature of the water rises by dT, then dQ = dT, the specific heat being unity. ∴
dS =
dT . T
Integrating, we get S =
∫
dT = loge T + A, T
where A is the constant of integration. If we take S = 0 at 0 °C (T= 273 K), then A = – loge 273. Hence, for 1 gm of water, we have Swater = loge T - loge 273 = loge
T . 273
Now, the amount of heat needed to convert 1 gm of water into steam at a temperature T is L, where L is the latent heat. Hence the increase in entropy will be L/T. Thus the total entropy of 1 gm of dry saturated steam at a temperature T will be Ssteam = Swater +
T L L = loge + . 273 T T
Physical Significance
193
Nernst Heat Theorem : Third Law of Thermodynamics : The second law of thermodynamics defines only difference in entropy between two equilibrium states connected by a reversible path. Such a path , however, exists if both the states lie on the same sheet of a p - V - T surface. If we consider two different substances, or metastable phases of the same substance, such a reversible path may not exist. Therefore, the second law does not uniquely determine even the difference in entropy. Nernst, in 1905, supplied a rule for unique determination of entropy. It states : The entropy of any system at absolute zero is a universal constant, which may be taken to be zero. This statement is known as ‘Nernst heat theorem’ or ‘third law of thermodynamics’ ; it means that 5 = 0 at T = 0, whatever be the values of any other parameters on which entropy S may depend. Thus, the third law renders the entropy of any state of any system unique. Two important consequences of the third law are : (i) Heat capacities of a system vanish at absolute zero, (ii) Coefficient of volume expansion of any substance vanishes at absolute zero. An alternative statement of the third law is the unattainability of absolute zero. A fundamental feature of all cooling processes is that the lower the temperature achieved, the harder it is to go still lower. It can be generalised in the following statement: It is impossible by any procedure, no matter how idealised, to reduce any system to the absolute zero of temperature in a finite number of operations. This statement implies that the entropy of a system cannot be reduced to zero.
Thermal Physics
194
PROBLEMS 1. One end of a copper rod is in contact with a heat reservoir at 127 °C and the other with a reservoir at 27°C. During a certain time-interval, 1200 cal of heat is conducted through the rod. We can look into the total entropy-change and whether the entropy of the rod change. Solution : The hot reservoir at a constant temperature of 127°C (= 400 K) gives up 1200 cal of heat to the cooler reservoir at a constant temperature of 27°C (= 300 K). The entropychanges of the reservoirs are ΔS (hot) =
−1200 Q = = – 3.0 cal/K. 400 T
(The heat given up is considered as negative.) 1200 Q = 300 = + 4.0 cal/K. T
and
ΔS (cold)
∴
ΔS total = – 3.0 + 4.0 = 1.0 cal/K.
=
The copper rod is in the steady state. It is neither absorbing nor giving up any heat. Hence there is no change in the entropy of the rod. 2. One mole of an ideal gas expands isothermally and reversibly to twice its original volume. Calculate the change in entropy of the gas, of the universe. We may see as what happens if the gas had undergone free expansion. Solution : The change in entropy of the gas in expanding from Vi to Vf (=2 Vi) is given by Vf ΔS = μ R loge V i
= 1 mole × 8.31 joule/(mole-K) × loge 2 = 1 × 8.31 × 0.693 = + 5.76 joule/K.
Physical Significance
195
This is an entropy increase. There is also an equal decrease in the entropy of the heat source which supplies heat to the gas for isothermal expansion. Thus, the entropy of the universe (gas-plus surroundings) remains unchanged. If the gas undergoes free expansion, then the entropy of the surroundings does not change because the expanding gas is thermally isolated from its surroundings. Now the entropy of the universe increases by 5.76 joule/K . 3. Three grams of nitrogen expand isothermally to twice the original volume. The change in entropy many be calculated. The molecular weight of nitrogen is 28. Given : R = 8.31 joule/(mole-K), / = 4.18 joule/calorie and loge 2 = 0.693. Solution : 3 gram nitrogen contains 3/28 mole ( μ = 3/28). Now ΔS = μ R loge (Vf/Vi)
=
3 mole × 8.31 joule/(mole-K) × loge 2 28
=
3 × 8.31 × 0.693 = 0.617 joule/K 28
=
0.617 = 0.148 cal/K. 4.18
4. A 2-litre box is divided into two equal parts by a central barrier. One part is filled with hydrogen and the other with nitrogen at NTP. The barrier is removed and the gases mix. Let us find into volume of entropy increase due to mixing. Solution : At NTP 1 mole of gas occupies 22.4 litre. Hence 1 litre (each part of the box) contains (1/22.4) mole of gas. Now, the entropy-increase of each gas is Vf R loge2. ΔS = μ R loge V = 22.4 i
Thermal Physics
196
The total entropy-increase is twice this , i.e., ΔS =
2R loge2 = 0.0619 R. [ Q loge2 = 0.693] 22.4
Now R = 8.31 joule/(mole -K) = 1.99 cal/(mole-K). [Q 1 cal = 4.18 joule] ∴
ΔS = 0.0619 × 1.99 = 0.123cal/K.
5. An electric current of 10 amp is maintained for 1 sec in a resistor of 25 ohm while the temperature of the resistor is kept constant at 27°C. The entropy-change of the resistor and the entropy-change of the universe are calculated as follows. Solution : The heat dissipated in the resistor is Q = i2 Rt = (10)2 × 25 × 1 = 2500 joule . Since the resistor is maintained at 27°C (= 300 K), the heat Q is not absorbed by the resistor, but taken away by the surroundings. Hence the entropy-change of the resistor is zero. The entropy-change of the surroundings is O 2500 ΔS
(surroundings)
=
2500 Q = = 8.33 joule/K. 300 T
This is also the entropy-change of the universe. 6. The same current is maintained for the same time in the same resistor, but now thermally insulated. The initial temperature of the resistor is 27°C. If the resistor has a mass of 10 g and specific heat 0 84 joule/g-K, then entropy-change of the resistor, and also of the universe many be calculated as follows. Solution : The thermally-insulated resistor now absorbs the heat Q (= 2500 joule) dissipated in it and its temperature rises. If Tf is its final temperature, then Q = mc ΔT
Physical Significance or
197
2500 joule = 10 g × 0.84 joule/(g-K) × (Tf – 300)
∴
Tf – 300 = 298 K
or
Tf = 598 K.
We know that when the temperature of a substance of mass m and constant specific heat c changes from Ti to Tf, the entropy-change is Tf dQ Tf mcdT = ∫T ΔS = ∫Ti i T T Tf
T = mc [ log e T ]T = mc log e T i f
i
Tf
= 2.3026 mc log10 T . i Hence the entropy-change of the resistor is 598 ΔS (resistor) = 2.3026 × 10 g × 0.84 Joule/(g-K) × log10 300
= 2.3026 × 10 × 0.84 × log10 2 = 2.3026 × 10 × 0.84 × 0.3010 = 5.8 joule/K. Since the resistor is thermally insulated, there is no communication of heat between it and the surroundings, so that the entropy-change of the surroundings is zero. Hence ΔS (universe) = 5.8 joule/K.
7. 1 kg of water at 273 K is brought into contact with a heat-reservoir at 373 K. When the water has reached 373 K, let’s see what would be the entropy-change of the water of the heat reservoir and of the universe. Solution : The water (mass m = 1000 gm, specific heat c = 1 cal/gm-K) takes in heat from the reservoir and its temperature rises from an initial value Ti = 273 K to a final value Tf = 373 K. The change in its entropy is
Thermal Physics
198 ΔS (water) = 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 1000 × 1 × log10 (373/273) = 2.3026 × 1000 × 1 × 0.1354 = + 312 cal/K. The water has been placed in contact with a single reservoir at 373 K which forms the surroundings. In this case heat flows from the reservoir to water irreversibly. The quantity of heat flow is the heat required to raise the temperature of water from 273 K to 373 K. Thus. Q = mc (Tf – Ti) = 1000 × 1 × (373 - 273) = 105 cal. The reservoir gives up this heat at a temperature of 373 K. Hence the change in its entropy is ΔS (reservoir)
=
Q = T
−10 5 = – 268 cal/K . 373
The entropy of the reservoir thus decreases. The entropy-change of the universe is ΔS (universe) = ΔS (water) + ΔS (reservoir)
= 312 cal/K – 268 cal/K = + 44 cal/K. This is the increase in the entropy of the universe. 8. If the water had been heated from 273 to 373 K by first bringing it in contact with a reservoir at 323 K and then with a reservoir at 373 K, let us see the entropy-change of the universe. Solution : The water undergoes the same rise of temperature (although in two steps). Hence its net entropy-change will be same as before i.e. ΔS (water) = + 312 cal/K.
Physical Significance
199
Now, the first reservoir heats the water from 273 to 323 K so that the heat given up by this reservoir is Q1 = mc (Tf – Ti) = 1000 × 1 × (323 - 273) = 50,000 cal. This heat is given up at a temperature of 323 K. The second reservoir heats the water form 323 to 373 K so that the heat given up by this reservoir is Q2 = 1000 × 1 × (373 - 323) = 50,000 cal, but this heat is given up at a temperature of 373 K. Hence the total entropy-change of both the reservoirs is ΔS (reservoirs) = –
50, 000 50, 000 − 323 373
= – 155 – 134 = – 289 cal/K. The entropy-change of the universe in this case is therefore ΔS (universe) = 312 – 289 = + 23 cal/K.
Thus, in this case the change (increase) in the entropy of the universe is less than before. 9. The water being heated from 273 to 373 K with almost no change in entropy of the universe is explained as below. Solution : It is clear from above that larger is the number of reservoirs ranging from 273 to 373 K used for heating the water, lesser will be the change in the entropy of the universe. 10. When a body of mass 5 g is heated from 100 K to 300 K the change in entropy is calculated as below. The specific heat of the body is 0.1 cal g–1 deg–1. Solution : The change in entropy of a body (mass m, specific heat c) heated from Kelvin temperature Ti to Tf is given by ΔS = 2.3026 mc log10 (Tf/Ti),
Thermal Physics
200
= 2.3026 × 5 g × 0.1 cal g–1 deg–1 × log10 (300/100) = 2.3026 × 5 × 0.1 × 0.4771 = 0.55 cal/°C . The entropy increases. 11. 10 gram of oxygen is heated from 50 °C to 150 °C at constant volume. The change in entropy is found as follows. Given : Cv = 5 cal/(mole-K). Solution : In usual notations : ΔS = 2.3026 mc log10 (Tf/Ti).
Here m = 10g =
10 mole (molecular weight of oxygen is 32), 32
c = Cv= 5 cal/(mole-K), Ti = 50 + 273 = 323 K and Tf= 150 + 273 = 423 K . 10 cal 423 ∴ ΔS = 2.3026 × 32 mole × 5 mole − k × log10 323
= 2.3026 ×
10 × 5 × (log 423 - log 323) 32
= 2.3026 ×
10 × 5 × (2.6263 – 2.5092) 32
= 0.42 cal/K. The entropy increases. 12. For silver the molar specific heat at constant pressure in the range 100 K to 200 K is given by Cp = 0.076 T - 0.00026 T2 - 0.15 cal/(mole-K), where T is Kelvin temperature. If 2 moles of silver are heated from 100 K to 200 K, calculate the change in entropy. Solution : Let dQ be the heat taken in by μ moles of silver at temperature T for an infinitesimally small temperature-rise dT. If Cp be the molar specific heat at temperature T, then we have
Physical Significance
201
dQ = μ CpdT. The corresponding entropy-change is dS =
dQ dT = μ Cp . T T
The entropy-change for a temperature-rise from 100 K to 200 K is therefore ΔS =
∫
200
100
dS =
∫
200
100
μCp
dT T
2 200 ⎛ 0.076T − 0.00026T − 0.15 ⎞ 2 ⎜ ⎟ dT [ ∴ μ = 2] = ∫100 T ⎝ ⎠
200 ⎛ 0.15 ⎞ ⎟ dT = 2 ∫100 ⎜ 0.076 − 0.00026T − T ⎠ ⎝ 200
⎡ ⎤ 0.00026T 2 − 0.15 log e T ⎥ = 2 ⎢0.076T − 2 ⎣ ⎦ 100
= 2[0.076 (200 – 100) - 0.00013 (2002 – 1002) – 0.15 loge2] = 2 [(0.076 × 100) – (0.00013 × 30000) – (0.15 × 0.693)] = 2 [7.6 – 3.9 – 0.104] = 7.192 cal/K . 13. Two blocks of copper, each of mass 850 g, are put into thermal contact in an insulated box. Their initial temperatures are 52 °C and 12 ºC and the specific heat of copper is 0.1 cal (g-°C). Let us find the change in the entropy of the system and of the universe? Solution : Let the final equilibrium temperature of the blocks be t °C. The heat lost by the hotter block must be absorbed by the cooler one, that is, mc (52 – t) = mc (t – 12) ∴
t = 32°C.
Thermal Physics
202
Now, change in entropy of the hotter block in cooling from 52 °C (= 325 K) to 32 °C (= 305 K) = 2.3026 mc log10 (Tf/Ti) = 2.3026 × 850 × 0.1 × (log 305 – log 325) = 2.3026 × 850 × 0.1 × (2.4843 - 2.5119) = – 5.40 cal/°C. Again, change in entropy of the cooler block in heating from 12 °C (= 285 K) to 32 °C (= 305 K) = 2.3026 mc log10 (Tf/Ti) = 2.3026 × 850 × 0.1 × (log 305 – log 285) = 2.3026 × 850 × 0.1 × (2.4843 – 2.4548) = + 5.77cal/°C. ∴ net increase in the entropy of the system is
5.77 – 5.40 = 0.37 cal/°C . Because the process is carried out in an insulated box, the entropy-change of the surroundings is zero so that the entropychange of the universe is + 0.37 cal/°C, an increase. 14. The change in entropy when 40 g of water at 50 °C is mixed with 80 g of water at 20 °C is calculated as below. Solution : Let the temperature of the mixture be t °C. The specific heat of water c = 1 cal/(g-deg). By the principle of calorimetry, we have 40 × 1 × (50 – t) = 80 × 1 × (t – 20) ∴
t = 30°C.
Now, change, in entropy of 40 g of water in cooling from 50 °C (= 323 K) to 30 °C (= 303 K) = 2.3026 mc log10 (Tf/Ti) = 2.3026 × 40 × 1 × (log 303 – log 323)
Physical Significance
203
= 2.3026 × 40 × 1 × (2.4814 – 2.5092) = – 2.56 cal/K . The – sign indicates decrease in entropy. Again, change in entropy of 80 g of water in heating from 20 °C (= 293 K) to 30 °C (303 K) = 2.3026 mc log10 (Tf/Ti) = 2.3026 × 80 × 1 x (log 303 – log 293) = 2.3026 × 80 × 1 × (2.4814 – 2.4669) = + 2.68 cal/K. The + sign indicates increase in entropy. Therefore, the net increase in the entropy of the system is 2.68 – 2.56 = 0.12cal/K. 15. 10 gm of ice at 0 °C melts into water at the same temperature. The change in entropy is calculated below. The latent heat is 80 cal/g. Solution : If the ice is to be melted reversibly then it must be put in contact with a heat-reservoir whose temperature is higher than 0 °C by only an infinitesimally small amount. Then we can write the entropy-change of the ice-water system as ΔS =
∫
dQ . T
But the temperature remains constant at 0 °C (= 273 K). ∴
ΔS =
1 Q dQ = . ∫ T T
The heat which must be supplied to 10 g of ice to melt is 10 × 80 = 800 cal; and T = 273 K. 800 ΔS = 273 = 2.93cal/K,
which is positive. Thus the entropy of the system increases.
Thermal Physics
204
In the reversible melting there will be an exactly equal decrease in the entropy of the heat- reservoir so that the entropychange of the system plus surroundings is zero, a characteristic of the reversible process. 16. If, in the above problem, the ice is melted by dropping it into a large quantity of water at 20 °C , then what would be the change in the entropy of the water-reservoir and what would be the total entropy-change of the original ice and the water-reservoir. Solution : When the ice at 0 °C is melted by putting it in a large quantity of water at 20 °C, the melting of ice will be irreversible (heat will always flow from water to ice). In this case we shall have the following entropy-changes : Entropy-change of ice will be ΔS ice = + 2.93 cal/K . (Calculated in part a)
Entropy-change of water-reservoir which gives out 800 cal of heat to the ice (since it is ‘large’, its temperature may be considered to remain constant at 20 °C i.e. 293 K) will be ΔS reservoir =
−800 Q = = – 2.73 cal/K. 293 T
Therefore, the total entropy-change of the system plus surroundings will be ΔS total = ΔS ice + ΔS reservoir
= + 2.93 – 2.73 = + 0.20 cal/K . Thus the total entropy increases, as is the characteristic of an irreversible process. 17. The change in entropy when 10 g of ice at 0 °C is converted into water at 50 °C by heating is calculated as flows. The latent heat of ice is 80 cal/g and specific heat of water is 1 cal/(g-°C.)
Physical Significance
205
Solution : The change in entropy when a substance of mass m and latent heat L melts at a constant Kelvin temperature T is given by Q mL = ΔS = T T ∴ the change in entropy when 10 g of ice melts at 0 C
(= 273 K) is 10 × 80 = 2.93 cal/K. 273
Again, the change in entropy when 10 g of water is heated from 0 °C (= 273 K) to 50 °C (= 323 K) is 2.3026 mc log10 (Tf/Ti) = 2.3026 × 10 × 1 × (log 323 – log 273) = 2.3026 × 10 × (2.5092 – 2.4362) = 1.68 cal/K. ∴ total change in entropy = 2.93 + 1.68 = 4.61 cal/K.
This is entropy-increase. 18. The change in entropy when 1 g of water at 0 °C is converted into steam at 100 °C is found out as follows. Take specific heat of water to be constant at 1 cal/(g-°C) and latent heat of steam at 100 °C as 540 cal /g. Solution : The change in entropy when the temperature of a substance of mass m and specific heat c changes from Ti to Tf is Tf
ΔS = 2.3026 mc log10 T . i
Hence the increase in entropy when the temperature of 1 g of water rises from 0 °C (= 273 K) to 100 °C (= 373 K) is 2.3026 × 1 × 1 × log10
373 273
Thermal Physics
206
= 2.3026 × (2.5717 – 2.4362) = 2.3026 × 0.1355 = 0.312 cal/K . Now, the heat absorbed by a substance of mass m and latent heat L at temperature T is mL, so that the entropy change is mL/T. Hence, the increase in entropy when 1 gm of water at 100 °C (= 373 K) evaporates =
1 × 540 = 1.448 cal/K . 373
∴ total increase in entropy = 0.312 + 1.448 = 1.76 cal/K.
19. The increase in entropy when 1 g of ice at 0 °C changes into 1 gm of steam at 100 °C, assuming the specific heat of water as 1 cal/g-deg , latent heat of ice 80 cal/g and latent heat of steam 540 cal/g may be found. Solution : The amount of heat Q given to 1 g of ice at 0 °C (i.e. 273 K) to convert into water at the same temperature is L cal where L is the latent heat. Hence the increase in entropy is 80 Q L = = = 0.293 cal/K. 273 T T
Now, the increase in entropy when the temperature of 1 g of water (sp. heat c = 1 cal/g-deg) rises from Ti (= 0 °C = 273 K) to Tf (= 100 °C = 373 K) = 2.3026 mc log10 (Tf/Ti) = 2.3026 × 1 × 1 × (log 373 – log 273) = 2.3026 × (2.5717 – 2.4362) = 2.3026 × 0.1355 = 0.312 cal/K . Now, the heat required to convert 1 g of water at 100 °C (= 373 K) into steam is L cal where L is the latent heat of vaporisation. Therefore , the increase in entropy in vaporisation is
Physical Significance
207
540 L = = 1.448 cal/K. 373 T ∴ total increase in entropy = 0.293 + 0.312 + 1.448 = 2.053
cal/K. 20. The increase in entropy when 1 kg of ice at 0 °C is converted into steam at 100 °C is calculated below. Specific heat of water =10 kcal/(kg °C), latent heat of ice = 3 4 x 10s joule/kg, latent heat of steam = 22 68 x 10s joule/kg ,7 = 42 joule/cal. Solution : The increase in entropy when 1 kg of ice is melted into water at 0 °C (= 273 K) is 1kg × (3.4 × 10 5 joule / kg Q mL = = T T 273 K
= 1.245 × 103 joule/K =
1.245 × 10 3 = 0.296 × 103 cal/K. 4.2
The increase in entropy when 1 kg of water is heated from 0 °C (= 273 K) to 100 °C (= 373 K) is 2.3026 mc log10 (Tf/Ti) = 2.3026 × 1 kg × 1 kcal/(kg-K) × (log 373 – log 273) = 2.3026 × 1 × 1 × (2.5717 × 2.4362) = 0.312 kcal/K = 0.312 × 103 cal/K . Finally, the increase in entropy when 1 kg of water is vaporised to steam at 100 °C (= 373 K) is 1kg × (22.68 × 10 5 joule / kg Q mL = = T T 273 K
= 6.080 × 103 joule/K
Thermal Physics
208 =
6.080 × 10 3 = 1.448 × 103 cal/K. 4.2
∴ total increase in entropy = (0.296 + 0.312 + 1.448) × 103
= 2.056 × 103 cal/K. 21. The change in entropy upon the conversion of 10 g of ice at —20 °C into steam at 100 °C may be found. Given : sp. heat of ice =0.5 cal/g-K, latent heat of ice = 80 cal/g, latent heat of steam = 539 cal/g. Solution : Ice (mass = 10 gm, sp. heat c = 0.5 cal/g-K) is first heated from –20 °C (= 253 K) to 0 °C (= 273 K). The increase in its entropy is ΔS1 = 2.3026 mc log10 (Tf/Tt)
= 2.3026 × 10 × 0.5 × (log 273 – log 253) = 2.3026 × 10 × 0.5 × (2.4362 – 2.4031) = 0.381 cal/K. Ice now melts to water at 0 °C. The increase in entropy is ΔS2 =
10 × 80 mL = = 2.930 cal/K 273 T
The water (m = 10 g , c = 1 cal/g-deg) is now heated from 0 °C (= 273 K) to 100 °C (= 373 K). The increase in entropy is ΔS3 = 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 10 × 1 × (log 373 – log 273) = 2.3026 × 10 × 1 × (2.5717 – 24362) = 3.120 cal/K. Finally, the water is converted into steam at 100 °C (= 373 K). The increase in entropy is ΔS4 =
10 × 539 mL = = 14.450 cal/K. 373 T
Physical Significance
209
∴ total increase in entropy
= ΔS1 + ΔS2 + ΔS3 + ΔS4 = 0.381 + 2.930 + 3.120 + 14.450 = 20.88 cal/K. 22. 10 g of steam at 100 °C is converted into water at the same temperature. The change in entropy is computed. The latent heat of steam is 540 cal/g. Solution : The steam (mass m = 10 g, latent heat L =540 cal/g) condenses into water at 100 °C (= 373K), giving out heat mL. The change in entropy is ΔS =
10 × 540 Q −mL = =– = –14.48 cal/K. 373 T T
The entropy of the steam decreases . The entropy of the universe, however, increases because the entropy of the coldreservoir which absorbs the heat given out by the steam increases by an amount greater than 14.48 cal/K. 23. 10 g of water at 20 °C are converted into ice at -10 °C at constant pressure. Assuming specific heat of ice to be 0.5 cal/(g-°C), and latent heat 80 cal/g, let us find the change in entropy. Solution : The water (mass m = 10 g, specific heat c = 1 cal/g-deg) first cools from 20 °C (= 293 K) to 0 °C (273 K). The change in its entropy is ΔS1 = 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 10 × 1 × log10 (273/293) = 2.3026 × 10 × 1 × (log 273 – log 293) = 2.3026 × 10 × 1 × (2.4362 – 2.4669) = – 0.707 cal/K . The water now freezes into ice at constant temperature of 273 K , giving out heat mL . The change in entropy is
Thermal Physics
210 ΔS2 =
10 × 80 −mL =– = – 2.930 cal/K. 273 T
Finally, the ice so formed (mass m = 10 g, specific heat c = 0.5 cal/g-deg) cools from 0°C ( = 273 K) to –10 °C (= 263 K). The change in entropy is ΔS3 = 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 10 × 0.5 × log10 (263/273) = 2.3026 × 10 × 0.5 × (log 263 – log 273) = 2.3026 × 10 × 0.5 × (2.4200 – 2.4362) = –0186 cal/K. ∴ total change in entropy = ΔS1 + ΔS2 + ΔS3
= – 0.707 – 2.930 – 0186 = – 3.823 cal/K. The entropy of water decreases. In the actual (irreversible) process, the entropy of the universe (water + surroundings) increases. 24. 0.1 kg of steam at 100 °C is converted into ice at –10 °C. The change in entropy is calculated. Given : specific heat of ice = 0.5 cal/(g-°C), specific heat of water = 1.0 cal/ (g-°C), latent heat of steam = 540 cal/g, latent heat of ice = 80 cal/g. Solution : The change in entropy of the steam (m = 100 g, L = 540 cal/g) in condensing into water at 100 °C (= 373 K) is ΔS1 =
100 × 540 Q −mL = =– = –144.8 cal/K. 373 T T
The change in entropy of water (m = 100 g , c = 1.0 cal/g-°C) in cooling from 100 °C (= 373 K) to 0 °C (= 273 K) is ΔS2 = 2.3026 mc log10 (Tf /Ti)
= 2.3026 × 100 × 1.0 × (log 273 – log 373) = 2.3026 × 100 × 1.0 × (2.4362 – 2.5717) = – 31.2 cal/K.
Physical Significance
211
The change in entropy of water (m = 100 g , L = 80 cal/g) in freezing into ice at constant temperature of 0 °C (= 273 K) is ΔS3 =
100 × 80 Q −mL = =– = – 29.3 cal/K. 273 T T
The change in entropy of ice (m = 100 g , c = 0.5 cal/g-°C) in cooling from 0 °C (= 273 K) to - 10 °C (=263 K) is ΔS4 = 2.3026 mc log10 (Tf/Ti)
= 2.3026 × 100 × 0.5 × (log 263 – log 273) = 2.3026 × 100 × 0.5 × (2.4200 – 2.4362) = –1.86 cal/K. ∴ total change in entropy = ΔS1 + ΔS2 + ΔS3 + ΔS4
= – 144.8 – 31.2 – 29.3 – 1.86 = – 207 cal/K. The entropy of the system decreases . In the actual process the entropy of the universe would increase. 25. The change in entropy when 1 g of tin is heated from 127 °C to 313 °C is calculated as follows. Given : melting point of tin = 232 °C, latent heat of fusion of tin = 14 cal / g, sp. heat of solid tin = 0 055 cal/(g-°C), sp. heat of molten tin = 0 064 cal/(g-°C). Solution : On heating the tin, its temperature first rises from 127 °C to 232 °C, then it melts at constant temperature, and then the temperature of the molten tin rises from 232 °C to 313 °C. Now, the change in entropy when the temperature of 1 g of tin (specific heat c = 0.055) rises from Ti = 127 °C (= 400 K) to Tf= 232 °C (= 505 K) = 2.3026 mc log10
Tf Ti
= 2.3026 × 1 × 0.055 × log10
505 400
Thermal Physics
212
= 2.3026 × 0.055 × (log10 505 – log10 400) = 2.3026 × 0.055 × (2.7033 – 2.6021) = 2.3026 × 0.055 × 0.1012 = 0.0128 cal/K. The increase in entropy when 1 g of tin melts at 505 K is 1 × 14 Q mL = = = 0.0277 cal/K. 505 T T
Finally, the increase in entropy when the temperature of the molten tin rises from 232 °C (= 505 K) to 313 °C (= 586 K) Tf
= 2.3026 mc log10 T i = 2.3026 × 1 × 0.064 × log10
586 505
= 2.3026 × 0.064 × (log10 586 – log10505) = 2.3026 × 0.064 × (2.7679 – 2.7033) = 2.3026 × 0.064 × 0.0646 = 00095 cal/K. ∴ total increase in entropy
= 0.0128 + 0.0277 + 0.0095 = 0.050 cal/K. 26. The change in entropy of 1 g of nitrogen when its temperature rises from 40 °C to 80 °C while its volume increases fourfold is calculated. For nitrogen cv = 0.18 cal/(g-°C) and molecular weight M = 28g/mol. Take R = 2 .0 cal/mole-°C. Solution : When (i moles of a gas pass from an initial state pi Vi, Ti to a final state pf, Vf, Tf, the change in entropy in terms of temperatures and volumes is given by Tf Vf ΔS = μCv log e T + μR log e V , i i
where R is universal gas constant.
Physical Significance
213
1 g of gas has
1 mol, where M is molecular weight. M
Therefore, for 1 g of gas, we have
Tf R Vf Cv ΔS = M log e T + M log e V . i i
Here
cal Cv = Cv = 0.18 g − °C , M
R = 2.0 = so
cal and mol − °C
2.0cal / mol − °C R = 2.8 g / mol M cal = 0.0714 g − °C ; Tf Ti
∴
=
Vf 80 + 273 353 = and V = 4. 40 + 273 313 i
353 ΔS = 0.18 loge 313 + 0.0714 loge 4
= 2.3026 [0.18 × (log 353 – log 313) + 00714 log 4] = 2.3026 [0.18 × (2.5478 – 2.4955) + 0.0714 × 0.6020] = 2.3026 [0.00941 + 0.04298] = 0.12cal/”C. 27. When 10 g of hydrogen is heated from 27 °C to 327 °C, its volume increases to four times the initial volume. The change in entropy is calculated. For hydrogen : Cv = 5 cal/mol-°C and molecular weight M = 2. Take R = 8 31 joule/ mole-°C.
Thermal Physics
214
Solution : For μ moles of a gas ,the change in entropy is Tf Vf ΔS = μCv log e T + μR log e V . i i
The molecular weight of hydrogen is 2. It means that 1 mole of hydrogen has 2 g gas. Thus 10 g of hydrogen contains 5 moles ( μ = 5). Now, Cv = 5 cal/mole–°C, R = 8.31 joule/mole-°C = = 2.0 cal/mole-°C, ∴
Tf
=
Ti
8.31 4.18
Vf 327 + 273 600 = = and = 4. Vi 27 + 273 300
ΔS = 5 × 5 × loge2 + 5 × 20 × loge4
= 2.3026 (25 log10 2 + 10 log
10
4)
= 2.3026 [(25 × 0.3010) + (10 × 0.6020)] = 2.3026 [7.525 + 6.020] = 31 cal/°C. 28. The change in entropy per g of oxygen gas during a process in which its temperature changes from 10 °C to 50 °C and the pressure changes from 76 cm to 100 cm of mercury is calculated Cv = 70 cal/mole –°C, R = 2.0 cal/mole –°C. Solution : For 1 g of a gas the change in entropy in terms of temperature and pressure is given by Tf R Pf Cp ΔS = M log e T − M log e P i i
The molecular weight M of oxygen is 32. Thus 7.0 273 + 50 2.0 100 ΔS = 32 log e 273 + 10 − 32 log e 76
=
1 [7 loge 1.141 – 2 loge 1.316] 32
=
2.3026 [7 × 0.05728 – 2 × 0.1192] 32
Physical Significance
215 =
2.3026 [0.4010 – 0.2384] 32
=
2.3026 × 0.1626 = 0.0117cal/°C. 32
29. Two moles of an ideal gas occupy a volume of 10 litres at a pressure of 4 atmospheres. It is first heated at constant volume until its pressure increases to 8 atmospheres, and then at constant pressure until its volume becomes 40 litres. The change in entropy is calculated as follows. Given : Cv = 3 cal/ (mole-deg) and R = 2 cal/(mole-deg). Solution : The change in entropy of μ moles of an ideal gas in terms of its pressure and volume is given by pf Vf ΔS = μCv log e p + μCp log e V . i i
In the first heating; Vf/ Vi = 1 and pf/pi = 8/4 = 2. Also μ = 2,Cv = 3 cal/(mole-deg) and Cp = Cv + R = 5 cal/(mole-deg). ∴
(ΔS)1 = 2 × 3 × loge2
= 6 × 0.6931
[∴ loge 2 = 0.6931]
= 4.1586 cal/deg. In the second heating ; pf/pi = 1, Vf/Vi = 40/10 = 4 . ∴
(ΔS)2 = 2 × 5 × loge 4
= 10 × (2 × 0.6931) = 13.862 cal/deg. The total entropy-change is therefore ΔS = (ΔS)1 + (ΔS)2
= 4.1586 + 13.862 = 18.020 cal/deg. 30. We can defend or refute the claim that with temperature of the surroundings 27 °C, 12000 kilocalories of heat available from a reservoir at 627 °C is more useful than 14000 kilocalories of heat available at 327 °C.
Thermal Physics
216
Heat energy is available for conversion into work only when it is let down from a higher to a lower temperature. If a working substance in Carnot engine takes in heat Q1 at higher temperature T 1 and gives up heat Q 2 at lower temperature T2, then the heat available for useful work ⎛ Q2 ⎞ = Q1 – Q2 = Q1 ⎜ 1 − Q ⎟ ⎝ 1 ⎠ ⎛ T2 ⎞ = Q1 ⎜ 1 − T ⎟ ⎝ 1 ⎠
⎡ Q2 T2 ⎤ = ⎥ ⎢∴ ⎣ Q1 T1 ⎦
In the first case, Q1 = 12000 kcal, T1 = 627 + 273 = 900 K and T2 = 27 + 273 = 300 K. Thus ⎛ 300 ⎞ ⎟ = 8000 kcal. useful energy = 12000 ⎜ 1 − ⎝ 900 ⎠
In the second case, Q1 = 14000 kcal, T1 = 327 + 273 = 600 K and T2 = 27 + 273 = 300 K. Thus ⎛ 300 ⎞ ⎟ = 7000 kcal. useful energy = 14000 ⎜⎝ 1 − 600 ⎠
Hence the claim is correct.
Changing State
217
9 Changing State Clausius–Clapeyron’s Equation A substance can exist in three states–solid, liquid and gas. Out of these three, only two states can generally coexist in equilibrium. Whenever there is a change of state, either a solid changes into liquid or liquid into vapour, the temperature remains constant as far as the change takes place. This temperature, although depends upon pressure, is characteristic of each substance. When the change is from solid to liquid state, the characteristic temperature is called the melting point of the solid and when it is from liquid to vapour state, the temperature is called the boiling point of the liquid. The melting point or boiling point have got a specific value at a specific pressure and vice versa. It is possible to obtain a relation showing how the melting and boiling points vary with pressure, by applying the second law of thermodynamics. The relation thus obtained is known as Clausius Clapeyron equation or the first latent heat equation.
218
Thermal Physics
Let ABCD and EFGH represent the two isothermals at infinitely close temperatures T and (T + dT) respectively. Referring to the Fig. below, the parts AB and EF correspond to the liquid state of the substance. At B and F, substance is purely in the liquid state. Along BC and FG, the change of state is in progress and the liquid and vapour states coexist in equilibrium. At C and G the substance is purely in vapour state. From C to D and G to H, the substance is in the vapour state. Let P and (P + dP) be the saturated vapour pressures of the liquid at temperatures T and (T + dT) respectively. Let V1 and V2 be the volumes of the substance at F and G respectively. Let us draw two adiabatics from F and G meeting the lower isothermal at M and N respectively. Let us suppose that 1 gm. of the substance is taken round a reversible Carnot cycle FGMNF, allowing it to expand isothermally along FG, adiabatically along GN, compressing it isothermally along NM and adiabatically along MF. The amount of heat Q1 absorbed along FG is equal to the latent heat of vaporization (L + dL) at temperature (T + dT), as substance changes completely from liquid state at F to the vapour state at G. Also the quantity of heat Q2 rejected along the isothermal compression NM, is L, the latent heat at temperature T. Here latent heat is supposed to vary with temperature.
Changing State
219
Applying the principle of Carnot’s reversible cycle Q2 Q1 T1 Q1 or = = T2 Q2 T2 T1
or
Q1 − Q2 T1 − T2 = Q1 T2
We have, here Q1 = L + dL, Q2 = L, T1 = T + dT and T2 = T ∴
L + dL − L T + dT − T = L T dL dT = L T L dL = dT T
or or
... (1)
The amount of heat converted into work during the cycle FGMNF, Q1 – Q2 = L + dL – L = dL
... (2)
But the work done during the Carnot cycle is given by the area FGMNF, which may be treated as a parallelogram. Hence, dL (in work units)= Area FGMNF = FG × Perpendicular distance between FG and NM = (V2 – V1) × dP
... (3)
where V2 and V1 are the specific volumes (i.e. masses per unit volume) in vapour and liquid states respectively, dP expresses the difference of pressure between FG and NM. Substituting this value of dL in (1), dP(V2 – V1) =
or
L dT T
dP L = dT T (V2 − V1 )
... (4)
Thermal Physics
220
This is called the Clapeyron’s latent heat equation and holds for both the changes of state, i.e., from liquid to vapour and solid to liquid. In the latter case L will represent the latent heat of fusion, V1 and V2 the volume occupied by 1 gm. of substance in solid and liquid states respectively. It may be noted that L in eq. (4) is to be expressed in work units i.e. in ergs/gm. or Joules/kg. Applications Effect of Pressure on Boiling Points of Liquids: When a liquid boils i.e. changes from liquid state to gaseous state, there is an increase in its volume so that V2 > V1 or the quantity (V2 – V1). is positive. Hence from eq. (4) above, dP/dT is a positive quantity which means the boiling point of a liquid rises with increase in pressure or vice versa. Thus a liquid will boil at lower temperature under reduced pressure. Hence water will boil at a temperature less than 100°C, when the atmospheric pressure is less than 76cm. of mercury–the normal pressure. Effect of Pressure on Melting Points of Solids: When a solid melts, there may be an increase in volume as in the case of certain substances like wax and sulphur or there may be a decrease in volume as in the case of certain substances like ice, gallium and bismuth. (a) When V2 > V1 (As for wax and sulphur)
dP is a – ve dT
quantity. This means that melting point of such substances rises with increase in pressure. (b) When V2 < V1 (As for ice, gallium and bismuth), (V2 – V1) is a negative quantity. Hence dP/dT is also negative which means that melting point of such substances decreases with increase in pressure. Thus, ice will melt at a temperature lower than 0°C when the pressure is higher than the normal pressure, 76 cm. of mercury.
Changing State
221
Second Latent Heat Equation The second latent heat equation, or the equation of Clausius, gives the variation of latent heat of a substance with temperature and connects it with the specific heat of the substance in the two states.
Let C1 denote the specific heat of a liquid in contact with its vapours and C2 the specific heat of saturated vapours in contact with its liquid. Referring to the Fig. above, let us consider that 1 gm. of the substance is taken round the cycle BFGCB. The quantity of heat absorbed by the substance (liquid) in passing from B to F, when its temperature rises by dT is C1 dT. In passing along FG, when the substance changes from liquid to vapour at constant temperature T + dT, it absorbs a quantity of heat L + dL. In passing from G to C, the temperature of the substance (vapours) falls by dT and hence it gives out a quantity of heat C2 dT, while in passing along CB, when it condenses from vapour to liquid at constant temperature T, gives out a quantity of heat L. Hence the net amount of heat absorbed during the cycle is C1 dT + L + dL – C2 dT – L = (C1 – C2) dT + dL.
Thermal Physics
222
This must be equal to the work done which is equal to the area of the cycle or the area FGNMF in the limiting case and hence L = dP (V2 − V1 ) = dT (Proved in § above) T L
( C1 − C2 ) dT + dL = T dT
∴
or
( C1 − C2 ) dT =
or
C2 − C1 =
L dT − dL T dL L − . dT T
This is the latent heat equation of Clausius.
PROBLEMS 1. Calculate the change in the boiling point of water when the pressure is increased from 1 to 2 atmospheres. Given: Specific volume of water and steam at 100°C are 1 and 1601 cm3 /gm. Latent heat of steam = 540 cal /gm. Boiling point of water at 1 atmospheric pressure = 100°C. Solution: From Clausius–Clapeyron equation dP L = dT T (V2 − V1 ) dT =
or
Given that
T (V2 − V1 ) dP L
dP = 2 – 1 = 1 atmosphere = 106 dynes/cm2
T = Boiling point of water = 100°C = 373K V2 – V1 = 1601– 1 = 1600 cm3/gm L = 540 cal/gm = 540 x 4.2 x 107 ergs/gm ∴
Change in boiling point of water
Changing State
223 dT =
373 × 1600 × 106 540 × 4.2 × 107
= 26.6K = 26.6°C 2. Calculate the latent heat of ice given that change of pressure by one atmosphere changes the melting point of ice by 0.0074°C and when one gm. of ice melts its volume changes by 0.0907 c.c. Solution:
From Clausius Clapeyron equation dP L T= dT T (V2 − V1 ) L=
or
dP.T (V2 − V1 ) dT
Here, dP = 1 atmosphere = 1.1013 × 106 dynes/cm2 T = 0°C = 273K, V2 – V1 = 0.0907 c.c. dT = 0.0074°C = 0.0074 K L=
1.013 × 106 × 273 × 0.0907 0.0074
= 3.39 × 109 ergs/gm. =
= 81.1 cal/gm.
3.39 × 10 9 cal/gm 4.18 × 107
3. Calculate the change in the boiling point of water when the pressure is increased from 1.0 to 10 atmospheres. Given specific volume of steam = 1677 c.c. /gm., latent heat of steam = 540 cal./gm., boiling point of water at one atmosphere pressure = 100°C = 373K and pressure of one atmosphere = 1.0 × 106 dynes/cm2. Solution:
From Clausius Clapeyron equation dP L = dT T (V2 − V1 )
We have
dT =
T ( V2 − V1 ) dP L
Thermal Physics
224 Given that dP = 10 – 1 = 9 atoms. = 9 × 1 × 106 dynes/cm2
T = Boiling point of water = 100°C = 373K V2 = 1677 c.c./gm., V2 = 1 c.c./gm. V2 – V2 = 1676 c.c./gm. and
L = 540 cal./gm. = 540 × 4.2 × 107 ergs/gm.
Therefore, change in boiling point of water dT =
373 × 1676 × 9 × 106 540 × 4.2 × 107
= 248.07°C = 248.07 K 4. Calculate the change in the melting point of wax for a pressure of 50 atmospheres from the following data Melting point = 64°C, volume at 0°C = 1 cc; volume of the solid at the melting point = 1.161 cc; volume of the liquid at the melting point = 1.166 c.c, density of the solid at 64°C = 0.96gm. /c.c. latent heat = 97 cal. gm. Solution: Here the melting point (64°C) is given to be at one atmosphere pressure and we are required to find the change in the melting point for a pressure of 50 atmosphere, i.e. for a pressure change of (50 – 1) = 49 atmosphere. ∴
dP = 49 × 76 × 13.6 × 981 dynes/cm2 T = 64°C = 64 + 273 = 337 K, L = 97 cal. = 97 × 4.2 × 107 ergs.
Now mass of the solid at melting point = Volume x density = 1.161 × 0.96 gm. ∴
Specific volume of the solid at melting point V1 =
Volume 1.161 = = 1.0417 c.c./gm mass 1.161 × 0.96
Changing State
225
and specific volume of the liquid at melting point V = ∴
Volume 1.166 = = 1.0461c .c ./gm mass 1.161 × 0.96
V2 – V1 = 1.0461 –1.0417 = 0.0044 c.c./gm.
Now Clausius Clapeyron equation is L dT = T (V − V ) dP 2 1
So that dT = =
dP × T (V2 − V1 ) L 49 × 76 × 13.6 × 981 × 337 × 0.0044 97 × 4.2 × 107
= 0.018K = 0.018°C. 5. Calculate the depression in the melting point of ice by one atmospheric increase of pressure, given that the latent heat of fusion of ice L = 3.4 × 105 joule/kg. Specific volume of 1 kg. of ice and water at 0°C are 1.091 × 10–3 m2 and 1.000 × 10–3 m3 respectively. 1 atmospheric pressure = 105 N/m2. Solution: Given dP = l atmosphere = 105 N/m2 T = 0°C = 27311. L = 3.4 x 105 joule/kg. and specific volume of 1 kg. of ice V1 = 1.091 × 10–3 m3 Specific volume of 1 kg. of water V2 = 1.000 × 10–3 m3 ∴
V2 – V1 = (1.000 × 10–3 – 1.091 × 10–3) m3 = – 0.091 × 10–3
Now we have from Clapeyron equation
Thermal Physics
226 L dP = T (V − V ) dT 2 1 dT =
=
dP × T (V2 − V1 ) L 10 5 × 273 ( −0.091 × 10 −3 ) 3.4 × 10 5
= – 0.0073 k = – 0.0073°C. Hence the melting point of ice will be depressed by 0.0073°C per atmosphere increase of pressure. 6. Calculate the depression of the melting point of ice (L = 80 Cals.) per atmosphere increase of pressure, if ratio of the densities of ice and water at 0°C is 10/11. Solution:
From Clausius–Clapeyron eqn.
T(V2 − V1 )dP L T = 27311, L = 80 cal. = 80 × 4.2 × 107 ergs. dT =
Here
dP = 1 atmosphere = 1.03 x 106 dynes/cm2 and and
V2 Specific volume of water = V1 Specific volume of ice =
But ∴ ∴
Hence
Den s ity of ice 10 = Den s ity of water 11
V2 = 1.0 cm3/4gm. V1 =
11 × 1.0 = 1.1cm3 / gm. 10
V2 – V1 = 1.0–1.1 = –0.1cm3/gm. dT =
273 × ( −0.1) × 1.013 × 106 80 × 4.2 × 107
= – 0.0082°C. Thus, melting point of ice will decrease by 0.0082°C per atmosphere increase of pressure.
Changing State
227
7. Calculate the pressure required to make water freeze at –1°C. Change of specific volume when 1 gm of water freezes into ice = .091 cc., J = 4.2 x 107 ergs/cal., 1 atmosphere 106 dynes/ cm2 and the latent heat of ice = 80 cal. /gm. Solution: Given dT = – 1°C, V2 – V1 = –0.091 c.c. L = 80 cal. = 80 x 4.2 x 107 ergs, T = 0 + 273 = 273 K Applying Clapeyron’s latent heat equation dP L = dT T(V2 − V1 ) or dP
80 × 4.2 × 107 × ( −1) L dT = dynes/cm 2 . T(V2 − V1 ) 273 × ( −0.091) =
80 × 4.2 × 107 atoms. 273 × −0.091 × 106
= 135.2 Atoms. Hence to lower the melting of ice by 1°C, the pressure must be raised by 135.2 atmospheres. Now melting point of ice at atmospheric pressure is 0°C. Hence pressure required to make ice freeze at – 1°C = 135.2 + 1 = 136.2 Atoms. 8. Calculate the change in vapour pressure of water as the boiling temperature changes from 100°C to 103°C. Given latent heat of steam = 540 cal. /gm. and specific volume of steam = 1670 cc. /gm. Solution: The Clapeyrons’ equation describing the rate of change of vapour pressure with temperature is
or
dP L = dT T (V2 − V1 ) dP L × dT = dT T (V2 − V1 )
Thermal Physics
228 Here
L = 540 cal. = 540 × 4.2 × 107 ergs., T = 100 + 273 = 373K
dT = (103 – 100) = 3°, V2 = 1670 c.c./gm., V1 = 1 c.c./gm. ∴
Change in vapour pressure dP = =
540 × 4.2 × 107 × 3 dynes/cm2 373 × (1670 − 1) 540 × 4.2 × 107 × 3 atmosphere 373 × 1669 × 76 × 13.6 × 981
= 0.17091 atmospheres. 9. Water boils at 100.5°C and 99.5°C when the atmospheric pressures are respectively 77.371 and 74.650 cm. of mercury. Calculate the volume of 1 gm. steam at 100°C the latent heat being 537 cal. /gm. Solution:
Given that dT = 100.5–99.5 = 1° L = 537 cal. = 537 × 4.2 × 107 ergs. T = 100°C = 373 K dP = (77.371 – 74.650) = 2.721 cm. of Hg = 2.721 × 13.6 × 981 dynes/cm2.
Now Clapeyron’s latent heat equation is
or
Since
dP L = dT T (V2 − V1 ) L dT V2 − V1 = T dP 537 × 4.2 × 107 × 1 = = 1658 c .c . 373 × 2.721 × 13.6 × 981
V1 = 1 c.c.
Hence the volume of 1 gm. of steam at 100°C V2 = 1658 + 1 = 1659 c.c.
Changing State
229
10. A Carnot’s engine having 100 gm. water–steam as working substance has, at the beginning of the stroke, volume 104 c.c. and pressure 788 mm. (B.P. = 101°C). After a complete isothermal change from water into steam, the volume is 167, 404 c.c. and the pressure is then lowered adiabatically to 733.7 mm. (B.P.° = 90°C). If the engine is working between 99°C and 101°C, calculate the latent heat of steam. Solution:Volume of 100 gm of water at the beginning of stroke = 104 c.c 104 = 1.04c.c./ gm. 100 Similarly specific volume of steam after isothermal change ∴
Specific volume V1 =
167404 = 1674c.c./ gm 100 = 1674.04 – 1.04 = 1673 c.c.
V2 = ∴
V 2 – V1
Change in pressure dP = 781 – 733.7 = 54.3 mm. = 5.43 cm. of Hg = 5.43 × 13.6 × 981 dynes cm
–2
101 + 99 100o C = 373K 2 dT = 101 – 99 = 2°.
Mean temperature T = and
Now according to Clapeyron’s latent heat equation
or
dP L = dT T (V2 − V1 ) T (V2 − V1 ) × dP L= dT 373 × 1673 × 5.43 × 13.6 × 981 = ergs. 2 373 × 1673 × 5.43 × 13.6 × 981 = cal ./gm. 2 × 4.2 × 107
= 540.0 cal./gm.
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230
11. Calculate the specific heat of saturated steam at 100°C from the following data: L at 90°C = 545.25 cals. L at 100°C = 539.30 cals. L at 110°C = 533.17 cals. Specific heat of water at 100°C = 1.013 cals/gm. Solution: Given C1 = 1.013, C2 = ? dL L at 110o C − L at 90o C 533.17 − 545.25 = = dT 20 20 12.08 =− = − 0.604 cal ./gm. 20 dL L − Hence from C2 − C1 = dT T dL L C2 = C1 + − dT T 539.30 C2 = 1.013 − 0.604 − 373
= – 1.037 cals./gm. 12. If L = 800 –.705 T, show that the specific heat of steam is negative. Solution: Given that L = 800 – 0.705 T ∴
dL = – 0.705. dT
Now at normal B.P. of water, i.e. at 100°C = 373 K, we shall have L = 800 – .795 × 373 = 537. Also specific heat of water in liquid state, C1 = 1 Hence from second latent heat equation, dL L − dT T 537 = 1 + ( −.705) − = − 1.14. 337
C2 = C1 +
Thus, the specific heat of saturated water vapour at 100°C is negative.
Changing State
231
13. The latent heat of water diminishes by 0.695 cal. /gm. for each degree centigrade rise in temperature at temperature 100°C and latent of water vapour at 100°C, is 540 cal. /gm. calculate the specific heat for saturated steam. Solution: By second latent heat eqn., the specific heat C2 of saturated steam is given by C2 = C1 +
Here
dL L − dT T
dL = −0.695 cal ./gm.o C dT
T = 100°C = 373 K L = 540 cal./gm. and
C1 = specific heat in liquid state (water) = 1.0 cal./gm. C2 = 1 + (−0.695) −
540 373
= 1 – .695 – 1.448 = – 1.143 cal/gm.°C. 14. Generally the specific volume of a liquid is much less as compared to specific volume of its vapour. Assuming that the vapour obeys ideal gas equation, prove that P = constant e–(L/RT). Solution : Clausius Clapeyron equation gives dP L = dT T(V2 − V1 )
... (i)
Given specific volume of liquid V1 << specific volume of vapour V2 or
V2– V1 = V2 = V (say)
Thermal Physics
232 Hence eq. (i) becomes dP L = dT TV But vapour obeys perfect gas equation
... (ii)
PV = RT RT P Substituting value of V in eq. (ii), ∴
V=
dP = dT
or
L LP = 2 RT RT ⎛ ⎞ T⎜ ⎟ ⎝ P ⎠ dP L = .dT dT RT 2
Integrating log e P = − or
L + Constant RT
P = e[ −( L / RT ) + Constant]
= e Constant × e − L/ RT = Constant × e( − L/ RT ) .
Triple Point (i) The boiling point of water increases with increase in pressure and vice versa. The curve AB represents the relation between pressure and temperature and is called the steam line or vaporization line Fig. The liquid and vapour are in stable equilibrium together only along the line AB. At all points above AB the substance is all liquid and below it there exists vapour only. If at a point M, pressure is raised keeping temperature constant, boiling point will consequently increase and all vapour will condense into liquid. Similarly if at M, pressure is decreased, all the liquid will vaporize and only vapour will remain.
Changing State
233
(ii) The melting point of ice decreases with increase in pressure i.e. ice melts below 0°C at a pressure higher than the atmospheric pressure. The pressure temperature relationship can be represented by a curve CD or C’D’ which is called the ice line or fusion line. The curve CD, which slopes to the left, is for ice type substances whose melting point is raised with increase in pressure; while C’D’, which slopes to the right, is for wax type substances whole melting point is raised with increase in pressure. The substance is entirely solid on the left of the curve while entirely liquid on the right. The curve represents the equilibrium between the solid and liquid state. (iii) When the pressure on ice is raised, evaporation from its surface slows down. The equilibrium between the solid and vapour states of a substance can be represented by a curve EF called the Hoar Forst line or sublimation line. Above the curve EF, the substance is all solid and below it all vapour. These three curves, when plotted on the same diagram, are found to meet in a single point O as shown in Fig. This point is called the triple point. At the triple point, the pressure and
Thermal Physics
234
temperature are such that the solid, liquid and vapour states of the substance can exist simultaneously in equilibrium.
To Show that there is Only one Triple Point : Suppose that the three curves do not meet at a point but intersect enclosing an area ACF shown shaded in Fig. Then according to ice line CD, the substance must be entirely solid in the shaded area as it is to the left of CD. According to the steam line AB the substance must be entirely liquid as it is above AB and according to hoar frost line EF the substance in the shaded portion must be entirely vapour as it is below EF. But these three conclusions contradict one another and hence the shaded triangle ACF cannot exist. Thus, the three curves should meet in a single point O called the triple point. It should be noted that the triple point of water is not fixed but is different for different allotropic forms of ice.
PROBLEMS 1. The coordinates of the triple point of water are t = 0.0075°C and p = 0.0060 atmosphere. Calculate the slope of the ice line in atoms. /°C. What is the physical significance of the negative sign, of the slope ? Solution: Referring to Figures, let O be the triple point of water. Then the coordinates of point O are
Changing State
235
P = 0.0060 atoms, t = 0.0075°C. Now we know that the freezing point of water into ice at 1 atmosphere pressure is 0°C. Hence the coordinates of point D on the ice line are P’ = 1 atoms., t’ = 0°C Hence for ice line OD dP = P’ – P = 1– 0.006 = 0.994 atoms. and
dT = dt = t’ – t = 0 – 0.0075 = 0.0075°C
Therefore, the slope of ice line is dP dP 0.994 =− atoms/o C = dT dt 0.0075
= – 132.5 atoms./°C. Negative sign with the slope dP/dT indicates that the melting point of ice decreases with an increase in pressure. 2. Calculate the pressure and temperature of the triple point of water. Given that the lowering of melting point of ice per atmosphere increase of pressure is 0.0072°C and the saturated vapour pressure at 0°C = 460 m. m. while at 1 °C = 4.94 m. m. Solution: Let the triple point correspond to a temperature toC and pressure p mm. of mercury. Saturated vapour pressure at 0°C = 4.60 mm. Saturated vapour pressure at 1°C = 4.94 mm. Increase in saturated vapour pressure for 1°C rise in ∴ temperature = 4.94 – 4.60 = 0.34 mm. Hence increase in vapour pressure for t°C = 0.34t Therefore, saturated vapour pressure at triple point t°C is given by p = (4.60 + 0.34 t) mm.
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236
Now the melting point of ice at 760 m.m. (1 atmosphere) pressure is 0°C. But at the triple point the pressure is (4.60 + 0.34t). Hence Decrease in pressure = 760 – (4.60 + 0.34t) = (755.4 – 0.34t) mm. It is given that a decrease in pressure of 760 mm. (1 atoms.) will increase the melting point of ice by 0.0072°C. Therefore by a decrease of (755.4 – 0.34t) mm. in pressure. Increase in melting point = or
0.0072 × (755.4 − 0.34t) 760
⎡ 0.0072 × (755.4 − 0.34t) ⎤ Melting point = ⎢0 + ⎥⎦ 760 ⎣
But this is the temperature t at a triple point. Thus t=
or
0.0072 × (755.4 − 0.34t) 760
760t = 5.4388 – 0.002448t
or 760.002448t = 5.4388 t =
5.4388 760.002448
= 0.007156°C. Substituting this value of t in the expression for p, we get p = (4.60 – 0.34 × 0.007156) = 4.60243 mm. of Hg. 3. The vapour pressure P (in mm. of mercury) of solid ammonia is given by 3754 log e P = 23.03 − T while that of liquid ammonia is given by
Changing State
237 loge P = 19.49 −
3063 T
where T is in K. Calculate the triple point of ammonia. Solution: At triple point, the vapour pressure of the substance in each of the three states is identical. Hence equating the vapour pressure of solid ammonia log e P = 23.03 −
3754 T
... (i)
3063 T
... (ii)
with that of liquid ammonia log e P = 19.49 −
We have 23.03 −
or
3754 3063 = 19.49 − T T
1 ( 3754 − 3063) = 23.03 − 19.49 T 691 = 3.54 T
or ∴
T = 195.2K
Substituting the value of T in eq. (i), we get log e P = 23.03 −
or
loge P = 3.8
or
2.303 log10 P = 3.8
or ∴
log10 P =
3754 195.2
3.8 = 1.652 2.303
P = 44.87 mm Hg.
Thus the coordinates of triples point of ammonia are T = 195.2K and P = 44.87 m.m. of Hg.
Thermal Physics
238
Thermodynamical Potentials and Their Relations with Thermodynamical Variables : The thermodynamic state of a homogeneous system may be represented by means of certain selected variables, such as pressure P, volume V, temperature T and entropy S. Out of these four variables, any two may vary independently and when known enable the others to be determined. Thus there are only two independent variables and the others may be considered as their functions. There exists certain relations between these thermodynamical variables. The first and second law of thermodynamics provide two relations which may respectively be stated as, dQ = dU + PdV and
dQ = TdS
Combining the two laws, we have TdS = dU + PdV or
dU = TdS – PdV.
This expresses the change in internal energy of the system in terms of four thermodynamical variables. However for a complete knowledge of the system, certain other relations are required and for this purpose we introduce some functions of the variables P, V, T and S, known as thermodynamical potentials or’ the thermodynamic functions. There are four principal thermodynamic potentials and we shall discuss them one by one. Internal or Intrinsic Energy: We have seen already that according to first law of thermodynamics, there is a certain function, U, of the variables which characterises the system and this function may be called the intrinsic energy or the internal energy. When the system passes from one state to another, the change in the internal energy is independent of the route followed between the two states. The internal energy of a system is defined by the equation,
Changing State
239 dU = dQ – dW
where dW is the external work done and may be replaced by PdV while dQ may be put equal to TdS by second law of thermodynamics. Thus dU = TdS – PdV Taking partial differentials of the intrinsic energy with respect to variables S, and V, we get ⎛ ∂U ⎞ ⎜ ⎟ = T ⎝ ∂S ⎠V
and
⎛ ∂U ⎞ ⎜ ⎟ = −P ⎝ ∂V ⎠S
These are the relations connecting the internal energy U with the thermodynamical variables S, V, T and P. Now since dU is a perfect differential*, we must have ∂ ⎛ ∂U ⎞ ∂ ⎛ ∂U ⎞ ⎜ ⎟ = ⎜ ⎟ ∂V ⎝ ∂S ⎠V ∂S ⎝ ∂V ⎠S ∴
⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = −⎜ ⎟ V ∂ ⎝ ⎠S ⎝ ∂S ⎠V
…(i)
This result is the first thermodynamical relation of Maxwell. Helmholtz’s Function F: The first and second law of thermodynamics combined together provide dU = TdS – dW If we consider processes in which the system exchanges heat with the surrounding and is thereby maintained at a constant temperature T, then TdS = d (TS) and we can write dU = d (TS)–dW or
d(U – TS) = – dW
or
dF = – dW
where F = U – TS is known as the Helmholtz free energy or more appropriately the work function (because eq. dF = – dW shows that for reversible changes the work done by the system
Thermal Physics
240
is equal to the decrease in this function F). Thus the Helmholtz free energy is defined by the equation, F = U – TS, where the value of F depends only on the state of the substance and dF is a perfect differential. Now dF = dU – d (TS) = dU – TdS – SdT. But
dU = TdS – PdV
∴
dF = TdS – PdV – TdS – SdT = – PdV – SdT.
Here T and V are independent variables. Taking partial differentials of F ⎛ ∂F ⎞ ⎛ ∂F ⎞ ⎜ ⎟ = − P and ⎜ ⎟ = −S ⎝ ∂V ⎠T ⎝ ∂T ⎠V
and since dF is perfect differential ∂ ⎛ ∂F ⎞ ∂ ⎛ ∂F ⎞ ⎜ ⎟ = ⎜ ⎟ ∂V ⎝ ∂T ⎠V ∂T ⎝ ∂V ⎠T ⎛ ∂S ⎞ ⎛ ∂P ⎞ ⎜ ⎟ = ⎜ ⎟ ⎝ ∂V ⎠T ⎝ ∂T ⎠V
…(ii)
This is the second thermodynamical relation of Maxwell. Enthalpy or Total Heat H: Enthalpy is an extensive thermodynamical property and is of particular significance. It is mathematically defined as H = U + PV. The differentiation of which yields dH = dU + d (PV) = (TdS – PdV) + PdV + VdP = TdS + VdP.
Changing State
241
At a constant pressure (dP = 0), dH = TdS = dQ, the quantity of heat given to the system from an external source. This explains the name of heat function given to it. It also shows that for an isobaric process, change in enthalpy is equal to the heat given to the system. Now taking partial differentials of H treating S and P as independent variables. ⎛ ∂H ⎞ ⎛ ∂H ⎞ ⎟ = V. ⎜ ⎟ = T and ⎜ ⎝ ∂P ⎠S ⎝ ∂S ⎠ P
And since dH is a perfect differential ∂ ⎛ ∂H ⎞ ∂ ⎛ ∂H ⎞ ⎜ ⎟ = ⎜ ⎟ ∂P ⎝ ∂S ⎠P ∂S ⎝ ∂P ⎠S
or
⎛ ∂T ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = ⎜ ⎟ ⎝ ∂P ⎠S ⎝ ∂S ⎠ P
…(iii)
This is the third thermodynamical relation of Maxwell. Gibb’s Potential G : From the definition of enthalpy, we have H = U + PV. or
dH = dU + PdV + VdP
which on putting
dU = TdS – PdV becomes
dH = TdS – PdV + PdV + VdP = TdS + VdP So that if the process be isothermal [TdS = d (TS)] as well as isobaric (dP = O), we get d (H – TS) = O or dG = O i.e. G = H – TS = constant. The function
G = H – TS = U + PV – TS = U – TS + PV = F + PV
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242
is known as the Thermodynamic potential at constant pressure or Gibb’s function. It is obvious that the function G remains constant if a thermodynamic process remains isothermal as well as isobaric. Thus the fundamental equation defining Gibb’s function G is G = U – TS + PV or
dG = dU – d (TS) + d (PV) = dU – TdS – SdT + PdV + VdP = – PdV – SdT + PdV + VdP (Q dU = TdS – PdV) = VdP – SdT.
Here the independent variables are P and T. Taking partial differentials ⎛ ∂G ⎞ ⎛ ∂G ⎞ ⎜ ⎟ = V and ⎜ ⎟ = −S ⎝ ∂P ⎠T ⎝ ∂T ⎠ P
and since dG is a perfect differential ∂ ⎛ ∂G ⎞ ∂ ⎛ ∂G ⎞ ⎜ ⎟ = ⎜ ⎟ ∂T ⎝ ∂P ⎠T ∂P ⎝ ∂T ⎠P
or
⎛ ∂V ⎞ ⎛ ∂S ⎞ ⎜ ⎟ = −⎜ ⎟ ⎝ ∂T ⎠ P ⎝ ∂P ⎠T
…(iv)
This is the fourth thermodynamical relation of Maxwell. Thus we see that the four thermodynamical potential U (S, V) ; F (T, V) ; H (S, P) and G (T, P) lead us to four thermodynamical relations – known as Maxwell’s relations. The four functions can be expressed in terms of measurable quantities P, V, T and S as follows : dU = TdS – PdV, dF = – PdV – SdT dH = TdS + VdP and
dG = VdP – SdT.
Changing State
243
The four quantities U (S, V), H (S, P), F (T, V) and G (T, P) are called thermodynamic potentials because the thermodynamic variables S, T, P and V can be derived from them by their differentiations with respect to the independent variables associated with them in the same manner as the components of a force are deduced from a force potential. These have already been derived in the equations above but are once again listed below : ⎛ ∂G ⎞ ⎛ ∂F ⎞ ⎟ = −⎜ ⎟ S = −⎜ ∂ T ⎝ ⎠P ⎝ ∂T ⎠V
⎛ ∂U ⎞ ⎛ ∂H ⎞ T= ⎜ ⎟ =⎜ ⎟ ⎝ ∂S ⎠V ⎝ ∂S ⎠P ⎛ ∂U ⎞ ⎛ ∂F ⎞ ⎟ = −⎜ ⎟ P = −⎜ V ∂ ⎝ ⎠S ⎝ ∂V ⎠T ⎛ ∂H ⎞ ⎛ ∂G ⎞ ⎟ =⎜ ⎟ V= ⎜ ⎝ ∂P ⎠S ⎝ ∂P ⎠T
The above equations give the values of thermodynamical variables in terms of thermodynamical potentials.
