The entropy change on going from state 1 to state 2 is given by Eq. (3.21) as S S2 S1 兰21dqrev/T, where T is the absolute temperature. For a reversible process, we can apply (3.21) directly to calculate S. For an irreversible process pr, we cannot in-tegrate dqpr/T to obtain S because dS equals dq/T only for reversible processes. For an irreversible process, dS is not necessarily equal to dqirrev/T. However, S is a state function, and therefore S depends only on the initial and final states. We can there-fore find S for an irreversible process that goes from state 1 to state 2 if we can conceive of a reversible process that goes from 1 to 2. We then calculate S for this reversible change from 1 to 2, and this is the same as S for the irreversible change from 1 to 2 (Fig. 3.6).
In summary, to calculate S for any process; (a) Identify the initial and final states 1 and 2. (b) Devise a convenient reversible path from 1 to 2. (c) Calculate S from
S 兰21dqrev/T.
Let us calculate S for some processes. Note that, as before, all state functions refer to the system, and S means Ssyst. Equation (3.21) gives Ssyst and does not include any entropy changes that may occur in the surroundings.
1. Cyclic process. Since S is a state function, S 0 for every cyclic process.
2. Reversible adiabatic process. Here dqrev 0; therefore
(3.23) Two of the four steps of a Carnot cycle are reversible adiabatic processes.
3. Reversible phase change at constant T and P. At constant T, (3.21) gives (3.24) where, since T is constant, we took 1/T outside the integral. qrevis the latent heat of the transition. Since P is constant, qrev qP H [Eq. (2.46)]. Therefore
(3.25) Since H qP is positive for reversible melting of solids and vaporization of liquids, S is positive for these processes.
¢S ¢H
T rev. phase change at const. T and P
¢S
冮
12dqTrev 1T冮
12dqrevqTrev¢S 0 rev. ad. proc.
Section 3.4 Calculation of Entropy Changes 87
Irrev.
Rev.
1
2
Figure 3.6
Reversible and irreversible paths from state 1 to 2. Since S is a state function, S is the same for each path.
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Chapter 3
The Second Law of Thermodynamics 88
EXAMPLE 3.1
S for a phase changeFind S for the melting of 5.0 g of ice (heat of fusion 79.7 cal/g) at 0°C and 1 atm. Find S for the reverse process.
The melting is reversible and Eq. (3.25) gives
(3.26) For the freezing of 5.0 g of liquid water at 0°C and 1 atm, qrevis negative, and
S 6.1 J/K.
Exercise
The heat of vaporization of water at 100°C is 40.66 kJ/mol. Find S when 5.00 g of water vapor condenses to liquid at 100°C and 1 atm. (Answer:30.2 J/K.)
4. Reversible isothermal process. Here T is constant, and S 兰21T1 dqrev T1兰21dqrev qrev/T. Thus
(3.27) Examples include a reversible phase change (case 3 in this list) and two of the four steps of a Carnot cycle.
5. Constant-pressure heating with no phase change. First, suppose the heating is done reversibly. At constant pressure (provided no phase change occurs), dqrev dqP CPdT [Eq. (2.51)]. The relation S 兰21dqrev/T [Eq. (3.21)] becomes
(3.28)
If CPis essentially constant over the temperature range, then S CPln (T2/T1).
EXAMPLE 3.2
S for heating at constant PThe specific heat capacity cPof water is nearly constant at 1.00 cal/(g °C) in the temperature range 25°C to 75°C at 1 atm (Fig. 2.15). (a) Find S when 100 g of water is reversibly heated from 25°C to 50°C at 1 atm. (b) Without doing the cal-culation, state whether S for heating 100 g of water from 50°C to 75°C at 1 atm will be greater than, equal to, or less than S for the 25°C to 50°C heating.
(a) The system’s heat capacity is CP mcP (100 g)[1.00 cal/(g °C)] 100 cal/K. (A temperature change of one degree Celsius equals a change of one kelvin.) For the heating process, (3.28) with CPconstant gives
(b) Since CP is constant, the reversible heat required for each of the processes with T 25°C is the same. For the 50°C to 75°C change, each in-finitesimal element of heat dqrev flows in at a higher temperature than for the 25°C to 50°C change. Because of the 1/T factor in dS dqrev/T, each dqrev
1100 cal>K2 ln 323 K
298 K 8.06 cal>K 33.7 J>K
¢S
冮
TT21
dqrev T
冮
TT21
CP
T dT CP ln T2 T1
¢S
冮
TT21
CP
T dT const. P, no phase change
¢S qrev>T rev. isotherm. proc.
¢S ¢H
T 179.7 cal>g2 15.0 g2
273 K 1.46 cal>K 6.1 J>K
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produces a smaller increase in entropy for the higher-temperature process, and
S is smaller for the 50°C to 75°C heating. The higher the temperature, the smaller the entropy change produced by a given amount of reversible heat.
Exercise
Find S when 100 g of water is reversibly heated from 50°C to 75°C at 1 atm.
(Answer: 31.2 J/K.)
Now suppose we heat the water irreversibly from 25°C to 50°C at 1 atm (say, by using a bunsen-burner flame). The initial and final states are the same as for the re-versible heating. Hence the integral on the right side of (3.28) gives S for the irre-versible heating. Note that S in (3.28) depends only on T1, T2, and the value of P (since CPwill depend somewhat on P); that is, S depends only on the initial and final states. Thus S for heating 100 g of water from 25°C to 50°C at 1 atm is 33.7 J/K, whether the heating is done reversibly or irreversibly. For irreversible heating with a bunsen burner, portions of the system nearer the burner will be at higher temperatures than portions farther from the burner, and no single value of T can be assigned to the system during the heating. Despite this, we can imagine doing the heating reversibly and apply (3.28) to find S, provided the initial and final states are equilibrium states.
Likewise, if we carry out the change of state by stirring at constant pressure as Joule did, rather than by heating, we can still use (3.28).
To heat a system reversibly, we surround it with a large constant-temperature bath that is at the same temperature as the system, and we heat the bath infinitely slowly.
Since the temperature of the system and the temperature of its surroundings differ only infinitesimally during the process, the process is reversible.
6. Reversible change of state of a perfect gas. From the first law and Sec. 2.8, we have for a reversible process in a perfect gas
(3.29)
(3.30) If T2 T1, the first integral is positive, so increasing the temperature of a perfect gas increases its entropy. If V2 V1, the second term is positive, so increasing the volume of a perfect gas increases its entropy. If the temperature change is not large, it may be a good approximation to take CVconstant, in which case S ⬇ CVln (T2/T1) nR ln (V2/V1). A mistake students sometimes make in using (3.30) is to write ln (V2/V1) ln (P1/P2), forgetting that T is changing. The correct expression is ln (V2/V1) ln (P1T2/P2T1).
7. Irreversible change of state of a perfect gas. Let n moles of a perfect gas at P1, V1, T1 irreversibly change its state to P2, V2, T2. We can readily conceive of a reversible process to carry out this same change in state. For example, we might (a) put the gas (enclosed in a cylinder fitted with a frictionless piston) in a large constant-temperature bath at temperature T1and infinitely slowly change the pres-sure on the piston until the gas reaches volume V2; (b) then remove the gas from contact with the bath, hold the volume fixed at V2, and reversibly heat or cool the gas until its temperature reaches T2. Since S is a state function, S for this
¢S
冮
TT21
CV1T2
T dT nR ln V2
V1 perf. gas ¢S
冮
12CV1T2 dTT nR冮
12 dVVdS dqrev>T CV dT>T nR dV>V
dqrev dU dwrev CV dT P dV CVdT nRT dV>V
Section 3.4 Calculation of Entropy Changes 89 lev38627_ch03.qxd 2/29/08 3:12 PM Page 89
Chapter 3
The Second Law of Thermodynamics 90
reversible change from state 1 to state 2 is the same as S for the irreversible change from state 1 to state 2, even though q is not necessarily the same for the two processes. Therefore Eq. (3.30) gives S for the irreversible change. Note that the value of the right side of (3.30) depends only on T2, V2, and T1, V1, the state functions of the final and initial states.
EXAMPLE 3.3
S for expansion into a vacuumLet n moles of a perfect gas undergo an adiabatic free expansion into a vacuum (the Joule experiment). (a) Express S in terms of the initial and final tempera-tures and volumes. (b) Calculate Smif V2 2V1.
(a) The initial state is T1, V1, and the final state is T1, V2, where V2 V1. T is constant because mJ (T/V)Uis zero for a perfect gas. Although the process is adiabatic (q 0), S is not zero because the process is irreversible.
Equation (3.30) gives S nR ln(V2/V1), since the temperature integral in (3.30) is zero when T2 T1. (b) If the original container and the evacuated container are of equal volume, then V2 2V1and S nR ln 2. We have
Exercise
Find S when 24 mg of N2(g) at 89 torr and 22°C expands adiabatically into vacuum to a final pressure of 34 torr. Assume perfect-gas behavior. (Answer:
6.9 mJ/K.)
8. General change of state from (P1, T1) to (P2, T2). In paragraph 5 we considered
S for a change in temperature at constant pressure. Here we also need to know how S varies with pressure. This will be discussed in Sec. 4.5.
9. Irreversible phase change. Consider the transformation of 1 mole of supercooled liquid water (Sec. 7.4) at 10°C and 1 atm to 1 mole of ice at10°C and 1 atm.
This transformation is irreversible. Intermediate states consist of mixtures of water and ice at 10°C, and these are not equilibrium states. Moreover, with-drawal of an infinitesimal amount of heat from the ice at 10°C will not cause any of the ice to go back to supercooled water at 10°C. To find S, we use the following reversible path (Fig. 3.7). We first reversibly warm the supercooled liq-uid to 0°C and 1 atm (paragraph 5). We then reversibly freeze it at 0°C and 1 atm (paragraph 3). Finally, we reversibly cool the ice to 10°C and 1 atm (para-graph 5). S for the irreversible transformation at 10°C equals the sum of the entropy changes for the three reversible steps, since the irreversible process and the reversible process each connect the same two states. Numerical calculations are left as a problem (Prob. 3.14).
10. Mixing of different inert perfect gases at constant P and T. Let naand nbmoles of the inert perfect gases a and b, each at the same initial P and T, mix (Fig. 3.8).
By inert gases, we mean that no chemical reaction occurs on mixing. Since the
¢S>n ¢Sm R ln 2 38.314 J>1mol K2 4 10.6932 5.76 J>1mol K2
Liquid water –10°C, 1 atm
Liquid water 0°C, 1 atm
Ice –10°C, 1 atm
0°C, 1 atmIce
Figure 3.7
Irreversible and reversible paths from liquid water to ice at 10°C and 1 atm.
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gases are perfect, there are no intermolecular interactions either before or after the partition is removed. Therefore the total internal energy is unchanged on mixing, and T is unchanged on mixing.
The mixing is irreversible. To find S, we must find a way to carry out this change of state reversibly. This can be done in two steps. In step 1, we put each gas in a constant-temperature bath and reversibly and isothermally expand each gas separately to a volume equal to the final volume V. Note that step 1 is not adiabatic.
Instead, heat flows into each gas to balance the work done by each gas. Since S is extensive, S for step 1 is the sum of S for each gas, and Eq. (3.30) gives
(3.31) Step 2 is a reversible isothermal mixing of the expanded gases. This can be done as follows. We suppose it possible to obtain two semipermeable membranes, one permeable to gas a only and one permeable to gas b only. For example, heated palladium is permeable to hydrogen but not to oxygen or nitrogen. We set up the unmixed state of the two gases as shown in Fig. 3.9a. We assume the absence of friction. We then move the two coupled membranes slowly to the left. Figure 3.9b shows an intermediate state of the system.
Since the membranes move slowly, membrane equilibrium exists, meaning that the partial pressures of gas a on each side of the membrane permeable to a are equal, and similarly for gas b. The gas pressure in region I of Fig. 3.9b is Pa and in region III is Pb. Because of membrane equilibrium at each semipermeable membrane, the partial pressure of gas a in region II is Pa, and that of gas b in re-gion II is Pb. The total pressure in region II is thus Pa Pb. The total force to the right on the two movable coupled membranes is due to gas pressure in regions I and III and equals (Pa Pb) A, where A is the area of each membrane. The total force to the left on these membranes is due to gas pressure in region II and equals (Pa Pb) A. These two forces are equal. Hence any intermediate state is an equi-librium state, and only an infinitesimal force is needed to move the membranes.
Since we pass through equilibrium states and exert only infinitesimal forces, step 2 is reversible. The final state (Fig. 3.9c) is the desired mixture.
¢S1 ¢Sa ¢Sb naR ln 1V>Va2 nbR ln 1V>Vb2
Section 3.4 Calculation of Entropy Changes 91
(c) V
Vacuum
(b)
I II III
Vacuum
Permeable to a only
Permeable to b only
V V
(a)
Impermeable
Figure 3.9
Reversible isothermal mixing of perfect gases. The system is in a constant-temperature bath (not shown). (To make the figure fit the page, the sizes of the boxes don’t match those in Fig. 3.8, but they should match.)
Figure 3.8
Mixing of perfect gases at constant T and P.
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Chapter 3
The Second Law of Thermodynamics 92
The internal energy of a perfect gas depends only on T. Since T is constant for step 2, U is zero for step 2. Since only an infinitesimal force was exerted on the membranes, w 0 for step 2. Therefore q U w 0 for step 2. Step 2 is adiabatic as well as reversible. Hence Eq. (3.23) gives S2 0 for the reversible mixing of two perfect gases.
S for the irreversible mixing of Fig. 3.8 equals S1 S2, so Eq. (3.31) gives
(3.32) The ideal-gas law PV nRT gives V (na nb)RT/P and Va naRT/P, so V/Va (na nb)/na 1/xa. Similarly, V/Vb 1/xb. Substituting into (3.32) and using ln (1/xa) ln 1 ln xa ln xa, we get
(3.33) where mix stands for mixing and xaand xbare the mole fractions of the gases in the mixture. Note that mixS is positive for perfect gases.
The term “entropy of mixing” for mixS in (3.33) is perhaps misleading, in that the entropy change comes entirely from the volume change of each gas (step 1) and is zero for the reversible mixing (step 2). Because S is zero for step 2, the entropy of the mixture in Fig. 3.9c equals the entropy of the system in Fig. 3.9a.
In other words, the entropy of a perfect gas mixture is equal to the sum of the entropies each pure gas would have if it alone occupied the volume of the mixture at the temperature of the mixture. Note that Eq. (3.33) can be obtained by adding the results of applying (3.30) with T2 T1to each gas.
Equation (3.33) applies only when a and b are different gases. If they are identical, then the “mixing” at constant T and P corresponds to no change in state and S 0.
The preceding examples show that the following processes increase the entropy of a substance: heating, melting a solid, vaporizing a liquid or solid, increasing the vol-ume of a gas (including the case of mixing of gases).
Summary
To calculate S ⬅ S2 S1, we devise a reversible path from state 1 to state 2 and we use S 兰21(1/T) dqrev. If T is constant, then S qrev/T. If T is not constant, we use an expression for dqrev to evaluate the integral; for example, dqrev CP dT for a constant-pressure process or dqrev dU dwrev CVdT (nRT/V) dV for a perfect gas.
Desperate students will memorize Eqs. (3.23), (3.24), (3.25), (3.27), (3.28), (3.30), (3.32), and (3.33). But such desperate behavior commonly leads to confu-sion, error, and failure, because it is hard to keep so many equations in memory and it is hard to keep straight which equation goes with which situation. Instead, learn only the equation S 兰12(1/T) dqrev, and start with this starred equation to find S for a reversible process. To evaluate this integral, we either do something with 1/T or do something with dqrev. For a reversible isothermal process (including reversible phase changes), we take 1/T outside the integral. For constant-pressure heating with-out a phase change, we use CP dqP/dT to write dq CP dT and then integrate (CP/T). For a process in an ideal gas, we use the first law to write dqdU dw CVdT P dV, substitute this into the S equation, use PV nRT to express P/T as a function of V, and integrate. The equation (3.32) for the mixing of perfect gases at constant T and P can be found by adding the entropy changes for the volume change of each perfect gas.
¢mixS naR ln xa nbR ln xb perf. gases, const. T, P
¢S naR ln 1V>Va2 nbR ln 1V>Vb2
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