Page 111 - Physical Chemistry

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Chapter 3 The internal energy of a perfect gas depends only on T. Since T is constant for
The Second Law of Thermodynamics 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 S 0 for the reversible
2
mixing of two perfect gases.
S for the irreversible mixing of Fig. 3.8 equals S S , so Eq. (3.31)
1
2
gives
¢S n R ln 1V>V 2 n R ln 1V>V 2 (3.32)
a
a
b
b
The ideal-gas law PV nRT gives V (n n )RT/P and V n RT/P, so V/V
b
a
a
a
a
(n n )/n 1/x . Similarly, V/V 1/x . Substituting into (3.32) and using
b
a
b
a
a
b
ln (1/x ) ln 1 ln x ln x , we get
a
a
a
¢ mix S n R ln x n R ln x perf. gases, const. T, P (3.33)
a
a
b
b
where mix stands for mixing and x and x are the mole fractions of the gases in
b
a
the mixture. Note that mix S is positive for perfect gases.
The term “entropy of mixing” for mix S 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 T T to each gas.
2
1
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 S S , we devise a reversible path from state 1 to state 2 and
1
2
2
we use S (1/T) dq . If T is constant, then S q /T. If T is not constant,
rev
rev
1
we use an expression for dq rev to evaluate the integral; for example, dq rev C dT
P
for a constant-pressure process or dq rev dU dw rev C dT (nRT/V) dV for a
V
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
2
only the equation S (1/T) dq , and start with this starred equation to find S
rev
1
for a reversible process. To evaluate this integral, we either do something with 1/T
or do something with dq . For a reversible isothermal process (including reversible
rev
phase changes), we take 1/T outside the integral. For constant-pressure heating with-
out a phase change, we use C dq /dT to write dq C dT and then integrate
P
P
P
(C /T). For a process in an ideal gas, we use the first law to write dq dU dw
P
C dT PdV, substitute this into the S equation, use PV nRT to express P/T as
V
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.