Page 109 - Physical Chemistry

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Chapter 3 reversible change from state 1 to state 2 is the same as S for the irreversible
The Second Law of Thermodynamics
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 T , V , and T , V , the state
2 2 1 1
functions of the final and initial states.

EXAMPLE 3.3 S for expansion into a vacuum

Let 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 S if V 2V .
m 2 1
(a) The initial state is T , V , and the final state is T , V , where V V .
1 1 1 2 2 1
T is constant because m ( T/ V) is zero for a perfect gas. Although the
J U
process is adiabatic (q 0), S is not zero because the process is irreversible.
Equation (3.30) gives S nR ln(V /V ), since the temperature integral in (3.30)
2 1
is zero when T T . (b) If the original container and the evacuated container
2 1
are of equal volume, then V 2V and S nR ln 2. We have
2 1
¢S>n ¢S R ln 2 38.314 J>1mol K2410.6932 5.76 J>1mol K2
m
Exercise
Find S when 24 mg of N (g) at 89 torr and 22°C expands adiabatically into
2
vacuum to a final pressure of 34 torr. Assume perfect-gas behavior. (Answer:
6.9 mJ/K.)

8. General change of state from (P , T ) to (P , T ). In paragraph 5 we considered
1 1 2 2
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 at 10°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 n and n moles
a b
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

Liquid water Ice
–10°C, 1 atm –10°C, 1 atm
Figure 3.7

Irreversible and reversible paths
Ice
from liquid water to ice at 10°C Liquid water 0°C, 1 atm
0°C, 1 atm
and 1 atm.