Page 110 - Physical Chemistry

lev38627_ch03.qxd 2/29/08 3:12 PM Page 91

91
Section 3.4
Calculation of Entropy Changes

Figure 3.8
Mixing of perfect gases at
constant T and P.

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

¢S ¢S ¢S n R ln 1V>V 2 n R ln 1V>V 2 (3.31)
a
1
b
b
b
a
a
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 P a
and in region III is P . Because of membrane equilibrium at each semipermeable
b
membrane, the partial pressure of gas a in region II is P , and that of gas b in re-
a
gion II is P . The total pressure in region II is thus P P . The total force to the
b
a
b
right on the two movable coupled membranes is due to gas pressure in regions I
and III and equals (P P )A, where A is the area of each membrane. The total
a
b
force to the left on these membranes is due to gas pressure in region II and equals
(P P )A. These two forces are equal. Hence any intermediate state is an equi-
a
b
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.
Permeable to Permeable to Impermeable
a only b only
I II III
Figure 3.9
Vacuum Vacuum Reversible isothermal mixing of
perfect gases. The system is in a
constant-temperature bath (not
shown). (To make the figure fit the
V V V
page, the sizes of the boxes don’t
match those in Fig. 3.8, but they
(a) (b) (c) should match.)