Page 234 - Physical Chemistry

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Equation (7.14) applies to both open and closed systems. A quick way to obtain (7.14) Section 7.3
is to divide dG SdT VdP by n. Although dG SdT VdP applies to a The Clapeyron Equation
closed system, G is an intensive property and is unaffected by a change in system size.
m
Use of (7.14) in (7.13) gives
b
b
a
a
S dT V dP S dT V dP (7.15)
m
m
m
m
where dT and dP are the infinitesimal changes in T and P on going from point 1 to
point 2 along the a-b equilibrium line. Rewriting (7.15), we have
b
a
b
a
1V V 2 dP 1S S 2 dT (7.16)
m
m
m
m

a
dP S S b m ¢S m ¢S
m
(7.17)*
a
dT V V b ¢V m ¢V
m
m
where S and V are the entropy and volume changes for the phase transition b → a.
For the transition a → b, S and V are each reversed in sign, and their quotient is
unchanged, so it doesn’t matter which phase we call a.
For a reversible (equilibrium) phase change, we have S H/T, Eq. (3.25).
Equation (7.17) becomes
dP ¢H m ¢H
one component two-phase equilib. (7.18)*
dT T ¢V m T ¢V
Equation (7.18) is the Clapeyron equation, also called the Clausius–Clapeyron equa-
tion. Its derivation involved no approximations, and (7.18) is an exact result for a one-
component system.
For a liquid-to-vapor transition, both H and V are positive; hence dP/dT is pos-
itive. The liquid–vapor line on a one-component P-T phase diagram has positive slope.
The same is true of the solid–vapor line. For a solid-to-liquid transition, H is virtu-
ally always positive; V is usually positive but is negative in a few cases, for example,
H O, Ga, and Bi. Because of the volume decrease for the melting of ice, the solid–
2
liquid equilibrium line slopes to the left in the water P-T diagram (Fig. 7.1). For nearly
all other substances, the solid–liquid line has positive slope (as in Fig. 7.3). The fact
that the melting point of ice is lowered by a pressure increase is in accord with Le
Châtelier’s principle (Sec. 6.6), which predicts that a pressure increase will shift the
equilibrium to the side with the smaller volume. Liquid water has a smaller volume
than the same mass of ice.
For melting, V is much smaller than for sublimation or vaporization. Hence the
m
Clapeyron equation (7.18) shows that the solid–liquid equilibrium line on a P-versus-
T phase diagram will have a much steeper slope than the solid–vapor or liquid–vapor
lines (Fig. 7.1).
Liquid–Vapor and Solid–Vapor Equilibrium
For phase equilibrium between a gas and a liquid or solid, V m,gas is much greater than
V m,liq or V m,solid unless T is near the critical temperature, in which case the vapor and
liquid densities are close (Fig. 7.2). Thus, when one of the phases is a gas, V
m
V m,gas V m,liq or solid V m,gas . If the vapor is assumed to behave approximately ideally,
then V m,gas RT/P. These two approximations give V RT/P, and the Clapeyron
m
equation (7.18) becomes
dP>dT P ¢H >RT 2
m
d ln P ¢H m
solid–gas or liq.–gas equilib. not near T (7.19)*
dT RT 2 c