Page 229 - Physical chemistry understanding our chemical world

196 PHASE EQUILIBRIA

by 1.2 K if the pressure is increased fivefold. Calculate V m for the wax as
it melts. Take H (melt) = 8.064 kJ mol −1 .

Justification Box 5.1

Consider two phases (call them 1 and 2) that reside together in thermodynamic equi-
librium. We can apply the Gibbs–Duhem equation (Equation 4.31) for each of the two
phases, 1 and 2.

For phase 1 : dG (1) = (V m(1) dp) − (S m(1) dT) (5.2)

For phase 2 : dG (2) = (V m(2) dp) − (S m(2) dT) (5.3)

where the subscripts ‘m’ imply molar quantities, i.e. per mole of substance in each phase.
Now, because equilibrium exists between the two phases 1 and 2, the dG term in
each equation must be the same. If they were different, then the change from phase 1 to
phase 2 (G (2) − G (1) ) would not be zero at all points; but at equilibrium, the value of
G will be zero, which occurs when G (2) = G (1) . In fact, along the line of the phase
boundary we say dG (1) = dG (2) .
In consequence, we may equate the two equations, saying:

(V m(1) × dp) − (S m(1) × dT) = (V m(2) × dp) − (S m(2) × dT)

Factorizing will group together the two S and V terms to yield

(S m(2) − S m(1) ) dT = (V m(2) − V m(1) ) dp

which, after a little rearranging, becomes

dp S m(1→2)
= (5.4)
dT V m(1→2)

O
Finally, since G = H O − T S O (Equation (4.21)), and since G = 0at
O
equilibrium:
H O
m
= S m(1→2)
T
Inserting this relationship into Equation (5.4) yields the Clapeyron equation in its famil-
iar form.
O
dp H m(1→2)
=
dT T V m(1→2)
In fact, Equation (5.4) is also called the Clapeyron equation. This equation holds for
phase changes between any two phases and, at heart, quantitatively defines the phase