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204 PHASE EQUILIBRIA

Assumption 3: V m is the molar change in volume during the phase change. The
3
value of V m = V m(g) − V m(l) ,where V m(l) is typically 20 cm mol −1 and V m(g) is
−1
3
3
22.4dm mol −1 (at s.t.p.), i.e. 22 400 cm mol . In response to the vast discrepancy
between V m(g) and V m(l) , we assume that V m ≈ V m(g) ,i.e. that V m(l) is negligible by
comparison. This third approximation is generally good, and will only break down at
very low temperatures.
First, we rewrite the Clapeyron equation in response to approximation 2:

dp H m O
=
dT TV m(g)
Next, since we assume the gas is ideal, we can substitute for the V m term via the
ideal-gas equation, and say V m = RT ÷ p:

dp H m O p
= ×
dT T RT
Next, we multiply together the two T terms, rearrange and separate the variables, to
give:
1 H m O 1
dp = × dT
p R T 2

Integrating with the limits p 2 at T 2 and p 1 at T 1 gives
O
We place the ‘ H m ÷ R’
term outside the right- p 2 1 H m O T 2 1
hand integral because p dp = R T 2 dT
its value is constant. p 1 T 1
Subsequent integration yields

1
H m O T 2
[ln p] p 2 =− ×
R T
p 1
T 1
Next, we insert limits:
H m O 1 1
ln p 2 − ln p 1 =− −
R T 2 T 1

And, finally, we group together the two logarithmic terms to yield the Clausius–Clap-
eyron equation:
O
p 2 H 1 1
ln =− −
p 1 R T 2 T 1