#### 3.D.2.a. Vaporization Curve

Consider the coexistence curve between the
liquid and gaseous phases of a pure substance.
This is indicated as the vaporization curve between the triple A and
critical point C in Figure 3.4.

If we neglect the volume changes of the
liquid phase, we have

Δ
v
lg
=
v
g
−
v
l
≃
v
g
=
RT
P

where the last equality assumes the vapor
to be an ideal gas.

The Clausius-Clapeyron equation thus
simplifies to

(
dP
dT
)
coex
=
Δ
h
lg
TΔ
v
lg
≃
PΔ
h
lg
R
T
2
(3.15)

where
Δ
h
lg
is the molar latent heat of vaporization. Treating
Δ
h
lg
as a constant, we get, along the vaporization
curve,

∫
dP
P
=
Δ
h
lg
R
∫
dT
T
2

lnP=−
Δ
h
lg
RT
+const

P=
P
∞
exp(
−
Δ
h
lg
RT
)
(3.16)

where
P
∞
is the pressure for
T→∞
.

#### Exercise 3.2

Find the molar heat capacity
c
coex
along the vaporization curve.

**Answer**

c
coex
=T
(
∂s
∂T
)
coex

With *T*,
*P* as independent
variables, we have

ds=
(
∂s
∂T
)
P
dT+
(
∂s
∂P
)
T
dP

On the vaporization curve, we have

dP=
(
dP
dT
)
coex
dT≃
PΔ
h
lg
R
T
2
dT

so that

(
ds
)
coex
=[
(
∂s
∂T
)
P
+
(
∂s
∂P
)
T
(
∂P
∂T
)
coex
]dT

≃[
(
∂s
∂T
)
P
+
(
∂s
∂P
)
T
PΔ
h
lg
R
T
2
]dT

Thus

c
coex
=
c
P
+
(
∂s
∂P
)
T
PΔ
h
lg
RT

Using

(
∂s
∂P
)
T
=−
(
∂v
∂T
)
P
=−
R
P

where the last equality is valid for an
ideal gas, we have

c
coex
=
c
P
−
Δ
h
lg
T
(2)

which means for
T<
Δ
h
lg
c
P
,
we have
c
coex
<0
so that the vapor gives off heat as *T* is raised.