### 3.D.4. The van der Waals Equation

The (molar) van der Waals equation

(
P+
a
v
2
)(
v−b
)=RT
(2.12)

can be written in various forms such as

P=−
a
v
2
+
RT
v−b
(3.41a)

or

v
3
−(
b+
RT
P
)
v
2
+
a
P
v−
ab
P
=0
(3.41)

An isotherm with
T<
T
C
in the *v-P*
plane is shown in Fig.3.9.

In order to simplify the notations, we
shall replace all quantities *x* with
their reduced values
x
¯
=
x
x
C
,
where the subscript *C* denotes value
at the critical point. Using the fact
that the isotherm
T=
T
C
has an inflection point at the critical point,
we get

(
∂P
∂v
)
T=
T
C
=
2a
v
C
3
−
R
T
C
(
v
C
−b
)
2
=0
(3.42)

(
∂
2
P
∂
v
2
)
T=
T
C
=−
6a
v
C
4
+
2R
T
C
(
v
C
−b
)
3
=0
(3.42a)

Eliminating
T
C
gives

4a
v
C
3
(
v
C
−b
)
=
6a
v
C
4
Þ
v
C
=3b

so that (3.42) becomes

T
C
=
2a
R
v
C
3
(
v
C
−b
)
2
=
8a
27bR

which turns (3.41a) into

P
C
=−
a
9
b
2
+
4a
27
b
2
=
a
27
b
2

Eq(2.12) thus becomes

(
a
P
¯
27
b
2
+
a
9
b
2
v
¯
2
)(
3b
v
¯
−b
)=
8a
27b
T
¯

Þ
(
P
¯
+
3
v
¯
2
)(
3
v
¯
−1
)=8
T
¯
(3.44)

P
¯
=−
3
v
¯
2
+
8
T
¯
3
v
¯
−1
(3.44a)

v
¯
3
−
1
3
(
1+
8
T
¯
P
¯
)
v
¯
2
+
3
P
¯
v
¯
−
1
P
¯
=0
(3.44b)

Note that the independence of
these equations on *a* and *b* is another example of the law of
corresponding states.

#### Maxwell Construction

An isotherm with
T<
T
C
in the *v-P*
plane is shown in Fig.3.9.

Now, for equilibrium states, an isotherm
v(
P
)
must be single- valued except at transition
points. Thus, if one tries to bring the
system quasi-statically and isothermally from A to I, the actual path followed must
be one similar to ABCEGHI. The task is
to determine the position of the vertical line segment CEG, which indicates a
coexistence of phases [cf. Fig.3.7].

Note that one implication of the above
discussion is that the states on the segment DEF are not in equilibrium. This is indeed the case since the slope of
the curve there is positive so that

κ
T
=−
1
v
(
∂v
∂P
)
T
<0
Þ mechanical instability

Consider now the molar Gibbs free energy
g(
T,P
)
along the isotherm in Fig.3.9.

Since
T=const
,
we have
dg=vdP
or

g
X
=
g
A
+
∫
P
A
P
X
v(
P
)dP
(3.46)

=
g
A
+I(
X,A
)

where
I(
X,A
)=
∫
P
A
P
X
v(
P
)dP
. The values of *g* as *X* moves from A to I
are shown in Fig.3.10. The salient
features are as follows:

1.
From A to D:
g
X
=
g
A
+I(
X,A
)
increases monotonically from
g
A
to
g
D
=
g
A
+I(
D,A
)
.

2.
From D to F:
g
X
=
g
D
+I(
X,D
)
=
g
D
−I(
D,X
)
decreases monotonically from
g
D
to
g
F
=
g
D
−I(
D,F
)
.

3.
From F to I:
g
X
=
g
F
+I(
X,F
)
increases monotonically from
g
F
to
g
I
=
g
F
+I(
I,F
)
.

As shown in Fig.3.10, the segments AD and
FI intersects in the *g-P* plane. Since the equilibrium state has the lowest
energy, the quasi-static isothermal path in the *g-P* plane runs as A → intersect → I . Referring back to the *v-P* plane, we see that C and G must be the position of this
intersect on the segments AD and FI, respectively.

Now, the condition
g
C
=
g
G
means

g
A
+I(
C,A
)=
g
F
+I(
G,F
)

=
g
D
−I(
D,F
)+I(
G,F
)

=
g
A
+I(
D,A
)−I(
D,F
)+I(
G,F
)

Þ
I(
C,A
)−I(
D,A
)=−I(
D,F
)+I(
G,F
)
(a)

Using

I(
C,A
)−I(
D,A
)=−I(
D,C
)

and

I(
D,F
)=I(
D,E
)+I(
E,F
)

eq(a) becomes

I(
D,C
)−I(
D,E
)=I(
E,F
)−I(
G,F
)
(3.48)

i.e.,
Area 2=Area 1
[see shaded areas in
Fig.3.9]

This is called the **Maxwell construction**.