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

The (molar) van der Waals equation

( P+ a v 2 )( vb )=RT                                      (2.12)

can be written in various forms such as

P= a v 2 + RT vb                                               (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.