### 3.H.3. Exercise 3.4

Compute α, β, δ and γ for a van der Waals fluid.

For convenience, we set

ε= T T C T C = T ¯ 1            ω= v v C v C = v ¯ 1             π= P P C P C = P ¯ 1

so that the reduced van der Waals equation

( P ¯ + 3 v ¯ 2 )( 3 v ¯ 1 )=8 T ¯

can be rewritten as

( π+1+ 3 ( ω+1 ) 2 )( 3ω+2 )=8( ε+1 )                      (1)

Solving for π gives

π=1 3 ( ω+1 ) 2 + 8( ε+1 ) 3ω+2

Using

( ω+1 ) 2 ( 3ω+2 )=27ω8 ω 2 3 ω 3

3( 3ω+2 )=9ω6

8 ( ω+1 ) 2 ( ε+1 )=8+16ω+8 ω 2 +8ε+16εω+8ε ω 2

we have

π= 8ε+16εω+8ε ω 2 3 ω 3 2+7ω+8 ω 2 +3 ω 3                             (2)

##### The Degree of the Critical Isotherm δ

Rewriting (3.84) in the present notation gives

π | ρ ¯ 1 | δ   = | 1 v ¯ 1 | δ ω δ              at ε=0

Eq(2) thus gives

π= 3 ω 3 2+7ω+8 ω 2 +3 ω 3 ω 3

so that δ=3 .

##### The Isothermal Compressibility Exponent γ

Rewriting (3.87) in the present notation gives

κ T ε γ           at     ω=0

where

κ T = 1 v ( v P ) T ( w π ) ε = ( π ω ) ε 1

Using (2), we have

( π ω ) ε = 16ε+16εω9 ω 2 2+7ω+8 ω 2 +3 ω 3 ( 8ε+16εω+8ε ω 2 3 ω 3 )( 7+16ω+9 ω 2 ) ( 2+7ω+8 ω 2 +3 ω 3 ) 2

=8ε14ε=6ε              for ω=0

Hence, κ T ε 1  so that γ=1 .

##### The Degree of the Coexistence Curve β

Rewriting (3.85) in the present notation gives

ρ ¯ l ρ ¯ g = 1 v ¯ l 1 v ¯ g = v ¯ g v ¯ l v ¯ l v ¯ g = ω g ω l v ¯ l v ¯ g = A β ( ε ) β

or

ω g ω l ( ε ) β                      (a)

along the coexistence curve.  Now, the intercepts (points C and G in Fig.3.9) of an isotherm T 0 < T C  with the coexistence curve is given by the Maxwell construction as

C G vdP =0                          (6)

where the integral is along the isotherm T= T 0 .  In terms of the present notation, we have

C G ( 1+ω )dπ =0                        (6a)

where π is given by (2) with ε= ε 0 = T 0 T C 1 .  In the vicinity of the critical point, (2) can be written as

π=( 4ε+8εω+4ε ω 2 3 2 ω 3 ) ( 1+ 7 2 ω+4 ω 2 + 3 2 ω 3 ) 1

=( 4ε+8εω+4ε ω 2 3 2 ω 3 )   [ 1 7 2 ω+( 49 4 4 ) ω 2 +( 343 8 3 2 ) ω 3 + ]

4ε+8εω+4ε ω 2 3 2 ω 3 7 2 ω( 4ε+8εω )+ 33 4 ω 2 ( 4ε )+

4ε6εω+9ε ω 2 3 2 ω 3 +                  (5)

where only terms ε n ω m  with n+m3  are kept.  For fixed ε= ε 0 , we have

dπ( 6 ε 0 +18 ε 0 ω 9 2 ω 2 + )dω                        (5a)

so that (6a) becomes

C G ( 1+ω )( 6 ε 0 +18 ε 0 ω 9 2 ω 2 + )dω =0

Þ            C G ( 6 ε 0 +12 ε 0 ω 9 2 ω 2 + )dω =0

[ 6 ε 0 ω+6 ε 0 ω 2 3 2 ω 3 + ] C G =0

6 ε 0 ( ω g ω l )+6 ε 0 ( ω g 2 ω l 2 ) 3 2 ( ω g 3 ω l 3 )+=0              (9)

where, by definition, ω= ω g  at G and ω= ω l  at C [see Fig.3.8-9].

Now, both G and C are at the same temperature and pressure so that (5) gives

4 ε 0 6 ε 0 ω g +9 ε 0 ω g 2 3 2 ω g 3 +=4 ε 0 6 ε 0 ω l +9 ε 0 ω l 2 3 2 ω l 3 +

Þ            6 ε 0 ( ω g ω l )+9 ε 0 ( ω g 2 ω l 2 ) 3 2 ( ω g 3 ω l 3 )+=0              (8)

This is compatible with (9) only if

9 ε 0 ( ω g 2 ω l 2 )=6 ε 0 ( ω g 2 ω l 2 )

Þ            ω g 2 = ω l 2                             (8a)

Since ω g = v g v C v C 0  and ω l = v l v C v C 0 , eq(8a) implies

ω g = ω l 0                             (8b)

Either (8) or (9) then gives

12 ε 0 ω g 3 ω g 3 0

Þ            ω g =0               or             ω g 2 =4 ε 0

Thus, slightly away from the critical point, we have

ω g =2 ε 0                              (11)

where ε 0 <0  since T 0 < T C .  Comparing with condition (a) gives β= 1 2 .

##### The Heat Capacity Exponent α

Rewriting (3.86) in the present notation gives

c v { ( ε ) α' ( +ε ) α            for         ε<0 ε>0                 at     ω=0

Now, as one approaches the critical point along the isochore ω=0 , one gets [see (3.88)],

c v ={ x g c v g + x l c v l + x g T 0 κ T g ρ g 3 ( ρ g T ) coex 2 + x l T 0 κ T l ρ l 3 ( ρ l T ) coex 2 c v g            for           ε<0 ε>0

Very close to the critical point, the liquid and vapor phases merges so that

c v { c v g + [ T 0 κ T ρ 3 ( ρ T ) 2 ] coex c v g           for           ε 0 ε 0 +

Thus, there is a discontinuity at the critical point,

Δ= c v ( T C ) c v ( T C + )= [ T 0 κ T ρ 3 ( ρ T ) 2 ] crit                 with  v= v C

Now,

ρ T = ρ C T C ρ ¯ T ¯   = ρ C T C T ¯ ( 1 v ¯ )   = ρ C T C ε ( 1 ω+1 )

= ρ C T C ( ω+1 ) 2 ω ε   ρ C T C ( ω+1 ) 2 2 ω

where eq(11) was used to obtain the last equality.  Thus

1 ρ 3 ( ρ T ) 2 ( ω+1 ) 3 ρ C 3 [ ρ C T C ( ω+1 ) 2 2 ω ] 2   = 4 T C 2 ρ C ( ω+1 ) ω 2

Similarly,

κ T = 1 v ( v P ) T   = 1 P C v ¯ ( v ¯ P ¯ ) T

= 1 P C ( ω+1 ) ( ω π ) ε   = 1 P C ( ω+1 ) ( π ω ) ε 1

Using (5a), we have

( π ω ) ε 6ε+18εω 9 2 ω 2 +

so that

1 κ T P C ( 1+ω )( 6ε18εω+ 9 2 ω 2 + )

Putting everything together, we have

Δ [ T C ( ε+1 ) 4 T C 2 ρ C ( ω+1 ) ω 2 P C ( 1+ω )( 6ε18εω+ 9 2 ω 2 + ) ] crit

where crit means ω=2 ε  with ε=0 .  Thus

Δ 4 P C T C ρ C ( 6 ε ω 2 18 ε ω + 9 2 + )

4 P C T C ρ C ( 3 2 +9 ε + 9 2 + )

Using P C T C ρ C = 3 8 R , we have

Δ 9 2 R                             (14)

Since the jump is finite, we have α=α'=0 .