3.H.2. The Critical Exponents For Pure PVT Systems

There are 4 critical exponents of interest.

(a)    Degree of the Critical Isotherm, δ

P P C P C 0 A δ | ρ ρ C ρ C | δ sign( ρ ρ C )                 T= T C               (3.84)

where the subscript 0 denotes properties of the ideal gas.

Experimentally, 6>δ4 .

(b)    Degree of the Coexistence Curve, β

ρ l ρ g ρ C = A β ( ε ) β          T< T C               (3.85)

where ρ l ρ g  is the order parameter.  Experimentally, β0.34 .

(c)    Critical Exponent of the Heat Capacity, α, α'

C V ={ A α ' ( ε ) α' A α ( +ε ) α      for  T< T C T> T C       ρ= ρ C          (3.86)

Experimentally, αα'0.1 .

(d)    Isothermal Compressibility, γ

κ T κ T 0 ={ A γ ' ( ε ) γ' A γ ( +ε ) γ      for  T< T C T> T C                            (3.87)

Experimentally, γ'1.2  and γ1.3 .

Exponent Inequalities

Using

        ( P v ) T = 1 v κ T = ρ m κ T and          v T = T ( m ρ )= m ρ 2 ρ T

where ρ the mass density and m is the molar weight, we can rewrite eq(3.40) as

        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               (3.88)

where c now denotes specific heat (molar heat capacity /m).  Since all terms on the right are positive, we have

        c v x g T 0 κ T g ρ g 3 ( ρ g T ) coex 2                         (3.89)

Now, when the critical point is approached from below, we have

        c v ( ε ) α'                                      [from (3.86)]

        ρ ρ C    

        κ T ( ε ) γ'                                     [from (3.87)]

        [ ( ρ l ρ g ) T ] coex ( ε ) β1                       [from (3.85)]

Taking the square of the last gives

        ( ρ l T ρ g T ) coex 2 ( ε ) 2β2

On the other hand, ( ρ l T + ρ g T ) coex 2  is expected to be regular, [see (3.26)], so that

        1 4 [ ( ρ l T ρ g T ) 2 + ( ρ l T + ρ g T ) 2 ] coex   = 1 2 [ ( ρ l T ) 2 + ( ρ g T ) 2 ] coex  

( ρ T ) coex 2 ( ε ) 2β2

Putting everything into (3.89), we have

        c v = A 1 ( ε ) α' A 2 κ T ( ρ T ) coex 2 = A 3 ( ε ) γ'+2β2                 (3.90)

where the Aj's are constants.  Since 0<ε<1 , we have

ln( ε )=| ln( ε ) |<0

Thus, the singular part of the logarithm of (3.90) gives

        α'| ln( ε ) |( γ'+2β2 )| ln( ε ) |

Þ            α'+γ'+2β2                         (3.92)

which is known as the Rushbrook inequality.