Partial Molal Properties Using Mixture State Equations

5. Partial Molal Properties Using Mixture State Equations

Any partial molal property can be determined if a mixture equation of state can be developed.

a. Kay’s rule Since v = V/N, for a mixture, VW equation of state assumes the form

P=N R T/(V – N b 2 m 2 )–N a m /V , where

(153) The pseudo–critical temperature and pressure are functions of composition. The partial molal

a m = (27/64) R 2 T 2 cm /P cm and b m = (1/8) R T cm /P cm .

volume

(154) can be obtained by differentiating Eq (152). It is noted that the pseudo–critical temperature

ˆv i =( ∂V/∂N i ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K

T cm and pressure P cm are also functions of composition. See example below. q. Example 19

A two–component mixture consisting of 40% carbon dioxide (1) and 60% acetylene (2) on molal basis at 320 K and 100 bar is well–described by Kay’s rule and the VW equation of state. Obtain an expression for the partial molal volume ˆv CO2 and deter-

mine ˆv CO 2 . Solution

Using Kay’s rule, for a 2 component mixture

T cm =X 1 T c1 +X 2 T c2 , and P cm =X 1 P c1 +X 2 P c2 , where

X 1 =N 1 /N, X 2 =N 2 /N, and N = N 1 +N 2 .

The total differential of Eq. (152)

dP = N R dT/(V – b m N) + dN R T/(V – b m N) – (N R T/(V – b m N) 2 )(dV– d b m N– b m dN)

2 2 2 2 – 2NdN 3 a

m /V –N d a m /V + (2N a m /V )dV.

(A) If we keep T, P, and N 2 are held constant, then dN = dN 1 . Consequently, Eq. (A) as-

sumes the form,

0 = 0 + dN

1 R T/(V – b m N) – (N R T/(V – b m N) )(dV – d b m N– b m dN 1 )

(B) 0= 2 R T/(V – b

2 2 2 2 – 2N dN 3

1 a m /V –N d a m /V + (2N a m /V ) dV, i.e.,

m N) – (N R T/(V – b m N) ) (dV/dN 1 – (d b m /dN 1 )N – b m )

(C) The partial molal volume ˆv 1 = dV/dN 1 = (2N a m /V 2 – R T/(V – b m N) – (N R T/(V – b m N) 2 )

– 2N a /V 2 –N 2 (d a /dN )/V 2 + (2N m 2 m 1 a m /V 3 )dV/dN 1 .

((d b 2 2 2 3 m 2 /dN 1 )N+ b m )+N (d a m /dN 1 )/V )/((2N a m /V ) – (N R T/(V– b m N) ). (D)

Differentiating Eq. (153) with respect to N 1 ,

2 2 2 (d 2 a m /dN 1 ) N 2 =(54/64) R T cm (dT cm /dN 1 )/P cm –(27/64) R T cm (dP cm /dN 1 )/P cm , (E) (d 2 b

m /dN 1 ) N 2 = (1/8) R (dT cm /dN 1 )/P cm – (1/8) R T cm (dP cm /dN 1 )/P cm , (F) (dT cm /dN 1 ) N 2 =(1/N)T c,1 –(N 1 /N 2 )T c,1 –(N 2 /N 2 )T c,2 =(1/N)(T c,1 –X 1 T c,1 –X 2 T c,2 ), and(G) (dP cm /dN 1 )

N 2 = (1/N)P c,1 –(N 1 /N )P c,1 –(N 2 /N )P c,2 = (1/N)(P c,1 –X 1 P c,1 –X 2 P c,2 ). (H) In case of infinite dilution, as N 1 →0, T cm →T c2 and P cm →P c2 , so that (dT cm /dN 1 ) N 2 = → (T c1 –T c,2 )/N, and (dP cm /dN 1 ) N 2 → (P c1 –P c,2 )/N, i.e.,

2 ˆv 2

1 = ((2 a m / v )–( R T/( v – b m )) – ( R T/( v – b m ) )(N(d b m /dN 1 )+ b m )

2 3 + (Nd 2 a m /dN 1 )/V ))/((2 a m / v )–( R T/( v – b m ) ), (cf. Eq. (D)). (I) As X 1 →0, N 1 →0, b m →b 2 and v → v 2 . Similarly Eqs. (E) and (F) can be rewritten as