Relations between Partial Molal and Pure Properties

5. Relations between Partial Molal and Pure Properties

We have discussed the partial molal volume and now focus on other partial molal properties.

a. Partial Molal Enthalpy and Gibbs function Since the enthalpy H(T, P, N 1 ,N 2 , ... .) = U(T, P, N 1 ,N 2 , ... .) + PV(T, P, N 1 ,N 2 , ... .), the partial molal enthalpy ˆh i =( ∂H/∂N i ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K can be expressed as

ˆh i =( ∂U/∂N i ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K + P( ∂V/∂N i ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K , i.e., (26) ˆh i = ˆu i +P ˆv , where

(27) ˆu i =( ∂(N u )/ ∂N i ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K = N( ∂ u / ∂N i ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K + u .

Similarly, since G = H – TS, ˆg i =( ∂H/∂N i ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K – T( ∂S/∂N i ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K , i.e.,

(28) ˆg i = ˆh i –T ˆs i .

(29) Likewise,

ˆa i = ˆu i –T ˆs i . (30)

b. Differentials of Partial Molal Properties Applying the Gibbs–Duhem equation

(12c) in terms of the Gibbs energy, i.e., B = G, ∂G/∂T = –S, and ∂G/∂P = V, so that

( ∂B/∂T) P,N dT + ( ∂B/∂T) T,N dP – Σ k d ˆb k N k =0

–S dT + V dP – Σ k d ˆg k N k = 0.

In terms of intensive properties, this relation may be written in the form – Σ k ˆs k N k dT + Σ k ˆv k N k dP – Σ k d ˆg k N k =– Σ k N k (d ˆg k + ˆs k dT– ˆv k dP) = 0, i.e.,

For arbitrary N k >0,

d ˆg k =– ˆs k dT + ˆv k dP. (31)

Differentiating Eq. (29) and using (31) to eliminate dg ˆ k ,

d ˆh k = Td ˆs k + ˆv k dP. (32a) Dividing Eq. (32) by dT at constant pressure, the partial molal specific heat ˆc pk = ∂ ˆh k / ∂T = T ∂ ˆs k / ∂T.

(32b) Subtracting the term d(Pv k ) from Eq. (32a), we obtain the relation

d ˆu k =Td ˆs k –Pd ˆv k . (33a) Similarly

ˆc vk =( ∂ ˆu k / ∂T) = T (∂ ˆs k / ∂T). (33b) These relations for partial molal properties are similar to those for pure substances.

Maxwell’s relations can be likewise derived. Subtracting d(T ˆs k ) from Eq. (33a), we obtain the relation

d ˆa k =– ˆs k dT – P d ˆv k . (34) This implies that ˆs k =– ∂ ˆa k / ∂T, P = – ∂ ˆa k / ∂ ˆv k . These expressions are similar to those for pure

properties. Maxwell’s relations can be likewise derived. Furthermore, from Eq. (32a) ∂ ˆh k /dP = T ∂ ˆs k /dP + ˆv k .

(35) Using the Maxwell’s relations we can show that ∂ ˆh k /dP = – T ∂ ˆv k /dT + ˆv k . For the entropy, the G–D relation is

(36) Since S = Σ k ˆs k N k and ∂ ˆs k / ∂T = ˆc pk /T, we may use Maxwell’s relations to simplify the second

( ∂S/∂T) P,N dT + ( ∂S/∂P) T,N dP – Σ k d ˆs k N k =0

term, i.e., ( ∂S/∂P) T,N = –( ∂V/∂T) P , where V = Σ k Ν k ˆv k so that

Σ k N k ( ˆc p,k /T)dT – Σ k (N k ∂ ˆv k / ∂T) dP – Σ k d ˆs k N k = 0, or

Σ k N k (( ˆc p,k /T)dT – ( ∂ ˆv k / ∂T)dP – d ˆs k ) = 0. Therefore,

d ˆs k =( ˆc p,k /T)dT – ( ∂ ˆv k / ∂T) dP. (37) Using Eqs. (37) in Eq. (32a),

d ˆh k = ˆc p,k dT + ( ˆv k –T( ∂ ˆv k / ∂T)) dP, (38) which is again similar to the corresponding expression for a pure substance. Likewise,

d ˆu k = ˆc v,k dT + (T( ∂P/∂T) – P) d ˆv k (39) Equations (37) to (39) are similar to the corresponding expressions for a pure substance. Thus

if state equations are available for mixtures, ˆu k , ˆh k and ˆs k can be determined.

i. Remarks Maxwell’s relations can be obtained using Eqs. (31)–(34). These relations are similar

to those for pure components. Consider the derivative ( ∂/∂N i ( ∂V/∂T) P, N N N 1 , 2 ,..., ji ≠ ,..., N K ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K . Switching or-

der of differentiation, the expression equals the term

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

Likewise, ( ∂/∂N i ( ∂S/∂T) P, N N N 1 , 2 ,..., ji ≠ ,..., N K ) T, P, N N N 1 , 2 ,..., ji ≠ ,..., N K =T ∂ ˆs i / ∂T = ˆc pi ,

(40) which is again an expression that is similar to that for a pure substance.