Physical Chemistry Third Edition

1262 D Some Derivations of Formulas and Methods

Let a functionSbe defined by

SS(T,V)f(V)−T+C (D-4)

whereCis a constant. Equation (D-4) applies for all values ofTandV, not just values

on the curve. Sincefis a function ofV,Sis a function ofTandVfor our closed

system, and is therefore a state function. Now we need to show thatdSvanishes on

the reversible adiabat. For reversible adiabatic processesTis equal tof(V), andSis

equal to the constantC. Therefore, for reversible adiabatic processes

dS 0 (reversible adiabatic processes) (D-5)

Since reversible adiabatic processes cannot lead away from the curve,dqrevvanishes

only on the curve. Sincef(V) represents a unique curve,dSvanishes only on the curve,

and we can write

dSydqrev (D-6)

whereyis a function that is nonzero in the vicinity of the curve. SinceSis a function,

dSis exact andyis an integrating factor. We have shown in Chapter 3 thaty 1 /Tis

a valid integrating factor.

D.2 Proof That the Liquid and Vapor Curves Are

Tangent at an Azeotrope

To show this fact we write the Gibbs–Duhem relation, Eq. (4.6-10), for the liquid phase

containing two components at constant temperature and pressure. Using Eq. (6.3-6) for

the activity and dividing byRT, we obtain the following version of the Gibbs–Duhem

relation:

x 1 d[ln(a 1 )]+x 2 d[ln(a 2 )]0(TandPconstant) (D-7)

wherex 1 andx 2 are the mole fractions in the liquid anda 1 anda 2 are the activities in

the liquid.

We assume that an ideal gas phase is at equilibrium with the solution. Using con-

vention I, we equate the chemical potential of the solvent in the two phases:

μ◦ 1 (I)+RTln(a 1 )μ◦ 1 (gas)+RTln(P 1 /P◦) (D-8)

For an infinitesimal equilibrium change in state at constantTandP

RTd[ln(a 1 )]RTd[ln(P 1 /P◦)] (D-9)

When Eq. (D-9) and the analogous equation for substance 2 are substituted in Eq. (D-8),

we obtain for the vapor at equilibrium with the solution

x 1 d[ln(P 1 /P◦)]+x 2 d[ln(P 2 /P◦)] 0 (D-10)

x 1

1

P 1

dP 1 +x 2

1

P 2

dP 2  0 (D-11)