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)