Problems and Solutions on Thermodynamics and Statistical Mechanics

130 Problem8 8 Sdutiom on Thermodynamics 8 Statidicd Mechanics

Solution:

x << 1. Thus the mole chemical potential of the solution is

(a) We assume the mole concentration of the solute in the solution is

1.11 (P, T) = 1.1; (P, T) - XRT ,

where py(p,T) is the mole chemical potential of the pure solvent. If the
mole chemical potential of the vapor phase is &(p,T), the equilibrium
vapor pressure of the solvent, PO, is determined by

1.1; (Po, To) = 1.1; (Po, To).

When the external pressure (the total pressure) is p, the condition of

equilibrium of vapor and liquid is

Making use of Taylor’s theorem, we have from the above two equations

Using the thermodynamic relation dp = -SdT + Vdp, we can write the

above as

P - Po = [(S2 - S,)(T - To) - XRT]/(V2 - Vl) ,

or

PO = P - [L(T - To)/T - zRT]/(V2 - VI) ,

where V is the mole volume, S is the mole entropy, and L is the latent

heat, L = T(S2 - S1).

(b) The triple point of water is the temperature TO at which ice, water

and vapor are in equilibrium. The ice point is the temperature T at which

pure ice and air-saturated water are in equilibrium at 1 atm. Utilizing the

result in (a) we have

T - To = T(V2 - Vi)(p - po)/L + xRT2/L ,

where V, and V1 are respectively the mole volumes of ice and water. From

V2 > Vl and L < 0, we know the ice point is lower than the triple point.