Question

A and B are for and ideal solution. Pure A has a vapor pressure of 84.3 torr, while pure B has a vapor pressure of 41.2 torr. The mole fraction of A in the liquid phase is equal to 0.32.

a). Calculate PA,PB and Ptotal.

b). If a portion of the gas phase is removed and condensed in a seperate container, what will PA,PB, and Ptotal of the new system be?

Answer 1

Raoult's Law for a solution of two volatiles is this:

P_total = P°_A χ_A + P°_B χ_B

a)

Where P_total = total pressure

P°_A = Partial pressure of pure A ,  χ_{A =} mole fraction of A in the liquid phase

P°_B = Partial pressure of pure B ,  χ_{B =} mole fraction of B in the liquid phase

Partial pressure of A = P_{A =} P°_A χ_A  = 84.3 torr x 0.32 = 26.976 = 27.0 torr

χ_{B = 1 -} χ_A = 1- 0.32 = 0.68

Partial pressure of B = P_{B =} P°_B χ_B  = 41.2 torr x 0.68 = 28.016 = 28.0 torr

P_total = 27.0 torr + 28.0 torr = 55.0 torr

P_A = 27.0 torr

P_B =  28.0 torr

b)

Vapor mole fraction of A = P_A / P_total = 26.976 torr / 54.992 torr = 0.49

Vapor mole fraction of B = P_B/ P_total = 28.016 torr / 54.992 torr = 0.51

So the new system the mole fraction of A and B are 0.49 and 0.51 respectively.

Partial pressure of A = P_{A =} P°_A χ_A  = 84.3 torr x 0.49 = 41.307 = 41.3 torr

Partial pressure of B = P_{B =} P°_B χ_B  = 41.2 torr x 0.51 = 21.012 = 21.0torr

P_total = 41.3 torr + 21.0 torr = 62.3 torr