Question

At 300K, the vapor pressure of pure liquid A and liquid B are 37.33 kPa and 22.66 kPa respectively. If we mix 2 mol liquid A with 2 mol liquid B, the total vapor pressure above the solution is 50.66 kPa and the molar fraction of vapor A is 0.60. Assume the vapor is an ideal gas. Calculate (a) the activity of A and B in the solution (b) the Gibbs energy change of mixing ΔmixG (c) If the solution is an ideal solution, what’s the number of ΔmixGid?

Answer 1

a) the activity of a substance is given as a = x. r where a is activity and x is its mole fraction and r is the activity coeficient.

as the vapor is ideal, r is approximately uynity and Raoult's law is appplicable

Thus activity of a and B = their mole fractions in solution = 0.5 each.

b) we habe delta g mix = nRT [x₁ logx₁ + x₂ logx₂] where n is total mles and x is the molefraction of each compoent in solution.

Since the gas is an idel solution, we can use this relation to gas solution. The given solution is not ideal solution , it is not following Raoult's law. [ p₁x₁ + p₂x₂is not equal to total pressure]

Here n = 1 and x1 = 0.60 and x2 = 0.40

thus delta G mix = 1x8.314 x300 [0.6 log0.6 + 0.4 loh0.4]= - 728.755 J

c) we habe delta g mix = nRT [x₁ logx₁ + x₂ logx₂] where n is total mles and x is the molefraction of each compoent in solution,

as the solution is ideal here n= 4 moles X of A = 0.5 and x of B = 0.5 r= 8.314 J, t = 300K

thus delta G mix = 4x 8.314x 300[ 0.5log0.5 +0.5log0.5] = - 3003.3 J or -3.0kJ