Question

At 300 K, the vapor pressure of pure liquid A and B is 200 mmHg and 450 mmHg, respectively. If the total pressure of the mixture A and B is 350 mmHg and the assumption that the solution is ideal, calculate: (a) Mole fraction of A and B in liquid phase and vapor phase, (b) The activity and activity coefficient of A and B in the solution, and (c) The mixing free energy, Agm, and the mixing entropy, Asm, of the solution if the total mole of A and B in the solution is 5.0 mole.

Answer 1

The vapor pressure of pure liquids A and B: P_A⁰ = 200 mmHg, P_B⁰ = 450 mmHg

The total pressure (P) = 350 mmHg

The mole fractions of A and B in the liquid phase (x_A and x_B) can be calculated according to Raoult's law:

P = P_A⁰ * x_A + P_B⁰ * (1-x_A) (Note: x_B = 1 - x_A)

i.e. 350 = 200x_A + 450 - 450x_A

i.e. x_A = 0.4

Therefore, x_B = 1-0.4 = 0.6

Now, the mole fractions of A and B in the vapor phase (y_A and y_B) can be calculated according to Dalton's law:

y_A= P_A/P = 200*0.4/350 = 0.23

And y_B = 1-0.23 = 0.77