Need Help with an Ideal Solution Chem Problem with Vapor Pressure and Partial Pressures?

Question

Two liquids A and B form an ideal solution. A solution of A and B is prepared at 30 degrees(C) n which the mole fraction of A is 0.256. In the vapor phase in equilibrium with this solution, the mole fraction of A is 0.318 and the total pressure is 673 torr.

a. Which component A or B is more volatile?

b. Calculate the partial pressure of A and B int he vapor phase.

c. Calculate the vapor phase of pure liquid A and pure liquid B at 30 degrees(C)

Any direction on what to do would be simply fantastic.

Answer 1

The whole problem deals with Raoult's law. It states that the partial pressure of a component equals i liquid phase mole fraction times the vapor pressure of the pure component:

e.g. for A

p(A) = x(A) ∙ p°(A)

For ideal gas mixtures the partial pressure of a component is equal to mole fraction times total pressure

p(A) = y(A) ∙ p

Hence:

y(A) ∙ p = x(A) ∙ p°(A)

That's all U need to to answer the problem.

(a)

A is more volatile.

You can reason this without calculation from considering the ratio of the mole fractions for each component. Vapor phase mole fraction of A is greater than its liquid phase mole fraction. Since we've got a binary mixture and the mole fractions sum up to unity, the vapor phase mole fractions of B is smaller than its liquid phase mole fraction.

Formally:

x(A)+x(B) = 1 => x(B) = 1-x(A) = 1-0.256 = 0.744

vapor

y(A)+y(B) = 1 => y(B) = 1-y(A) = 1-0.318 = 0.682

So A tends more to the gas phase than B. And that's the original meaning of more volatile.

A more formal reasoning would be, that the more volatile component has the higher vaopr pressure. You can easily show that the vapor pressure of A is greater. Just solve equation above for it:

y(A) ∙ p = x(A) ∙ p°(A)

=>

p°(A) =  p ∙ y(A) / x(A) = 673torr ∙ 0.318 / 0.256 = 836tor

Similarlyly for B:

p°(B) =  p ∙ y(B) / x(B) = 673torr ∙ 0.682 / 0.744 = 617torr

(b)

p(A) = y(A) ∙ p = 0.318 ∙ 673torr = 214torr

p(B) = y(B) ∙ p = 0.682 ∙ 673torr = 459torr

(c)

For pure components partial pressure in gas phase equals the vapor pressure of the pure component. As calculated in a.

p(A) = p°(A) = 836torr

p(B) = p°(B) = 617torr

Note:

The fact, that p°(A) is greater than standard atmospheric pressure of 760torr,  indicates that the normal boiling point of A is below 30°CNeverthelessee the could be establishsh a vaporequilibriumum at 30°C with pure A, but only if you pressurize to a level above p°(A).