# Liquification of ideal solution of two liquids from vapor phase [closed]

A and B form an ideal solution. In a cylinder piston arrangement, $$\pu{2.0 mol}$$ of vapor of liquid A and $$\pu{3 mol}$$ of vapor of liquid B are taken at $$\pu{300 torr}$$ and $$T~\pu{K}$$. At what pressure $$30\%$$ of the total amount of substance of vapor will it liquefy? Given $$\ce{Pa^{.}}=\pu{300 torr}$$, $$\ce{Pb^{.}}=\pu{600 torr}$$, ($$\sqrt{48.25}=6.95$$). Kindly provide a conceptual and brief solution with an explanation. The answer comes out to be $$\pu{445 torr}$$. This question is from a preparatory-test for The indian jee Advanced exam, And i think it can be a discussion. As i put my input into it already, i dont think this should be a homework-question, But instead, it raises a thought and creates new perspectives

• You first. What have you done so far? Commented Aug 12, 2023 at 10:54
• i tried raoult`s law, daltons law. Didnt get answer. I actually dont know what to do with external 300 torr presurre given in question.And Should i consider that after 30% of vapours are liqufied, now equillibrium between external pressure and internal pressure is established?? Commented Aug 12, 2023 at 11:38
• No. The 300 torr initial pressure is increased to a higher final pressure. You need to find out what that final pressure will have to be to end up with 30% liquid at the same temperature T. I think that the initial pressure of 300 torr means that the initial vapor is not saturated. Commented Aug 12, 2023 at 11:44
• got that..How can i proceed with equations now. Because thats where i lack now. My equations earlier had no use of given numerical estimation, which indicates i might be at wrong path. Can u please write them? Commented Aug 12, 2023 at 11:53
• Please edit in your attempt of solving the exercise. Please also explain what ist actually given. Commented Aug 12, 2023 at 14:07

• $$L$$ is the amount of substance of liquid
• $$V$$ is the amount of substance of vapor
• $$x$$ is the mole fraction A in the liquid
• $$y$$ is the mole fraction of A in the vapor

Total amount of substance: $$L+V=5\tag{1}$$

Amount of substance of A in liquid plus vapor: $$Lx+Vy=2\tag{2}$$

Raoult's Law: $$y=\frac{300x}{300x+600(1-x)}=\frac{x}{2-x}\tag{3}$$

Fraction of total amount of substance that is liquid: $$\frac{L}{L+V}=0.3\tag{4}$$

So, from Eqns. 1 and 4, we have $$L = 1.5$$ and $$V = 3.5$$. Substituting these results and Eqn. 3 into Eqn. 2 gives: $$1.5x+\frac{3.5x}{2-x}=2,$$ which leads to the quadratic equation $$1.5x^2-8.5x+4=0.$$

Solving this quadratic equation for $$x$$ using the quadratic formula:
$$x=\frac{8.5-\sqrt{48.25}}{3}=0.517$$ This gives $$P=300x+600(1-x)=300(2-x)=\pu{445 torr}.$$

• you Are a GOAT ...... I was stuck for almost a week.... Commented Aug 13, 2023 at 16:09
• thanks martin too.✌🏻 Commented Aug 13, 2023 at 16:19