-3
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ You first. What have you done so far? $\endgroup$ Commented Aug 12, 2023 at 10:54
  • $\begingroup$ 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?? $\endgroup$
    – om Makadia
    Commented Aug 12, 2023 at 11:38
  • $\begingroup$ 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. $\endgroup$ Commented Aug 12, 2023 at 11:44
  • $\begingroup$ 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? $\endgroup$
    – om Makadia
    Commented Aug 12, 2023 at 11:53
  • $\begingroup$ Please edit in your attempt of solving the exercise. Please also explain what ist actually given. $\endgroup$ Commented Aug 12, 2023 at 14:07

1 Answer 1

1
$\begingroup$
  • $L$ 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 $$P=300x+600(1-x)=300(2-x)=\pu{445 torr}.$$

$\endgroup$
2
  • $\begingroup$ you Are a GOAT ...... I was stuck for almost a week.... $\endgroup$
    – om Makadia
    Commented Aug 13, 2023 at 16:09
  • $\begingroup$ thanks martin too.✌🏻 $\endgroup$
    – om Makadia
    Commented Aug 13, 2023 at 16:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.