Question

A mixture of two miscible volatile ideal liquids P and Q (obeying Raoult’s law) is kept

in a vessel (molar ratio of P and Q in the mixture is m). At a suitable temperature T,

the vapour above the liquid is condensed in another vessel. The liquid obtained on con-

densation is allowed to evaporate and establish equilibrium with its vapour. The vapour

is then condensed in another vessel. The process of such evaporation and condensation

is repeated for n times. If the ratio of the vapour pressure of pure P to that of pure Q is

p, the molar ratio of P and Q in the condensed liquid obtained after n

th cycle (for finite

n > 1) is​

Answer 1

Answer:

See below.

Explanation:

We are given that there are two miscible liquids which form an ideal solution. Hence, by Raoult's law, the vapour pressures of both the liquids will be directly proportional to their mole fractions.

$\dfrac{p_{1}}{p_{2}}=\dfrac{\dfrac{n_{1}}{n_{1}+n_{2}}}{\dfrac{n_{2}}{n_{1}+n_{2}}}=\dfrac{n_{1}}{n_{2}}$

Now, we collect the vapour and condense it. Condensation would yield a liquid which is similar to the previous one. Just like steam, when cooled ( condensed ), forms water again.

Now this liquid ( which was condensed ) is again vapourized to some extent. Again the vapour pressure will be proportional to the mole fraction of the component.

Repeating this cycle again and again, Raoult's law is not violated. Hence,

$\dfrac{p_{1}}{p_{2}}=\dfrac{n_{1}}{n_{2}}=m$