Question

The vapour pressures of pure liquids A and B are 450 and 700 mm Hg respectively at 350K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Answer 1

To solve this question, you must recall the Raoult’s law and Dalton’s law of partial pressure. Raoult’s Law states that in a solution, the partial vapour pressure of each volatile compound is directly proportional to its mole fraction. Dalton’s law of partial pressure states that the total pressure of a mixture of a number of non-reacting gases is equal to the sum of pressures exerted by individual gases. We shall substitute the values in the equations given to find the composition of vapour phase.

Formula used: ${{\text{P}}_A}{\text{ = P}}_A^{\text{0}}{{\text{X}}_A}$ and ${{\text{P}}_B}{\text{ = P}}_B^{\text{0}}{{\text{X}}_B}$
Where, ${{\text{P}}_A}$ is partial vapour pressure of A
${{\text{P}}_B}$ is the partial vapour pressure of B
And ${X_A},{X_B}$ are mole fractions of A and B in liquid phase respectively.
From Dalton’s law, we have ${{\text{P}}_A}{\text{ = P}}_A^{\text{0}}{{\text{Y}}_A}$ and ${{\text{P}}_B}{\text{ = P}}_B^{\text{0}}{{\text{Y}}_B}$.
Where, ${Y_A},{Y_B}$ are mole fraction of A and B in vapour phase respectively.

Complete step by step solution:
We are given in the question, ${\text{P}}_N^{\text{0}} = 450{\text{ mm Hg}}$ and ${\text{P}}_B^{\text{0}} = 700{\text{ mm Hg}}$
The total pressure of the gaseous solution is given as, ${P_T} = 600{\text{ mm Hg}}$
We can write ${P_T}$ using the Raoult’s law as, ${P_T} = {\text{P}}_A^{\text{0}}{{\text{X}}_A}{\text{ + P}}_B^{\text{0}}{{\text{X}}_B} = {\text{P}}_A^{\text{0}}{{\text{X}}_A} + (1 - {X_A})$
$\Rightarrow 600 = 450 \times {X_A} + 700\left( {1 - {X_A}} \right)$
$\Rightarrow 600 = 700 - 250{X_A}$
Solving this, we get:
$\therefore {X_A} = 0.40$ and ${X_B} = 0.60$
For finding the composition of the mixture in vapour phase:
${{\text{P}}_A}{\text{ = P}}_A^{\text{0}}{{\text{Y}}_A} = 450 \times 0.40 = 180{\text{ mm Hg}}$ and ${{\text{P}}_B}{\text{ = P}}_B^{\text{0}}{{\text{Y}}_B} = 700 \times 0.60 = 420{\text{ mm Hg}}$
Mole fraction of gas A in the vapour phase can be given by,
${Y_A} = \dfrac{{{P_A}}}{{{P_A} + {P_B}}} = \dfrac{{180}}{{180 + 420}}$
$\therefore {Y_A} = 0.30$
And the mole fraction of gas B in the vapour phase can be given by,
${Y_B} = \dfrac{{{P_B}}}{{{P_A} + {P_B}}} = \dfrac{{420}}{{180 + 420}}$
$\therefore {Y_B} = 0.70$

Note: Raoult’s law is given for an ideal mixture. A solution that shows a deviation from Raoult's law over the entire range of various compositions of the mixture, the solution is said to be a non- ideal or real solution. In a real solution, the pair of liquids that constitute the mixture do not have a uniformity throughout attractive forces, which may be either adhesive or cohesive. Thus, they show a deviation from Raoult's Law.