Question

The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. 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

$p=p_A+p_B\\ \Rightarrow p=p_A^{\star }x_A+p_B^{\star }x_B\\ \Rightarrow p=p_A^{\star }x_A+p_B^{\star }(1-x_A)\\ \Rightarrow x_A(p_A^{\star }-p_B^{\star })=p-p_B^{\star }\\ \Rightarrow x_A=\dfrac{p_B^{\star }-p}{p_B^{\star }-p_A^{\star }}\\ \Rightarrow x_A=\dfrac{700-600}{700-450}\\ =0.4\\ x_B=1-0.4=0.6\\ y_A=\dfrac{p_A}{p}\\ =\dfrac{p_A^{\star }x_A}{p}\\ =\dfrac{450\times 0.4}{600}\\ =0.3\\ y_B=1-0.3=0.7$

Answer 2

Answer :

Given :-

Vapour pressure of pure liquid A, $ρ°_{A}$ = 450 mm of Hg

Vapour pressure of pure liquid A, $ρ°_{B}$ = 700 mm of Hg

Total vapour pressure, $ρ_{total}$ = 600 mm of Hg

By formula,

$ρ_{A} = ρ°_{A} χ_{A} , ρ_{B} = ρ°_{B} χ_{B}$

⇒ $ρ_{A} = 450 χ_{A} and ρ_{B} = 700(1 - χ_{A})$

∵ $ρ_{A} + ρ_{B} = 600$

⇒ 450 $χ_{A} + 700 - 700χ_{A} = 600$

⇒ -250$χ_{A} = -100$

⇒ $χ_{A} = \frac{100}{250} = 0.4$ and $χ_{B} = 1 - 0.4 = 0.6$ in liquid phase.

In vapour phase,

$χ_{A} = \frac{ρ_{A}}{ρ_{A} + ρ_{B}} = \frac{ρ°_{A}χ_{A}}{ρ°_{A}χ_{A} + ρ°_{B}χ_{B}}$

= $\frac{450 × 0.4 }{450 × 0.4 + 700 × 0.6}$

=$\frac{180}{180 + 420}$

= $\frac{180}{600}$ = $\frac{3}{10}$

⇒ $χ_{A} = 0.3$ and $χ_{B} = 0.7$ in vapour phase.

[∵ $χ_{A} + χ_{B} = 1$]