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

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.

Solution:

It is given that:

$p_{\mathrm{A}}^{0}=450 \mathrm{~mm}$ of $\mathrm{Hg}$

$p_{\mathrm{B}}^{0}=700 \mathrm{~mm}$ of $\mathrm{Hg}$

ptotal = 600 mm of Hg

From Raoult’s law, we have:

$p_{\mathrm{A}}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}$

$p_{\mathrm{B}}=p_{\mathrm{B}}^{0} x_{\mathrm{B}}=p_{\mathrm{B}}^{0}\left(1-x_{\mathrm{A}}\right)$ Therefore, total pressure, $p_{\text {tocal }}=p_{\mathrm{A}}+p_{\mathrm{B}}$

$\Rightarrow p_{\text {tocal }}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}+p_{\mathrm{B}}^{0}\left(1-x_{\mathrm{A}}\right)$

$\Rightarrow p_{\text {total }}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}+p_{\mathrm{B}}^{0}-p_{\mathrm{B}}^{0} x_{\mathrm{A}}$

$\Rightarrow p_{\text {total }}=\left(p_{\AA}^{0}-p_{\mathrm{B}}^{0}\right) x_{\mathrm{A}}+p_{\mathrm{B}}^{0}$

$\Rightarrow 600=(450-700) x_{\mathrm{A}}+700$

$\Rightarrow-100=-250 x_{\mathrm{A}}$

$\Rightarrow x_{\mathrm{A}}=0.4$

Therefore, $x_{\mathrm{B}}=1-x_{\mathrm{A}}$

= 1 − 0.4

= 0.6

Now, $p_{\mathrm{A}}=p_{\mathrm{A}}^{0} x_{\mathrm{A}}$

= 700 × 0.6

= 420 mm of Hg

Now, in the vapour phase:

Mole fraction of liquid $\mathrm{A}=\frac{p_{\mathrm{A}}}{p_{\mathrm{A}}+p_{\mathrm{B}}}$

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

$=\frac{180}{600}$

= 0.30

And, mole fraction of liquid B = 1 − 0.30

= 0.70