Question

A gaseous mixture contains equal masses of $O_2$ and $C{H}_4$ . If the total pressure of the mixture is 15 atm then the partial pressure of methane in the mixture is

Answer 1

Step 1:

Given that,

$w_{O_2}=w_{C{H}_4}=1gm$

$M_{O_2}=32gm{mol}^{-1}$ $M_{C{H}_4}=16gm{mol}^{-1}$

$n_{O_2}=\frac{1}{32}{n}_{C{H}_4}=\frac{1}{16}$

$P_T=15atm$

Step 2:

$p_{C{H}_4}=X_{C{H}_4}\times 15=\left(\frac{\frac{1}{16}}{\frac{1}{16}+\frac{1}{32}}\right)\times 15=\left\{\frac{1}{1+\frac{1}{2}}\right\}15$

$=\frac{2}{3}\times 15=10atm$

Hence, the partial pressure of methane in the mixture is 10 atm.