A x : 1 molar mixture of $\mathrm{He}$ and  ${\mathrm{CH}}_4$is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. If the composition of the mixture effusing out initially is 8 : 1. then calculate the value of x?

(ii) Molar ratio of $\mathrm{He}$ and ${\mathrm{CH}}_4$ is x : 1 

$\therefore$ Partial pressure ratio of $\mathrm{He}$ and ${\mathrm{CH}}_4$ is 

$\left(\frac{x}{x+1}\right)\times 20:\left(\frac{1}{x+1}\right)\times 20$                                                                                    ($\therefore$total of pressure= 20 bar)

$\frac{n_{\mathrm{He}}^{\mathrm{\prime }}}{n^{\mathrm{\prime }}{\mathrm{CH}}_4}=\sqrt{\frac{{\mathrm{M}}_{{\mathrm{CH}}_4}}{{\mathrm{M}}_{\mathrm{He}}}}\times \frac{{\mathrm{P}}_{\mathrm{He}}}{{\mathrm{P}}_{{\mathrm{CH}}_4}}$

                     ($\therefore$ time of diffusion for both is same )

$\sqrt{\frac{16}{4}}\times \frac{x}{1}=8:1$

$\begin{array}{l}\Rightarrow \frac{x}{1}\sqrt{\frac{16}{4}}=\frac{8}{1}\\ \Rightarrow x=4\end{array}$

The composition of mixture initially gone out for  $\mathrm{He}$ and ${\mathrm{CH}}_4$    is 8 :1.