Question

A $4:1$ molar mixture of $He$ and $C{H}_4$ is contained in vessel at $20$ bar pressure. Due to a hole in the vessel, the gas mixture leakes out. What is the composition of mixture effusing out initially?

• A. $33.3\%He,66.7\%$ $C{H}_4$
• B. $66.7\%He,33.3\%$ $C{H}_4$
• C. $40\%He,60\%$ $C{H}_4$
• D. $60\%He,40\%$ $C{H}_4$

Answer 1

Answer: (B)

$66.7\%He,33.3\%$ $C{H}_4$

According to the formula for rate of diffusion,
$\frac{r_{He}}{r_{C{H}_4}}=\sqrt{\frac{M_{C{H}_4}}{M_{He}}}\times \frac{P_{He}}{P_{C{H}_4}}$
$P\propto moles$
$\frac{r_{He}}{r_{C{H}_4}}=\sqrt{\frac{16}{4}}\times \frac{4}{1}=2\times 4=8:1$
Molar composition effusing $=8:1$
Weight composition effusing $=8\times 4:1\times 16=32:16$
% of $He=\frac{32}{48}\times 100=66.7$%
% $C{H}_4=100-66.7=33.3$%