A mixture of He and CH_4 in a molar ratio of | vedclass

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(A) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_{He}}{

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$8$

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(A) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{He}}}$.Substituting the molar masses ($M_{CH_4} = 16 \ g/mol$,$M_{He} = 4 \ g/mol$): $\frac{r_{He}}{r_{CH_4}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2$.This means the rate of effusion of $He$ is $2$ times that of $CH_4$.The composition of the effusing mixture is given by the ratio of the rates of effusion multiplied by their initial molar ratio in the container: $\frac{	ext{Rate}_{He}}{	ext{Rate}_{CH_4}} = \frac{r_{He}}{r_{CH_4}} 	imes \frac{n_{He}}{n_{CH_4}} = 2 	imes \frac{4}{1} = \frac{8}{1}$.Thus,the initial composition of the effusing gas mixture is $8:1$.

Column-$I$	Column-$II$
$A$. Hydrogen gas $(P=200 \ atm, T=273 \ K)$	$p$. Compressibility factor $Z \neq 1$
$B$. Hydrogen gas $(P \approx 0, T=273 \ K)$	$q$. Attractive forces are dominant
$C$. $CO_2$ $(P=1 \ atm, T=273 \ K)$	$r$. $PV=nRT$
$D$. Real gas with very large molar volume	$s$. $P(V_m-b)=RT$

$A$ mixture of $He$ and $CH_4$ in a molar ratio of $4:1$ is present in a container at a pressure of $20 \ bar$. Due to a hole in the container,the gas mixture leaks out. What will be the composition of the gas mixture effusing out initially (in $: 1$)?

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