A 4 : 1 molar mixture of He and CH_4 is kept | vedclass

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(A) According to Graham's Law of Effusion,the rate of effusion of a gas is directly proportional to its partial pressure and inversely proportional to

Vedclass · Accepted Answer

$8 : 1$

Vedclass · Answer

(A) According to Graham's Law of Effusion,the rate of effusion of a gas is directly proportional to its partial pressure and inversely proportional to the square root of its molar mass.$\frac{r_{He}}{r_{CH_4}} = \frac{p_{He}}{p_{CH_4}} 	imes \sqrt{\frac{M_{CH_4}}{M_{He}}}$Given the molar ratio $n_{He} : n_{CH_4} = 4 : 1$,the partial pressures are proportional to the mole fractions,so $\frac{p_{He}}{p_{CH_4}} = \frac{4}{1}$.The molar masses are $M_{He} = 4 \ g/mol$ and $M_{CH_4} = 16 \ g/mol$.Substituting these values:$\frac{r_{He}}{r_{CH_4}} = \frac{4}{1} 	imes \sqrt{\frac{16}{4}} = 4 	imes \sqrt{4} = 4 	imes 2 = 8$.Thus,the ratio of the rates of effusion is $8 : 1$.

$A$ $4 : 1$ molar mixture of $He$ and $CH_4$ is kept in a vessel at $20 \ bar$ pressure. Due to a hole in the vessel,the gas mixture leaks out. What is the composition of the mixture effusing out initially?

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