A x : 1 molar mixture of He and CH_4 is conta | vedclass

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(B) 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

$4$

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(B) 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: $\frac{r_{He}}{r_{CH_4}} = \frac{P_{He}}{P_{CH_4}} \sqrt{\frac{M_{CH_4}}{M_{He}}}$.Given the molar ratio in the vessel is $x : 1$,the partial pressures are proportional to the mole fractions: $P_{He} = \frac{x}{x+1} 	imes 20 \ bar$ and $P_{CH_4} = \frac{1}{x+1} 	imes 20 \ bar$.The molar masses are $M_{He} = 4 \ g/mol$ and $M_{CH_4} = 16 \ g/mol$.Substituting these into the effusion equation: $\frac{8}{1} = \left( \frac{x}{1} ight) \sqrt{\frac{16}{4}}$.$8 = x 	imes \sqrt{4} = x 	imes 2$.Therefore,$x = \frac{8}{2} = 4$.

$A$ $x : 1$ molar mixture of $He$ and $CH_4$ 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 the value of $x$ will be

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