In a container,0.5 mol each of H_2, SO_2,and | vedclass

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(A) According to Graham's Law of Diffusion,the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass $

Vedclass · Accepted Answer

$P_{SO_2} > P_{CH_4} > P_{H_2}$

Vedclass · Answer

(A) According to Graham's Law of Diffusion,the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass $(Rate \propto \frac{1}{\sqrt{M}})$.Since the gases are in a container,if there is any leakage or diffusion,the gas with the lowest molar mass will diffuse the fastest.The molar masses are: $M(H_2) = 2 \ g/mol$,$M(CH_4) = 16 \ g/mol$,and $M(SO_2) = 64 \ g/mol$.Since $M(H_2)  Rate(CH_4) > Rate(SO_2)$.After $3$ hours,the gas that diffuses the fastest will have the lowest amount remaining in the container.Therefore,the amount remaining follows the order: $n(SO_2) > n(CH_4) > n(H_2)$.Since partial pressure is directly proportional to the number of moles $(P \propto n)$,the order of partial pressures will be $P_{SO_2} > P_{CH_4} > P_{H_2}$.

In a container,$0.5 \ mol$ each of $H_2, SO_2$,and $CH_4$ are taken. The order of their partial pressures after $3$ hours will be:

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