0.5 text{ mole} of each of H_{2},SO_{2},and C | vedclass

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(B) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass. The molar masses are: $H_

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$pSO_{2} > pCH_{4} > pH_{2}$

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(B) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass. The molar masses are: $H_{2} = 2 	ext{ g/mol}$,$CH_{4} = 16 	ext{ g/mol}$,$SO_{2} = 64 	ext{ g/mol}$. Since $M(H_{2})  r(CH_{4}) > r(SO_{2})$. Therefore,the amount of gas remaining in the container after time $t$ will be in the reverse order of their diffusion rates. The amount left is in the order: $SO_{2} > CH_{4} > H_{2}$. Since partial pressure $p$ is directly proportional to the number of moles $n$ $(p = \frac{nRT}{V})$,the order of partial pressures is $pSO_{2} > pCH_{4} > pH_{2}$.

$0.5 \text{ mole}$ of each of $H_{2}$,$SO_{2}$,and $CH_{4}$ are kept in a container. $A$ hole was made in the container. After $3 \text{ h}$,the order of partial pressures in the container will be

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