0.5 mol each of H_2, SO_2 and CH_4 are prese | vedclass

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(A) According to Graham's Law of Effusion,the rate of effusion $r \propto \frac{1}{\sqrt{M}}$.Calculating molar masses: $M(H_2) = 2 \ g/mol$,$M(CH_4)

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) According to Graham's Law of Effusion,the rate of effusion $r \propto \frac{1}{\sqrt{M}}$.Calculating molar masses: $M(H_2) = 2 \ g/mol$,$M(CH_4) = 16 \ g/mol$,$M(SO_2) = 64 \ g/mol$.Since the rate of effusion is inversely proportional to the square root of molar mass,the order of effusion rates is $r_{H_2} > r_{CH_4} > r_{SO_2}$.This means that $H_2$ effuses out the fastest,leaving the least amount behind,while $SO_2$ effuses the slowest,leaving the most amount behind.Since partial pressure is directly proportional to the number of moles remaining,the order of partial pressures will be $P_{SO_2} > P_{CH_4} > P_{H_2}$.

$0.5 \ mol$ each of $H_2, SO_2$ and $CH_4$ are present in a container. If a hole is made in the container,what will be the order of partial pressures of the gases remaining in the container after $3 \ hours$?

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