Diffusion of CH_{4(g)} and O_{2(g)} occurs un | vedclass

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(A) According to Graham's Law of Diffusion, under similar conditions of pressure and temperature, the rate of diffusion $(r)$ of a gas is inversely pr

Vedclass · Accepted Answer

$1.414$

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(A) According to Graham's Law of Diffusion, under similar conditions of pressure and temperature, the rate of diffusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$Therefore, the ratio of the rates of diffusion of $CH_{4}$ and $O_{2}$ is given by: $\frac{r_{CH_{4}}}{r_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{CH_{4}}}}$Given molar masses: $M_{CH_{4}} = 16 \ g/mol$ and $M_{O_{2}} = 32 \ g/mol$.Substituting the values: $\frac{r_{CH_{4}}}{r_{O_{2}}} = \sqrt{\frac{32}{16}} = \sqrt{2} \approx 1.414$.

Diffusion of $CH_{4(g)}$ and $O_{2(g)}$ occurs under similar conditions, then the ratio of their rates of diffusion is

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