Calculate the ratio of the effusion of CO and | vedclass

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(B) According to Graham's law of effusion,the rate of effusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$: $r

Vedclass · Accepted Answer

$1: 1$

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(B) According to Graham's law of effusion,the rate of effusion $(r)$ of a gas is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.For two gases $CO$ and $N_2$ at the same temperature and pressure,the ratio of their rates of effusion is given by: $\frac{r_{CO}}{r_{N_2}} = \sqrt{\frac{M_{N_2}}{M_{CO}}}$.The molar mass of $CO$ is $12 + 16 = 28 \ g/mol$.The molar mass of $N_2$ is $14 	imes 2 = 28 \ g/mol$.Substituting these values: $\frac{r_{CO}}{r_{N_2}} = \sqrt{\frac{28}{28}} = \sqrt{1} = 1$.Thus,the ratio is $1: 1$.

Calculate the ratio of the effusion of $CO$ and $N_2$,when temperature and pressure gradients are held constant?

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