The same mass of CH_4 and H_2 is taken in a c | vedclass

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(A) Let the mass of both gases be $m \ g$.Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{m}{16}$.Number of moles of $H_2$ $(n_{H_2})$ = $\frac{m}{2}$

Vedclass · Accepted Answer

$8/9$

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(A) Let the mass of both gases be $m \ g$.Number of moles of $CH_4$ $(n_{CH_4})$ = $\frac{m}{16}$.Number of moles of $H_2$ $(n_{H_2})$ = $\frac{m}{2}$.The partial pressure of a gas is proportional to its mole fraction $(x_i = \frac{n_i}{n_{total}})$.Mole fraction of $H_2$ $(x_{H_2})$ = $\frac{n_{H_2}}{n_{H_2} + n_{CH_4}} = \frac{\frac{m}{2}}{\frac{m}{2} + \frac{m}{16}}$.$x_{H_2} = \frac{\frac{m}{2}}{\frac{8m + m}{16}} = \frac{\frac{m}{2}}{\frac{9m}{16}} = \frac{m}{2} 	imes \frac{16}{9m} = \frac{8}{9}$.Therefore,the partial pressure caused by $H_2$ is $\frac{8}{9}$ of the total pressure.

The same mass of $CH_4$ and $H_2$ is taken in a container. The partial pressure caused by $H_2$ is:

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