16 g each of H_2,He,and O_2 are present in a | vedclass

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(C) The pressure $P$ of a gas mixture is given by $P = \frac{nRT}{V}$. Since $R, T,$ and $V$ are constant,$P \propto n_{total}$.In the first container

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$3.6$

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(C) The pressure $P$ of a gas mixture is given by $P = \frac{nRT}{V}$. Since $R, T,$ and $V$ are constant,$P \propto n_{total}$.In the first container,the total moles $n_1 = n_{H_2} + n_{He} + n_{O_2} = \frac{16}{2} + \frac{16}{4} + \frac{16}{32} = 8 + 4 + 0.5 = 12.5 \ mol$.Given $P_1 = 10 \ atm$,so $10 \propto 12.5$.In the second container,the total moles $n_2 = n_{He} + n_{O_2} = \frac{16}{4} + \frac{16}{32} = 4 + 0.5 = 4.5 \ mol$.Therefore,the new pressure $P_2 = \frac{n_2}{n_1} 	imes P_1 = \frac{4.5}{12.5} 	imes 10 = 0.36 	imes 10 = 3.6 \ atm$.

$16 \ g$ each of $H_2$,$He$,and $O_2$ are present in a container exerting $10 \ atm$ pressure at $T \ (K)$. The pressure in $atm$ exerted by $16 \ g$ each of $He$ and $O_2$ in the second container of same volume and temperature is

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