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(B) $1$. Initial state: Each part has a volume $V = 0.5 \, L$. $2$. For $H_2$: Moles $n_1 = \frac{2 \, g}{2 \, g/mol} = 1 \, mol$. $3$. For $CH_4$: Mo

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$2P$

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(B) $1$. Initial state: Each part has a volume $V = 0.5 \, L$. $2$. For $H_2$: Moles $n_1 = \frac{2 \, g}{2 \, g/mol} = 1 \, mol$. $3$. For $CH_4$: Moles $n_2 = \frac{16 \, g}{16 \, g/mol} = 1 \, mol$. $4$. Using ideal gas law $PV = nRT$,since $n_1 = n_2 = 1 \, mol$ and $V$ and $T$ are the same for both,the pressure in each part is $P = \frac{nRT}{V}$. $5$. After removing the partition,the total volume becomes $V_{total} = 1 \, L$ and total moles $n_{total} = n_1 + n_2 = 1 + 1 = 2 \, mol$. $6$. The new total pressure $P_{total} = \frac{n_{total}RT}{V_{total}} = \frac{2RT}{1} = 2 	imes \frac{RT}{V_{part}} = 2P$.

$A$ box of $1 \, L$ capacity is divided into two equal parts by a thin partition,which are filled with $2 \, g$ of $H_2$ and $16 \, g$ of $CH_4$ respectively. The temperature of each part is $T$. The pressure in each part is $P$. If the partition is removed,the total pressure will be:

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