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(A) For container $X$,the initial state is $P_1 V = n_1 RT$. After expansion to $2V$ with pressure change $\Delta P$,the final state is $(P_1 - \Delta

Vedclass · Accepted Answer

$5$

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(A) For container $X$,the initial state is $P_1 V = n_1 RT$. After expansion to $2V$ with pressure change $\Delta P$,the final state is $(P_1 - \Delta P) 2V = n_1 RT$.Equating the two: $P_1 V = 2P_1 V - 2\Delta P V$,which gives $P_1 V = 2\Delta P V$.Thus,$n_1 = \frac{2\Delta P V}{RT}$.For container $Y$,the initial state is $P_2 V = n_2 RT$. After expansion to $2V$ with pressure change $0.2 \Delta P$,the final state is $(P_2 - 0.2 \Delta P) 2V = n_2 RT$.Equating the two: $P_2 V = 2P_2 V - 0.4 \Delta P V$,which gives $P_2 V = 0.4 \Delta P V$.Thus,$n_2 = \frac{0.4 \Delta P V}{RT}$.The ratio of masses is equal to the ratio of moles: $\frac{w_X}{w_Y} = \frac{n_1}{n_2} = \frac{2\Delta P V / RT}{0.4 \Delta P V / RT} = \frac{2}{0.4} = 5$.

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