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(A) Let the initial number of moles be $n$. The initial state is given by $P_{1}V_{1} = nRT_{1}$,where $T_{1} = 250\, K$.When $25\%$ of the diatomic m

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$5$

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(A) Let the initial number of moles be $n$. The initial state is given by $P_{1}V_{1} = nRT_{1}$,where $T_{1} = 250\, K$.When $25\%$ of the diatomic molecules $(X_{2})$ dissociate,the reaction is $X_{2} ightarrow 2X$. If $25\%$ of $n$ moles dissociate,the number of moles of $X_{2}$ remaining is $0.75n$,and the number of moles of $X$ produced is $2 	imes 0.25n = 0.5n$.The total number of moles in the final state is $n' = 0.75n + 0.5n = 1.25n = \frac{5n}{4}$.The final state is given by $P_{2}V_{2} = n'RT_{2}$,where $V_{2} = 2V_{1}$ and $T_{2} = 2000\, K$.Substituting the values: $P_{2}(2V_{1}) = \left(\frac{5n}{4}ight)R(2000)$.Dividing the two equations: $\frac{P_{2}(2V_{1})}{P_{1}V_{1}} = \frac{(5n/4)R(2000)}{n R(250)}$.$2 \frac{P_{2}}{P_{1}} = \frac{5}{4} 	imes \frac{2000}{250} = \frac{5}{4} 	imes 8 = 10$.Therefore,$\frac{P_{2}}{P_{1}} = \frac{10}{2} = 5$.

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