For the reaction PCl_5 rightleftharpoons PCl_ | vedclass

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(A) For the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the degree of dissociation $\alpha$ is given by $\alpha \propto \frac{1}{\sqrt{P}}$.Since

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$4$

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(A) For the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the degree of dissociation $\alpha$ is given by $\alpha \propto \frac{1}{\sqrt{P}}$.Since $PV = nRT$,at constant temperature,$P \propto \frac{1}{V}$.Substituting this into the relation,we get $\alpha \propto \sqrt{V}$.Given that the volume increases to $16$ times the initial volume $(V_2 = 16V_1)$,the new degree of dissociation $\alpha_2$ is:$\frac{\alpha_2}{\alpha_1} = \sqrt{\frac{V_2}{V_1}} = \sqrt{\frac{16V_1}{V_1}} = \sqrt{16} = 4$.Therefore,the degree of dissociation becomes $4$ times the initial value.

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