If $\alpha$ is neglible as compared to 1 for the following reaction ${\text{XY}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{\hspace{0.17em}XY}}_{\left(\text{g}\right)}{\text{\hspace{0.17em}+\hspace{0.17em}Y}}_{\left(\text{g}\right)}$. The degree of

dissociation $\left(\alpha \right)$for the above reaction is            

${\text{XY}}_{\text{2}\left(\text{g}\right)}\rightleftharpoons {\text{\hspace{0.17em}XY}}_{\left(\text{g}\right)}{\text{\hspace{0.17em}+\hspace{0.17em}Y}}_{\left(\text{g}\right)}\text{\hspace{0.17em};\hspace{0.17em}1\hspace{0.17em}}-\text{\hspace{0.17em}α}$

If P is the equilibrium pressure, then

${\text{P}}_{{\text{xy}}_{\text{2}}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{1\hspace{0.17em}}-\text{\hspace{0.17em}α}}{\text{1\hspace{0.17em}+\hspace{0.17em}a}}\text{.\hspace{0.17em}P\hspace{0.17em}\hspace{0.17em}}[Totalmoles:1-\alpha +\alpha =1+\alpha ]$

${\text{P}}_{\text{xy}}{\text{\hspace{0.17em}=\hspace{0.17em}P}}_{\text{y}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{α}}{\text{1\hspace{0.17em}+\hspace{0.17em}α}}\text{.\hspace{0.17em}P}$

${\text{Hence,\hspace{0.17em}K}}_{\text{p}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{P}}_{\text{xy}}{\text{\hspace{0.17em}×\hspace{0.17em}P}}_{\text{y}}}{{\text{P}}_{{\text{xy}}_{\text{2}}}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\frac{\text{α}}{\text{1\hspace{0.17em}+\hspace{0.17em}α}}\text{.\hspace{0.17em}P\hspace{0.17em}×\hspace{0.17em}}\frac{\text{α}}{\text{1\hspace{0.17em}+\hspace{0.17em}α}}}{\frac{\text{1\hspace{0.17em}-α}}{\text{1\hspace{0.17em}+\hspace{0.17em}α}}\text{.\hspace{0.17em}P}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{α}}^{\text{2}}\text{P}}{\text{1\hspace{0.17em}}-{\text{\hspace{0.17em}α}}^{\text{2}}}{\text{\hspace{0.17em}=\hspace{0.17em}α}}^{\text{2}}\text{P}[\because \text{\hspace{0.17em}α\hspace{0.17em}<<}]$

$\text{or\hspace{0.17em}α\hspace{0.17em}=\hspace{0.17em}}{\left({\text{K}}_{\text{p}}\text{/2}\right)}^{\text{1/2}}\text{\hspace{0.17em}or\hspace{0.17em}α\hspace{0.17em}}\propto \frac{\text{1}}{{\text{P}}^{\text{1/2}}}$

Hence, (d) is correct.

Since, V is inversely proportional to P

So, $\text{α\hspace{0.17em}}\propto {\text{\hspace{0.17em}V}}^{\text{1/2}}\text{\hspace{0.17em};\hspace{0.17em}hence(a)\hspace{0.17em}is\hspace{0.17em}correct.}$

$\text{Alternatively,\hspace{0.17em}}\left[{\text{XY}}_{\text{2}}\right]\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{1\hspace{0.17em}}-\text{\hspace{0.17em}α}}{\text{V}}\text{\hspace{0.17em};\hspace{0.17em}}\left[\text{XY}\right]\text{\hspace{0.17em}=\hspace{0.17em}}\left[\text{Y}\right]\text{\hspace{0.17em}=}\frac{\text{α}}{\text{V}}$

${\text{Hence,\hspace{0.17em}K}}_{\text{C}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\left[\text{XY}\right]\left[\text{Y}\right]}{\left[{\text{XY}}_{\text{2}}\right]}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{α\hspace{0.17em}/Y\hspace{0.17em}×\hspace{0.17em}α\hspace{0.17em}}/\text{V}}{(\text{1\hspace{0.17em}}-\text{\hspace{0.17em}α})\text{/\hspace{0.17em}V}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{α}}^{\text{2}}}{\text{V}}\text{\hspace{0.17em}or\hspace{0.17em}α\hspace{0.17em}=\hspace{0.17em}}{\left({\text{K}}_{\text{C}}\text{V}\right)}^{\text{1/2}}\text{\hspace{0.17em}or\hspace{0.17em}α\hspace{0.17em}}\propto {\text{\hspace{0.17em}V}}^{\text{1/2}}$