Consider the following equilibrium in a closed container.
${\text{PCl}}_5$ gas at a certain pressure is introduced in the container at 27°C. However , the total pressure at equilibrium at 207°C was found to be double the initial value. The % dissociation of ${\text{PCl}}_5$  at 207° C is :
 

            ${\text{PCl}}_{\text{5}}\rightleftharpoons {\text{PCl}}_{\text{3}}+{\text{Cl}}_{\text{2}}$
$\begin{array}{cccc}t=0& P& 0& \begin{array}{cc}0& (at27°C)\end{array}\end{array}$
$\begin{array}{cccc}t=0& \frac{480P}{300}& 0& \begin{array}{cc}0& (at207°C)\end{array}\end{array}$
$\begin{array}{cccc}t=eq.& \frac{480P}{300}-P\text{'}& P\text{'}& \begin{array}{cc}P\text{'}& (at207°C)\end{array}\end{array}$
Total pressure at eq. at  $207°C=\frac{480P}{300}-P\text{'}+2P\text{'}=2P$
(given)
$\therefore \frac{8P}{5}+P\text{'}=2P$
$\therefore P\text{'}=\frac{2P}{5}$
% dissociation of ${\text{PCl}}_{\text{5}}$  at $207°C=\frac{P\text{'}}{\left(\frac{480P}{300}\right)}\times 100$ 
$\text{=}\frac{2P/5}{8P/5}\times 100=25\%$