Pure PCl₅ is introduced into an evacuated chamber and comes to equilibrium at 247^oC and 2.0 atm. The equilibrium gaseous mixture contains 40% chlorine by volume. Calculate ${\mathrm{K}}_{\mathrm{p}}\text{ at }{247}^{\circ }\mathrm{C}$ for the reaction  ${\mathrm{PCl}}_5\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{PCl}}_3\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right)$

The gaseous mixture contains 40% Cl₂ and 40% PCl₃,  since they are produced in 1 : 1 mole ratio. The ${\mathrm{PCl}}_5\%\text{ is }20$ 

For ideal gases, mole % = volume%
$$\begin{gathered} {\mathrm{P}}_{{\mathrm{Cl}}_2}={\mathrm{P}}_{{\mathrm{PCl}}_3} \\ {\mathrm{P}}_{{\mathrm{PCl}}_3}=\frac{{\mathrm{n}}_{{\mathrm{PCl}}_3}}{\mathrm{n}}\times \mathrm{P}=\frac{0.4}{1}\times 0.2= 0.80 \mathrm{atm} \\ {\mathrm{P}}_{{\mathrm{PCl}}_5}=2\times 0.2=0.40 \mathrm{atm} \\ {\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{P}}_{{\mathrm{PCl}}_3}·{\mathrm{P}}_{{\mathrm{Cl}}_2}}{{\mathrm{P}}_{{\mathrm{PCl}}_5}} \\ =\frac{0.80\times 0.80}{0.40}=1.6 \mathrm{atm} \end{gathered}$$