The mole % of Cl₂ at equilibrium for the reaction.
$\begin{array}{l}{\mathrm{PCl}}_5\rightleftharpoons {\mathrm{PCl}}_3+{\mathrm{Cl}}_2\\ \left(\mathrm{g}\right)\left(\mathrm{g}\right)\left(\mathrm{g}\right);{\mathrm{k}}_{\mathrm{p}}=0.2\end{array}$ 

Under total pressure of 2 atm is _________

$\begin{array}{l}{\mathrm{PCl}}_5\rightleftharpoons {\mathrm{PCl}}_3+{\mathrm{Cl}}_2\\ \begin{array}{lll}{\mathrm{P}}_0& 0& 0\end{array}\\ \begin{array}{lll}{\mathrm{P}}_0(1-\mathrm{\alpha })& {\mathrm{P}}_0\mathrm{\alpha }& {\mathrm{P}}_0\mathrm{\alpha }\end{array}\\ \end{array}$
${\mathrm{k}}_{\mathrm{p}}=\frac{{\left({\mathrm{P}}_0\mathrm{\alpha }\right)}^2}{{\mathrm{P}}_0(1-\mathrm{\alpha })}={\mathrm{P}}_0\frac{{\mathrm{\alpha }}^2}{1-\mathrm{\alpha }}$
${\mathrm{x}}_{{\mathrm{cl}}_2}=\frac{{\mathrm{PCl}}_2}{{\mathrm{P}}_{\mathrm{tot}}}=\frac{{\mathrm{P}}_0\mathrm{\alpha }}{{\mathrm{P}}_0(1+\mathrm{\alpha })}=\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}$
$\begin{array}{l}\text{ mole }\mathrm{\%}\mathrm{y}{\mathrm{Cl}}_2={\mathrm{x}}_{{\mathrm{cl}}_2}\times 100=23\mathrm{\%}\\ 0.2={\mathrm{P}}_0\cdot \frac{{\mathrm{\alpha }}^2}{1-\mathrm{\alpha }}-1\\ {\mathrm{p}}_0(1+\mathrm{\alpha })=2\\ 0.1=\frac{{\mathrm{\alpha }}^2}{1-{\mathrm{\alpha }}^2}\Rightarrow \frac{1-{\mathrm{\alpha }}^2}{{\mathrm{\alpha }}^2}10\end{array}$
$\begin{array}{l}1-\mathrm{\alpha }=10{\mathrm{\alpha }}^2\\ 1/{\mathrm{\alpha }}^2=1\\ \mathrm{\alpha }=\frac{1}{\sqrt{11}}=0.3\\ {\mathrm{K}}_{\mathrm{p}}=0.2={\mathrm{P}}_0\frac{{\mathrm{\alpha }}^2}{1-\mathrm{\alpha }}\\ {\mathrm{P}}_{\text{tot }}=2={\mathrm{p}}_0(1+\mathrm{\alpha })\\ {\mathrm{x}}_{{\mathrm{cl}}_2}=\frac{0.3}{1.3}\times 100=23\mathrm{\%}\end{array}$