$\begin{aligned} & \mathrm{PCl}_5=5 \mathrm{~mole} \\ & \mathrm{Ar}=4 \mathrm{~mole} \\ & \mathrm{P}_{\text {Total }}=\frac{9 \times 0.82 \times 610}{100}=4.5 \mathrm{~atm} \\ & \mathrm{P}_{\mathrm{PCl}_5}=\frac{5 \times 4.5}{9}=2.5 ; \mathrm{P}_{\mathrm{Ar}}=\frac{4 \times 4.5}{9}=21 \\ & \mathrm{PCl}_5 \rightleftharpoons \mathrm{PCl}_3+\mathrm{Cl}_2 \\ & 2.5-\mathrm{P} \quad \mathrm{P} \quad \mathrm{P} \\ & \mathrm{P}_{\text {total }}=2.5-\mathrm{P}+\mathrm{P}+\mathrm{P}+\mathrm{P}_{\text {AAL }}=6 \\ & \mathrm{P}=1.5\end{aligned}$
$K_p=\frac{1.5 \times 1.5}{1}=2.25$