Under what pressure an eqiumolar mixture of pcl3 and cl2 be placed at 250 degree centigrade in order to obtain pcl5 at 1 atm

Q. Under what pressure must an equimolar mixture of $PC{l}_3$ and $C{l}_2$ be placed at ${250}^{\circ }C$ in order to obtain $PC{l}_5$ at $1atm$?
($K_p$ for dissociation of $PC{l}_5=1.78)$.

Q. In the reaction $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$ the partial pressure of $PC{l}_3,C{l}_2$ and $PC{l}_5$ are $0.3,0.2$ and $0.6atm$ respectively at equilibrium. If partial pressure of $PC{l}_3$ and $C{l}_2$ was increased twice, what will be the partial pressure of $PC{l}_5$ is in $atm$ at new equilibrium condition:

Q. In an evacuated closed isolated chamber at ${250}^{o}C$. $.02$ mole $P{Cl}_5$ and $.01$ mole ${Cl}_2$ are mixed $(P{Cl}_5\rightleftharpoons P{Cl}_3+{Cl}_2)$. At equilibrium density of mixture was $2.49g/L$ and pressure was $1$ atm. The number of total moles at equilibrium will be approximately:

Q. In the reaction $P{Cl}_5\rightleftharpoons P{Cl}_3+{Cl}_2$ the partial pressure of $P{Cl}_3,{Cl}_2$ and $P{Cl}_5$ are 0.3,0.2 and 0.6 $atm$ respectively.If partial pressure of $P{Cl}_3$ and ${Cl}_2$ was increased twice,what will be the partial pressure of $P{Cl}_5$ in $atm$ :

Q. The degree of dissociation is $0.4$ at $400K$ and $1atm$ for the gaseous reaction.
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$. Assuming ideal behaviour of all the gases. Calculate the density of the equilibrium mixture at $400K$ and $1$ atm Pressure.