Chemical equilibrium between gases

For the equilibrium: $PCl_5(g)$ $PCl_3(g)$ $ + $ $Cl_2(g)$ At a given temperature, there are: $K_p$ $=$ $2.25\;atm$ Pure phosphorus pentachloride is introduced into an empty closed enclosure of constant volume $V$ and it can be seen that, at equilibrium, the partial pressure of this gas is $0.25 \;atm$. 1) Calculate the partial pressures of the other two gases: The equilibrium constant $K_p$ is defined by: ....$K_p$ $=$ $\frac{p_{PCl_3}p_{Cl_2}}{p_{PCl_5}}$ Let x be the partial pressure of $PC1_3$ at equilibrium. Since each mole $PCl_5$ that dissociated produced one mole of $PCl_3$ and one mole of $Cl_2$, these two gases will be present .... at equilibrium, so the partial pressure of $Cl_2$ at equilibrium is equal to .... and the expression of the constant $K_p$ becomes: ....$2.25$ $=$ $\frac{x\cdot x}{0.25} $

and so:

$x=\sqrt{2.25\cdot 0.25}$      $=$ $0.75$   atm

At equilibrium: $p_{PCl_3}$ $=$ $ p_{Cl_2}$ $=$ $0.75$   atm 2) Let's determine the initial pressure of $PCl_5$: At constant volume and temperature, the partial pressures are proportional to the number of moles. The reasoning with the partial pressures can be done in the same way as with the moles: If $0.75$ atm of $PCl_3$ have appeared, then by looking at $PCl_5$, we can say that .... $0.75$ atm $PCl_5$ will have disappeared. The initial pressure of $ PCl_5 $ was thus ....$0.25$   (the actual pressure) + $0.75$ $=$ $1\;atm$