Will the total equilibrium pressure increase in gas phase reaction PCl5(g) = PCl3(g) + Cl2(g) after chlorine is removed? Will the total pressure be greater than the initial equilibrium pressure if all $\ce{Cl2}$ is removed and the following reaction is allowed to re-equilibrate?
$$\ce{PCl5(g) &lt;=&gt; PCl3(g) + Cl2(g)}$$
Assume constant temperature and volume.

I know that when $\ce{Cl2}$ is removed, the reaction will go forward and I also know expressions of $K_\text{p}$ and $K_\text{c}$.  I also know Le Chateliers Principle. I have thought about it for a long time but have not made any progress.
First it is necessary to understand that due to the stoichiometry of the reaction, if you move the equilibrium to the products side, pressure will be increased. This is based on the fact that 1 mole of gas produces 2 moles of gas. In other words the number of molecules in the system is increased. By removing all of the $\ce{Cl2}$ you force the reaction to move to the products side (Le Chatelier's Principle). Thus more $\ce{PCl3}$ will be produced along with $\ce{Cl2}$. Eventually you will end up with more atoms inside the system and that will lead to increase of the pressure.