How will the equilibrium partial pressure of carbon dioxide be affected? \begin{equation}\ce{CaCO3(s) -&gt; CaO(s) + CO2(g)}\quad\Delta H^\circ = +556\ \mathrm{kJ/mol} \tag{endothermic}\end{equation}

How will the equilibrium partial pressure of  $\ce{CO2}$ be affected if the volume is decreased?

So, I thought of course the moles will increase when you decrease the volume hence $pV = nRT$, thus $p = \frac{nRT}{V}$. Therefore, $p$ is inversely proportional to $V$. But, the right answer is NO CHANGE, can anyone help me to understand why?
The equilibrium constant for the reaction is 

$$K=\frac{P_{\ce{CO2}}}{P^\circ}=e^{-\Delta G^\circ/RT}$$

Changing the volume changes neither $\Delta G^\circ$ or $T$ (at least this is implicitly implied), so $K$ and, hence, $P_{\ce{CO2}}$ does not change.

For an ideal gas 

$$P_{\ce{CO2}}=\frac{n_{\ce{CO2}}RT}{V}$$

So if $V$ decreases then $n_{\ce{CO2}}$ must also decrease in order for $K$ to remain constant. So, when the equilibrium is perturbed by the volume-decrease the equilibrium shifts to the left to re-establish equilibrium, in accord with Le Chatelier's principle.