Does entropy really not increase here? 
Two vessels separated by a partition have equal volume $V_0$ and equal temperature $T_0$. They both contain the same ideal gas, and the particles are indistinguishable. The left vessel has pressure $P_0$ and the right vessel has pressure $2P_0$. After the partition is removed and the system equilibrates, what is the net change in entropy?

At first my intuition tells me that $\Delta S=0$ since, for the whole system, $dQ=0$ and $dW=0$ throughout the entire process. Moreover, if we look at only one vessel - say, the left one - the change in entropy is given by:
$$\begin{align}\Delta S_1&=\frac{\Delta E_1}{T_0}\\
&=\frac{\Delta (c_V NT)}{T_0}\\
&=\frac{c_V T_0\Delta (N)}{T_0}\\
&=c_V\frac{N_0}{2} \,\,\,\,\textrm{(from ideal gas formula)}
\end{align}$$
If we do it for the right vessel, we simply get $\Delta S_2 = -\Delta S_1$, so once again $\Delta S=0$. But the accepted answer to this Physics SE question says otherwise. Is there something I'm missing?

[EDIT] To show where I got $\Delta N$, notice that if the left vessel has $N_0$ moles of gas at first, the right vessel has $2N_0$ moles of gas.
$$N_2=\frac{P_2T_0}{V_0}=\frac{2P_0T_0}{V_0}=2N_0$$
So in the end, the total system will have $N_0+2N_0=3N_0$ moles of gas. Since the volumes are equal, in the end each will have $\frac{3}{2}N_0$ moles of gas.
 A: The change in entropy is certainly not zero.  It is greater than zero for this spontaneous process.  Just because the Q in an irreversible process is zero does not mean that the entropy change is zero.  The entropy change is the integral of dQ/T only for a reversible path. 
I get $\frac{3}{2}P_0$ for the final pressure.  The initial number of moles in the left container is $\frac{P_0V_0}{RT_0}$ and the initial number of moles in the right container is $\frac{2P_0V_0}{RT_0}$.  If the initial moles in the left container goes from $P_0$ to $1.5P_0$ (compression) at temperature $T_0$, what is its change in entropy?  If the initial moles in the right container goes from $2P_0$ to $1.5P_0$ (expansion) at temperature $T_0$, what is its change in entropy?  What is the total change in entropy for the process?
You can check your result against my final answer of $$\Delta S=\frac{P_0V_0}{T_0}\ln{(32/27)}$$
EDIT:  If you increase the pressure on an ideal gas isothermally and reversibly, then dU=0.  So, $$TdS=PdV=d(PV)-VdP=-VdP=-\frac{nRT}{P}dP$$Integrating, we get:$$\Delta S=-nR\ln(P_2/P_1)$$
