Entropy change in sudden expansion

Question

For an irreversible sudden expansion from $V$ to $2V$, no heat is added during the expansion. However, the entropy changes by $N\log2$. I'm not sure how there can be a change in entropy without any heat added, since $dS = \frac{dQ}{T} = 0$. Of course, integration can yield $\Delta S = C$, where $C$ is a constant, but I'm not sure if this is the correct mathematical and physical way of thinking about this. Thank you for any and all help.

Answer 1

Your original equation is incorrect. The entropy change is not $\Delta S=\int{\frac{dq}{T}}$. The correct equation is$$\Delta S=\int{\frac{dq_{rev}}{T}}$$ where $dq_{rev}$ is the heat flow for an alternate reversible process between the same two end states. For such a reversible path, the heat flow will not be zero.