Removing and replacing a barrier between two gas halves should conserve entropy — resolves an apparent violation of the 2nd law via indistinguishability.

Same Gas

Before barrier removal: $S=2N{k}_B\ln V$
After removal (naive): $S=2N{k}_B\ln (2V)=2N{k}_B\ln V+2N{k}_B\ln 2$ ← entropy increased?
Fix: particles are indistinguishable, so microstates $=V^N/N!$

After removal (corrected):

Same — no entropy change.

Two Different Gases

Now the two sides have distinguishable particles (blue vs red). Mixing increases entropy:

Replacing the barrier leaves $N/2$ blue and $N/2$ red on each side — not the original state. The number of ways to arrange this mixed partition adds back:

So entropy stays elevated — consistent with the 2nd law.
Key point: entropy change from mixing depends on whether the particles are distinguishable. Same gas → no entropy of mixing. Different gases → $\mathrm{\Delta }S=2N{k}_B\ln 2$.