# Mixing entropy for ideal gases is decreasing

Consider two ideal (microcanonical) gases with particle number $$N_{1/2}$$, energy $$E_{1/2}$$ in a box with volume $$V$$. Also let's assume $$\frac{E_1}{N_1} = \frac{E_2}{N_2}$$.

Let's call the corresponding phase space volume $$\Gamma(E_{1/2},N_{1/2},V)$$. Now let's mix these two gases:

Then in my opinion it must hold that:

$$\Gamma(E_1+E_2,N_1+N_2,V) \geq \Gamma(E_1,N_1,V)\cdot\Gamma(E_2,N_2,V)$$.

Since on the rhs we're only taking the possibility of $$N_1$$ particels having the combined energy $$E_1$$, and $$N_2$$ particles having the combined energy $$E_2$$ into account. For $$N_1 + N_2$$ particles having the combined energy $$E_1 + E_2$$, there are of course more options than the latter.

From this follows:

$$S(E_1+E_2,N_1+N_2,V) \geq S(E_1,N_1,V)+S(E_2,N_2,V)$$.

Now let $$N_1$$ and $$N_2$$ be indistinguishable.

Then the Sackur–Tetrode equation tells us that ($$\lambda = \frac{E_1}{N_1} = \frac{E_2}{N_2} = \frac{E_1+E_2}{N_1+N_2}$$):

$$S(E_1 + E_2, N_1 + N_2, V) - S(E_1,N_1,V) - S(E_2,N_2,V) = k [(N_1+N_2)\text{ln}(\frac{V}{N_1+N_2}\lambda^{3/2}) - N_1\text{ln}(\frac{V}{N_1}\lambda^{3/2}) - N_2\text{ln}(\frac{V}{N_2}\lambda^{3/2}) = k[N_1\text{ln}(\frac{N_1}{N_1+N_2})+N_2\text{ln}(\frac{N_2}{N_1+N_2})]\leq 0.$$

How can i make sense of this?

You are not comparing the same things. The entropy of the Sackur-Tetrode equation is: $$S = \ln\frac{\Gamma}{N!}$$ while your subadditivity argument uses: $$S = \ln\Gamma$$ Physically, your final result makes sense. You are mixing two volumes of the same gas with the same temperature but different densities. When you combine them, the temperature still does not change but they need to equilibrate to a new intermediate density. This lowers the entropy of the low density gas and increases the entropy for the high entropy gas. Your calculation shows that the former wins over the latter.