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I'm confused about the entropy change if two gases, initially separated, are mixed together in a rigid box. I use the following

$$\Delta S_1= n_1 c_{v,1} \mathrm{ln}\left( \frac{T_f}{T_{i,1}}\right) + n_1 R \mathrm{ln}\left(\frac{V_1+V_2}{V_1}\right)\tag{A}$$

$$\Delta S_2= n_2 c_{v,2} \mathrm{ln}\left( \frac{T_f}{T_{i,2}}\right) + n_2 R \mathrm{ln}\left(\frac{V_1+V_2}{V_2}\right)\tag{B}$$

And $\Delta S=\Delta S_1+\Delta S_2$. I'm ok with this.

But I read in a textbook that I can use formulas $(\text{A})$ and $(\text{B})$ only if the two gases are different.

In the case where I have the mixture of the same gas, I must use

$$\Delta S_1= n_1 c_{v,1} \mathrm{ln}\left( \frac{T_f}{T_{i,1}}\right) - n_1 R \mathrm{ln}\left(\frac{p_f}{p_{i,1}}\right)\tag{C}$$

$$\Delta S_2= n_2 c_{v,2} \mathrm{ln}\left( \frac{T_f}{T_{i,2}}\right) - n_2 R \mathrm{ln}\left(\frac{p_f}{p_{i,2}}\right)\tag{D}$$

Otherwise, the change in entropy would be zero.

Is that correct or is there something I am missing?

If that is true, then how can that be? For instance, $(\text{A})$ and $(\text{C})$ should be equivalent as $S$ is a state function. I do not see how can there be a difference in those formulas for calculating $S$ and the reason why two of them give the correct result and the others do not.

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    $\begingroup$ check Gibbs paradox en.wikipedia.org/wiki/Gibbs_paradox $\endgroup$
    – hyportnex
    Jun 8, 2016 at 22:20
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    $\begingroup$ Let's simplify this a little. Suppose the two gases were at the same temperature and pressure to start with. Would there be a difference in the entropy change if the gases were different compared to if they are the same? $\endgroup$
    – Chet Miller
    Jun 8, 2016 at 22:25
  • $\begingroup$ @ChesterMiller Thanks for the answer! Apparently $(A)$ and $(B)$ seems to contain "something more" than $(C)$ and $(D)$: the terms of entropy of mixing, as calculated here chemwiki.ucdavis.edu/Core/Physical_Chemistry/Thermodynamics/…. Infact, just using the equation of state for gas $1$, indicating with $\Delta S_{1,A}$ the change in entropy of gas $1$ calculated with $(A)$ and with $\Delta S_{1,C}$ the change in entropy of gas $1$ calculated with $(C)$ we have $$\Delta S_{1,C}=\Delta S_{1,A}-n_1 R ln(\frac{n_1+n_2}{n_1})$$ $\endgroup$
    – Sørën
    Jul 16, 2016 at 20:14
  • $\begingroup$ If this is correct, then it is right to use $(B)$ in the case in which I have in the two parts the same gas at same temperature and pressure, otherwise the entropy change would be non zero (that's a basic of Gibbs paradox, as far as I understood). But in the opposite case, in which I have two different gases but with same pressure and temperature, which of the two formulas should I use? My guess would be $(A)$, which gives exactly just the term of mixing, which instead is missed in $(C)$, so in that case, for gas $1$ it would hold that $$\Delta S_1=n_1 R ln (\frac{n_1+n_2}{n_1})$$ $\endgroup$
    – Sørën
    Jul 16, 2016 at 20:15
  • $\begingroup$ Would this be correct in that case? And is the fact that $(C)$ (which would give $\Delta S_1=0$) is wrong due to Gibbs paradox again? $\endgroup$
    – Sørën
    Jul 16, 2016 at 20:17

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Entropy is function of state multiplicity. If you have the same gas, you wouldn't change state multiplicity and thus cannot change its entropy. That's different if two gases are different.

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