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.

  • 2
    $\begingroup$ check Gibbs paradox en.wikipedia.org/wiki/Gibbs_paradox $\endgroup$
    – hyportnex
    Commented Jun 8, 2016 at 22:20
  • 1
    $\begingroup$ 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? $\endgroup$ Commented 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
    Commented 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
    Commented 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
    Commented Jul 16, 2016 at 20:17

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.