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Suppose that we have two containers containing an ideal gas and both have energy $E_1 = E_2 = E$, $V_1 = V_2 = V$ and $n_1 = n_2 = n$ which implies that they also have the same temperature (assuming that both contain the same kind of gas particles, e.g. diatomic).

The entropy of the combined system is:

$$ S = \ln(\Omega(E)) \quad \text{where} \ \ \Omega(E) =\Omega_1(E_1) \cdot \Omega_2(E_2)$$

We now place the containers in thermal contact (only heat can be exchanged) and since no macroscopic change takes place (systems have same temperatures) we are in equilibrium, which in turn means that according to the second law of thermodynamics:

$$\Delta S = S' -S=0$$

But the final entropy is:

$$ S' =\ln\left(\sum_{\epsilon=0}^{2E} \Omega_1(2E-\epsilon) \cdot \Omega_2(\epsilon)\right)$$

which is greater than $S$ in contrast with the second law of thermodynamics.

Should be the second law be better written as:

$$\Delta \langle S \rangle \geq 0$$

?

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  • $\begingroup$ If you put angle brackets or not depends on the thermodynamic ensemble. $\endgroup$
    – Mauricio
    Aug 16, 2022 at 17:32
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    $\begingroup$ Try to evaluate the difference in the two entropy values in terms of $E_1,V_1,N_1$. I expect it will turn out to be negligible when compared to $2S_1$. In thermodynamic limit, contribution to extensive quantities due to interactions on 2D boundary can be neglected. $\endgroup$ Aug 16, 2022 at 19:15

2 Answers 2

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Here there is the full formal proof.

We have a trivial upper and lower bound of the sum $\sum_{\epsilon=0}^{2E} \Omega_1(2E-\epsilon) \cdot \Omega_2(\epsilon)$ (where all terms are positive) by using the maximum of the sum terms, say the term for $\epsilon = \bar \epsilon$: $$ \Omega_1(2E-\bar\epsilon) \cdot \Omega_2(\bar\epsilon)\leq\sum_{\epsilon=0}^{2E} \Omega_1(2E-\epsilon) \cdot \Omega_2(\epsilon) \leq \sum_{\epsilon=0}^{2E} \Omega_1(2E-\bar \epsilon) \cdot \Omega_2(\bar\epsilon)=N_{2E} \Omega_1(2E-\bar \epsilon) \cdot \Omega_2(\bar\epsilon) $$ where $N_{2E}$ is the number of terms of the sum. For a fixed spacing between the levels $\epsilon$, say $\delta \epsilon$, $N_{2E} = \frac{2E}{\delta \epsilon}+1$. Therefore, $$ \log \left(\Omega_1(2E-\bar\epsilon) \cdot \Omega_2(\bar\epsilon)\right)\leq S'\leq \log N_{2E} +\log \left(\Omega_1(2E-\bar\epsilon) \cdot \Omega_2(\bar\epsilon)\right). $$ For large systems (thermodynamic limit), $\log N_{2E}$ becomes negligible with respect to the other term, and $S'$ asymptotically coincides with the value of its maximum term.

Since the two systems are indistinguishable, $\Omega_1=\Omega_2$, and the maximum term corresponds to $\bar \epsilon = E$, with entropy $$ S'(2E) = S(2E)=2S(E). $$ Once again, we see that Statistical Mechanics requires the thermodynamic limit to agree with Thermodynamics.

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  • $\begingroup$ Nice proof! But I have a doubt: since $N_{2E}$ is an extensive quantity, how can you be so sure that it is negligible in the thermodynamic limit? $\endgroup$
    – Javi
    Aug 17, 2022 at 8:23
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    $\begingroup$ @Javi $N_{2E}$ is extensive. Its logarithm is not. This is at variance with the number of states that increases exponentially with the size, giving an extensive entropy. $\endgroup$ Aug 17, 2022 at 10:22
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The process that you describe is indeed reversible. As suggested in the book by Pathria, you can check that the change of entropy is zero in an ideal gas using the Sackur-Tetrode equation.

In addition, to show that $$\sum_{\epsilon=0}^{2E} \Omega_1(2E-\epsilon) \cdot \Omega_2(\epsilon)\approx \Omega_1(E) \cdot \Omega_2(E),$$ I guess that one could approximate the sum by just one of its terms. This would be the discrete version of the saddle-point approximation. One can do such a thing in the thermodynamic limit ($N\rightarrow\infty$), when the distribution of energies can be taken as a delta function.

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