Consider first a closed system divided into two subsystems A and B, separated by a fixed, adiabatic and impermeabler wall. These have a different amount of the same ideal gas, $n_{i,A}$ and $n_{i,B}$, $n_{i,A} > n_{i,B}$. Their temperatures and volumes are the same. Pressure is different, as per they different amounts of gas. Now change the wall so that it is permeable. Gas from subsystem A will quickly diffuse into subsystem B, until equilibrium is achieved. The entropy of each subsystem is straightforward to determine:

$$S_{A,B}=n_{A,B}[c_v\ln T+ R\ln(V/n_{A,B})+k]$$

Using this information, it's easy to determine that $\Delta S_A<0$ and $\Delta S_B > 0$ (knowing that $n_{f,A}=n_{f,B}=n_T/2$). In other words, the entropy of subsystem A decreases and the entropy of subsystem B increases. It's also possible to the determine that the overall entropy change is an increase. However, what I'm trying to point out here is that the entropy of subsystem A decreases.

Now consider a similar system, but this time with only 10 particles we can place either in the right or in the left (let's say that left is A and right is B). Originally, we have 7 particles in A and 3 in B. We let the system evolve until there are 5 particles in each side: equilibrium is achieved. The Botzmann postulate for entropy is $S=k_B \log W$. We can caclulate the mutiplicity with the binomial coefficient. It's easy to see that, now, both entropies increase (and obviously the total entropy of the system increases) because the macrostate with 5 particles in one side has more microstates than either the one with 3 or 7 (in fact, both entropies increase the same amount).

So, what's going on? How come in the purely thermodynamical version, one entropy decreases, while in the statistical version, both increase?

  • $\begingroup$ Sorry, how exactly are you calculating the entropy change for the 10-atom case? It seems like you're assuming that there are precisely 10 slots for atoms on each side; is that correct? If so, this condition that $n_{i,A}+n_{i_B}$ slots are available is not present in the first $S_{A,B}$ calculation, nor does it seem to apply to a container of an ideal gas. $\endgroup$ Oct 17, 2022 at 19:23
  • $\begingroup$ @Chemomechanics Yes, I am assuming that there are only 10 slots available on each side. Is this limitation what changes the result from the second to the first? It does make sense, as if you calculate the binomial coefficient for an available number of slots higher than 10 (say, 1000), the number of microstates goes down for subsystem A as its 7 particles become 5, and it increases for subsystem B. $\endgroup$
    – agaminon
    Oct 18, 2022 at 19:31

1 Answer 1


Here is the correct calcualtion of entropy. Assume that each box is a lattice with $L$ sites, each with volume $v_0$ so that the volume of each box is $V=L v_0$. Put $N_1$ particles in box 1 and $N-N_1$ in box 2. The entropies are:

$$ S_1 = k_B \ln\frac{L!}{N_1! (L-N_1)!}, \quad S_1 = k_B \ln\frac{L!}{(N-N_1)! (L-N+N_1)!} $$

To compare the above result with the ideal gas calculation we must operate in the dilute limit $N/L\to 0$. In this limit particles truly become point masses as the ideal gas requires. Using the Stirling formula $\ln x! = x\ln x - x$ the entropy of compartment 1 becomes $$ \frac{S_1}{k_B N_1} = -\ln\frac{N_1}{L} - \frac{L-N_1}{N_1}\ln\frac{L-N_1}{L} $$ With $N_1\ll L$ we have $L-N_1 \approx L$ and the entropy becomes $$ \frac{S_1}{k_B N_1} = -\ln\frac{N_1}{L} = \ln\frac{L}{N_1} $$ Now write the volume of the system as $V = v_0 L$ where $v_0$ is the volume of a lattice site, also the volume of a molecule. The entropy becomes $$ \frac{S_1}{k_B N_1} = \ln\frac{V}{N_1} - \ln v_0 \Rightarrow \boxed{ \frac{S_1}{k_B N_1}= \ln\frac{V}{N_1} + \text{const} }. $$ The last result gives the ideal gas entropy as a function of volume at fixed temperature.

Conclusion To obtain the ideal gas limit in the combinatorial model we must take the system in the limit of infinite dilution.

The progression to the ideal gas state is demonstrated below with $L=10$, $N=10, 8, 6$: as $N$ decreases we approach the ideal-gas behavior where $S_1$ decreases and $S_2$ increases. For true ideal gas behavior we need to use a much larger $L$ relative to $N$.

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  • $\begingroup$ Your situation is different as you are making calculations using more openings than particles. If you run the calculations using $L=10=N$, then things should agree with me. Of course, whether it's appropiate to consider that there are $L=N$ openings only, is another question. $\endgroup$
    – agaminon
    Oct 18, 2022 at 19:45
  • $\begingroup$ I see now what you mean. However, w cannot have an ideal gas when the lattice is full. The ideal gas limit represents poient particles and this condition is approached when $N\ll L$. I updated the answer to demonstrate the relationship between the combinatorial solution and the ideal gas. $\endgroup$
    – Themis
    Oct 18, 2022 at 23:02
  • $\begingroup$ Fantastic, this answer gets to the problem perfectly. $\endgroup$
    – agaminon
    Oct 18, 2022 at 23:16

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