# How do you correct a partition function to account for indistinguishability of particles in general?

Consider a container of ideal gas divided in half by a partition where each half is at the same temperature, pressure, volume, and number of moles. In my statistical mechanics class we have gone over the Gibbs paradox / mixing paradox and so I know that, if the particles are indistinguishable, the entropy increase from removing the partition and allowing the particles to mix is zero.

But what if the two gases in the two halves of the container are distinguishable — for example of they are two different ideal gases? (They are still at the same temperaure, pressure, volume, number of moles as each other.)

Every resource that I have read on the Gibbs paradox says that if you have a partition function for a system where all the particles are considered distinguishable, then the way to get from that to a partition function for the same system but where the particles are indistinguishable is to divide the partition function by $$N!$$, the number of permutations of particles which would leave the system in an indistinguishable state.

Applying that strategy to my hypothetical situation in which we start with two distinguishable gases in each half of the container, the partition function of the system of two mixed distinguishable gases $$Z_2$$ should be related to the partition function of the system if the gases were indistinguishable $$Z_1$$ by $$Z_1 = Z_2 / A$$ where $$A$$ is the number of permutations of particles that interchange some number of left-side particles with some number of right-side particles without interchanging left-side particles with left-side particles or right-side particles with right-side particles. By counting, the number of such permutations is

$$A = \sum_{n=0}^N \binom{N}{n}^2 n!$$

where $$N$$ is the number of particles that started on each side of the container. Since the partition function of a system where the particles are distinguishable is multipled by $$A$$, the entropy of mixing should be $$k \ln(A)$$.

However, calculating the entropy of mixing using regular old macroscopic thermodynamics or by using the Sackur-Tetrode equation shows that the entropy of mixing is $$k N \ln(4)$$, which is not equal to $$k \ln(A)$$.

Why is the entropy of mixing not $$k \ln(A)$$? What is wrong with my derivation?

• I think you should be considering two different indistinguishable gases. Different, i.e. distinguishable between each other, but indistinguishable within themselves. Then the end result should be $\frac{Z_1^{N_1}}{N_1!}\frac{Z_2^{N_2}}{N_2!}$ utilising the total combined volume. I do not know why you had such a weird exchange factor as $Z_1=Z_2/A$. I think your entropy of mixing is really $k(2N)\ln2$ because you assumed each gas has the same number of particles, $N$, leading to total number $2N$. Commented May 5, 2023 at 10:44
• Note that Gibbs style correction is an approximation. Later you will learn that real gases are either Boson or Fermion, and you have to take into account their proper ways to do the indistinguishability, and Gibbs style correction is but an interpolation between the two for the Maxwell-Boltzmann limiting case. Commented May 5, 2023 at 10:45

Yes. And if we have a mixture we divide by the product $$N_1! N_2! \cdots$$. For example, the partition function of binary ideal gas mixture of point particles is $$Q(T,V,N_1,N_2) = \frac{1}{N_1! N_2!} \left(\frac{V}{\Lambda_1}\right)^{N_1} \left(\frac{V}{\Lambda_2}\right)^{N_2}$$ with $$\Lambda_i = \left(\frac{h}{2\pi m_i k_B T}\right)^{1/2}$$