As mentioned in Partition Function and Density of States, we reach a paradox when mixing the many non-interacting classical particles. In classical statistical mechanics, the free energy as derived in (9) is

(13)$A = -k_B T N ( \ln V - 3 \ln \lambda_T )$

Imagine we have two systems, one has $$N_1$$ particles and volume $$V_1$$ while the other has $$N_2$$ particles and volume $$V_2$$. Now we mix the two systems. Our physics intuition would tell us that the free energy of this new system should be $$A = A_1 + A_2$$ since they are non-interacting particles. However, the free energy shown in (13) tells us that

$A_{\text{mixed}} = \cdots - k_B T \ln (V_1^{N_1} V_2^{N2})$

which is different from the result we expect, i.e.,

$A = \cdots - k_B T (N_1 + N_2) \ln (V_1 + V_2) = \cdots - k_B T \ln \left( (V_1 + V_2)^{N_1+N_2} \right)$

That is, free energy becomes neither intensive nor extensive in our theory.

To make the free energy extensive, we could choose to divide volume by the particle number. Then a new term will appear in the epression for free energy, i.e., $$N\ln N$$. On the other hand, we have $$\ln N! = N\ln N -N$$ from Sterling approximation. In large systems, we can define the free energy in the following way

$A = - k_B T N ( \ln(V/N!) - 3 \ln \lambda)$

which is equivalent to

$Z_N = \frac{Z_1^N}{N!}$

This definition “solves” the Gibbs mixing paradox. The explanation of this modification requires quantum mechanics.

Warning

We can’t just pull out some results from statistical mechanics and apply them to a small system of a few particles. In stat mech we use a lot of approximations like Sterling approximation, many of which are only valid when particle number is huge.

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