In my lectures, we derived that the expected number of particles in each side of a connected two-compartment system is proportional to the volume of each side. We did this by dividing each side into small volumes and using the principle of each a priori probabilities. However we assumed that the perticles are indistinguishable, which gave an important factor of $\frac{1}{N_1!N_2!}$ towards the number of distinct microstates giving rise to a macrostate. However, I am not at all sure why we have assumed the particles are indistinguishable. My confusion stems from two points:

  • I thought that classically, it was assumed that the particles are distinguishable, and that is one of the assumptions of which Maxwell-Boltzmann statistics is based. Granted, we are not deriving the Maxwell-Boltzmann distribution, however both the Maxwell-Boltmann distribution and the result derived here fit well with experiemnt and are considered within the classical regime. I feel like we should be able to use the same assumptions for both.

  • Secondly, on an intuitive level, I don't see that it should make a difference whether the particles are distinguishable or not to this result. It feels to me like in either case, the most probable distribution should be that which has a umber of particles proportional to the volume on either side....


1 Answer 1

  1. Maxwell-Bolztmann statistics is an approximation of Fermi-Dirac and Bose-Einstein statistics when quantum effects are negligible. However, it predicts a non-physical result: the entropy increases when you remove the partition in your example (called Gibbs paradox). The paradox can be resolved if we put in a factor of $$\frac{1}{N!}$$ by hand.

  2. It doesn't matter for the particle distribution in different volumes because this is the result of maximizing entropy when you already have the partition in the system.


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