Confusion about the entropy of mixing

[sha1000]

TL;DR Summary

I'm seeking clarity on a seeming discrepancy between thermodynamics and statistical mechanics concerning the calculation of entropy when two identical boxes are combined. While thermodynamics suggests the combined entropy remains the same, a statistical mechanics viewpoint indicates a significant increase in entropy due to the increase in potential microstates. Can anyone help resolve this?

Hello everyone,

I am seeking some clarification regarding a question related to thermodynamics and statistical mechanics. My understanding is that when we combine two identical boxes with the same ideal gas by removing the wall between them, the resulting system's entropy stays the same. Essentially, the total entropy of the new system is the summation of the entropies of the original two boxes (i.e., Stot = S1 + S2 = 2S1 or 2S2).

However, from the standpoint of statistical mechanics, it appears that this entropy increase might not be as straightforward. Let's consider that we have N1 particles in a volume V1, which results in an entropy of S1. If we duplicate this system with a partition in place, we can simply double the entropy. But, if we remove the partition, we're left with 2N1 particles in a volume of 2V1. My confusion arises from the fact that when calculating the number of microstates in this new system, the entropy seems to increase significantly due to the doubled number of particles in the doubled volume.

Could anyone shed some light on this apparent discrepancy between these two views?

Thank you in advance for your help!

 

[Bystander]

https://www.google.com/search?q=gib...0i10i512l9.25128j0j7&sourceid=chrome&ie=UTF-8

 

[sha1000]

https://www.google.com/search?q=gibbs+paradox+in+statistical+mechanics&rlz=1C1CHBD_en-GBUS1019US1019&oq=gibb's+paradox&aqs=chrome.2.69i57j0i10i512l9.25128j0j7&sourceid=chrome&ie=UTF-8

Thank you for your reply. I'm glad to know that there is real basis for my confusion and that my train of thought is correct.

 

[vanhees71]

TL;DR Summary: I'm seeking clarity on a seeming discrepancy between thermodynamics and statistical mechanics concerning the calculation of entropy when two identical boxes are combined. While thermodynamics suggests the combined entropy remains the same, a statistical mechanics viewpoint indicates a significant increase in entropy due to the increase in potential microstates. Can anyone help resolve this?

Hello everyone,

I am seeking some clarification regarding a question related to thermodynamics and statistical mechanics. My understanding is that when we combine two identical boxes with the same ideal gas by removing the wall between them, the resulting system's entropy stays the same. Essentially, the total entropy of the new system is the summation of the entropies of the original two boxes (i.e., Stot = S1 + S2 = 2S1 or 2S2).

However, from the standpoint of statistical mechanics, it appears that this entropy increase might not be as straightforward. Let's consider that we have N1 particles in a volume V1, which results in an entropy of S1. If we duplicate this system with a partition in place, we can simply double the entropy. But, if we remove the partition, we're left with 2N1 particles in a volume of 2V1. My confusion arises from the fact that when calculating the number of microstates in this new system, the entropy seems to increase significantly due to the doubled number of particles in the doubled volume.

Could anyone shed some light on this apparent discrepancy between these two views?

Thank you in advance for your help!

If the gases in the two parts of the box are of identical particles (atoms/molecules), then the entropy doesn't change. If they are not identical there's mixing entropy. You get this right within statistical mechanics, using quantum theory. Anyway, quantum statistical mechanics is simpler than classical. If you know enough quantum theory, it's thus easier to learn statistical physics starting from quantum many-body theory and, for equilibrium, the maximum-entropy principle and take the classical limit to get the results of classical statistics. The same also holds for off-equilibrium statistical mechanics, where you can derive the Boltzmann(-Uehling-Uhlenbeck) equation via the Kadanoff-Baym equations of quantum many-body theory.