Partition function of mixture of two gases

[millerm]

TL;DR Summary

Determine the partition function of a mixture of 2 gases

I have a question about statistical physics. Suppose we have a closed container with two compartments, each with volume V , in thermal contact with a heat bath at temperature T, and we discuss the problem from the perspective of a canonic ensemble. At a certain moment the separating wall is replaced by a membrane that is permeable for particles of gas A, but not for particles of gas B. What is the total partition function of the final system as a function of $N_A^R$ ?

My approach was to first consider a mixture between the particles in container B, $N_B$ and some of the particles from container A, let's say $N_A^R$, and calculate the partition function: $Z_1=Z(N_B)Z(N_A^R)$. The partition function of compartment A is: $Z_2=Z(N_A-N_A^R)$. And next take the partition function of two subsystems in contact with a thermal reservoir: $Z_{tot}=Z_1Z_2$. Is this approach correct? If yes, what do you take for the volume of $Z(N_A^R)$ ?

I hope someone can help me with this.

 

[AndyW]

Hello! Your approach is mostly correct, but there are a few things to consider.

First, the total partition function of the final system as a function of $N_A^R$ can be written as $Z_{tot}=Z_1(N_B)Z_2(N_A-N_A^R)$, since the partition functions of the two subsystems are multiplied together.

Next, for the volume of $Z(N_A^R)$, you can consider it as the volume of the compartment A that is now occupied by the particles of gas A. So, it would be $V-N_A^R$ since $N_A^R$ particles have been removed from the original volume V.

However, there is one more thing to consider. In your approach, you have assumed that the particles from compartment A are mixed with the particles from compartment B. But, in reality, the particles of gas A are only permeable through the membrane and will not mix with the particles of gas B. So, the partition function for compartment A should be modified to take into account this constraint. This can be done by considering the partition function of a single particle of gas A in compartment A, which would be $Z_1^{single}=e^{-\beta \epsilon_A}$, where $\epsilon_A$ is the energy of a single particle of gas A. Then, the partition function for compartment A would be $Z_2=Z_1^{single}^{N_A^R}$.

Overall, your approach is on the right track, but it would need some modifications to account for the constraint of the particles not mixing. I hope this helps!