Effect on the partition function upon the mixing of two ideal gases

Question

Consider a box that is separated into two compartments by a thin wall. Each compartment has a volume V and temperature T. The first compartment contains N atoms of ideal monatomic gas A and the second compartment contains N atoms of ideal monatomic gas B. Assume that the electronic partition functions of both gases are equal to 1. The molecular partition function for each component is given by $$q_i = \frac{V}{Λ_i^3}$$

Firstly I am asked to write the total initial canonical partition function which is given in the question as $Q_{initial}=Q_AQ_B$. The algebra is fine, but why would the two separate gases have a total canonical function?

I was then asked to show that after the gases mix (and do not react) that the following is true: $$\frac{Q_{mixed}}{Q_{initial}}=4^N$$.

I got that $Q_{initial}=\frac{q_A^Nq_b^N}{(N!)^2}$. Therefore, the only reasonable expression for the mixed function is: $$Q_{mixed}=\frac{(2q_A)^N(2q_b)^N}{(N!)^2}$$

However, I cannot see why the individual molecular partition functions will have doubled. Especially since A and B are distinct gases.

Answer 1

The fact that $Q_{\mathrm{initial}} = Q_{A}Q_{B}$ implies that the two subsystems, respectively consisting of gas A and gas B, are independent. Because $\log Q_{\mathrm{initial}} = \log Q_{A} + \log Q_{B}$, it follows that various extensive thermodynamic quantities (e.g., internal energy, free energies, entropy) of the total system are simply given by the sum of the corresponding quantities of the subsystems.

The partition functions are doubled because each gas occupies double the initial volume after mixing, i.e., $V \rightarrow 2V$.