Can a system have varying partition function?

Question

Follow up from my previous question,

I am not sure if a partition function for a system should be constant or not. If not, what are examples of constrained systems with partition function that varies with the system's state?

Here is a particular example that came to mind:

Consider a closed rigid container of temperature $T$, the total number of particles $N$ and volume $V$ in contact with a reservoir of temperature $T$. This container is divided into two sections $A$ and $B$, separated by a barrier that does not allow particles to pass through. The volumes of two sections $V_A$ and $V_B$ can vary as long as $V_A+V_B=V$. Section $A$ is filled with ideal gas of one kind and section $B$ is filled with ideal gas of another kind.

The partition function for section $A$ is

$$Z_A=e^{N_A} \left(\frac{V_A}{N_A}\right)^{N_A} \left(\frac{2\pi m}{h^2 \beta}\right)^{\frac32 N_A}$$

The partition function for section $B$ is

$$Z_B=e^{N_B} \left(\frac{V_B}{N_B}\right)^{N_B} \left(\frac{2\pi m}{h^2 \beta}\right)^{\frac32 N_B}$$

The partition function for the container is

$$Z=Z_AZ_B=e^{N_A+N_B}\left(\frac{V_A}{N_A}\right)^{N_A}\left(\frac{V_B}{N_B}\right)^{N_B}\left(\frac{2\pi m}{h^2 \beta}\right)^{\frac32 (N_A+N_B)}=e^{N}\left(\frac{V_A}{N_A}\right)^{N_A}\left(\frac{V_B}{N_B}\right)^{N_B}\left(\frac{2\pi m}{h^2 \beta}\right)^{\frac32 N}$$

Since the barrier does not allow particle flow, in order to maintain chemical equilibrium, assume that $N_A+N_B=\frac{N}{2}$,

$$Z=e^{N}V_A^{\frac{N}{2}}V_B^{\frac{N}{2}}\left(\frac{2}{N}\right)^{N}\left(\frac{2\pi m}{h^2 \beta}\right)^{\frac32 N}$$

It seems like the partition function $Z$ varies with values of $V_A$ and $V_B$ and is maximized when $V_A=V_B=\frac{V}{2}$.

Does my example qualify as a constrained system with varying partition function?

Answer 1

The partition function of an equilibrium system is a function of $V$, $T$, $N$: $$Z = Z(T,V,N)$$ For multiple systems that do not interact, the total partition function is the product of the individual functions: $$ Z(T,V_A,V_B,N_A N_B) = Z(T,V_A,N_A) \cdot Z(T,V_B,N_B) $$ or more compactly $$ Z = Z_A\cdot Z_B $$ under the conditions $$ T=\text{same in all parts},\quad V_A+V_B=\text{const},\quad N_A+N_B=\text{const} $$ Notice that $Z$ of the total system is a function of five variables while the partition function of each part is a function of three.

With $N_A=N_B=N/2$ fixed you have shown that the condition $V_A=V_B=V/2$ minimizes the free energy of the two-part system. Indeed, your condition amounts to $P_A=P_B$, i.e., the two parts in this case are in mechanical equilibrium with each other.

To summarize: The partition function of the two-part system depends not only on the total volume and total number of particles in the entire system, but also on how we partition these properties between the two parts. An arbitrary partitioning of will not produce equilibrium between the parts. Equilibrium is obtained when the partition function (or its log, which is proportional to minus free energy) is maximized. At the maximum the partition function is still a function (not a constant) that depends on three variables, temperature, total volume and total number of particles.

Answer 2

This might clarify what is varying. In your formulation, the partition function \begin{equation} Z(V_A) = e^N [V_A(V - V_A)]^{N/2} \left ( \frac{2}{N} \right)^N \left ( \frac{2\pi m}{h^2\beta} \right )^{N/2} \end{equation} describes a different system for each value of $V_A$. Forcing this to be one system might be artificial but there's an easy way to do that. We can declare that changes in $V_A$ instead take us to different microstates. Then there are more microstates to sum over and the partition function becomes \begin{equation} \mathcal{Z} = \int_0^V Z(V_A) dV_A \end{equation} which is constant.