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For a canonical ensemble, I wonder if the partition function for a system $Z(T,V,N)$ is constant or not. If it is constant, then does it follow that the system's Helmholtz energy:

$$F=-k_BT\ln Z$$

is constant as well?

If Helmholtz energy is constant, then how it is possible for a system to increase or decrease its Helmholtz energy?

I understand that as the system evolves to its most-likely microstate, which has maximum entropy and lowest Helmholtz energy, its Helmholtz energy should decrease. But it cannot happen if Helmholtz energy is constant.

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For an unconstrained system at equilibrium at fixed temperature ($T$), volume ($V$), and number of particles ($N$), Helmholtz's free energy is a constant and there is no way to make it changing. It is a consequence of being at equilibrium.

When we say that the free energy satisfies a minimum principle, we mean that, if the system is at equilibrium under some internal constraint, and such a constraint is relaxed, the new equilibrium system will have a lower free energy. However, as soon there is the internal constraint, this implies that the free energy is a function non only of $T,V,$ and $N$, but also of some additional variables representing the constraint.

For example, if the starting system is a fluid made by two subsystems of the same substance in two subvolumes $V_1$, and $V_2$, with $V_1+V_2=V$, containing $N_1$ and $N_2$ particles, and a movable wall allows to change the two subvolumes, the minimum principle of Helmholtz's free energy implies that after releasing the constraint, the new equilibrium state will correspond to the minimum of the function $ F(T,V,N;V_1) $ as a function of $V_1$ at fixed $T,V,$ and $N$. By the way, the final state is the state such that the two subsystems have the same pressure.

Notice, that after the constraint removal, the actual transition towards the new equilibrium state may go through non-equilibrium states. Instead, $F(T,V,N;V_1)$ for values of $V_1$ intermediate between the initial and the final one, represents the free energy along a reversible quasi-static process.

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    $\begingroup$ Thank you. 1. For an unconstrained system, there is a single microstate when there is single equilibrium state, right? 2. For a constrained system, if free energy is changing, then its partition function also changes? If this system has two components A and B, then its partition function is $Z(T,V,N)=Z_AZ_B$, assuming that particles are distinguishable? $\endgroup$
    – Ray Siplao
    Mar 30, 2023 at 15:44
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    $\begingroup$ @RaySiplao 1. No. A single equilibrium state (macrostate) usually corresponds to many microstates (the only exception is $T=0$. 2. Yes. Free energy and partition functions are in a one-to-one correspondence. $\endgroup$ Mar 30, 2023 at 21:58

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