After asking this question, I am still confused about some facts regarding Helmholtz energy and thermal equilibrium.

Based on my understanding so far, for a system in contact with surroundings which is a reservoir of constant temperature $T$, Helmholtz energy $F$ measures the system’s capability to do any work on surroundings under constant constant temperature.

I wonder if the change in Helmholtz energy $\Delta F=\Delta U-T\Delta S$ assumes that the system is in thermal equilibrium with the surroundings or the system can have any arbitrary well-defined temperature. I think it is the latter.

According to the inequality $W_{by}\le-\Delta F$, as the system does work on surroundings, its $F$ decreases. The system will keep doing work until its $F$ reaches the minimum value which only happens when the system is in thermal equilibrium with the surroundings. In this state, the system is unable to do any further work.

In the linked question, Chet Miller wrote (emphasis mine):

What the equation $W\leq(-\Delta F)$ means is that, for all possible processes, both reversible and irreversible, between the same pair of initial and final thermodynamic equilibrium states, subject to the constraint that all heat transfer takes place with an ideal constant temperature reservoir at the same temperature T as the initial (and final) states, the maximum amount of work that the system can do on the surroundings is the same for the subset of all reversible paths and is equal to $-\Delta F$

I really don't understand why the initial and final states should be in thermal equilibrium. After all in thermal equilibrium, the system's $F$ is already minimized, so why the system should do work to decrease its $F$ further?

  • $\begingroup$ How would you propose to ascertain the value of the Helmholtz free energy in a system experiencing an irreversible process, which typically features significant spatial non-uniformities in temperature, pressure, specific volume, and viscous stresses? $\endgroup$ Mar 10, 2023 at 13:03

2 Answers 2


There are two separate issues that need to be addressed. First, how to understand the minimization of the Helmholtz energy; and second, how the Helmholtz energy relates to work. I will address each separately.

Minimization of $A(T,V,n)$

We consider a system with uniform temperature $T$, fixed total volume $V$ and fixed number of particle $n_i$ of type $i=1,2\cdots K$ (the system may be multicomponent). At equilibrium the Helmhotlz energy of this system is $$A(T, V_A+V_B, n_{Ai}+n_{Bi}) \tag{1}$$ This system has uniform temperature $T$, uniform pressure $P$ and uniform chemical potential $\mu_{i}$ for all components.

Next we divide this system arbitrarily into two parts: part $A$ whose state is $(T,V_A, n_{Ai})$, and part $B$ whose state is $(T,V_B=V-V_A, n_{Bi}=n_i-n_{Ai}$. Each part is in internal equilibrium, i.e., compartment $A$ has uniform $T$, $P_A$, $\mu_{Ai}$, and compartment $B$ has uniform $T$, $P_B$, $\mu_{Bi}$.

enter image description here

For arbitrary partitioning, $P_A\neq P_B$ and $\mu_{Ai}\neq \mu_{iB}$, which means that generally the parts are not in equilibrium with each other. As a consequence of the second law we must have $$ \tag{2} A(T, V_A, n_{Ai}) + A(T, V_B, n_{Bi}) \geq A(T, V_A+V_B, n_{Ai}+n_{Bi}) $$ and in more condensed form $$ \tag{3} A_A + A_B - A \geq 0 \Rightarrow \Delta A = A - (A_A+A_B) \leq 0 $$ The equal sign applies if and only if $P_{A}=P_{B}$ and $\mu_{Ai}=\mu_{Bi}$, i.e., when the parts are in equilibrium with respect to each other. In other words:

the equilibrium state has the lowest Helmholtz energy among all possible ways to partition $V$ and $n_i$ into any number of parts under uniform temperature

Notice that in this construction the parts are always in internal equilibrium. This is the way that equilibrium thermodynamics treats non equilibrium states, in this case a gradient in pressure and chemical potential.

Also notice that as we partition $V$ and $n_i$ at fixed $T$ the internal energy of the system will vary between different partitions. We assume the presence of a bath that can provide or absorb the required energy. This is the bath that ensures $T$ is the same in all parts at all times.

Helmholtz energy and work

The inherent tendency of systems to reach equilibrium is associated with the ability to produce work. A non equilibrium state is characterized by gradients which could be leveraged to produce work. This is a very general principle: a waterfall, a voltage difference, a temperature difference, all of such gradients can produce work – but only if we place a suitable machine in the path of the flow (of fluid, current, heat, etc). The difference of the Helmholtz energy between the final (equilibrium) state and the initial (non-equilibrium) state is the maximum amount of work that can be extracted: $$\tag{4} W = A - (A_A + A_B) = \Delta A \leq 0 $$ I want to correct a statement in the original post:

According to the inequality $W_{by}\le-\Delta F$, as the system does work on surroundings, its $F$ decreases. The system will keep doing work until its $F$ reaches the minimum value which only happens when the system is in thermal equilibrium with the surroundings. In this state, the system is unable to do any further work.

This is not true. Since the total volume is fixed, no work is done against the surroundings. To obtain work from the equilibration process we need to build some machine with weights, springs, semipermeable membranes and similar mechanical contraptions. If no such machinery is present the amount of work is zero, just like a waterfall with no turbine in its path to capture its energy before it is converted into internal energy at the bottom.

  • $\begingroup$ Wow thank you, I am starting to get it now. If you don't mind, I have some more questions: 1. So the maximum work that can be extracted from the system is $W=\Delta A\le0$ but what this work can be used for if it cannot be done against surroundings? 2. Is the work extracted $W$ spontaneous or not? I assume it is spontaneous because the process of reaching equilibrium is spontaneous? $\endgroup$
    – Jimmy Yang
    Mar 13, 2023 at 5:49
  • $\begingroup$ I see, I was confused in the first place because I didn't know there are two kinds of equilibrium states: thermal equilibrium state and 'equilibrium' state (minimum energy state). $\endgroup$
    – Jimmy Yang
    Mar 13, 2023 at 5:53
  • 1
    $\begingroup$ @JimmyYang (1) Yes, minimization od $A$ requires thermal as well mechanical and chemical equilibrium (2) For the calculation of $\Delta A$ (or any other property) all states must be in equilibrium. More specifically, each part must be in internal equilibrium, though parts need not be in equilibrium with each other. Equilibrium thermodynamics can only deal with states in internal equilibrium. $\endgroup$
    – Themis
    Mar 13, 2023 at 20:31
  • 1
    $\begingroup$ Absolutely, any of these types of work are included. Indeed, this is how we analyze electrochemical cells; in this case we would rather use the Gibbs energy $G=U+PV-TS$ since we typically study the problem under constant $T$ and $P$ (Helmholtz requires constant $T$ and $V$). $\endgroup$
    – Themis
    Mar 14, 2023 at 11:18
  • 1
    $\begingroup$ Here is how to think about free energy. It is analogous to the potential energy difference between the tob and the bottom of a waterfall: the maximum amount of work that can be priduced by the waterfall is $\Delta E_\text{pot}$, but this work will be produced only if we place a turbine in the path of the waterfall. $\endgroup$
    – Themis
    Mar 14, 2023 at 22:44

The question is about Helmholtz's free energy. However, the same question could be asked for any other thermodynamic potential (all satisfy a minimum condition, although under different external conditions) or for the fundamental equations derived from entropy (provided the minimum condition is changed into a maximum).

The important fact is that all fundamental equations are thermodynamic state functions, i.e., functions depending only on a bunch of thermodynamic variables and not on the huge number of microscopic degrees of freedom. Such a huge saving is allowed by thermal equilibrium. Without thermal equilibrium, the manifold of microscopic degrees of freedom should be taken into account in the description of the system.

As a consequence of the fact that fundamental equations are defined only for equilibrium states, minimum and maximum principles must be understood as the selection of the particular equilibrium state, among a set of equilibrium states corresponding to the same thermodynamic conditions, but different additional constraints.

Notice that we still can have inequalities connecting variation of free energy between two equilibrium states and the work or heat used to go from one state to the other with a non-equilibrium transformation (a non-quasi-static transformation). The reason is that heat or work are process-related quantities that in general do not allow a description in terms of thermodynamic variables of the system.

In summary, a change of free energy $\Delta F$, assumes, and can be written or evaluated, ony if the initial and final state are equilibrium states, each characterized by some value of volume, temperature, and number of particles (for simple fluid systems). Thermodynamics has no means to describe the time evolution of free energy. It can only say something about differences between free energies of equilibrium states.

  • $\begingroup$ Can you clarify what "equilibrium state" means here? Two interacting systems are in thermal equilibrium only if their equilibrium state matches? $\endgroup$
    – Jimmy Yang
    Mar 10, 2023 at 8:14
  • $\begingroup$ @JimmyYang apart from the external bath (if present) there is no other system than the thermodynamic system of interest. Equilibrium state refers to the thermodynamic state of the system of interest (no internal motions, no gradients if there is no external field etc.) $\endgroup$ Mar 10, 2023 at 12:37
  • $\begingroup$ So the 'equilibrium' state simply refers to the system's state, which is described by macroscopic quantities like $P$, $V$, $U$? It makes no reference to thermodynamic equilibrium? In that case, I find the word 'equilibrium' confusing here. $\endgroup$
    – Jimmy Yang
    Mar 10, 2023 at 21:46
  • $\begingroup$ @JimmyYang there is no difference.. $\endgroup$ Mar 11, 2023 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.