How a thermodynamical system relaxes to equilibrium?

For a system in contact with a reservoir with well-defined constant temperature $$T$$, its change in Helmholtz energy satisfies the following inequality:

$$\Delta F \le -W_{by}$$

where $$W_{by}$$ denotes the work done by the system.

I know that in thermal equilibrium, Helmholtz energy of the system is minimized. But I don’t understand what the system is doing to minimize its free energy.

Based on the inequality, the maximum work that can be done by the system is $$-\Delta F$$. I am not sure if it indicates that the system spontaneously do work to decrease its free energy.

As the system approaches thermal equilibrium, its entropy increases $$\Delta S>0$$ until it reaches its maximum. I wonder if the system also increases its internal energy $$\Delta U>0$$, so that at equilibrium $$\Delta F=\Delta U-T\Delta S=0$$? If so, what are possible reasons for the system to increase its internal energy?

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 $$-\Delta F$$; all other paths are irreversible, and do less work on the surroundings. The reversible paths are not spontaneous, and must be carefully controlled such that all changes along the path are quasi-static.

An example of this is expansion of an ideal gas in contact with an ideal constant temperature reservoir at the initial temperature of the gas. The maximum work and decrease in F for a reversible path are $$W=-\Delta F=nRT\ln{(V_2/V_1)}$$The work for an irreversible path, say where we drop the pressure from the initial value $$nRT/V_1$$ to the final pressure $$nRT/V_2$$ and hold it at this final value for the entire expansion, the work is $$W=nRT\left[1-\frac{V_1}{V_2}\right]$$Mathematically, this is less than for the reversible path.

For a non-ideal gas, the internal energy can change even if the temperature is constant. It can also change at constant temperature if there is a phase change or a chemical reaction.

• Thank you for the answer. I don't understand what you mean by "reversible paths are not spontaneous." So all the work done by the system is not spontaneous? What triggers the system to do work? Also, how did you derive $W=nRT\left[1-\frac{V_1}{V_2}\right]$? Mar 9, 2023 at 0:54
• A reversible process is constrained to do work very gradually in a highly controlled manner. In a reversible expansion, for example, work can be done by very gradually removing pebbles from atop a piston situated in a vertical cylinder, at a sequence of increasing elevations. This very gradually eases up on the gas pressure. For the irreversible work W, the force per unit area exerted by the gas on the piston is constant at $nRT/V_2$, and the volume change is $V_2-V_1$, so the work is $W=\frac{nRT}{V_2}(V_2-V_1)$. Mar 9, 2023 at 2:33
• So based on what you wrote, $W=-\Delta F$ is the maximum possible work the system can perform on surroundings while it undergoes a reversible process where initial and final states are in thermal equilibrium. But in thermal equilibrium, $F$ is already minimized. I don't see the point of the system doing further work to decrease its $F$? Mar 9, 2023 at 4:25
• I don't understand what you are saying, Mar 9, 2023 at 11:24
• In the reversible expansion, the gas is in thermal equilibrium with the reservoir over the entire process. In the irreversible expansion, the gas is in thermal equilibrium only before expansion and when the expansion ends; during the expansion, even though the gas temperature matches the reservoir temperature at the interface with the reservoir at all times, the gas temperature away from this interface is lower than the temperature at the interface, and the volume average gas temperature is less than the reservoir temperature.. Mar 10, 2023 at 21:21

The minimization of free energy is a mathematical "invention" that formalizes the equilibrium state. The system does nothing to minimize its free energy, it can only respond to physical forces of various kinds that drive the system to equilibrium. These are capable of producing work. The Helmholtz energy is the maximum work that could be extracted from them as the system approaches equilibrium under the condition that temperature, volume and number of moles are all constant.

With respect to the last question:

As the system approaches thermal equilibrium, its entropy increases $$\Delta S>0$$ until it reaches its maximum. I wonder if the system also increases its internal energy $$\Delta U>0$$, so that at equilibrium $$\Delta F=\Delta U-T\Delta S=0$$? If so, what are possible reasons for the system to increase its internal energy?

There are several issues with the above statements that make it difficult to give a direct answer:

• It is not true that "$$\Delta S>0$$ until it reaches its maximum". This statement applies to an isolated system (constant $$U$$, $$V$$ and $$N$$) but here we are dealing with a system at constant $$T$$, $$V$$ and $$N$$.

• The correct equilibrium condition is $$\Big(dF = dU - T dS = 0 \Big)_{T,V,N}$$ Notice the differentials. For a spontaneous process $$\Delta F$$ is negative, not zero.

• By first law, $$dU = dQ + dW$$. Therefore, internal energy may increase or decrease, depending on the process that takes place.

• Thanks for the answer. 1. So the system cannot spontaneously work itself into its equilibrium state? What are examples of physical forces that drive the system to equilibrium? 2. I don't understand the meaning of "work that could be extracted." It refers to work by the system to the reservoir? 3. For a spontaneous process $\Delta F$ is positive, not zero. I think you mean "For a spontaneous process $\Delta F$ is negative, not zero"? Mar 9, 2023 at 1:00
• 1. The approach to equilibrium is always spontaneous. For pressure equilibration the force is due to the pressure difference between the two sides of a wall. In thermal equilibration it is the higher momentum of the molecules on the hotter side of the wall. 2. Think of a waterfall: work can be extracted from the tendency of the water to reach a lower potential energy. This work will not be extracted unless we place a suitable machine (turbine) in the path of the water. 3. Yes -- thanks for catching. Mar 9, 2023 at 14:57

The more general equation for dF is $$dF=-SdT-PdV+\sum{\mu_idn_i}$$where the $$\mu$$'s are chemical potentials. The criterion for equilibrium of a closed system at constant temperature and volume is minimization of F (with respect to, say, chemical conversion): $$dF=\sum{\mu_idn_i}=0$$