# Why the insistence that the process be reversible for $dW_{RWS} = -dF$ to hold (Helmholtz free energy)?

In a discussion about the (change in the) Helmholtz potential being interpretable as the maximum available amount of work for a system in contact with a thermal reservoir (i.e. the free energy), Callen seems to insist this fact is true only for reversible processes. Why should this be? I understand that any quasistatic process performed in contact/equilibrium with a thermal reservoir (as is the case here) will necessarily be reversible (since heat exchange proceeds at the same temperature), but to my reading Callen's emphasis on reversible seems to suggest there's something extra here I am perhaps missing? After all, as far as I can tell we must have $$dW_{RWS} = -dU - dU^r$$ by pure energy conservation for any process.

• Again, Callen does not assume that quasistatic = reversible. Reversible is a strict subset of quasistatic, and they will only equal if you assume things like friction do not exist in the quasistatic limit. Commented Jul 2, 2023 at 5:59

This is a wrong understanding:

...any quasistatic process performed in contact/equilibrium with a thermal reservoir (as is the case here) will necessarily be reversible (since heat exchange proceeds at the same temperature).

Only the thermal exchanges within the reservoir are always reversible meaning that a thermal reservoir, by definition, can and may interchange an arbitrary amount of entropy always at the same fixed temperature, say $$T^0$$, AND then the reservoir's internal energy that is supplied with an entropy exchange $$\Delta S^0$$ is exactly $$\Delta U^0=T^0\Delta S^0$$. This says nothing about the processes that may occur in the finite system with which the reservoir exchanges entropy $$\Delta S^0$$ at temperature $$T^0$$. We only know that the system receives $$\Delta S^0$$ entropy and $$\Delta U^0$$ energy at $$T^0$$ temperature. The actual change of entropy and of energy within the system depends on what else it is also connected to, i.e., what Callen calls work reservoirs AND how the process takes place. Energy conservation holds in all processes but entropy is conserved only in a reversible process. Callen assumes that besides the thermal reservoir, each work reservoir is characterized by a single fixed intensive quantity representing a single work exchange and the corresponding energy transfer is always reversible within that reservoir. The work IN the system is not necessarily reversible, just the exchange with the respective reservoir is within the reservoir.

The obvious example is having a huge, ideally, infinite charge storage device, a battery, connected to a large resistance so the current is very small and has no effect on the battery itself. The battery supplies work, current at some FIXED voltage $$V_b$$ at some FIXED temperature, $$T_b$$, both assumed to be essentially unchanged if the current is small, and the resistance generates entropy at a constant rate $$\dot s = \frac{V_bI}{T_r}=\frac{V_b^2}{RT_r}$$ where $$T_r$$ is the temperature of the thermal reservoir, the heat sink, attached to the resistor $$R$$ and is absorbing the entropy generated by the electric that is a pure friction work done on the resistance.

In this example, $$T_b$$ is not necessarily the same as $$T_r$$. The battery temperature must be fixed so that we can say that $$V_b$$ is fixed by the temperature dependent electrochemistry of the battery, but from the point of view of the working system, the resistance, $$T_r$$ represents the thermal reservoir and $$V_b$$ the work reservoir.

• If I am understanding you correctly, the resistor is the (primary) system here, the battery is the RWS, and the thermal reservoir is the thermal reservoir (in terms of mapping to Callen's usage)?
– EE18
Commented Jul 2, 2023 at 15:13
• If my comment above is correct, is the resolution not that this is not a quasistatic process with respect to the resistor, no matter how small the current? I will have to think about this though, thank you for the example suggestion.
– EE18
Commented Jul 2, 2023 at 15:20
• The reason I made the assertion that quasistatic + entropy exchange at constant temperature implies reversible is because (I believe) quasistatic + adiabatic implies reversible, and I thought this was a natural generalization since the system + thermal reservoir are adiabatic as considered together?
– EE18
Commented Jul 2, 2023 at 15:23
• yes, that is almost exactly what I meant. What I left out, but now I admit to, is whether the battery itself is the RWS, or in fact there are two work sources at different potentials, one at the potential $V_1$ the other at say $V_2$ so that $V_1-V_2=12V$. I like the battery example because it is obvious (?) that no matter how slow, that is small, your current is the Joule dissipation is unavoidable and it can never be reversible. Commented Jul 2, 2023 at 15:23
• I know you believe that quasistatic + adiabatic = reversible but within the confines of regular English it is a false religion... This issue deserves a separate question for it cannot be answered in a comment. Commented Jul 2, 2023 at 15:27

My understanding of how this works is as follows: The first law of thermodynamics tells us that $$\Delta U=Q-W$$. For a system that can only exchange heat with an reservoir at temperature T, and for which the initial and final equilibrium states of the system are also at temperature T, the 2nd law of thermodynamics tells us that $$\Delta S=\frac{Q}{T}+\sigma$$where $$\sigma$$ is the amount of entropy generated (positive) due to irreversibility within the system. If we combine these two equations, we get $$\Delta F=\Delta U-T\Delta S=-W-T\sigma$$or$$W=-\Delta F-T\sigma$$So the maximum work in going from the initial state to the final state is $$W_{max}=-\Delta F$$, and occurs when there are no irreversibilities.

• Thank you for this answer here; I am going to think about it as well as the other example given by Hyportnex below. If you get the chance, I would greatly benefit/appreciate you taking a look at the example given by Hyportnex. Am I indeed to conclude that quasistatic + entropy exchange at constant temperature does not in fact imply reversibility?
– EE18
Commented Jul 2, 2023 at 15:22
• I agree that one source of irreversibility within a system is current flow through a resistor embedded within the system. More typically, irreversibility within a system is caused by transport processes proceeding at finite rates: 1. Heat conduction occurring at finite temperature gradients 2. Viscous dissipation occurring at finite velocity gradients 3. Molecular diffusion occurring at finite concentration gradients Another source or irreversibility is chemical reactions proceeding at finite reaction rates. Commented Jul 2, 2023 at 17:11
• Should my conclusion indeed then be that quasistatic + adiabatic does not necessarily imply reversible? I had held that as sacrosanct given (as mentioned below) that $dQ = TdS$ when quasistatic (assuming a system with no matter exchange) so that $dQ = 0 \implies dS = 0$. But perhaps I was wrong? Does Callen perhaps tacitly consider/examine a subset of (quasistatic?) processes in his book?
– EE18
Commented Jul 2, 2023 at 17:27
• It depends on what you mean by quasi static. Commented Jul 2, 2023 at 17:55
• By quasistatic I mean "essentially in an (local) equilibrium" at all points during the process. Operationally, by quasistatic I mean "able to use differentials" to describe changes in $U$ etc. for the given process; this is what I believe is meant by Callen in his Chapter 4.2.
– EE18
Commented Jul 2, 2023 at 17:59