Let's say we have a system in contact with a large reservoir. Let's also fix the number of particles in the system, as well its volume, so that the only thing that could be exchanged is energy. Now we know that any increase in entropy must be the sum of any increase in entropy in the system plus any increase in entropy in the reservoir:

$$dS_{total} = dS + dS_R.$$

Generally speaking, any change in entropy is due to the following factors:

$$dS = \frac{1}{T}dU + \frac{P}{T}dV + \frac{\mu}{T}dN.$$

But as we said above, we are fixing the volume and the number of particles, so in our case:

$$dS = \frac{1}{T}dU.$$

This is also true of the reservoir:

$$dS_R = \frac{1}{T_R}dU_R.$$

Putting this into our equation for the total entropy:

$$dS_{total} = dS + \frac{1}{T_R}dU_R.$$

We also know that any change in the energy of the system is a loss for the reservoir:

$$dU = -dU_R.$$

Up to here, I have no issues. However, the textbook I am looking at now makes the following assertion:

"...the temperature of the reservoir is the same as the temperature of the system,"

which means that $T_R = T$.

Substituting $T_R$ for $T$ and $U_R$ with $-U_R$:

$$dS_{total} = dS - \frac{1}{T}dU = -\frac{1}{T}(dU - TdS) = -\frac{1}{T}dF,$$

where F is the Helmholtz free energy, $U-TS$. In the other words, increasing total entropy involves lowering the system's Helmholtz free energy.

Using a similar argument, where the pressure, rather than the volume, is kept constant, it can be shown that $dS_{total} = -\frac{1}{T}dG$, where G is the Gibbs free energy, $U+PV-TS$.

Here's my dilemma: why do you assume that the temperature of the system is equal to that of the reservoir? Generally speaking, a system and its environment's temperatures wouldn't necessarily be the same.

Okay, so maybe the intent of all this is to imply that the equation only holds once the system has reached the same temperature as the reservoir. But if this were the case, would that really make sense? After all, since the system and the reservoir are the same temperature, there will be no net heat flow between the two. And in the first case, since the volume is fixed, there is no expansion of either the system or the reservoir. By the first law of thermodynamics, $U=W+Q$. We've already said that Q would have to be 0 and W cannot be an work that comes from a changing volume. This only leaves "other" work ("such as electrical work", whatever that even means). Then the equation would just say:

$$dS_{total} = dS + dS_R = \frac{1}{T}dU + \frac{1}{T}dU_R.$$

Since I don't really get what $W_{other}$ is, I can't say for sure that $dU_R = -dU$, but let's say it is so and that the reservoir is doing the "other" work on the system. Otherwise the assumption that $dU_R = -dU$ is wrong, which is specifically against the textbook's assumptions. So if we make this assumption then we get:

$$dS_{total} = \frac{1}{T}dU - \frac{1}{T}dU = 0.$$

So it's trivial: If they are the same temperature there is no heat flow, plus there is no expansion work done by constraint, and any "other" work done must increase one's entropy by the exact decrease in the other, since they are both at the same temperature.

So either $T_R=T$ and the answer is trivial, or $T_R\neq T$ which is not what the textbook said, making the textbook's whole derivation wrong.

Yet the textbook has to be correct so I must be wrong, but how?

  • $\begingroup$ If same temperatures of system and reservoir are assumed, and if reversible heat transfer is assumed, then no entropy is generated in heat exchange, so $dS_{total} = 0$ and $d(U-TS)=0$ always. Which author, book, section/page is this? $\endgroup$ Commented Jun 3, 2019 at 10:38
  • $\begingroup$ I think that what they are assuming is that the change takes place reversibly and very slowly, with a tiny temperature difference over a very long period of time, so that the heat transfer is finite while the temperature driving force is infinitesimal. Usually, as in Denbigh, the system is not analyzed this way. They usually assume that the system boundary is kept at the initial temperature of the system, and the system re-equilibrates to that temperature in the end. So, in the irreversible case, entropy would be generated within the system. $\endgroup$ Commented Jun 3, 2019 at 12:09
  • $\begingroup$ @janLalinsky The textbook is Thermal Physics by Daniel Schroeder, section 5.2: Free Energy as a Force toward Equilibrium. $\endgroup$
    – Israel
    Commented Jun 3, 2019 at 15:06

1 Answer 1


The exposition in Thermal Physics by Schroeder, section 5.2, is somewhat botched, but the idea is sound. What is actually assumed in these kinds of argument is that the system has the same temperature as the reservoir just before the process, and just after the process. Temperature of the system during the process is immaterial. It may not even be defined.

If temperature of the system was the same all the time, one would be tempted to additionally assume that the whole process is reversible and use the relation $$ dS = \frac{dQ}{T}. $$ Then $S_{total}$ would be constant during this process. But this is not what is being studied.

What is being studied is any irreversible process, which starts and ends at the same temperature as that of reservoir $T_R$. Thus a better way to write down the result would be

$$ \Delta S_{total} = -\frac{1}{T_R}\Delta(U-TS) $$

with deltas instead of differentials, to stress the fact that values at endpoints of a possibly long-running process are considered.

I wrote $U-TS$ instead of $F$, because for some systems, $F$ is more complicated than just $U-TS$, such as for mix of liquid water and ice, see e.g. this answer by GiorgioP:

Entropy as a function of temperature: is temperature well defined?

  • $\begingroup$ Interesting...I have two followup questions: 1. Are you then saying that in a situation where I take a can of soda and put it in the ocean, the can will not necessarily lower its Hemholtz free energy since their temperatures don't match? 2. As I said in my question above, doesn't this severely limit the usefulness of result? If the temperature of the system already matches its environment, doesn't it lead to trivial conclusions like no change in entropy? $\endgroup$
    – Israel
    Commented Jun 3, 2019 at 22:13
  • $\begingroup$ 1) That is a more specific question, not addressed by this kind of theorem. But it can be analyzed in similar way, I recommend you try it; it turns out that whether $F$ decreases or not depends on whether the initial temperature of the can is higher or lower than that of the ocean. 2) It limits the applicability, but it isn't trivial. Entropy of the system can change even if temperature does not. Also, entropy of the super-system can increase if the process is irreversible, no matter that temperature at the beginning is the same as temperature in the end. $\endgroup$ Commented Jun 4, 2019 at 2:20

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