Why equation 5.29 in Schroeder's Thermal Physics is not equal to 0?

Question

In Schroeder's "An Introduction to Thermal Physics" in section 5.2 (page 161), Schroeder considers the case of a system that is in thermal contact with a reservoir that is at a constant temperature $T_\text{R}$. I'll quote the relevant section:

The total entropy of the universe can be written as $S_\text{total}=S+S_\text R$, where a subscript $\text R$ indicates a property of the reservoir, while a quantity without a subscript refers to the system alone... let's consider a small change in the total entropy: $$\text dS_\text{total}=\text dS+\text dS_\text{R}\tag{5.26}\label{5.26}$$ I would like to write this quantity entirely in terms of system variables. To do so, I'll apply the thermodynamic identity, in the form $$\text dS=\frac1T\ \text dU+\frac PT\ \text dV-\frac\mu T\ \text dN\tag{5.27}\label{5.27}$$ to the reservoir. First, I'll assume that $V$ and $N$ for the reservoir are fixed - only energy travels in and out of the system. Then $\text dS_\text R=\text dU_\text R/T$, so equation \eqref{5.26} can be written $$\text dS_\text{total}=\text dS+\frac1T_\text R\text dU_\text R\tag{5.28}\label{5.28}$$ But the temperature of the reservoir is the same as the temperature of the system, while the change $\text dU_\text R$ in the reservoir's energy is minus the change $\text dU$ in the system's energy. Therefore, $$\text dS_\text{total}=\text dS-\frac1T\text dU=-\frac1T(\text dU-T\ \text dS)=-\frac1T\ \text dF\tag{5.29}\label{5.29}$$ Aha! Under these conditions (fixed $T$, $V$, and $N$), an increase in the total entropy of the universe is the same thing as a decrease in the Helmholtz free energy of the system. So we can forget about the reservoir, and just remember that the system will do whatever it can to minimize its Helmholtz free energy.

My question is, if $\text dV$ and $\text dN$ are also $0$ for the system, then shouldn't the thermodynamic identity for the system then be $\text dS=\text dU/T$, leading to $\text dS-\text dU/T$ in equation $\eqref{5.29}$ being equal to $0$?

Answer 1

This confusion is caused by the schizophrenia of traditional thermodynamic teaching. On the one hand it says that "heat" is only energy transported in motion like work but it is not work, and tries to conflate reversible thermo-statics with irreversible processes happening in time.

Schroeder is talking about $dS_{tot}=dS+\frac{1}{T_R}dU_R$ and goes on to show that also $dS_{tot}=-\frac{1}{T}dF$ when he is assuming a pure thermal irreversible process because his side condition was that $dN=0;dV=0,$ and then there is only thermal interaction.

Your question as to why that is not zero since the Gibbs equation under no work condition says $dS=\frac{dU}{T}$. That is because while it true that $dS_{tot}=dS+\frac{1}{T_R}dU_R$ but $dU=-dU_R$ with $T<T_R$ for "heat" be absorbed spontaneously by the system. Instead in an irreversible "no-work" process $dS_{tot} > 0 $, "the" 2nd law, showing the internally produced entropy must always be positive.

At the end of the process $T=T_R$ but not when it starts. Now the only thing you can say is that during the process $dU-T_R dS = dU-TdS-(T_R-T)dS=-(T_R-T)dS\le 0$ or $dF \le 0$ as a result of the irreversibility at every stage, so by the end you must still have $dF\le 0.$ Gibbs called the quantity $dA= (T-T_R)dS-(p-p_R)dV+(\mu-\mu_R)dN+... \le 0 $ the availability , nowadays it is more popular to call it exergy, the intensive constants $T_R, p_R, \mu_R..$ represent various work reservoirs, thermal, volume, chemical species, etc., with which the system may exchange $S, V, N, ...$ extensive quantities.

Answer 2

My question is, if $\text dV$ and $\text dN$ are also $0$ for the system, then shouldn't the thermodynamic identity for the system then be $\text dS=\text dU/T$, leading to $\text dS-\text dU/T$ in equation $\eqref{5.29}$ being equal to $0$?

Yes, unless the system undergoes some type of spontaneous process, not involving a change in temperature or volume, which then generates entropy making $dF\lt 0$.

There is a discussion of this in the Wikipedia article on Helmholtz Free Energy. See the section Minimum Free Energy and Maximum Work Principles.

Hope this helps.

Answer 3

I think you are correct. Schroeder here appears to be motivating the idea of free energy, but it seems silly to me to do that in the very circumscribed special case when there is simultaneously no work done on the system, no work done on the reservoir, AND no temperature difference between the two. In that case not only would $dU = -dU_R$ and $dF = 0$, but there would be no heat flow either, so $dU = dU_R = 0$. Since free energy is precisely a measure of how much work might ideally be made available in a process, it's odd to bring it up specifically in a situation where the constraints prevent any work from being done.