In his $Thermodynamics$, Fermi proves beautifully the following (rephrased):

For a system undergoing a cyclic process, $$\oint {\delta Q\over T}\leq 0,$$ and for a reversible cyclic process, it is an equality.

Then he states the following without any proof:

”... and $\oint {\delta Q\over T}= 0$ which is valid only for reversible cycles.”

Question: How does he conclude the converse of the implication as stated above? In other words, how to prove that the Clausius integral evaluating to zero implies the cycle to be reversible? (Without introducing the concept of entropy.)

Edit: Well, you may use entropy, if you can’t prove without it.

It is mentioned on page $48$ of his classy $Thermodynamics$.

  • $\begingroup$ What do mean by the parenthetical statement? $\endgroup$
    – Bob D
    Dec 7 '19 at 10:06
  • $\begingroup$ @BobD first parentheses mean I slightly modified the wording. The second asks to prove what is asked without using entropy. $\endgroup$
    – Atom
    Dec 7 '19 at 10:18
  • $\begingroup$ So you want to prove it without referring to entropy? $\endgroup$
    – Bob D
    Dec 7 '19 at 10:26
  • $\begingroup$ @BobD yes...... $\endgroup$
    – Atom
    Dec 7 '19 at 10:41
  • $\begingroup$ You are again unclear what you are asking and I don’t intend to waste any more of my time $\endgroup$
    – Bob D
    Dec 7 '19 at 15:35

When you reverse a reversible process then $T>0$ still holds but $\delta Q$ becomes $-\delta Q$ and the Clausius inequality must hold for these heat exchanges: $\int \frac{-\delta Q}{T} \le 0$.
Compare this with the original inequality and the two can hold simultaneously if and only if $\int_{rev} \frac{\delta Q}{T} = 0$ for all reversible cycles.

To your question in the comment:

$\int_{irrev} \frac{\delta Q}{T} \ne 0$ is a separate assumption from the Clausius inequality postulate having $\le$ from which one gets that $\int_{irrev} \frac{\delta Q}{T} < 0$ always. In fact the amount missing from $=$ is a measure of the irreversibility of the process. Now it is conceivable mathematically that there are special contours (ie. cyclic processes) in the domain of states such that in general $<$ but for a specific strange contour you get $=$ while the process is still irreversible.

Note that even if such a contour existed it could not be extended "sideways" to any finite width and still having $=0$ because within that finite width "tube" around the contour you would have equality and thus $T$ be an integrating factor and have entropy well defined from the ratio of exchanged heat $\delta Q$ and $T$; so it must have zero measure and zero measure contours cannot be physical because of fluctuations that can move it into the $<$ territory. (I suspect something like this could happen at some phase transition boundaries, eg, magnetic hysteresis is always irreversible but phase transition itself is not necessarily but am not sure...)

  • $\begingroup$ What is ask is whether there is any irreversible cycle for which the integral evaluates to zero. $\endgroup$
    – Atom
    Dec 7 '19 at 12:18
  • $\begingroup$ Or in other words, is it only for reversible cycles when the integral is zero? $\endgroup$
    – Atom
    Dec 7 '19 at 12:21
  • $\begingroup$ I am getting what you're saying, and loving it. But I'm not following why "around that contour ... entropy well defined ..." make the "tube" of measure zero. Please explain! $\endgroup$
    – Atom
    Dec 7 '19 at 19:46
  • $\begingroup$ if $=$ holds in a finite domain (not something that is infinitely thin, i.e., a 1-D curve in an k-D, k>1, domain) then $T$ is an integrating factor and everything you have heard about reversibility, entropy, etc., will also hold. (Think of differentiability in 2D: it is not enough that the function is differentiable along a line, say, you need differentiability in all directions.) $\endgroup$
    – hyportnex
    Dec 7 '19 at 19:52

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