I found a similar question but I could not solve my doubt. So, if you consider this question to be a double, I am sorry.

Consider a system $S$ that undergoes a cyclic transformation and the $n$ sources from which it receives heat have temperatures $T_1, T_2... T_n$. Let $Q_i$ be the heat received/given by the $i$-th source. After deriving Clausius inequality for all cycles:

\begin{equation}\tag{1}\sum\limits_{i=1}^n\frac{Q_i}{T_i}\leqslant0 \end{equation}

I've seen some books (Fermi, Thermodynamics is an example) doing what follows:

If the cycle is reversible, we can consider the inverse cycle, and the only difference will be the opposite sign of the heats. So:


In order to have both this inequality and the $(1)$ satisfied, for a reversible cycle we must have:

\begin{equation}\tag{3}\sum\limits_{i=1}^n\frac{Q_i}{T_i}=0 \end{equation}

Okay, there it is. From the equation $(3)$ we can conclude that for reversible cycles equality signs holds.

But Fermi also concluded that the equality holds only in that case. We have proved that the equation $(3)$ is true in the case of a reversible cycle, but we haven't proved that the equality cannot hold in any other case, so how do we conclude that \begin{equation} \sum\limits_{i=1}^n\frac{Q_i}{T_i}<0 \end{equation} for non-reversible cycle?

Am I missing something or does that book take it somehow for granted? Please, note I am asking for a theoretical and mathematical explanation of this conclusion. Thanks in advance.

  • $\begingroup$ see physics.stackexchange.com/questions/518206/… $\endgroup$
    – hyportnex
    Jul 7 '20 at 18:40
  • $\begingroup$ I read the answer and I'm not sure of what it means. So can we have irreversibile processes where the equality sign holds? $\endgroup$
    – Feynman_00
    Jul 7 '20 at 19:02
  • $\begingroup$ no, for an irreversible process it is "<" $\endgroup$
    – hyportnex
    Jul 7 '20 at 19:04
  • $\begingroup$ Ok, like Fermi said. So \begin{equation}reversible\iff\sum\limits_{i=1}^n\frac{Q_i}{T_i}=0\end{equation} Still, he just proved it is "=" for reversible cycle but not the opposite. In other words, he proved: \begin{equation}reversible\implies\sum\limits_{i=1}^n\frac{Q_i}{T_i}=0\end{equation} but he did not prove that \begin{equation}\sum\limits_{i=1}^n\frac{Q_i}{T_i}=0\implies reversible\end{equation} $\endgroup$
    – Feynman_00
    Jul 7 '20 at 19:14
  • $\begingroup$ no he did not because as I said before it is a separate assumption $\endgroup$
    – hyportnex
    Jul 7 '20 at 19:31

In an irreversible process, entropy is generated within the system, so the total entropy change in each step is the sum of the entropy exchange with the surroundings at $T_i$ plus the (positive) entropy generated $\sigma_i$: $$\Delta S_i=\frac{Q_i}{T_i}+\sigma_i$$. So, if we add up the entropy changes for the entire cycle, we obtain: $$\Delta S=\sum{\Delta S_i}=0=\sum{\frac{Q_i}{T_i}}+\sum{\sigma_i}$$ But, $\sum{\sigma_i}\gt0$. Therefore,$$\sum{\frac{Q_i}{T_i}}\lt0$$


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