# Theoretical proof of Clausius inequality (Fermi)

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:

$$$$\tag{1}\sum\limits_{i=1}^n\frac{Q_i}{T_i}\leqslant0$$$$

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:

$$$$\tag{2}\sum\limits_{i=1}^n\frac{-Q_i}{T_i}\leqslant0\iff\sum\limits_{i=1}^n\frac{Q_i}{T_i}\geqslant0$$$$

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

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

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 $$$$\sum\limits_{i=1}^n\frac{Q_i}{T_i}<0$$$$ 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.

• Jul 7 '20 at 18:40
• I read the answer and I'm not sure of what it means. So can we have irreversibile processes where the equality sign holds? Jul 7 '20 at 19:02
• no, for an irreversible process it is "<" Jul 7 '20 at 19:04
• Ok, like Fermi said. So $$reversible\iff\sum\limits_{i=1}^n\frac{Q_i}{T_i}=0$$ Still, he just proved it is "=" for reversible cycle but not the opposite. In other words, he proved: $$reversible\implies\sum\limits_{i=1}^n\frac{Q_i}{T_i}=0$$ but he did not prove that $$\sum\limits_{i=1}^n\frac{Q_i}{T_i}=0\implies reversible$$ Jul 7 '20 at 19:14
• no he did not because as I said before it is a separate assumption 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,$$\sum{\frac{Q_i}{T_i}}\lt0$$