0
$\begingroup$

I'm having trouble trying to perform a general proof of Clausius' inequality. I am familiar with Fermi's (and other authors') proofs, where they imagine that the heat given/taken from the system S under consideration, in any of its infinitesimal steps, is provided by tiny Carnot engines/refrigerators (see e.g. this post: 2$^\text{nd}$ law of thermodynamics: equivalence of statements). enter image description here

However, these Carnot engines appear quite artificially in the proof. They are certainly not physically justifiable by any means, even if mathematically the proof is valid.

So I tried to generalize and imagine that these tiny engines/refrigerators are actually irreversible. Then one could physically argue that all heat given to the system comes from some cold reservoir somewhere, and does all kinds of hocus-pocus before part of it is transferred to S, and this would be represented by an irreversible cycle.

Problem is, I can't make it work. Here's what I've tried. If $C_i$ is a general irreversible machine, then $\delta Q_{i,0}+T_0\dfrac{\delta Q_i}{T_i}\leq 0$ (this comes from Carnot's theorem). But $\delta Q_i$, being the heat rejected by $C_i$, is negative, while equal (in modulus) to $\delta Q_i^S$, i.e. the heat absorbed by S in that infinitesimal part of its cycle. So $\delta Q_i^S = -\delta Q_i$. Likewise the heat released by the reservoir is $\delta Q_{i,0}^{res} = -\delta Q_{i,0}$, so one has $ \delta Q_{i,0}^{res} + T_0 \dfrac{\delta Q_i^S}{T_i} \geq 0$. Integrating,

$T_0\displaystyle\int \dfrac{\delta Q}{T} \geq -Q^{res}$.

From the 2nd Law, it can't be $Q^{res}>0$ otherwise we would have built a miraculous engine converting all heat into net work. So we have $Q^{res}\leq 0 \implies -Q^{res}\geq 0$ and therefore $T_0\displaystyle\int \dfrac{\delta Q}{T} \geq 0$, which is precisely the opposite of what we wanted to prove.

Any help in finding the flaw in the argument would be greatly appreciated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.