In my thermodynamics class, we've seen the Clausius inequality derived for a Carnot cycle, and then extended to any cycle. For the Carnot Cycle, we have that it's the most efficient possible cycle between two heat reservoirs, with an efficiency of 1 minus the ratio of the temperature of the cold reservoir to that of the hot one. Any other cycle between those two reservoirs must be less than or equal to the Carnot efficiency, with an efficiency of 1 minus the heat absorbed from the cold reservoir divided by the heat taken in from the hot one. Let $\eta_{\text {irr }}$ be the efficiency of any irreversible cycle and $\eta_{\text {rev }}$ be the efficiency of the Carnot cycle (as far as I understand, the only possible reversible cycle between two reservoirs):
$$\eta_{\text {rev }}=1-\frac{T_L}{T_H}$$
$$\eta_{\text {irr }}=1-\frac{Q_L}{Q_H}$$
$$1-\frac{Q_L}{Q_H}\leq 1-\frac{T_L}{T_H}$$
Rearranging our equation, a bit, we can derive the Clausius inequalities for a Carnot-type cycle.
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Reversible cycle: $$\quad \frac{Q_H}{T_H}-\frac{Q_L}{T_L}=0$$
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Irreversible cycle: $$\frac{Q_H}{T_H}-\frac{Q_L}{T_L}<0$$
Somehow, Clausius extended this to any cycle, claiming that:
- Reversible cycle: $$\oint \frac{\delta Q}{T}=0$$
- Irreversible cycle: $$\oint \frac{\delta Q}{T}<0$$
How do we derive that general version of the Clausius inequality that applies to any cycle?
Best Answer
The inequality for irreversible cycles follows from Carnot's theorem which states that no cyclic process can be more efficient than a Carnot cycle. The basic idea for deriving the equality for reversible processes is that the loop integral over an arbitrary reversible cycle can be expressed as the sum of loop integrals over Carnot cycles, which are all $0$.
We can get an expression for $\frac{\delta q}{T}$ in terms of exact differentials from the 1st law of thermodynamics: $$\mathrm dU = \delta q - p\mathrm dV\implies \frac{\delta q}{T} = \frac1T\mathrm dU + \frac pT \mathrm dV$$
The details of the expression don't matter, what's important is that it's in the form $F\mathrm dx + G\mathrm dy$ so that we can use Green's theorem to turn the integral over a cyclic process $\gamma_{rev}$ into an integral over the area it encloses in phase space (i.e. on a P-V diagram):
$$\oint_{\gamma_{rev}} \frac{\delta q}{T} = \oint_{\gamma_{rev}} (F\mathrm dx + G\mathrm dy) = \iint_{A_{rev}} \left(\frac{\partial G}{\partial x} - \frac{\partial F}{\partial y}\right)\mathrm dx\mathrm dy$$ Again the details don't matter and I'll denote $H = \frac{\partial G}{\partial x} - \frac{\partial F}{\partial y}$. The area being integrated over can be broken up into smaller areas and the total integral will be the sum of the integrals over the small areas: $$A = B \cup C \implies \iint_A H\mathrm dx\mathrm dy = \iint_{B}H\mathrm dx\mathrm dy + \iint_{C}H\mathrm dx\mathrm dy$$
The key to this derivation is that a Carnot cycle can be made arbitrarily small, and so we can express the enclosed area $A_{rev}$ as the combination of many tiny Carnot cycle areas $A_{rev} = A_{carnot1}\cup A_{carnot2}\cup ...$ so that the integral becomes $$\iint_{A_{rev}} H\mathrm dx\mathrm dy = \iint_{A_{carnot1}} H\mathrm dx\mathrm dy + \iint_{A_{carnot2}} H\mathrm dx\mathrm dy + ...$$ Then we can undo Green's theorem to get $$\oint_{\gamma_{rev}}\frac{\delta q}{T} = \oint_{\gamma_{carnot1}}\frac{\delta q}{T} + \oint_{\gamma_{carnot2}}\frac{\delta q}{T}$$ But we know that this integral over a Carnot cycle is $0$, so $$\oint_{\gamma_{rev}}\frac{\delta q}{T} = 0$$ for any reversible cyclic process $\gamma_{rev}$.