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.

  • Reversible cycle: $$\quad \frac{Q_H}{T_H}-\frac{Q_L}{T_L}=0$$

  • 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?


3 Answers 3


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 $\gamma_{rev}$.

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    $\begingroup$ Thank you! I haven't taken or used multivariable calculus in 3 years haha, so I'm going to have to go back to it to try and understand this...I'll keep you updated if I have any questions! $\endgroup$ Nov 19, 2023 at 2:57

Personally, I find the Clausius inequality easiest to think of without considering cycles at all. Thermodynamic processes obey a succinct relationship between the reversible and irreversible work done on or by a system, \begin{gather*} \delta w - \delta w_{rev} \geq 0 \end{gather*} This is just a way of mathematically stating that the most work that can be done in a thermodynamic process is for a reversible process. If we then write out the first law in its differential form, \begin{align*} dE &= \delta q + \delta w \\ &= \delta q_{rev} + \delta w_{rev} \end{align*} where in the latter we have noted that the first law holds for any process, reversible or irreversible. Then, we can say, \begin{align*} \delta w - \delta w_{rev} &= \delta q_{rev} - \delta q \\ \implies \delta q_{rev} - \delta q &\geq 0 \\ \implies \delta q_{rev} &\geq \delta q \\ \implies dS &\geq \frac{\delta q}{T} \end{align*} where in the second line we have used the work inequality and in the last line we have used the Clausius definition of entropy. This is the Clausius inequality you were seeking, where the equality holds only for a reversible process. There are profound implications on the possible efficiency of heat engines and other mechanical devices that stem from this inequality, which is essentially where you were trying to start from.


How do we get here, to the Clausius inequality for any cycle?

The Clausius inequality applies to any heat engine cycle, where $T$ is the temperature at the point of heat entry but which, unlike the Carnot Cycle, can vary over the course of the entire cycle. The inequality applies only to the heat engine and not the heat engine plus surroundings, where the total change in entropy for an irreversible cycle is greater than zero.

In order for the total change in entropy of any heat engine cycle to be zero over the cycle, any entropy transferred to the heat engine, $S_{IN}$, plus any entropy generated by the heat engine due to irreversible processes, $S_{GEN}$, must be transferred to the environment, $S_{OUT}$, or

$S_{IN} + S_{GEN} = S_{OUT}$

Since, for an irreversible cycle, more entropy is leaving the heat engine ($S_{IN} + S_{GEN}$), than is transferred into the heat engine ($S_{IN}$), the difference being the entropy generated by the irreversible process(es), the change in entropy of the heat engine over the cycle is

$$\oint_{engine} \frac{\delta Q}{T}\lt 0$$

Hope this helps.

  • $\begingroup$ Thank you! I'm in the process of understanding entropy, where understanding the Clausius inequality is a step towards that...so seeing entropy in the explanation confuses me a little. I think Clausius didn't understand entropy until after developing inequality as well, right? $\endgroup$ Nov 19, 2023 at 2:55
  • $\begingroup$ @joshuaronis Per Wikipedia article it appears it was the same time. He coined the term entropy to describe the quantity Q/T. But you don't need to know about entropy to understand the theorem. I will update my answer to show you why. $\endgroup$
    – Bob D
    Nov 19, 2023 at 13:59

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