How to derive the Clausius Inequality?

Question

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?

Answer 1

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 $\gamma_{rev}$.

Answer 2

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.

Answer 3

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.

Answer 4

I'll throw one more answer into the mix here. I enjoyed reading the others. This is from Enrico Fermi's 1936 book, "Thermodynamics".

Chapter IV specifically aims to produce Clausius' Theorem, and I'll give my understanding and summary of it here:

The Carnot cycle is a useful system to study because it provides us the basic relation between the heat added, $Q_H$, removed, $Q_C$, and the temperatures during those isothermal transitions, $T_C$ and $T_H$.

$$\frac{Q_H}{T_H}=-\frac{Q_C}{T_C} \tag{1}$$

Fermi creates a thought experiment involving $n$ Carnot engines to derive Clausius' Inequality.

He also makes an interesting observation which is often overlooked: In the Carnot cycle we implicitly assigned a negative number to $Q_C$, indicating that it is really "negative" heat added to the engine. Although this derivation of the Carnot cycle is the most natural, the negative sign causes some issues in Fermi's argument. We'll come back to that.

The basic layout of the thought experiment is that each of the Carnot engines operate between some temperature, $T_i$, and a common system, $T_0$. They are all in thermal contact with the common system and have the capacity to perfectly isolate from it, as Carnot engines do.

We can easily rewrite the equation for the first Carnot engine as:

$$\frac{Q_{1,0}}{T_0}=\frac{Q_1}{T_1} \tag{2}$$

Notice the signs are now a bit different (actually this matches what OP wrote). The notation $Q_{1,0}$ is the net heat flow from system 0 (the common system) to system 1 (a positive number), and $Q_1$ is now the heat flow from system 1 to system 0 (also positive). All $Q$'s are positive numbers now and we only need to keep track of the flow direction.

For any particular Carnot engine, it receives the following heat from the common system:

$$Q_{i,0}=T_0\frac{Q_i}{T_i}$$

Imagine we go through a complete cycle of the whole system, meaning each of the Carnot engines complete their cycle.

We can write an expression for the total amount of heat that the common system transferred to the Carnot engines:

$$Q_0=\sum_i Q_{i,0}=T_0\sum_i \frac{Q_i}{T_i} \tag{3}$$

If there is a net positive heat transfer from the common system to the Carnot engines, i.e., $(3)$ is positive, then we have a violation of Kelvin's Postulate. In other words, the common system can't lose heat at the end of a full cycle because it is at a constant temperature, $T_0$, throughout, and all Carnot cycles return to their initial $P,V$ state.

Since it is only $Q_0 > 0$ that creates a contradiction, then $Q_0\leq0$ is what remains. Extend this to a continuous distribution of sources and you get Clausius' Inequality (ignore the degenerate case of $T_0=0$).

Also try to keep in mind this is just a demonstration of Clausius' Experiment within the framework of Carnot Engines. We know that the inequality is much more generally true.

Edit: in my above representation of the Carnot cycle, I started with $Q_C$ being a negative number. Of course, this is an arbitrary choice, but it is consistent with the fact that when making an isothermal transition where heat is removed from the system, the volume decreases and therefore $Q_C=NkT_C\ln\frac{V_2}{V_1} < 0$.