In thermodynamics, a reversible process can be identified by a curve in the $pV$ plane, that is, a sufficiently regular function $\gamma :[a,b]\rightarrow \mathbb{R^2}$.

A cycle, then, will simply be a closed curve, i.e., a curve such that $\gamma (a)=\gamma (b)$. It is well known that for such processes $$\oint_\gamma \frac{dQ}{T}=0$$ So far so good.

But let's instead consider an irreversible cycle; specifically, one that is neither quasi-static: there is, then, no corresponding closed curve we can associate to such a process, because the system may not have a well-defined temperature of pressure at all times. But it is equally well known that for this kind of cycle $$\oint \frac{dQ_{irr.}}{T}<0$$

Mathematically, the integral of a differential form must be calculated along some line (hence, "line integral"). It simply doesn't make sense to calculate $\int d\omega$ along no curve! But in the case of irreversible paths there is no corresponding curve along which we can integrate, therefore what does $\oint \frac{dQ_{irr.}}{T}<0$ even mean?

  • $\begingroup$ When you specify a thermodynamic process, you're specifying a path in $P-V-T$ space (or whatever your thermodynamic variables are). So you integrate along that path. Each path has a specific $\delta Q$ associated with it. $\endgroup$ Commented Aug 26, 2017 at 20:36
  • $\begingroup$ @JahanClaes well, not really. If the process is irreversible, or at least non-quasi-static, temperature or pressure may not be the same everywhere, hence your argument doesn't hold for such processes. $\endgroup$
    – Adrian
    Commented Aug 26, 2017 at 20:42
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    $\begingroup$ You're confusing irreversible and quasi-static processes. If I put ice in some hot water, the process is irreversible, but the water always has a well-defined temperature. This is the kind of process the Clausius inequality is about. $\endgroup$
    – knzhou
    Commented Aug 26, 2017 at 20:56
  • $\begingroup$ The Clausius inequality indeed assumes that the temperature of the external reservoir that acts as heat source/sink is well-defined at all times. $\endgroup$
    – Christoph
    Commented Aug 26, 2017 at 21:04
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    $\begingroup$ In the Clausius inequality, T is supposed to be taken as the temperature at the interface between the system and the surroundings where the heat transfer is occurring. Since the surroundings are usually assumed to be represented by an ideal infinite reservoir (or reservoirs), T in the Clausius inequality is taken as the temperature of the reservoir (or reservoirs). This being the case, the integral in your equation is well-defined, and your equation is satisfied. $\endgroup$ Commented Aug 26, 2017 at 21:30

1 Answer 1


But in the case of irreversible paths there is no corresponding curve along which we can integrate, therefore what does $\oint \frac{dQ_{irr.}}{T}<0$ even mean?

The Clausius inequality you ask about is more accurately written in this way:

$$ \oint \frac{dQ_{irr.}}{T_{reservoir}} < 0. $$

That is, the temperature in the denominator is actually temperature of the reservoir in thermal contact with the system, not necessarily equal to temperature of the system itself.

The integral does not refer to integration in space of thermodynamical states of the system. Instead, it refers to the following theoretical construction (the simplest one where the system exchanges heat with at most one reservoir at a time; it could be extended to more complicated situations but that is not necessary here).

1) It is assumed there is a physical process where the system studied can possibly get into non-equilibrium states, but at the end it ends up in the same equilibrium state it started with. Let us introduce real number $t$ that tracks states of the system, at the beginning $t=0$ and at the end $t=t_{max}$.

2) The system accepts and gives off energy by heat transfer only with bodies that have temperature defined at all times. Those bodies are called reservoirs.

3) Let $Q(t)$ be net heat accepted by the system in time interval $[0;t]$ and let $T_{reservoir}(t)$ be temperature of that reservoir that is in contact with the body at time $t$, at time $t$.

4) Then it can be shown, based on the second law of thermodynamics, that the sum of reduced heats over the whole cycle is less than or equal to zero:

$$ \int_0^{t_{max}} \frac{dQ/dt}{T_{reservoir}(t)}\,dt \leq 0. $$

Because the values of time variable $t$ and $t_{max}$ do not play much role in the derivation, it is customary to rewrite the integral using the loop integral symbol $\oint$, understanding that the integration is over the actual progression of (possibly non-equilibrium) states that begins and ends in the same equilibrium state.

The simplification of writing the integral over monotonously increasing real variable $t$ into a general form, free of any auxiliary variable, is quite reasonable. After all, the value of the integral can be often calculated even without knowledge of functions $Q(t), T(t)$, for example by splitting into several integrals and using substitution method.

However, the ubiquitous omission of the subscript "reservoir" from "T" in the integral as given in many study documents is, I think, a significant mistake. If this point is not stressed enough by the teacher, students, when studying from their notes, are bound to confuse $T$ with temperature of the system. In hindsight it seems better to always write the subscript making a difference between the system, and the reservoir.

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    $\begingroup$ In item #2, it is not necessary that the system be in contact with one reservoir at a time. The development can be extended to a surface integral over the boundary of the system in which the local heat flux divided by the local boundary (reservoir) temperature is integrated over the boundary and over time. $\endgroup$ Commented Aug 26, 2017 at 22:06
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    $\begingroup$ @ChesterMiller, you are right, of course, but I do not want to complicate the answer with this possibility. I have edited the answer to mention that it considers the simple case. $\endgroup$ Commented Aug 26, 2017 at 22:13
  • $\begingroup$ Thanks, this was actually the answer I was expecting. I wasn't troubled by the $T$, really, because my textbook makes it clear that it refers to the temperature of the reservoire and not of the system. My issue was rather with the path over which to integrate. So it all boils down to a neglected time dependence, apparently, and hence that is not a true line integral/integral of a differential form? Anyway, standard thermo textbooks don't explain things which such a careful approach, where did you study these topics?Could you suggest some references? $\endgroup$
    – Adrian
    Commented Aug 27, 2017 at 15:13
  • $\begingroup$ @Nicol, indeed, the integral in the Clausius inequality is primarily not a line integral in multidimensional space of thermodynamic states $\mathbf X$, but is just an ordinary onedimensional integral, a limit of a sum of reduced heats $\Delta Q_k/T_{reservoir,k}$.that get transferred in the course of the process that ends up in the same state it began with. $\endgroup$ Commented Aug 27, 2017 at 23:20
  • $\begingroup$ However, if the system is always in some (quasi) equilibrium state at all stages of the process, so it always has a representative point in the space of thermodynamic states, the integral may be rewrittable as line integral in the space of thermodynamic states: $\oint \frac{dU-dW}{T}$ = $\oint \sum_i \frac{\frac{\partial U}{\partial X_i}-F_i}{T(\mathbf X)}dX_i$ where $F_i$ is "force" in the expression $dW = \sum_i F_i dX_i$. $\endgroup$ Commented Aug 28, 2017 at 10:07

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