In thermodynamics, how can $\oint \frac{dQ}{T}$ make sense for an irreversible process?

Question

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?

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.