# Clausius inequality leading to absurd result

Background: After deriving Clausius inequality, the author of this book derives the following relation:

Consider the cycle shown in the figure in which leg $$A \rightarrow B$$ is irreversible. In the equation $$0>\oint\frac{\mathrm{d}Q}{T}=\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d}Q}{T}+ \int_{B \operatorname{rev}}^{A} \frac{\mathrm{d}Q}{T}$$ the second term on the right-hand side of this equation is given by $$S(A)-S(B)$$ because it is taken over a reversible path. When we move this quantity to the left-hand side, we find that $$S(B)-S(A)>\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}.$$ Thus the difference in entropy between the points is greater than the integral of $$\mathrm{d} Q / T$$ over an irreversible change. Problem: Entropy is a state function so $$\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}= {\Delta} S.$$ By the inequality derived we have $${\Delta} S>{\Delta} S$$ which is absurd.

Considering the result $$\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T} for an infinitesimal path, we get $$\mathrm{d}S\ge\frac{\mathrm{d}Q}{T}$$ where the equality holds only for a reversible process (by the definition of entropy).

This means that in your expression $$\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T},$$ $$\frac{\mathrm{d} Q}{T}$$ is not equal to $$\mathrm{d}S$$ because the process is irreversible. Instead, you have $$\frac{\mathrm{d} Q}{T}<\mathrm{d}S$$ and so $$\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}<\int_{A \operatorname{irrev}}^{B} \mathrm{d}S=S(B)-S(A)$$ which is the original result.

• So $\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}\not= {\Delta} S$ for irreversible processes? Oct 8, 2021 at 9:44
• Yes, that's right Oct 8, 2021 at 9:58
• But on the next page, the author writes that "we used reversible paths to calculate the change in entropy. But because entropy is a state function, the same result obtains for any transformation, reversible or irreversible between the same state". So isn't there any contradiction here? Oct 8, 2021 at 10:03
• There is no contradiction. The fact that entropy is a state function means that $\int_{A \operatorname{irrev}}^{B} \mathrm{d}S=\int_{A \operatorname{rev}}^{B} \mathrm{d}S=S(B)-S(A)$ which I used in my answer. The point is that $\frac{\mathrm{d} Q}{T}\neq\mathrm{d}S$ for an irreversible process so $\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}\neq {\Delta} S$. Oct 8, 2021 at 10:12
• Thank you for answering my queries. Good day. Oct 8, 2021 at 13:30

For the irreversible path between the same two end states, dQ is different than dQ for the reversible path, and in the integral of dQ/T for the irreversible path, you are supposed to use the temperature at the boundary interface between the system and surroundings $$T_B$$. So for the irreversible path, you should be using $$\int{\frac{dQ_{irrev}}{T_B}}$$So the two integrals are nothing like one-another. The correct form of the inequality should read: $$\Delta S\geq \int{\frac{dQ_{irrev}}{T_B}}$$

• Chet, I like this notation because it clearly shows that for the irreversible path we cannot equate the ratio of heat absorbed/surrendered to the isothermal temperature to the increment of entropy. Nice. Oct 8, 2021 at 19:16

I find the following from Enrico Fermi's book to be the most explicit derivation that shows:

$$dS \geq \frac{dQ}{T}$$

Looking at a closed loop integral of the ratio of heat absorbed (or surrendered, depending on sign) to the heat bath temperature along each isotherm in a cycle (reversible or not),

$$\oint \frac{dQ}{T}=\int_{A}^{B} \frac{dQ}{T}+\int_{B}^{A}\frac{dQ}{T} \leq0$$

We can take the forward part of the cycle $$(A\rightarrow B)$$ as an irreversible transformation, and the return part of the cycle $$(B\rightarrow A)$$ as a reversible transformation. It is valid to do this because even irreversible cycles behave the same along the forward part of the cycle. We are just saying that as a limiting case, our cycle behaves reversibly along the return path:

$$(TdS=dQ)_{B \rightarrow A}$$ Therefore,

$$\int_{B}^{A}dS=\int_{B}^{A}\frac{dQ}{T}=S(A)-S(B)$$

$$\int_{A}^{B}\frac{dQ}{T} + S(A)-S(B) = \int_{A}^{B}\frac{dQ}{T}-[S(B)-S(A)] \leq 0$$

Conversely,

$$S(B)-S(A) \geq \int_{A}^{B}\frac{dQ}{T}$$

$$dS \geq \frac{dQ}{T}$$