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. enter image description here

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.


3 Answers 3


Considering the result $$\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}<S(B)-S(A)$$ 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.

  • $\begingroup$ So $\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}\not= {\Delta} S$ for irreversible processes? $\endgroup$
    – Osmium
    Commented Oct 8, 2021 at 9:44
  • $\begingroup$ Yes, that's right $\endgroup$ Commented Oct 8, 2021 at 9:58
  • $\begingroup$ 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? $\endgroup$
    – Osmium
    Commented Oct 8, 2021 at 10:03
  • 1
    $\begingroup$ 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 $\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}\neq {\Delta} S$. $\endgroup$ Commented Oct 8, 2021 at 10:12
  • $\begingroup$ Thank you for answering my queries. Good day. $\endgroup$
    – Osmium
    Commented 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: $$\Delta S\geq \int{\frac{dQ_{irrev}}{T_B}}$$

  • $\begingroup$ Thank you for your answer and your very precious time. $\endgroup$
    – Osmium
    Commented Oct 8, 2021 at 13:28
  • 1
    $\begingroup$ 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. $\endgroup$
    – michael b
    Commented 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,


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


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

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


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