According to the mathematical definition of entropy, it is only defined for reversible processes only. Then how can it be defined for irreversible processes? Please explain clearly.

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    $\begingroup$ The classical definition of entropy is only for variation of entropy ($\Delta S$), not absolute entropy. The $\Delta S$ of a system from one state to another is the integral of $dQ/T$ for a reversible process that would take the system from the initial state to the final one. The actual process that happened might not have been reversible, but you just pick a reversible one to calculate the $\Delta S$ (all choices will give the same result). $\endgroup$
    – Wood
    Nov 20, 2016 at 5:39
  • $\begingroup$ Also, your question is good, but the title should be more specific. $\endgroup$
    – Wood
    Nov 20, 2016 at 5:41
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    $\begingroup$ Check this Phys.SE answer I wrote for some related post; in a nutshell, entropy is a state function; it is defined both for reversible as well as irreversible processes. $\endgroup$
    – user36790
    Nov 20, 2016 at 10:00

1 Answer 1


In thermodynamics, from Clausius' theorem, the definition of entropy change is indeed

$$\Delta S_{A \to B} = S(B)-S(A) = \left(\int_A^B \frac{\delta Q} T\right)_R$$

where the subscript $R$ means that the integral must be evaluated along a reversible path.

It doesn't matter which reversible path we choose: the entropy difference between the states $A$ and $B$ will be $\left(\int_A^B \delta Q/T\right)_R$.

Therefore, $\Delta S_{A\to B}$ is path-independent: it only depends on the states $A$ and $B$. To evaluate it, we just have to choose any reversible path connecting $A$ and $B$. When the change of a quantity only depends on the initial and final states, we say that that quantity is a state function.

Now let's say that we perform an irreversible transformation from $A$ to $B$: what is the entropy change? Easy: it is $\left(\int_A^B \delta Q/T\right)_R$, where the integral is evaluated along any reversible path connecting $A$ and $B$.

Notice that it would be wrong to say that

$$\Delta S_{A \to B} = \left(\int_A^B \frac{\delta Q} T\right)_I \ \ \text{(wrong!)}$$

where $I$ is our irreversible path. In fact, the other part of Clausius' theorem tells us that

$$\Delta S_{A \to B} > \left(\int_A^B \frac{\delta Q} T\right)_I \Rightarrow \left(\int_A^B \frac{\delta Q} T\right)_R > \left(\int_A^B \frac{\delta Q} T\right)_I $$

so if you (erroneously) computed $\Delta S_{A \to B}$ as $\left(\int_A^B \delta Q/T\right)_I$ (along the irreversible path) you would be underestimating it.

Summing up: to compute the entropy change for a generic (reversible or irreversible) transformation from $A$ to $B$, choose any reversible path connecting $A$ to $B$ and compute the intrgral $\left(\int_A^B \delta Q/T\right)_R$.

  • $\begingroup$ Very much the same thing I wanted to write; +1 for invoking the statement of Clausius' Inequality. $\endgroup$
    – user36790
    Nov 20, 2016 at 10:01
  • $\begingroup$ @MAFIA36790 Fact: 90% of the questions about (thermodynamic) entropy can be solved through a correct application of Clausius' Inequality :-) $\endgroup$
    – valerio
    Nov 20, 2016 at 10:07
  • $\begingroup$ Very much indeed; most of my answers on entropy here are basically just the invocation of Clausius Inequality; like the post I linked above; good answer @valer. $\endgroup$
    – user36790
    Nov 20, 2016 at 10:10
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    $\begingroup$ I think that I would also be worth mentioning that, when applying the Clausius inequality for an irreversible case, the T you use in the integration is the temperature over the portion of the interface between the system and surroundings through which the heat is flowing. $\endgroup$ Nov 20, 2016 at 23:39
  • $\begingroup$ Indeed, @ChesterMiller; it is the very structure of Clausius' Inequality that the $T$ in the denominator is not that of system but that of the reservoir. When the process is reversible, then only we can switch the temperature of the system. $\endgroup$
    – user36790
    Nov 21, 2016 at 14:39

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