# Entropy for irreversible processes

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.

• 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). – Wood Nov 20 '16 at 5:39
• Also, your question is good, but the title should be more specific. – Wood Nov 20 '16 at 5:41
• 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. – user36790 Nov 20 '16 at 10:00

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 $\left(\int_A^B \delta Q/T\right)_R$.

• Very much the same thing I wanted to write; +1 for invoking the statement of Clausius' Inequality. – user36790 Nov 20 '16 at 10:01
• @MAFIA36790 Fact: 90% of the questions about (thermodynamic) entropy can be solved through a correct application of Clausius' Inequality :-) – valerio Nov 20 '16 at 10:07
• 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. – user36790 Nov 20 '16 at 10:10
• 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. – Chet Miller Nov 20 '16 at 23:39
• 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. – user36790 Nov 21 '16 at 14:39