# What's the variation of entropy of an irreversible process?

I'm studying entropy, in particular the increase of entropy principle, and I have a doubt about the entropy variation of an irreversible transformation.

If I have two thermodynamic states, A and B, and I go from A to B through an irreversible transformation (1) or a reversible transformation (2), aren't the entropy variations equal for (1) and (2)? I assume this because entropy variation only depends on the initial and final states, am I right?

So why, if I'm considering a transformation in an isolated system, the variation of entropy for a reversible transformation is $$\Delta S=0$$, but for an irreversible one, it is $$\Delta S>0$$ ? Shouldn't they be the same?

Also, if I consider Clausius integrals of this transformation from A to B, passing from (1) or (2), I obtain:

$$\Delta S=\int \left(\frac {dQ}T\right)_2 > \int \left(\frac {dQ}T\right)_1$$

And the integral calculated along (1) doesn't correspond with $$\Delta S$$ from A to B along (1), right? $$\Delta S$$ of an irreversible transformation should always be calculated along a reversible transformation from A to B.

So overall, why is $$\Delta S>0$$ for an isolated and irreversible transformation?

I hope you can understand my doubt, I'm not studying these topics in English, and it was really difficult to express my question.

• To get to the same final state using a reversible path, the system can not be treated as isolated. You will have to allow heat transfer with the surroundings in the case of the alternate reversible path. Commented Apr 21, 2022 at 13:02
• @ChetMiller thank you so much, now I can finally understand
– Fede
Commented Apr 21, 2022 at 13:39

aren't the entropy variations equal for (1) and (2)? I assume this because entropy variation only depends on the initial and final states, am I right?

Yes you are correct.

So why, if I'm considering a transformation in an isolated system, the variation of entropy for a reversible transformation is $$\Delta S=0$$, but for an irreversible one, it is $$\Delta S>0$$ ? Shouldn't they be the same?

The change in entropy between the two equilibrium states for any process is defined for a reversible transfer of heat by the equation

$$\Delta S_{AB}= \int_A^B\frac{dQ_{rev}}{T}$$

Where $$T$$ is the equilibrium temperature of the system. This applies regardless of whether the actual process is reversible and whether or not there actually is heat transfer occurring.

But there are two potential components of the variation in entropy: (1) Entropy transfer across the boundary between the system and surroundings and (2) entropy generation as the result of the process being irreversible. So

$$\Delta S_{AB}= \int_A^B\frac{dQ}{T_B}+\sigma_{AB}$$

The first term on the right is the entropy transfer at the boundary where $$T_B$$ in the temperature at the boundary between the system and the surroundings where heat $$Q$$ crosses and $$\sigma_{AB}$$ is the entropy generated if the process is irreversible.

So combining the two equations we have

$$\Delta S_{AB}=\int_A^B\frac{dQ_{rev}}{T}=\int_A^B\frac{dQ}{T_B}+\sigma_{AB}$$

Now, for an isolated system which by definition involves no heat or work transfer, $$dQ/T_{B}=0$$. If the process is reversible, then $$\sigma_{AB}=0$$ and we are left with

$$\Delta S_{AB}=\int_A^B\frac{dQ_{rev}}{T}=0$$

On the other hand, if the process is irreversible $$\sigma_{AB}\ne0$$ and we have

$$\Delta S_{AB}=\int_A^B\frac{dQ_{rev}}{T}=\sigma_{AB}\gt0$$

Which tells us we can calculate the entropy generated during an irreversible process by assuming any convenient reversible process connecting states A and B that satisfies the first law and which need not bear any resemblance to the actual process. A typical example is the free expansion of an ideal gas in an isolated system.

Hope this helps.