# Why does the added heat have to be reversible in the formula for calculating entropy change?

Here Scott Lawson tells us that in the formula below:

$$\Delta S=\frac{\Delta Q}{T}$$

the heat added, $$\Delta Q$$, must be reversibly added. Rewriting the formula we get:

$$\Delta S=\frac{\Delta Q_{rev}}{T}\tag{1}$$

Mr. Lawson is correct according to this answer and also Wikipedia. Now, let us consider another formula:

$$dS=\frac{dQ}{T}\tag{2}$$

Now, here, I don't think it matters if the heat is added reversibly or irreversibly because the heat added is infinitesimal. Due to the addition of this infinitesimal heat, the infinitesimal increase in entropy is $$dS$$. So, I think both $$dQ$$ and $$dQ_{rev}$$ indicate the same thing here and $$dS$$ in both cases will be the same. Am I correct?

• "I don't think it matters if the heat is added reversibly or irreversibly because the heat added is infinitesimal." I don't see why this would be the case. One can have an infinitesimal entropy increase from the heating and a supplemental infinitesimal entropy increase from the irreversible nature of the heating. Dec 3, 2021 at 18:37
• @Chemomechanics Woah! So, even if the same amount of infinitesimal heat is added, the amount of entropy increase will be greater for the irreversible addition of heat than the reversible addition of heat? Dec 3, 2021 at 18:43

So, I think both $$dQ$$ and $$dQ_{rev}$$ indicate the same thing here and $$dS$$ in both cases will be the same. Am I correct?

For the system it is the same, but for the surroundings it won't be if the path is irreversible.

Although the change in entropy of the system, be it an infinitesimal or finite change, is the same for all paths because entropy is a state function, in order to calculate what the change is you need to do it for a reversible path. Which means if the process is irreversible you need to assume any convenient reversible path and use

$$dS_{sys}=\frac{\delta Q_{rev}}{T}$$

to calculate the finite change between two equilibrium states

$$\Delta S_{sys}=\int_1^2\frac{\delta Q_{rev}}{T}$$

Hope this helps.

• So, if $Q$ amount of heat is added to the system, infinitesimal or finite, reversibly or irreversibly, the entropy increase will be the same for the system in both cases. However, the entropy change of the surroundings in the two cases will be different. Am I correct sir? Dec 3, 2021 at 19:19
• Yes, the change in entropy of the surroundings will be different in such a way that, for the irreversible path, the total change in entropy (system + surroundings) will be greater than zero, whereas it will be zero for the reversible path. As an exercise, do it for the irreversible constant pressure heat addition we previously discussed. Just assume some initial and final temperature and keep in mind that $T$ in the formula is the temperature at the boundary between the system and surroundings, which in this case would be the single fixed temperature of the thermal reservoir. Dec 3, 2021 at 19:54

$$\Delta S \ge Q/T$$ is the appropriate formula. The equal sign is only for reversible processes. Imagine a thermally isolated system, so $$Q=0.$$ There are a host of processes that can increase the entropy of the system. For example, say I have a thermally isolated chamber with a partition in the middle. All of the atoms are initially on one side of the partition. Then I remove the partition so the atoms can explore the rest of the chamber. The entropy clearly went up, but I added no additional heat. This was an irreversible process.

Consider this: The dQ's and the T's for a reversible path are not the same as the dQ's and the T's for an irreversible path between the same two thermodynamic equilibrium end states. For example, in an adiabatic irreversible process, all the dQ's are zero (so the integral of dQ/T is zero), but, for any reversible path between the same two end states, the dQ's cannot be zero, and the integral of dQ/T is greater than zero (and equal to the entropy change for both the adiabatic reversible path and the irreversible path).