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

Question

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?

Answer 1

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.

Answer 2

$$\Delta S \ge Q/T$$ 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.

Answer 3

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).