Entropy change in a reversible and Irreversible path

Question

Let's consider 2 cases. First where a system is taken from state 1 to 2 in a reversible path. Second where the same system is taken from state 1 to 2 in an irreversible path. Can we say that Entropy change in both the cases will be same. This question is more like an extended version of this question here

Answer 1

Yes! That is what you exactly mean by a state function. The value of a state function does not depend on the path and process, it only depends on the state of the system which is completely defined by $n$ (number of moles), $P$ (Pressure), $V$ (Volume) and $T$ (Temperature). Thus the entropy change will remain the same regardless of the type of process.

Answer 2

As @FakeMod has already pointed out, the difference in the entropy of a system between two equilibrium states is the same for all paths, reversible or irreversible, since it is state function. I could be wrong, but I believe the reason why you are asking the question again here is you may be misunderstanding the comments and answers you got in connection with the link you supplied.

In the link the comment was made by @Chet Miller that the following equation

$$\Delta S_{sys}=\int\frac{dQ_{rev}}{T}$$

"is valid to obtain $\Delta S$ only for reversible paths. So it is not independent of path, since it cannot be used for irreversible paths. And, for reversible paths, the integral is the same for all of them, irrespective of the reversible path chosen".

This statement is not the same as saying the difference in entropy between two states is different for a reversible process and an irreversible process, providing they connect the same two states. In fact, if you wish to determine the change in entropy for an irreversible process between two equilibrium states you can use the integral by assuming any convenient reversible process between the two states and then apply the integral.

The methodology is laid out very nicely in @Chet Miller "Grandpa Chet's Entropy Recipe" Here's a link to the recipe: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

I'll quote an excerpt from it that I found particularly helpful in understanding how to use the integral. But I strongly recommend you check out the entire recipe.

Devise a reversible path between the same two thermodynamic equilibrium states (end points). This reversible path does not have to bear any resemblance whatsoever to the actual irreversible process path. For example, even if the actual irreversible process is adiabatic, the reversible path you devise does not have to be adiabatic...try to devise a path that is simple to work with (i.e., for which it is easy to apply step 4). For the selected reversible path, evaluate the integral of dq/T from the initial state to the final state, where dq is the incremental amount of heat added to the system along the sequence of changes comprising the reversible path. This will be your change of entropy S.

Hope Chet's recipe and this helps.

Answer 3

To complement the other answers: the change in entropy of the system will be the same in both cases, but that of the environment will generally not.

If you split $\Delta S_\mathrm{tot} = \Delta S_\mathrm{sys} + \Delta S_\mathrm{env}$, then the second law says $\Delta S_\mathrm{tot} \geq 0$, with $=$ for reversible process and $>$ for irreversible processes. As $\Delta S_\mathrm{sys} = S_2 - S_1$ is the same in both cases, you can infer:

• Reversible: $\Delta S_\mathrm{env,rev} = -\Delta S_\mathrm{sys}$

• Irreversible: $\Delta S_\mathrm{env,irrev} > -\Delta S_\mathrm{sys}$