How is entropy a state variable, and how can we measure entropy from irreversible processes with reversible ones, from a Khan Academy video

My questions :

1. How is entropy a state variable?

2. Why can we use a reversible process to measure an irreversible process's change in entropy if irreversible processes generate extra (unaccounted for in reversible process) thermal energy?

I try to go in detail about them below. I also have a third question, I don't think I worded it correctly but I don't have it clearly grasped. It's at the bottom

For question (1) :

In this video https://www.khanacademy.org/science/physics/thermodynamics/laws-of-thermodynamics/v/proof-s-or-entropy-is-a-valid-state-variable, Sal derived entropy for an isotherm as the integral of a $$PV$$ diagram from the beginning point of the isotherm to the end of the isotherm, as in an isotherm $$\int Pdv = W = Q$$, and after that calculation Sal divided by the temperature of whatever gave that heat. So that makes $$Q/T$$ = change in entropy.

However, this means that the path taken to get from point $$A$$ to point $$B$$ matters, as if the path changes, the integral value also changes, which means the heat ($$Q$$) in entropy changes. You can imagine a linear path from point $$A$$ to $$B$$, having an integral value of some number, but if the path from point $$A$$ curves up then back down to point $$B$$, then the integral definitely changes, and thus does entropy.

So since from this perspective change in entropy does depend on the path on the $$PV$$ diagram, how is entropy a state variable?

For Question (2) :

From the video right after, https://www.khanacademy.org/science/physics/thermodynamics/laws-of-thermodynamics/v/thermodynamic-entropy-definition-clarification Sal gives a clarification to the question about entropy as a state variable, that the derivation used above is only applicable for reversible processes. He shows that in an irreversible process, thermal energy would be generated.

At 13:28, he says :

So in an irreversible system, [change in entropy] wouldn't be a valid state variable.

He later says that entropy change in an irreversible process can be measured as though the path taken was a reversible process.

This seems kind of contradictory, as the thermal energy generated from an irreversible process is not accounted for in a path of reversible process. So even if entropy was a state variable, do we not care about this extra heat?

For my question I have no idea the answer but also no idea the question, I have heard you couldn't accurately tell entropy change in an irreversible process, all you could do was bound it from a reversible process. Of course, net change in entropy of system and surroundings would stay $$0$$ as per reversible process, but maybe for just system or just surrounding you could bound the entropy change? I really haven't researched this, if you can please just tell me right or wrong.

• @Andrew Oh ok I see. thanks i'm gonna delete my comments now Apr 10, 2022 at 23:20
• Just to clarify, he did not say that entropy is not a state variable for an irreversible process, which would be incorrect. He pointed at the expression $Q/T$ to indicate that $Q$ is not a state variable, and can't be used to calculate what happens in an irreversible process. Apr 10, 2022 at 23:23
• Irreversible processes are tricky because usually you can't directly compute what happens during the process. So instead, you use a clever but indirect argument. At the start of the irreversible process the system is in the state $S_1$, and at the end it is in the state $S_2$. Since state variables only depend on the state and not the path, you can compute how any state variables change between $S_1$ and $S_2$ by using any path you want that connects these states. In particular, you can use a reversible path, for which we can calculate what happens. Apr 10, 2022 at 23:28
• @Andrew Thanks for the clarification Apr 10, 2022 at 23:34
• No problem, sorry the original comment was more hostile. Apr 10, 2022 at 23:34

(1) There is only one isotherm that connects any two points $$A$$ and $$B$$ on a state diagram (otherwise, a given state could have two values of the temperature). The argument in the video is only valid for isotherms. In your question, you essentially want to argue that if you change the path from the one in the video, and apply the logic in the video, you will calculate a different $$\Delta S$$ between $$A$$ and $$B$$. However, you cannot change the path while keeping the path an isotherm. If you do change the path, you have to generalize the argument in the video to handle paths which are not non-isotherms. In the end, if you do things carefully, you will find that entropy is a valid state variable, and its value does not depend on the path.

(2) Irreversible processes are tricky because usually you can't directly compute what happens during the process. So instead, you use a clever but indirect argument. At the start of the irreversible process the system is in the state 𝑆1, and at the end it is in the state 𝑆2. Since state variables only depend on the state and not the path, you can compute how any state variables change between 𝑆1 and 𝑆2 by using any path you want that connects these states. In particular, you can use a reversible path, for which we can calculate what happens.

I have heard you couldn't accurately tell entropy change in an irreversible process, all you could do was bound it from a reversible process.

Yes, this is correct.

• Just a further question, is there a proof that entropy is a state variable like the one for isotherms, but generalize able for all cases? Or did physicists just declare entropy as some concept that happens to be a state variable and without a formula (irreversible), to go according to 2nd law of thermodynamics? Apr 10, 2022 at 23:45
• @dummy097 Of course there is a proof :-) That is something that should get covered in a thermodynamics course. The classic argument uses Carnot engines. Here is one online textbook with an argument: chem.libretexts.org/Bookshelves/… Apr 10, 2022 at 23:51
• Here's another argument: theochem.ru.nl/~pwormer/Knowino/knowino.org/wiki/… Apr 10, 2022 at 23:58
• One thing to keep in mind, is that if you actually found a flaw in the foundations of thermodynamics, not only would you become very famous academically, but you could probably make a lot of money. Apr 11, 2022 at 0:03
• It will be in my mind all day. And thanks a lot for the information :D Apr 11, 2022 at 0:07