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I found out that if we want to find the entropy change in an irreversible process let's say $\text{state A} \to \text{state B}$ , we find or make an reversible path for the same process and find the entropy change for that irreversible process. The thing I don't get is that, even if entropy is a state function, does it mean that the change in entropy for the irreversible process we needed to find is the same as the change in entropy for the reversible path we constructed? Or am I being mistaken? If its the latter, please provide me a solution for it. Thanks.

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    $\begingroup$ That's exactly right. $\endgroup$ – Buck Thorn Oct 1 '20 at 18:11
  • $\begingroup$ @BuckThorn I actually can't make sense of it, like how can it be the same? even though its a state function, I can't get why is it true. Thanks for your reply, I hope that you can elaborate a bit on it. $\endgroup$ – FinalBOSS Oct 1 '20 at 18:12
  • $\begingroup$ You make a reversible path between the same two end states as the irreversible process. The entropy change for the irreversible process will be the same as for the reversible path. $\endgroup$ – Chet Miller Oct 1 '20 at 20:00
  • $\begingroup$ I really can't get as to why it's true, I know it's a state function, but I really cannot make sense of why the entropy would be same for both the irreversible and reversible process. $\endgroup$ – FinalBOSS Oct 2 '20 at 4:10
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Quantities such as heat and work can vary depending on the path (how you get from an initial to a final state). In contrast a defining aspect of a state function is that it is independent of the path.

Entropy is a state function, as a corollary therefore a path-independent property. The experimental determination of entropy involves measuring the heat transferred as a function of temperature during a reversible process and computing $\int \frac{dq}{T}$. That is another defining property of the entropy. It is not trivial to wrap your head around why it should be so, but entropy can be thought of as helping quantify a limiting or ideal property, in particular the maximum work that might be obtained from a process, or the least amount of work required to bring about a non-spontaneous change. It defines processes in which there are no losses (losses being consistent with irreversibility).

As an example, at constant T and p the Gibbs free energy change (which is related to the limiting work) is related to the entropy as follows:

$$-\frac{\Delta G}{T}=\Delta S_{system}+\Delta S_{surrounding}$$

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  • $\begingroup$ oh alright, I was having this idea, but the thing was that I was just wrapping my head to see why is it really true. Thank you for your answer, appreciate it! $\endgroup$ – FinalBOSS Oct 1 '20 at 19:04
  • $\begingroup$ but why do you relate entropy as being thought of as “maximum potential to do work” ? $\endgroup$ – FinalBOSS Oct 1 '20 at 19:06
  • $\begingroup$ Hopefully my edit clarifies one way. $\endgroup$ – Buck Thorn Oct 1 '20 at 19:10
  • $\begingroup$ yeah, thanks @Buck Thorn $\endgroup$ – FinalBOSS Oct 2 '20 at 4:09
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The answer above is perfectly fine. There is another way to see it. It is very important to realize what is a "system" and a "surrounding", what changes when we talk about Entropy and other state functions? when we talk about Entropy, we say "The total change in Entropy of Universe is never negative, it can be either zero or positive". We usually define a system and a surrounding but take surrounding in discussion only when we talk about entropy. when we are saying any quantity is state function, we mean it with respect to the system. So, both of them are not same. Take for example, any process which is performed by two ways:

  1. PV work (irreversibly), say, total change in entropy is "x".
  2. Electrochemically (reversibly), total change in entropy of system is "y" (since the process is reversible, realize, that total change will be zero) if someone asks you to find the change in entropy of surrounding in an irreversible process. Then, the way around this problem is that you simply use third law. whether the process is done irreversibly or reversibly, the del S system is same. So, "y" is del S (system), x has to be positive. x-y is the answer! You can find such problems as well as detailed concept in Klotz and Rosenberg's book on chemical thermodynamics.
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  • $\begingroup$ "since the process is reversible, realize, that total change will be zero" this is wrong.. it's only true for the full cycle for reversible process $\endgroup$ – Buraian Oct 3 '20 at 10:49
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Entropy change for the system between two points can be determined simply from plugging the final and initial states into the state equation of entropy. The fundamental idea is that entropy change only depends on the state of the system and the state of a system is described by the state variables.

However, the total entropy change (sys + surr) depends on how the irreversible process was. Suppose the total entropy change in an irreversible process for the whole universe was some quantity $ \Delta S_{net}$, and entropy change for the system along any reversible path is $ \Delta S_{sys}$ the entropy change in surroundings is:

$$ N = \Delta S_{net} - \Delta S_{sys}$$

Note: A reversible process is characterized by $N=0$, that is there is no entropy production and $N>0$ if irreversible


More on this extra entropy concept

For an example of this using physical situation, see the answer by Chet Miller here

For details on the terms in the formula, see the answer by Chet Miller here

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  • $\begingroup$ Your answer and Shivani's are admittedly very useful (perhaps more than mine). However for a reversible process we do say that N=0. $\endgroup$ – Buck Thorn Oct 3 '20 at 11:25
  • $\begingroup$ Good point I"ll add it in $\endgroup$ – Buraian Oct 3 '20 at 11:26

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