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If a system goes from state A to B in a reversible process, then $(\Delta S)_{\text{universe}}=0$. However, my confusion starts if we take an irreversible route. For the system, even if we take the irreversible route, we can always calculate it using the reversible route from A to B, so the system entropy is a state function. However, the surroundings entropy change is not negative of the system's entropy change in an irreversible process, meaning that $(\Delta S)_{\text{universe}}\neq 0$, despite the process still being between states of the system A and B. This suggests to me that $(\Delta S)_{\text{universe}}$ is not a state function but depends on the path (reversible vs irreversible). I was under the impression that entropy is a state function.

Is this reasoning correct?

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    $\begingroup$ Paths are reversible only in the limit. So your premise is incorrect. $\endgroup$
    – Karsten
    May 5, 2023 at 3:14
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    $\begingroup$ A system going through a closed irreversible path increases surrounding entropy, because returning the system to initial state is possible only by having surrounding in other than initial state, with higher entropy. $\endgroup$
    – Poutnik
    May 5, 2023 at 4:25

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TL;DR: entropy, for reversible and irreversible processes, for system, surroundings and universe, is always a state function. You are not accurately considering the initial and final states of surroundings (and universe) for reversible and irreversible processes.

The details

In system going from state $A$ to $B$ reversibly, say the state of the surroundings changes from $C$ to $D$. In system going from state $A$ to $B$ irreversibly, considering the same initial state of the surroundings, $C$, the final state will differ from $D$ (say $D'$)

Entropy change of system and surroundings for reversible and irreversible processes

For the reversible process,

$$ \begin{align} (\Delta S)_\text{uni} &= (\Delta S)_\text{sys} + (\Delta S)_\text{surr}\\ &= S_B-S_A + S_D-S_C \tag{1} \end{align} $$

For the irreversible process,

$$ \begin{align} (\Delta S)_\text{uni} &= (\Delta S)_\text{sys} + (\Delta S)_\text{surr}\\ &= S_B-S_A + S_{D'}-S_C \tag{2} \end{align} $$

From equations (1) and (2), the difference between the entropy change of the universe when the system undergoes an irreversible and a reversible process is given by $S_{D'}-S_D > 0$, where the inequality is a consequence of the second law of thermodynamics.

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There is no reversible route from A to B if those are the states of the universe.

If A and B are system states, and you take different paths, the surrounding state will be distinct for distinct paths. So in one case the entropy of the universe might increase by a certain amount, and in the other case it increases by a different amount.

However, if the entire universe is changed on a reversible path, there is nothing outside of the universe that can drive that. So either the universe is at equilibrium, with no more increase in entropy, or things are still happening and entropy is increasing.

Bottom line: there is no way to go reversible from a state A to a state B of the entire universe.

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  • $\begingroup$ The universe can undergo a reversible change. As pointed out by OP, when a system undergoes a reversible change, so do the surroundings and, hence, so does the universe. $\endgroup$
    – ananta
    May 5, 2023 at 3:14
  • $\begingroup$ @ananta what would be an example of such a reversible process? And how would you change that process to make it irreversible? $\endgroup$
    – Karsten
    May 5, 2023 at 3:17
  • $\begingroup$ A quasi-static expansion/compression of the system is one example of a reversible process. To make this process irreversible, carry out the expansion/compression within a finite time-period (fast). $\endgroup$
    – ananta
    May 5, 2023 at 3:20
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    $\begingroup$ @ananta I interpreted A and B as the state of the universe. Going from A to B is either reversible or not, there is no choice. $\endgroup$
    – Karsten
    May 5, 2023 at 10:55
  • $\begingroup$ @Karsten Hello Karsten. System can go from $A$ to $B$ reversibly, but the surroundings can't go from $C$ to $D$ reversibly. Am I misunderstanding you? States can be referred to the system and the surroundings, separately, but we cannot talk of step $A$ of the "universe" and step $B$ of the "universe". Do you agree? $\endgroup$ May 5, 2023 at 14:18
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A State Function has a value for a given state of a system. This means that if something undergoes a process and eventually returns to the exact same conditions as it started, its state functions U, G, H, S, PV will be exactly the same. The energy terms U, H, G are conserved so these will remain constant in the universe. S is not conserved and must increase for anything that happens. Since S has not changed in our system it must have increased in the universe. The trick is to expand our immediate universe to determine where and how the entropy increase happens.

This can be difficult but if reduced to extremes is very simple: Everything on Earth depends on Energy from the Sun, Directed energy of higher energy photons of LOW entropy. To maintain an energy steady state High entropy energy must be emitted. This is low energy infrared photons with higher entropy.

An example: A large Oak tree is measured to determine the energy etc. of everything down to the tilt of its leaves in the Sun and the rotational states of all its molecules. The tree is cut down, all its roots removed and used for lumber furniture and fuel. One of its acorns is planted in the exact same spot and with much TLC Grows to an exact clone of the tree down to the rotational states of the molecules. Its state functions are the same as the original tree! but the entropy increase of the universe needed to accomplish this is not only beyond all the supercomputers but beyond imagination. In the end one is content to simply have a new oak tree.

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  • $\begingroup$ The Gibbs energy is not conserved because its entropy component is not conserved. $\endgroup$
    – Karsten
    May 5, 2023 at 20:58

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