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When entropy is a state function and the entropy change for a reversible process is

$$\mathrm dS_\mathrm{rev} = \frac{\delta Q_\mathrm{rev}}{T} = 0,$$

how can an irreversible process be calculated by the same formula? $\delta Q_\mathrm{irr}$ is not a state function, therefore $\mathrm dS_\mathrm{irr}$ cannot be a state function.

Another description I have found was that the irreversible path should be seen as divided in a infinite amount of reversible steps. But an infinite times zero is also zero.

Can you help me please?

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    $\begingroup$ But the entropy change for a reversible path is 0. How can a reversible path be not zero if this is the definition? If State A has a entropy of 5 J/K and State B has a value of 20 J/K then we would need a entropy change of 15 J/K but a reversible heat transfer would result in 0 entropy change. $\endgroup$
    – Martin sssssssss
    Commented Jan 3, 2020 at 19:51
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    $\begingroup$ Entropy change of the universe is zero for the reversible process, but the entropy changes of both the system and surroundings can be nonzero, as long as equal magnitude and opposite in sign. $\endgroup$
    – Andrew
    Commented Jan 3, 2020 at 20:21
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    $\begingroup$ I took a liberty to improve formatting a little and tweaked the appearance of the symbols (please check whether the intended meaning is preserved). Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also, regarding "an infinite times zero is also zero", I wouldn't be so sure about that. $\endgroup$
    – andselisk ♦
    Commented Jan 3, 2020 at 23:57
  • $\begingroup$ Thank you, andselisk. I will use the codes for the next thread. $\endgroup$
    – Anna Dapont
    Commented Jan 5, 2020 at 10:33

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Entropy is indeed a state function, and thus depends only on the state of the system. Hence it doesn't matter how you get from state A to state B, the entropy change will be the same. The analogy would be that it doesn't matter which path you use to get from the base of a mountain to the summit, your elevation change will be the same. This is because altitude is a state function: your altitude depends only on how high you are (the state of your system), not how you got there.

But, since $dS= \frac{\text{đ}q_{rev}}{T}$, one can calculate the entropy change by integrating along a path connecting the two states, but that path must be reversible. I.e., it doesn't matter if the way the system changed from A to B was reversible or irreversible. $\Delta S$ will be the same. However, to calculate $\Delta S$ for that change from A to B, you need to integrate along some reversible path that connects A to B.

Also, it is possible to sum the effects of an infinite number of infinitely small steps. That's what happens when we do an integration. [More precisely, in an integration we calculate the sum in the limit as the step size goes to zero and the number of steps goes to infinity.] For a graphical representation of this, take a look at the graphics in Wikipedia's entry on Reimann sums (https://en.wikipedia.org/wiki/Riemann_sum), and imagine what happens in the limit as the bar width goes to zero.

The comments by Andrew, and Chet Miller, are both useful additions. Further expanding on Andrew's comment: The system's net change isn't affected by the path. This means that, to distinguish between the result of making a change in a system by a reversible vs. irreversible path, we need to look at differences in effects upon the surroundings.

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    $\begingroup$ Perhaps worth elaborating that the irreversible paths result in different amounts of entropy change of the surroundings even if the change to the system is the same as for the reversible path. $\endgroup$
    – Andrew
    Commented Jan 3, 2020 at 20:17
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    $\begingroup$ It might be worth mentioning that there are an infinite number of reversible paths between A and B, and all of them calculate exactly the same value for the entropy change. And if you integrate dq/T for an irreversible path, it will come out lower than for all reversible paths. $\endgroup$
    – Chet Miller
    Commented Jan 4, 2020 at 0:33
  • $\begingroup$ Could it also be that we look at the surroundings which has not changed but the state of the system has changed. Therefore we know that it was an irreversible process. To be sure,if an reversible isothermal expansion (against piston) increases the systems volume and therefore the entropy of the system. But because the entropy of the surroundingsa change for the exact same amount we have 0 entropy change for the universe and know that this was a reversible process. $\endgroup$
    – Anna Dapont
    Commented Jan 5, 2020 at 10:25
  • $\begingroup$ Attachment: the entropy of the surroundings DECREASED for the exact same amount. $\endgroup$
    – Anna Dapont
    Commented Jan 5, 2020 at 10:32
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    $\begingroup$ @AnnaDapont It sounds like you've got it right in your comments: In a reversible process, $\Delta S_{sys}$ is equal and opposite to $\Delta S_{surr}$ , so $\Delta S_{sys}+\Delta S_{surr}=\Delta S_{univ}=0$ . If we make the same change in the system, but instead do it irreversibly, $\Delta S_{sys}$ is still the same (because S is a state function), but $\Delta S_{surr}$ is more positive (or less negative) than in the reversible case, thus causing $\Delta S_{univ}>0$. All real processes are irreversible, and thus have $\Delta S_{univ}>0$. [NB: sys=system, surr=surroundings, univ=universe] $\endgroup$
    – theorist
    Commented Jan 5, 2020 at 23:53

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