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Pick a Carnot Cycle (being $T_1<T_2$), it is reversible, therefore $\Delta S_{univ, cycle}=0$.

The same result is obtained via the sum of all entropies associated with its transformation, which means: $\Delta S_{univ, cycle} = \Delta S_{gas+ambient,AB} + \Delta S_{gas+ambient,BC}+\Delta S_{gas+ambient,CD} +\Delta S_{gas+ambient,DA}$

Make the first adiabatic expansion irreversible, like in the picture: enter image description here

The same equation applies, however, since entropy is a state function, $S_{gas, cycle}=0$, $\Delta S_{univ, cycle} = \Delta S_{ambient,AB} + \Delta S_{ambient,BC}+\Delta S_{ambient,CD} +\Delta S_{ambient,DA}$

The adiabatic processes, BC and DA don't account for a change in entropy, as $Q_{exchanged} =0$, so $\Delta S_{amb,BC} = \int\limits_{B}^{C} \frac {dQ} T$ goes to $0$, same for DA

Therefore the total entropy gets to $\Delta S_{univ, cycle} = \Delta S_{ambient,AB} + \Delta S_{ambient,CD} = \Delta S_{univ, irreversible processes} = S_{gas+ambient,BC} = S_{gas,BC} $

How can the entropy also become not dependent on the irreversible process? Since it only depends on the isothermal transformations, how can different "degrees of irreversibility in BC" not affect the net entropy change in the universe?

PS: does this relate to the fact that entropy is defined as function of state of any reversible process? It always comes up like magic in my calculations and I can't explain why.

EDIT: Possibly a duplicate of this, but I'd actually prefer to know how this relates to cycle-wide calculations

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  • $\begingroup$ Cannot cycle expansion stage is reversible. $\endgroup$
    – user115350
    Commented Jun 11, 2017 at 15:38

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If the Carnot cycle is irreversible, the entropy change of the system is still zero (per cycle), but the entropy change of the reservoirs is not zero, and the entropy change of the universe is not zero. The entropy that is generated within the system due to irreversibility is transferred to the reservoirs during the "isothermal portions" of cycle.

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  • $\begingroup$ I don't think it is true that "the entropy that is generated within the system due to irreversibility is transferred to the reservoirs during the isothermal portions of cycle". As is pointed out in the question, entropy is a state function and so if the irreversible portion has the same start and end points as in the reversible path, the system must have the same entropy at these points. The irreversible path simply generates entropy in the surroundings. The statement that reversible changes generate no entropy is a statement about the entropy of the universe not the system. $\endgroup$
    – By Symmetry
    Commented Jun 11, 2017 at 16:38
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    $\begingroup$ @BySymmetry: I guess we have a disagreement between experts. I stand by what I said. If the process that the system experiences is irreversible, and the system passes through a cycle, then the change in entropy of the system over the cycle must be zero. The only way this can happen is if the entropy generated within the system during the cycle is transferred to the surroundings. The entropy generation is not taking place within the surroundings, since the surroundings are considered to be comprised of ideal reservoirs (featuring no entropy generation) $\endgroup$
    – Chet Miller
    Commented Jun 12, 2017 at 2:43
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The equation $$ \mathrm{d}S = \frac{\mathrm{d}Q_\mathrm{rev}}{T} $$ Is valid only along reversible paths. Along an irreversible path, therefore, the fact that no heat is transferred does not tell you that there was no change in entropy. This is made more explicit by Clausius' inequality $$ \mathrm{d}S \ge \frac{\mathrm{d}Q}{T} $$ which relates the change in entropy to the heat transferred along an arbitrary path.

In short the entropy increase does occur on the irreversible step and is not somehow shifted into the unchanged isothermal steps.

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  • $\begingroup$ Ok, but I really can't get my self around the fact that, I can actually calculate it using any (reversible or irreversible) transformation between my coordinates. In this case, I can get the deltaS from B to C using a reversible path, or even "cheating" (as above) by using the other, reversible, transformations in the cycle $\endgroup$
    – Massimo Pesavento
    Commented Jun 11, 2017 at 17:30
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    $\begingroup$ The point is that you say as $\Delta Q_{BC} = 0$, this implies $\Delta S_{amb,BC} = 0$. But this is not true $\Delta S_{amb,BC} \ge \int\frac{\mathrm{d}Q}{T} =0$, so the entropy of the surroundings is increasing, in accordance with the second law. This means that the surroundings are not, returning to their initial state after an irreversible cycle, i.e. the fact the the system is performing a cycle does imply that the surroundings are as well. The assumption is, however, that the surroundings are so large that any finite change in their state can be neglected. $\endgroup$
    – By Symmetry
    Commented Jun 11, 2017 at 17:42
  • $\begingroup$ @MassimoPesavento Here is how to think about entropy: Two systems, $A$ and $B$ interact by undergoing a process. The fact that entropy is a state function means that we can calculate the entropy change in each system from the knowledge of their end states, i.e., we can calculate $\Delta S_{AA'}$ and $\Delta S_{BB'}$. If the process is reversible, then $\Delta S_{AA'}+\Delta S_{BB'} = 0$; if it isn't, then $\Delta S_{AA'}+\Delta S_{BB'} > 0$. To put it differently, just because entropy is state function does not mean that it should also be conserved. $\endgroup$
    – Themis
    Commented Aug 15, 2019 at 17:01

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