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The change in entropy of the Carnot and reversible cycles is said to be 0. Several other loops are supposed to have a non-negative change in entropy.

This presents 2 problems which I cannot reconcile. 1) Entropy is supposed to be a state function so shouldn't any loop on the PV plane bring the system to its initial state and thus its initial entropy? 2) Any loop on the PV plane is reversible given enough heat sources. Shouldn't all reversible cycles, not just the Carnot, have a zero change in entropy. Wouldn't this mean that any loop on the PV plane would have a zero change of entropy?

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    $\begingroup$ Entropy is a state function, but the total entropy change to get a system from one state to another is not unique, it depends on the way the system movers from one state to the other. Along the way, of course, something had to have happened to external temperature baths, and that could have happened in many possible ways. A Carnot process is, after all, a model for a periodic machine that moves heat from one temperature bath to another while either producing or using mechanical energy. It's only unique because it is optimal in the sense of entropy production. $\endgroup$
    – CuriousOne
    Commented Dec 28, 2014 at 18:22

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1) Entropy is supposed to be a state function so shouldn't any loop on the PV plane bring the system to its initial state and thus its initial entropy?

Pair of values $P,V$ may not be enough to specify the state of the system, there may be other thermodynamics state variables. If so, it is possible system returns to the same values $P,V$, but it is in a different thermodynamic state because the additional quantities have different value.

But when the pair of values $P,V$ is sufficient to describe thermodynamic state, like it is for a simple homogeneous system such as gas far from condensation, its entropy is function of $P,V$ only and the system indeed gets the same entropy after the values of $P,V$ are restored.

This is true also for any irreversible cycle that has a point where the state is thermodynamic equilibrium state with definite $P,V$. If the the system gets back to such state, it does not matter whether this happened reversibly or not - the entropy returns to the same value characteristic for the equilibrium state $P,V$.

The claim that entropy increases when irreversible cycle is performed means that the entropy of (system + its environment (heat reservoir)) increases. That way, the entropy of the system may return to its original value, while the total entropy still increases.

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1) Entropy is supposed to be a state function so shouldn't any loop on the PV plane bring the system to its initial state and thus its initial entropy?

Yes, for a reversible process. For an irreversible process where heat has been lost along the way you do not reach the exact same state (less internal energy).

2) Any loop on the PV plane is reversible given enough heat sources.

While that is true for a part of the system, you need to consider the full closed system to see the total entropy change. A part of a system can easily have entropy changes that are zero or negative.

When ice is melting in a water bath (assuming isolated system):

  • The entropy of the water is decreasing, $\Delta S_{water} < 0$.
  • The entropy of the ice block is increasing, $\Delta S_{ice} > 0$.
  • The whole system will all in all experience an increase in entropy, $\Delta S_{system} = \Delta S_{water} +\Delta S_{ice} >0 $.
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  • $\begingroup$ I do not quite understand the explanation for the first statement. If you have a loop on the PV plane, then the cycle must have a change in internal energy of zero as it is a state function. The internal energy should be the same after each cycle? $\endgroup$
    – curiousgeorge
    Commented Dec 28, 2014 at 19:17
  • $\begingroup$ @Abhi I see what you mean. Yes, you will reach the same state in the pV diagram and entropy change will be zero so to say. But if not all processes along the way are reversible, then you need to add heat (as fuel e.g.) from somewhere else to reach that same state after each cycle. Entropy for the whole system will not remain constant, only for this cycle which is just a part of the whole system $\endgroup$
    – Steeven
    Commented Dec 29, 2014 at 13:01
  • $\begingroup$ I understand the first statement, but the second is still slightly confusing. The carnot cycle is the only cycle which is reversible between 2 heat sources. When you have several heat sources shouldn't there be many other cycles with 0 entropy? And when you have a very large number of heat sources, any loop will technically be reversible (a transfer of energy can be approximated as a series of several isothermal processes). So any loop should technically have a zero net change of entropy? $\endgroup$
    – curiousgeorge
    Commented Dec 29, 2014 at 19:45
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1) Entropy is supposed to be a state function so shouldn't any loop on the PV plane bring the system to its initial state and thus its initial entropy?

It is, but this is the PV plane, it is a projection of the total state space, so every PV point can correspond to different entropies. A reversible cycle will be closed in the PV plane, but not all closed cycles PV cycles are reversible.

Shouldn't all reversible cycles, not just the Carnot, have a zero change in entropy. Wouldn't this mean that any loop on the PV plane would have a zero change of entropy?

It would if you were in the SV or SP plane, but not in the PV plane.

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