# Irreversible processes don't account in entropy calculations?

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: 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

• Cannot cycle expansion stage is reversible. Jun 11 '17 at 15:38

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.
• 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. Jun 11 '17 at 17:42
• @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. Aug 15 '19 at 17:01