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We have defined $$\mathrm dS=\frac{\delta Q_\text{reversible}}{T}$$ and the second law of thermodynamics (Clausius inequality) says $$\mathrm dS\geq\frac{\delta Q}{T}$$ (Here $\delta Q$ is just the differential of heat regardless of it being reversible or irreversible) and the equality holds for it being a reversible process (as defined).

Now, $$\mathrm dS\geq\frac{\delta Q}{T}$$ If I put $\mathrm dS$ on the left as $\frac{\delta Q_\text{rev}}{T}$, I will get $\frac{\delta Q_\text{rev}}{T}\geq\frac{\delta Q}{T}$, or $\delta Q_\text{rev}\geq\delta Q$.

This implies that in a reversible process, more heat transfer takes place than in an irreversible one. Why is this so?

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  • $\begingroup$ because the internal irreversibilities of the system generate excess heat and that dissipative effect is always positive, so when a cycle is completed the internally generated heat is part of the total heat exchange. $\endgroup$ – hyportnex Jul 4 '16 at 23:42
  • $\begingroup$ expanding irreversibly (spontaneously) into a vacuum does no work, as the resisting pressure is zero, so no heat absorbed from surroundings. Expanding reversibly against pressure $ p$ results in work $\int pdV$ so heat is absorbed. $\endgroup$ – porphyrin Jul 6 '16 at 16:21
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In the Clausius inequality, the temperature in the denominator is supposed to be the temperature at the boundary of the system where the heat transfer is taking place. This important point is often omitted from many thermodynamics textbooks. In a reversible process, the temperature at the boundary is essentially equal to the temperature of the system, because the system temperature is uniform throughout. But, in an irreversible process, the temperature of the system is not uniform, and the temperature at the boundary can differ substantially from even the average temperature of the system. So the Clausius inequality does not always require that $\delta Q_{rev}\geq \delta Q$.

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  • $\begingroup$ So is the first answer in chemistry.stackexchange.com/questions/19735/… wrong? $\endgroup$ – Rick Jul 4 '16 at 13:08
  • $\begingroup$ the order got changed, i meant the one with 7 upvotes $\endgroup$ – Rick Jul 4 '16 at 14:30
  • $\begingroup$ The answer you referred to does not make the important distinction regarding the T in the Clausius inequality being the value at the boundary of the system where heat transfer is occurring. Otherwise, it seems pretty OK. $\endgroup$ – Chet Miller Jul 4 '16 at 14:55
  • $\begingroup$ Well, there he is saying "Therefore δqrev−δq=δw−δwrev≥0" which would mean δqrev≥δq. Morever, it say work in a reversible process is lesser than in a irreversible process. What about isothermal irreversible compression compared to a reversible one $\endgroup$ – Rick Jul 4 '16 at 15:10
  • $\begingroup$ Suppose you are operating your process in a cycle where $\Delta U$ and $\Delta S$ are both zero. Both the reversible process and the irreversible process start and end at the same state. The irreversible process uses a great big cycle with large W and large Q, while the reversible process uses a tiny little cycle with small W and small Q. You certainly can't say that Q for the reversible cycle is greater than the Q for the irreversible cycle. However, for the irreversible process cycle, the integral of $\delta Q/T$ at the heat transfer boundary(ies) is less than zero. $\endgroup$ – Chet Miller Jul 4 '16 at 22:21
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I try to explain as follows. To illustrate using the Clausius inequality consider a cyclic process. In the first step 1 to 2 the system is isolated and undergoes a spontaneous irreversible change and in the second step 2 to back to 1 the system interacts with the environment and returns by any reversible path. Because S is a state function $\Delta S = 0$ for this whole process. Thus $$ \Delta S = 0 > \int _1^2 \frac{\delta q_{irr}}T + \int _2^1 \frac{\delta q_{rev}}T$$ and the inequality is because of the irreversible step (note change in limits). The first integral is zero because the system is isolated, and the second integral is $S_1 - S_2 $ as entropy is a state function, thus $0> S_1 - S_2 $, which produces $$ \Delta S = S_2-S_1 > 0$$

thus we see that the entropy increases when the isolated system goes from state 1 to 2 by a general spontaneous (irreversible) process.

Thus you see that more heat is transferred in a reversible than irreversible process as you suggest. Hope this helps.

(From this if we consider the universe to be an isolated system and all naturally occurring processes are irreversible, one statement of the second law is that the entropy of the universe is constantly increasing. Clausius summarised the first two laws as ' The energy of the world is constant' and 'The entropy is tending to a maximum'). (See MCQuarrie & Simon, Physical Chemistry)

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