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Entropy is a state variable, i.e. the data about the initial and final states would suffice to define change in entropy . However, while applying the formula dS = integral of dQ/T, I have distinctly been told that dQ is to be calculated by assuming a hypothetical reversible process, irrespective of whether we are given a reversible or an irreversible process in the problem. Although heat depends on the thermodynamic path, S doesn't. So why should it matter whether the process is irreversible, whether Q is calculated for an irreversible process, as S doesn't depend on all this? (I know that the math won't work out and we'll end up with different values, but my question is a conceptual one.)

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For both the reversible and the irreversible paths of a system, the T in the equation is supposed to be the value at the interface between the system and surroundings through which dQ flows (see Fermi's Thermodynamics text). This integral is different between the reversible path and the irreversible path. The various integrals of dQ/T for all the possible irreversible paths are different from one another, but the integrals of dQ/T for all the reversible paths are the same, and greater than those for all the irreversible paths.

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  • $\begingroup$ "dQ for the reversible path is not equal to dQ for the irreversible path". So if I have a reversible and irreversible isochoric heat addition between the same two equilibrium states of an ideal gas, the heat added to the gas will not be the same or the reversible path and the irreversible path? No work is done and the change in internal energy is the same. $\endgroup$ – Bob D Jan 12 at 18:57
  • $\begingroup$ The total amounts of heat are the same, but the differential amounts divided by the boundary temperatures are not the same. That's what I was really trying to say (not very well). $\endgroup$ – Chet Miller Jan 12 at 19:10
  • $\begingroup$ Got it. It's the differential entropy not the differential heat that is different. For the irreversible heat transfer part of the total entropy change is generated whereas for the reversible heat transfer all of the entropy change is entropy transfer. $\endgroup$ – Bob D Jan 12 at 19:14
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    $\begingroup$ @BobD It is correct to say that entropy generated is not a property of the system (i.e., a function of. state), but depends on process path. It is also correct to say that the entropy exchanged between the system and its surroundings is not a property of the system, but depends on process path. However, the sum of the entropy generated plus the entropy exchanged is equal to the entropy change, which is a property of the system. It is sort of like work and heat. Neither work nor heat is a property of the system, but the difference is the internal energy, which is a property of the system. $\endgroup$ – Chet Miller Jan 13 at 14:46
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    $\begingroup$ Thanks Chet. You're analogy with the first law is a nugget that I am saving with all your other nuggets. Bob. $\endgroup$ – Bob D Jan 13 at 14:55

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