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Which of these statements is more accurate?

  1. Entropy is a state variable, whose change between points a and b we can calculate assuming a reversible path between those points. For a reversible path between points a and b, it is equal to $\frac{dQ}{T}$ integrated over that path. For an irreversible path from a to b, it is equal to $\frac{dQ}{T}$ integrated over that path plus some extra amount (the generated entropy).

  2. Entropy is a state variable only for reversible paths. All reversible paths from point a to point b have the same change in entropy. For a reversible path between points a and b, it is equal to $\frac{dQ}{T}$ integrated over that path. For an irreversible path from a to b, it is equal to $\frac{dQ}{T}$ integrated over a hypothetical reversible path plus some extra amount (the generated entropy). If the path is irreversible, there will be more entropy than if we had taken a reversible path.

joshuaronis
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  • What do you mean by “generated entropy?” The Clausius definition of entropy is just the reversible work differential over the temperature, and thus the change in entropy is the integral of the reversible heat differential over the temperature for the reversible path. – Matt Hanson Nov 21 '23 at 02:19
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    "Entropy is a state variable only for reversible paths.". That is untrue/ – Bob D Nov 21 '23 at 02:32
  • Option (1) is correct, Option (2) is incorrect. – Chet Miller Nov 21 '23 at 03:25
  • @ChetMiller of course, you and Bob_D are right but then we have the problem to define entropy within the Clausius framework for systems that may not have equilibrium states or reversible paths that can connect certain equilibrium states. What gives? – hyportnex Nov 21 '23 at 13:48
  • @Bob_D please, see, my comment above. – hyportnex Nov 21 '23 at 13:49
  • @hyportnex I don't know about Chet, but I'm not sure what you mean by "systems that may not have equilibrium states or reversible paths that can connect equilibrium states". Can you provide and example(s)? – Bob D Nov 21 '23 at 13:59
  • @hyportnex I agree with BobD. Please provide an example. – Chet Miller Nov 21 '23 at 14:02
  • @Bob_D a ferromagnet is never in a state of equilibrium, neither is a plastic body, see, e.g., Bridgman: The Thermodynamics of Plastic Deformation and Generalized Entropy https://doi.org/10.1103/RevModPhys.22.56 but despite the hysteretic cycles and lack of reversible paths entropy can be defined. – hyportnex Nov 21 '23 at 14:36
  • @ChetMiller https://i.imgur.com/qQTH8Xx.png https://i.imgur.com/YfZHv0o.png https://i.imgur.com/PHxNVdO.png https://i.imgur.com/s0anpxp.png https://i.imgur.com/rkO0Jaz.png https://i.imgur.com/31dTB0i.png https://i.imgur.com/NL3IgXG.png https://i.imgur.com/7m6E76f.png – hyportnex Nov 21 '23 at 15:10
  • So, for polymers, you're talking about configurational entropy, right. – Chet Miller Nov 21 '23 at 20:39
  • @ChetMiller I am not familiar with polymer physics, and cannot comment on that. My whole point was that the standard way of introducing entropy along the lines of Clausius $\Delta S=\int \frac{\delta Q_{rev}}{T}$ is inadequate to describe some of the most common materials and despite having been known for at least 100 years it is still being taught. But my problem with it is its incredibly unintuitive nature more than anything else; I have always hated it ever since I met it in college. Anyhow, I hope you are enjoying Bridgman (note: his $\tau$ is temperature and $T$ is tension/stress). – hyportnex Nov 22 '23 at 04:05
  • Concerning classical treatment entropy and entropy generation, it might help to see Transport Phenomena, Bird et at, Chapter 11, homework problem 11.D-1. – Chet Miller Nov 22 '23 at 11:18
  • @ChetMiller I am familiar with the Eckart article referenced in 11.D-1 of Bird, and is also referenced on page 1 of Bridgman (subtly reminding the reader of his, Bridgman's precedence...) , and that is exactly the kind of thermodynamics already outside the realm of Clausius style "classical* treatment. Both Bridgman and Eckart assume a constitutive relationship that underlies the inherent irreversible entropy production independent of whether the process is quasistatic or otherwise. There is no "$Q_{rev}$" in Eckart or Bridgman. – hyportnex Nov 22 '23 at 13:55
  • @Bob_D please see these https://physics.stackexchange.com/questions/769144/what-does-thermodynamically-consistent-mean/769179#769179 and https://physics.stackexchange.com/questions/787740/truesdells-formulation-of-second-law-for-homogeneous-systems-in-rational-thermo/787751#787751 – hyportnex Nov 23 '23 at 13:41

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