2
$\begingroup$

I am continuing to try to understand maximum work reversible processes (and a subset thereof -- Carnot cycles) better. I am here curious about the following system.

(1) Consider one mole of a gas (say, ideal, or van der Waals) which is expanded isothermally, at temperature $T_h$, from an initial volume $V_i$, to a final volume $V_f$. A thermal reservoir at temperature $T_c$, is available.

My question is about how I can know/prove that there exists a way to take the gas (the primary subsystem) reversibly with respect to a composite system consisting of the gas, the reservoir, a reversible work source (RWS), and some set of auxiliary systems (defined as undergoing no net change) from the initial state to the final state of the primary subsystem in this particular way (i.e. along the given isotherm).

I ask because in the context of the following conundrum given to us by Callen in his classic thermodynamics textbook:

(2) A system can be taken from state A to state B (where $S_B= S_A$) either (a) directly along the adiabat $S$ = constant, or (b) along the isochore AC and the isobar CB. The difference in the work done by the system is the area enclosed between the two paths in a P-V diagram. Does this contravene the statement that the work delivered to a reversible work source is the same for every reversible process?

The answer here -- I think -- is that there is no way to reversibly (with respect to the subsystem+surroundings --i.e. subsystem+whatever composite system you consider it as part of) take the primary subsystem to the final state as in b). Perhaps this answer is wrong though, and there is some separate reason that (2) above does not violate the so-called Maximum Work Theorem.

At any rate, the contrast between the two cases discussed above leads to my question: how can I know whether it is possible to take a (sub)system reversibly (again, with respect to some composite system containing the primary subsystem, a reversible heat source (the reservoir in the first example is a particular case), a reversible work source (RWS), and some set of auxiliary systems)? Is it always possible, or are there some changes of state for which this is not possible? If it is always possible, it would seem that my answer to the paradox in (2) is wrong, and I would appreciate any help in resolving it.

$\endgroup$
2
  • $\begingroup$ "Does this contravene the statement that the work delivered to a reversible work source is the same for every reversible process?" Are you quoting Callen here? $\endgroup$
    – Bob D
    Jun 23, 2023 at 21:31
  • $\begingroup$ Yes! See also my question here (physics.stackexchange.com/questions/766008/…). I am not so much asking as for an answer to that particular problem as using it to demonstrate that which I think Callen assumes -- that such a reversible process exists (in the context of situating the given system in some larger composite system). @BobD $\endgroup$
    – EE18
    Jun 23, 2023 at 21:40

1 Answer 1

2
$\begingroup$

The answer to this question is negative in the sense that not all systems allow for reversible processes at all, not even in principle. The most famous examples maybe hysteretic ferromagnetism and hysteretic plastic deformation along with all kinds of friction. There are many others. No matter how slow the process is there is irreversible entropy generation at every step along the way. So the issue of how to define entropy if you cannot have a reversible path between two equilibrium is germane. One way to handle the question is to dismiss the whole issue by saying that, for example, a ferromagnetic body is never in equilibrium and is not amenable to classical thermostatic methods. Not everybody is so pessimistic, for example Bridgman says that:

Physics, whether we choose to call it thermodynamics or not, should not be impotent in the face of any situation which can be completely characterized by measurements with macroscopic instruments. Yet many essentially irreversible processes can be completely characterized in terms of a few simple measurements. For instance, the irreversible process of heat conduction is exhaustively characterized in terms of a temperature gradient and the thermal conductivity of the material, and the irreversible generation of Joulean heat is fixed by current density and specific electrical resistance.

The method that Bridgman offers is one way to attack the problems of inherently irreversible processes thermodynamically, and being an experimentalist he is very strongly influenced by his own instrumentalist philosophy. But his way is just one among many and is not unique in the sense that there several "schools" of irreversible thermodynamics of which Callen discusses only one kind and whose origin is in the work of Onsager and Prigogine; moreover, Prigogine's way does not even address the hysteretic problem.

Summarizing, Clausius's classical definition of entropy between two equilibrium states hinges upon two assumption: (1). there are equilibrium states defined by the values of some extensive and intensive parameters and (2). these equilibrium states can always be connected by at least one path that is equilibrium at every step along the path. One can neither prove the existence of the equilibriums states nor the existence of the connecting reversible path. Only an experiment on the body can verify that such assumption is valid or not. In a textbook, such as Callen's, it is assumed that the system in question can exist in equilibrium and it is assumed that those states can always be connected with a reversible path; we define the system by its the caloric and thermic equations.

$\endgroup$
15
  • $\begingroup$ It seems you are using reversible to mean quasistatic, but perhaps I am misunderstanding you. At any rate, my question is whether it is possible to connect equilibrium states of a primary subsystem in a reversible (with respect to the universe; i.e. isentropic with respect to the universe) way in general (or at least in Callen's way of thinking). It seems maybe you are saying Callen would say yes? But then how does one resolve point (2) I raise above? $\endgroup$
    – EE18
    Jun 23, 2023 at 3:05
  • $\begingroup$ @EE18 it is important here, again, that the word reversible is to be used. Quasistatic alone is not enough. Hyportnex is correctly pointing out that if irreversible quasistatic processes cannot be avoided, then the assumption of reversible processes connecting every equilibrium state together is invalid. $\endgroup$ Jun 23, 2023 at 6:22
  • 1
    $\begingroup$ quasi-static IS NOT reversible. Quasi-static is infinitely slow, reversible is adiabatic and then is also isentropic for an isolated system. Confusing them or rather not distinguishing the concepts is a historical issue of either sloppy/bad writing or excluding certain processes from the possibility of thermodynamic description by declaring the impossibility. Callen excludes this problem by not talking about it; just turn your page to the appendix where he explicitly excludes ferromagnetic hysteresis from his analysis of magnetic work and energy of which he is really an expert. $\endgroup$
    – hyportnex
    Jun 23, 2023 at 8:14
  • $\begingroup$ Bridgman, who was one of the most famous and respected experimentalist and thermodynamicist of the 20th century, offered a procedure to try to extend thermodynamics to cover quasi-static but still irreversible processes. He surely was more than somewhat successful but that the issue is still not universally resolved is shown in the various forms of "schools" in existence. Thermodynamics of the undergraduate textbooks is thermo-statics of reversible processes, nature is more complicated. $\endgroup$
    – hyportnex
    Jun 23, 2023 at 8:23
  • 1
    $\begingroup$ sorry, above I meant to write in the 2nd sentence "...if reversible and is adiabatic then it is also isentropic". $\endgroup$
    – hyportnex
    Jun 23, 2023 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.