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Callen asks us to consider the following

A cylinder of length $L$ and cross-sectional area $A$ is divided into two equal-volume chambers by a piston, held at the midpoint of the cylinder by a setscrew. One chamber of the cylinder contains $N$ moles of a monatomic ideal gas at temperature $T_0$ • This same chamber contains a spring connected to the piston and to the end-wall of the cylinder; the unstretched length of the spring is $L/2$, so that it exerts no force on the piston when the piston is at its initial midpoint position. The force constant of the spring is $K_{spr}$; The other chamber of the cylinder is evacuated. The setscrew is suddenly removed. Find the volume and temperature of the gas when equilibrium is achieved. Assume the walls and the piston to be adiabatic and the heat capacities of the spring, piston, and walls to be negligible. Discuss the nature of the processes that lead to the final equilibrium state. If there were gas in each chamber of the cylinder the problem as stated would be indeterminate! Why?

I have completed the problem except for the bolded part. I suppose there are two ways of approaching this question depending on the interpretation. In interpretation (1), we assume that the initial gas pressures are identical. In interpretation (2), we do not have that the initial gas pressures are identical.

(1) Why should such a system be indeterminate? Denote the relevant forces by $F_{spr}, F_{g1}$, and $F_{g2}$. The initial state is surely the final (equilibrium) state once the setscrew is removed, since it is characterized by the equality $$F_{spr} + F_{g1} + F_{g2} = 0 + F_{g1} + F_{g2} = (P_1 - P_2)A = 0.$$ Is Callen saying that in general there could be other equilibria and that small fluctuations (which are outside the scope of classical thermodynamic theory, or so I thought?) could shift the system to other equilibria?

(2) Obviously the initial state is not an equilibrium state, but how can I see in this case that there is no (unique?) final equilibrium state?

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    $\begingroup$ "$(P_1-P_2)A=0$" doesn't provide an answer because those pressures are unknown. Think a little about where the heat from friction has gone when the piston finally comes to rest. (The answer is unambiguous when there's gas on only one side.) $\endgroup$ Commented May 27, 2023 at 19:20
  • $\begingroup$ Why doesn't it? I've specified in that interpretation (1) that $P_1=P_2$, so that in that line in my OP I am showing that we have zero net force on the piston. Am I missing something in your comment? @Chemomechanics $\endgroup$
    – EE18
    Commented May 27, 2023 at 19:43
  • $\begingroup$ @Chemomechanics I'm not sure what heat flow should have to do with anything; I think we are supposing that the piston is adiabatic? $\endgroup$
    – EE18
    Commented May 27, 2023 at 19:44
  • $\begingroup$ Presuming that the system with gas in both chambers is initially out of equilibrium, motion will occur. Agreed? Later, after the motion has died down from friction, where exactly is the associated thermal energy? $\endgroup$ Commented May 27, 2023 at 21:01
  • $\begingroup$ Assuming that the initial pressures are unequal, the final pressures must also be unequal. In addition, the final temperatures will be unequal. $\endgroup$ Commented May 28, 2023 at 0:11

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Regarding your question (1), I think it is related to the notorious conundrum (paradox?) of the adiabatic piston. Callen has two other problems related to it, and on page 51 he writes:

The case of a moveable adiabatic (rather than diathermal) wall presents a unique problem with subtleties that are best discussed after the formalism is developed more fully; we shall return to that case in Problem 2.7-3 and in Problem 5.1-2.

I suggest that you consult Gruber where this problem is analyzed in all its gory detail; it is a very readable work, I should add. Gruber and his coworkers subsequently published several other pieces at varying degree of complexity on the same subject.

Instead of rehashing what Gruber wrote, while I agree he does offer an effective solution to the problem, I discuss another approach to it because I think the issue really lies somewhere else. To me the real problem is in the aggressive (?) over-idealization of the concept of reversibility, constraints and reservoirs. I alluded this to you yesterday while referring to it as "singular perturbation".

To illustrate it I assume you know the so-called two-capacitor paradox. To remind you it is about connecting directly in parallel a charged capacitor with an uncharged capacitor of equal size and then noticing that from symmetry, in equilibrium, both the voltages and charges must be equal, so that either we have lost half the energy or we have lost half the charge. The usual explanation is that when you turn the switch on to connect the two capacitors, you get sparks, radiation etc. and that is not included in the energy balance. I think this explanation is completely backwards because there is no radiation in Kirchhoff's laws that describe abstract elements defined by constitutive relationships $i=C\frac{dv}{dt}, v=RI, v=L\frac{di}{dt}$ with impressed independent sources $v_s, i_s$, AND, of equal importance, a set of connection rules, namely all initial conditions must be such that both voltages and currents must be continuous and stay continuous throughout. The last requirement immediately disallows switching a charged capacitor on to a discharged one because their voltages are different and the resulting initial current is infinite, not continuous. Before somebody dismisses KVL and KIL as some approximation to the true Maxwell's equations, I would like to remind him/her that Kirchhoff's laws may be history's most tested theory of any theory, as the integrated circuits in digital computers with trillions of resistors, capacitors transistors testify to its absolutely phenomenal accuracy. These ICs are all designed using Kirchhoff's laws but they never connect one charged capacitor on to a discharged one with an ideal switch.

The moment you add to the switch an infinitesimal resistance just to show that it is not ideal, you will have fully restored both charge and energy conservation, and of course all design software would have that miniscule resistor added in it when it is needed. So the problem is in the unrealistic idealization and illicit couplings of the various abstract items. The most amazing result in the solution of the capacitor paradox is that it does not matter how large or how small that resistance of the switch or any of the connecting wires between the capacitors is as long as it is not zero: the dissipated energy in that resistance is always half the initial stored energy. The problem, the paradox, happens only when that connecting resistance is assumed to be exactly zero. This is the singular perturbation I was referring to.

Now what does this have to do with the adiabatic piston problem? There it is assumed that in a cylinder two fluids are separated by a moveable ideal frictionless, massless and adiabatic wall. Some claim that it will oscillate and never relax, some claim that temperatures may not equalize but pressures will, etc. I say that there is no real adiabatic wall, and there must be some amount of diabaticity leading to the equalization of temperatures and per force the equalization of pressures, as well, and problem solved.

In my view, what is missing is a clear set of connection principles/rules that would not allow a massless adiabatic wall to separate two thermodynamic fluids whose state depend on the hydrostatic pressure and temperature, exactly the parameters that are not allowed to be in communication by the idealized wall. There is no difficulty in the 1st part of your question because there the massless adiabatic wall is separating a mechanical system (elastic spring) from a thermodynamic system (gas). Because the spring has no temperature and it is characterized by its length, the mechanical properties by themselves are enough to specify the equilibrium.

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  • $\begingroup$ "I would like to remind him/her that Kirchhoff's laws may be history's most tested theory of any theory" It is incredibly weird to take those as evidence for KVL rather than Faraday's Law, of which KVL is a simplified limit. Adding inductance to KVL is a hack; it is automatically there in Faraday's Law. Anyway, yes, it is true that having no resistance is unphysical, but if you treat the problem fully, a tiny resistance leads to most of the energy leaving as radiation, whereas bigger resistance takes up more and more of the loss. i.e. Insisting that it takes up exactly half, is unphysical. $\endgroup$ Commented May 29, 2023 at 8:18
  • $\begingroup$ @naturallyInconsistent (1) It is not hacking how Faradays' law is added in KVL, but it is proper circuit element modeling within the realm of Kirchhoff's laws that is KVL/KCL together with connection (network) rules. Hacking and modeling are not the same as the failure of any reasonably complex circuit will show you. (2) the number of inductors are insignificant in digital circuits, (3) it does not matter that on a higher level a tiny inductance leads to more radiation than a smaller one, the KVL/KCL + connection rules within their realm satisfy both energy AND charge conservation... $\endgroup$
    – hyportnex
    Commented May 29, 2023 at 11:39
  • $\begingroup$ That is simply not true. KVL is only about electrostatic potential, and thus can never hope to cover the fact that inductance causes the integration of E field around a loop gives a non-zero result. The only way to do it properly is via Faraday's Law, and any attempt to coerce KVL to get it, is a hack. It is also totally unnecessary, because Faraday's Law is a drop-in replacement of KVL and will always explain the successes of KVL, with or without the hack. It is just the more general and more fundamental law to use, without the limitations that KVL has. $\endgroup$ Commented May 29, 2023 at 11:47
  • $\begingroup$ (4) the way an inductor is added to KVL is no more hacking than calling two charged plates a capacitor, if it is done badly (hacking) it does not work, (5) within KVL/KCL the "disappearing" energy is dissipation not radiation and if some of that radiation causes magnetic coupling between loops, and they do, most of the time it can be resolved by better modeling of the affected parts. This is true even with waveguides and with the system behavior of antennas that are always (!) analyzed and designed with Kirchhoff's equations. $\endgroup$
    – hyportnex
    Commented May 29, 2023 at 11:47
  • $\begingroup$ It is kind of ironic that you cannot see the mental contortions one has to go into, just to assert that "the 'disappearing' energy is dissipation not radiation" v.s. "if some of that radiation causes magnetic coupling between loops, [...] resolved by better modeling of the affected parts". There can only be one physical picture as to where the energy has gone, not "radiation if you model better, dissipation if you model badly". It would also explain why you could have so loudly stated that half the energy is used in a tiny resistor, as opposed to a smooth transition between radiation and that. $\endgroup$ Commented May 29, 2023 at 11:53

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