Is Mechanical Equilibrium Really Driven by Entropy Increase?

Question

It is a standard result in statistical mechanics that when two interacting systems are free to exchange energy and volume, then in the macrostate of maximum entropy the systems will have equal temperature and pressure. The second law of thermodynamics then suggests that in thermal and mechanical equilibrium, such interacting systems will always equilibrate so as to have equal temperature and pressure.

But consider two ideal gases separated by a partition, in perfect thermal contact but with different initial pressures. If the partition is suddenly allowed to move, then thinking of pressure in terms of force per unit area (and using the ideal gas law) we can exactly solve for the motion of the partition. Assuming there is some damping force present, it is certainly true that the system will eventually settle down to the state where the pressure on either side of the partition is equal, but the reason for this equilibrium seems to me to be purely mechanical, and completely unrelated to entropy increase and the second law of thermodynamics. In fact, it seems false to assume some sort of ergodic hypothesis for volume microstates here (which underlies the assumption that mechanical equilibrium occurs at a volume macrostate of maximal entropy), as it is certainly not the case that every volume microstate is equally likely at a given moment in time (thanks to our exact solution for the volume as a function of time).

So basically, I'm confused why mechanical equilibrium should have anything to do with entropy, given that the volume exchange process can, at least for simple examples, be completely understood using Newton's Laws. How do we reconcile these two viewpoints?

Answer 1

It is a standard result in statistical mechanics that when two interacting systems are free to exchange energy and volume, then in the macrostate of maximum entropy the systems will have equal temperature and pressure.

This is true if entropy exchange through heat transfer is allowed, but Callen, for example, in Thermodynamics and an Introduction to Thermostatistics takes care to note that an adiabatic frictionless partition would continue to oscillate from side to side, with similar reasoning to yours. (Of course, damping would ultimately eliminate this oscillation in the real world.) From problem 2.7-3:

"The hypothetical problem of equilibrium in a closed composite system with an internal moveable adiabatic wall is a unique indeterminate problem. Physically, release of the piston would lead it to perpetual oscillation in the absence of damping. With damping, the piston would eventually come to rest at such position that the pressures on either side would be equal, but the temperatures in each subsystem would then depend on the relative viscosity in each subsystem. The solution of this problem depends on dynamical considerations. Show that the application of the entropy maximum formalism is correspondingly indeterminate with respect to the temperatures but determinate with respect to pressures."

The problem is also discussed with care and solved in Müller's An Expedition to Continuum Theory with citations to the following reports:

• Crosignani B, Di Porto P, Segev M (1996) Approach to thermal equilibrium in a system with adiabatic constraints. Am J Phys 64:610–613 ("...while the final pressure turns out to depend in a unique way on initial temperatures and volumes, the two final temperature and volumes cannot be predicted [within the framework of elementary thermodynamics], since they depend upon parameters lying outside the domain of thermodynamic description.")

• Gislason EA (2010) A close examination of the motion of an adiabatic piston. Am J Phys 78(10):995–1001

The reference lists of these two papers will probably also be of interest to you, as they include various discussions of entropy and entropy paradoxes.

Answer 2

Assuming there is some damping force present, it is certainly true that the system will eventually settle down to the state where the pressure on either side of the partition is equal

Assume instead that there is some anti-damping force present, which accelerates the piston in the direction it is already moving. Then the piston will either oscillate with greater and greater amplitude or fly off to infinity. This is also purely mechanical.

The situation I have just described is clearly unphysical, whilst the case with ordinary damping is commonplace. Why is that? The reason is that the situation I have described would require drawing upon internal energy of an object and converting it directly into work, in violation of the second law. That is the factthat damping forces can exist and anti-damping forces cannot is a manifestation of thermodynamics.

Mechanics and thermodynamics provide different and complimentary information about the situation.

Answer 3

I used to find this notion mysterious, but after I started teaching thermo out of Lemons' "Mere Thermodynamics" it made sense to me. The notion is this:

When a system cannot increase its internal entropy, the Second Law of Thermodynamics (basically, that entropy increases until an equilibrium is reached) states that it will reach equilibrium by attaining a local minimum of energy.

The logic is actually fairly simple. If the system can't increase its entropy, then its energy isn't doing it any good, at least as far as increasing the universe's entropy. Consequently, it will give up its energy to its surroundings, to try to increase the total entropy. This is a useful principle to know, since it lets us solve for mechanical equilibrium without worrying about the quite complicated microscopic mechanisms of dissipation.

Lemons regards this analysis as not terribly useful, but to me it explains why essentially any macroscopic system comes to mechanical equilibrium. Consider a featureless particle moving in a 1D potential, $V(x)$. Subject to conservative mechanics, it might oscillate forever about a local minimum. Coupled dissipatively to an environment at a positive temperature, it will instead come to rest at that minimum, with the lost energy, divided by the environmental temperature, comprising the total increase in the universe's entropy. Transparently, the particle doesn't lose any entropy itself since it doesn't have any internal structure.

Answer 4

Change the picture you are describing by putting perfect springs inside the two boxes instead of a gas. This system can only exchange mechanical energy. If we start with the partition at a non equilibrium state and let it go, what happens? A purely mechanical treatment says that the partition will oscillate for ever. Even though the system passes a point of stable mechanical equilibrium it will not stop because it has momentum. We must introduce damping for the motion to eventually stop. Damping, however, means degrading the mechanical energy of the system into internal energy, and this in turn means entropy. To reach the equilibrium state and stop there, we need entropy.

Entropy itself does not cause anything, it's the myriad of forces acting on the microscopic constituents of matter. We can say with confidence, however, that when all mechanical motion stops the entropy of the universe is at a local maximum. This then allows us to calculate the final state simply by maximizing entropy under the constraints of the problem without having to worry about the forces that bring about the change.