We all know the classic scenario of free expansion. A contained gas expands into a vaccum and in the end we have $\Delta T = \Delta U = \Delta H = 0$ and $\Delta S = R \ln \frac{V_2}{V_1}$. This is fine for me, no problem here.

My problem is with the assumption of equilibrium at the final state. I'm thinking about the transient state. Since in this case there is unresisted expansion (i.e. the gas does accelerate), my concern is that, if the gas were inviscid ($\mu = 0$), it would never decelerate to reach equilibrium again. There'd be an eternal oscillation with waves of pressure.

I think that, as particles of the gas accelerate to expand, their internal energy becomes kinetic energy (thus cooling that region). This kinetic energy would only go back to being internal energy due to viscosity, thus reverting the system temperature to the original value.

Some friends argue, however, that kinetic energy is already contained in internal energy, thus the gas temperature wouldn't ever be altered anywhere. I disagree, though, because I think the only kinetic energy accounted for in the internal energy is that due to speeds which don't amount to any net displacement.

So, the questions are:

  1. In the classical scenario, is equilibrium indeed assumed in the final state? If not, I'm way off here.
  2. If so, would such process need viscosity to reach the assumed equilibrium? Even if the usual "ideal gas" is inviscid, I think this can be a "yes" assuming really low viscosity.
  3. Does the temperature of the gas vary locally due to acceleration of that region, and eventually back to the original value upon reaching equilibrium?
  • $\begingroup$ Part of the problem here, I think, is that the ideal gas is pathological to begin with. The molecules in the gas don't interact, so there's no way for an out-of-equilibrium ideal gas to evolve to equilibrium by redistributing energy. For a real gas where the molecules interact (even weakly), the temperature is not constant; the potential energy associated with the intermolecular forces necessarily changes due to the particles being on average farther apart. It might be worth looking up the Joule-Thomson effect on Wikipedia. $\endgroup$ – march Mar 7 '16 at 22:06
  • $\begingroup$ @march you gave me something to think about. I'm starting to think that the particles would never even accelerate in the first place. $\endgroup$ – André Chalella Mar 8 '16 at 22:25

I completely agree with your analysis of the situation.

  1. In the classical scenario, equilibrium is indeed reached in the final state.

  2. An (irreverisble) free expansion process needs viscosity to reach final equilibirum. If not the gas will oscillate forever. This will be the case also for a vertical adiabatic piston/cylinder arrangement in which the piston is suddenly released. If not for gas viscosity, the piston would oscillate forever. As ChE's we are taught that ideal gases are not inviscid, but, instead, exhibit finite viscosity even at very low pressures (where the viscosity becomes independent of pressure). See Transport Phenomena by Bird, Stewart, and Lightfoot.

  3. During the process of free expansion, temperature will vary spatially as different parts of the gas are expanded and compressed (as a result of losing and regaining kinetic energy), but as the system approaches the final state, heat conduction and viscous dissipation of kinetic energy will cause the gas to equilibrate uniformly at its initial temperature.


There's a point you raise in the question that's wrong. As you quote, 'in this case there is unresisted expansion (i.e. the gas does accelerate)', the gas is unresisted, but their speed remains the same. The gas particles are just colliding elastically with the walls and nothing accelerates or decelerates them.

  • $\begingroup$ Good remark, but this should really be a comment instead of an answer. $\endgroup$ – André Chalella Apr 9 '19 at 10:06
  • $\begingroup$ This is more a comment than an answer to OP's question. Also, it seems like you're talking about something that the OP wasn't. When they said "the gas does accelerate"; presumably they were talking about the bulk of the gas, analyzed as a fluid, not as a collection of molecules. Although on the molecular level, nothing may accelerate, on a macroscopic level, the gas would need to accelerate as it expands out of the opening, as it had 0 net velocity when in the pressurized container. $\endgroup$ – JMac Apr 9 '19 at 10:34
  • $\begingroup$ i thought by clarifying this misconception the OP's question might b resolved $\endgroup$ – user196075 Apr 9 '19 at 10:55

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