The question is: What would we observe if the tire being deflated was filled with helium instead of air?
I have encountered two contradicting theories about the possible outcome of the experiment:
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A: The observed cooling occurs as a result of adiabatic expansion of the compressed air. The air, as it escapes from the tire valve, does work on the atmosphere by pushing it away against atmospheric pressure to make room for the expanded air. The amount of work done, delta W, equals P(atm) * delta V (the extra volume that the expanded air takes). We assume delta Q = 0 (adiabatic process), and so this work is done at the expense of internal energy of the expanding air which is why it cools down. (Joule-Thomson effect is not a major factor here and only manifests as a slight deviation from the 'ideal gas' result.)
This mechanism (internal energy converted to work) will apply to helium as well, therefore, when deflated, a helium-filled tire will also cool down.
B: What we observe is indeed a kind of adiabatic expansion: free (unrestrained) adiabatic expansion, where no external work is extracted. There is no receding piston that would take away kinetic energy of the expanding air molecules. The tire begins at atmospheric temperature, so the 'tire molecules' have the same average kinetic energy as 'atmospheric molecules' and will not pass energy to them upon collision. The compressed air molecules do not 'push' the atmospheric molecules away by collisions; expansion occurs because the 'empty space' on the boundary is more likely to get occupied by one of the more densely occuring 'tire molecules' first.
In free adiabatic expansion, where no work is extracted (delta W = 0) and no heat transferred (delta Q = 0), the internal energy must remain the same (delta U = 0). The observed cooling occurs due to Joule-Thomson effect which predicts cooling of air by isenthalpic expansion at room temperature (because air is not ideal gas and so temperature does not directly correspond to internal energy).
For helium however, Joule-Thomson inversion temperature is far below room temperature, therefore, when deflated, a helium-filled tire will warm up instead.
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A's rebuttal: B is wrong. Free expansion (as in Joule expansion experiment) assumes expanding against vacuum, and only predicts that temperature of (ideal) gas will be the same after the gas reaches equilibrium. In our experiment, we vent the compressed air into the atmosphere and we do work on it. This work results in (generally inobservable) adiabatic heating of the rest of the atmosphere.
So the zero energy change would only be true if we included the entire atmosphere in the system, and waited until it reaches equilibrium, i.e., the cold valve stem is warmed up again by the surrounding air. That doesn't mean that there can't be work done between parts of the system (tire+air inside vs. rest of the atmosphere).
If our system includes only the tire and the air inside, then this is not a free expansion experiment, and the internal energy change of this system doesn't have to be zero.
As for the Joule-Thomson effect: Joule-Thomson experiment assumes isenthalpic expansion. B suggests that the internal energy of the tire air remains the same. But that would mean that there is positive enthalpy change, because the volume of the compressed air has increased (H = U + P(atm) * V). This invalidates the assumptions of a Joule-Thomson experiment.
B's rebuttal: A is wrong. No work is done by the compressed air on the atmosphere. Sure, a reversible, isentropic adiabatic process that allows gas to expand against pressure does positive work on the environment during expansion and always results in cooling; we could do that by venting air from the tire into a balloon. That would genuinely push the atmosphere away and thus do some work.
In our experiment though, the process is irreversible; the molecules get hopelessly mixed up and we will never be able to sort them out again. Entropy of the system is increased. Irreversible adiabatic expansion does not have to exert work on the environment, and in this case it doesn't.
The process is isenthalpic and is a valid Joule-Thomson experiment; volume of the compressed air does indeed increase, but the atmosphere is not blocked from permeating this extra volume, therefore the change in enthalpy is not justified.
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Who is right and who is wrong? Will a helium-filled tire cool down or warm up when deflated?
I don't know what the correct answer is; both theories sound credible and I don't have any helium at hand to actually do the experiment.
http://www.physicsforums.com/showthread.php?t=134458