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Joule's expansion experiment

  1. Sep 30, 2015 #1

    Geofleur

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    In his book, Theoretical Concepts in Physics, Malcolm Longair describes two experiments that Joule performed involving gas expansion (the discussion starts on pg. 217). In both experiments, there is a chamber containing gas, labeled A, and a chamber that is evacuated, labeled B. There is also a tube with a valve connecting A and B.

    Experiment 1: The whole setup is immersed in a bath of water and the valve opened. The gas spreads out evenly between A and B, with no temperature change of the water bath.

    Experiment 2: A and B are placed in their own separate water baths, and the valve is opened. The water surrounding A cools and that surrounding B warms.

    I understand why experiment 1 turns out the way it does, but I am not sure that I understand the result of experiment 2. Does the bath around B warm because gas molecules start bombarding the walls of the container, transferring kinetic energy to them? Does the bath around A cool because there are less molecules present to bombard its walls as some of the gas leaves? Is there a better way to think about this result?
     
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  3. Sep 30, 2015 #2
    Think of this like evaporation on a water surface. Why do you cool when you sweat? The water molecules leave your skin, taking kinetic energy with them, cooling the skin. This is basically what is happening at A, so yes you are correct.
     
  4. Sep 30, 2015 #3
    In the second experiment, the containers are not in thermal contact. Imagine that the process took place adiabatically, rather than in a constant temperature bath. The gas remaining in chamber A has been expanding and doing work to push the gas ahead of it into chamber B. So the gas in chamber A experiences expansion cooling. The gas entering chamber B is at the same temperature as that in chamber A at any given instant of time. But, in passing through the tubing, there is no change in the temperature of the gas because of the Joule-Thompson effect (enthalpy is independent of pressure). As gas enters chamber B, it compresses the gas that is already in there, and this causes compressional heating. The net effect is that the gas in chamber A would be hotter than originally, and the gas in chamber B would be colder than originally. The total internal energy of the gases in the two chambers would be unchanged, but the pressures would be equal, and lower than originally. The temperatures and pressures can be precisely calculated.
     
  5. Sep 30, 2015 #4

    Geofleur

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    @Chestermiller: If the gas were ideal, then we could ignore any collisions between gas molecules - so would the effects you mention disappear in that case? Also, the effect that Muqi and I were talking about would still be present, yes?
     
  6. Sep 30, 2015 #5
    Sorry Geofleur. I'm a continuum mechanics guy, so molecular explanations of this type are hard for me to relate to.

    Chet
     
  7. Oct 1, 2015 #6
    If I remembered well the effect is only due to variation with distance of inter-molecular potential energy. For an ideal gas there is no such potential energy (molecules exchange energy only through collisions) so the temperature doesn't vary.

    Edit. I think this is the experiment you are discussing:
    https://en.wikipedia.org/wiki/Joule–Thomson_effect

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  8. Oct 1, 2015 #7

    Philip Wood

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    Expansion of a gas into a vacuum is called 'Joule expansion" after James Prescott Joule. Joule-Kelvin (aka Joule-Thomson) expansion is rather different.

    When a gas expands into a vacuum it does no external work. So the heat flow into it/out of it is equal to the increase/decrease in its internal energy. This is true for both real and ideal gases.

    For an ideal gas (main requirement: no forces between molecules except during collisions) there is no change in the potential energy of the gas when it expands, so nothing to slow down the molecules in their random motions, so no change in gas temperature and no heat flow in or out.,

    For a real gas (attractive forces between molecules), when the gas expands the molecules become further apart on average, so the gas gains potential energy. At first (before heat has time to flow into the gas from the container walls), the gas cools as the molecules lose random kinetic energy such that the total internal energy of the gas is the same. But heat will gradually flow into the gas from the container walls, on account of the lower temperature of the gas.

    I thought that Joule's instruments were too insensitive to detect the consequent cooling of the surroundings when he did the (first) experiment with air at lowish pressure. Haven't a clue why container B should have warmed up in the second experiment you describe. I'd have thought both containers would haver cooled, assuming gas, container and water were in thermal equilibrium originally

    Await enlightenment..
     
  9. Oct 1, 2015 #8
    I need to expand on my discussion of the second experiment to make clearer what I was saying. The second experiment was very different from the first experiment, in that the first experiment involved unconstrained expansion into a vacuum, while the second experiment did not.

    The key features of the second experiment were (1) the thermal isolation of the chambers from each other and (2) the presence of the connecting tube with a constraining valve. The presence of the constraining valve prevents the gas from flowing freely between the two chambers. Although it is unstated in the description of the first post, the valve is cranked down to the point where it only allows the gas to escape very slowly from chamber A. As a result of this, the gas that remains in chamber A during the transition between the initial and final states does reversible work on the gas that it pushes out of the chamber. This causes the gas in chamber A to experience expansion cooling.

    I mentioned the Joule-Thompson effect in my first post. This occurs at the valve of the connecting tube. Gas enters the tube containing the value at a higher pressure and exits the tube at a lower pressure. But this is not a pressure change that would be consistent with the Bernoulli equation (for which mechanical energy is conserved). The pressure change is dominated by viscous frictional pressure loss in the valve. The associated viscous heating of the gas in passing through the value is exactly balanced by the expansion cooling of the gas so that, overall, there is no temperature change of the gas in passing through the valve. This is the mechanistic explanation of the Joule-Thompson effect, to wit, when a gas passes in steady flow through a constriction in a tube or through a porous plug, the change in enthalpy of the gas is zero, and the change in the temperature of the gas is zero.

    Now for chamber B. When the gas enters chamber B, because of the Joule Thompson effect in the valve, the temperature of the gas entering the chamber is the same as that currently present temperature in chamber A. The gas is entering chamber B very slowly, and the gas that is already present in the chamber (at any point during the change) is experiencing adiabatic reversible compression. Basically, the new gas that is entering the chamber is doing compressional heating work on the gas that is already in the chamber.

    Now, we know that, if the total system is adiabatic, the gas in chamber A will cool down in this experiment. And, we also know that, since no net work is done on the surroundings (overall, since the chambers are rigid) and no heat is exchanged with the surroundings, the overall change in internal energy of the gas in the two chambers is zero. But, since the gas in chamber A cools, the gas in chamber B must be hotter in the final state. So, even though the temperature of the gas entering chamber B from the connecting tube is cooler than the original temperature in the system, the compressional heating that occurs in chamber B is sufficient to, in the end, raise the temperature in chamber B above the original temperature.

    We can precisely calculate all this pretty easily.

    I hope this explanation makes sense. I'm very confident that it's correct.

    Chet
     
  10. Oct 1, 2015 #9

    Geofleur

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    Thanks a ton everyone - I feel like I understand much better now what's going on!
     
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