Cooling of an expanding gas

  1. I've been a bit confused about when/why an expanding gas cools.

    Here is what I've heard so far:

    1. Refrigeration systems work based on using heat that is used merely to break attraction of molecules of liquids at their boiling point (eg, freon released through an expansion valve gets cooler because particles break the attraction between the freon molecules, so some energy is lost to that "breaking" process, resulting in less KE in the particles involved and thus a lower temp). A common explanation of this, however, is that an expanding gas cools, not the boiling point bit.
    2. In Feynman's lectures he talks about how if you compress a gas, you are imparting energy onto the gas's particles and the KE and temperature rises (ie, a piston/compressor is "hitting" the particles faster). Then he says that expansion lowers the temperature. "So, under slow compression, a gas will increase in temperature, and under slow expansion it will decrease in temperature."
    3. The OP in this thread talks about how the air is measurably cooler after compressing, cooling, and then releasing it. https://www.physicsforums.com/showthread.php?t=549795
    4. The ideal gas law seems to *not* imply a temperature decrease in the above scenarios because pressure decreases as volume increases, so T should stay the same. But maybe I'm wrong about this because maybe p and V change at different rates which causes T to noticeably fluctuate.

    So, I'm mostly struggling to visualize 2 and 3 above. Why does Feynman say that expansion will decrease temperature? Also, why was the uncompressed air cooler? I'm trying to visualize those on a molecular level, but it's hard to see why things would get cooler in those cases.

    Thanks!
     
  2. jcsd
  3. To visualize 2: Imagine you are bouncing a tennis ball up and down on a racket. You move your racket upwards as it hits the ball and the ball bounces higher - it's kinetic energy increases. If you move your racket down you can "cushion" the impact and the ball loses kinetic energy.

    Similarly a gas particle hitting the wall of an expanding container loses energy. Particles losing kinetic energy is a temperature decrease.
     
  4. excellent visualization for that one. thanks!
     
  5. 3. Compress the air doing work on it with a piston and it gets hotter. Let if cool back to room temperature. Expand the air back to its original volume letting the air to perform work on the piston and it gets cooler.

    4. Yes p and V change in different ways and T drops in a expansion (As described by Feynman).

    The process described above works, but in real refrigerators other processes may be used. For instance

    1. When freon evaporates it absorbs latent heat from the environment cooling down.

    It's also possible to cool a non-ideal gas through the Joule–Thomson effect which allows a gas to cool during expansion even if no work is being done. (Some gasses will warm up instead of cool down)
     
  6. thanks!

    for 3, does this apply to a pressurized container that was opened in the free atmosphere? so you pressurize a bottle with a pump, detach the pump, let the container cool, and then open the lid. air rushes out and at least momentarily is cooler than the surrounding air. there is no piston for the escaping air to do work on. i believe that was the OPs experiment in the cited thread. why would the air flowing out be cooler?
     
  7. It is doing work on the air that it is displacing (to push the air back), just as if the air it is displacing were a piston.
     
  8. To complement Chestermiller's answer above, if you open the bottle of air in avacuum than it wouldn't cool because there would be nothing for the air to do work on.
     
  9. thank you for the help!

    might be more of an engineering question now, but couldnt you use pressurized gas as a form of refrigeration technology? or is the cooling effect not great enough and that is why we need to involve the phase change part (like in current refrigeration technology)?
     
  10. Yes and yes.
     
  11. Why wouldn't the air (taking it as a real gas) cool? I always thought in a Joule-Thomson process or free expansion to a vacuum, temperature always decrease as molecular KE is converted to PE due to a lower pressure at the exit/end state?

    Thanks
     
  12. In the Joule-Thompson process applied to an ideal gas, there is no change in temperature for the gas. Mechanistically, the expansion cooling occurring within the valve is exactly cancelled by the viscous heating in the valve.

    Chet
     
  13. I get that for an ideal gas enthalpy is pressure independent and temperature wouldn't change, but wouldn't a real gas always cool? Was the mechanism I am visualizing by which temperature decreases correct (particle KE converted to PE as P decreases at constant specific energy)?

    Thanks
     
  14. Well, we're talking about ideal gas. Also note that real gasses due to Joule-Thomson effect, somewhat paradoxically, sometimes warm up instead of cool down when they expand.
     
  15. Real gases often closely approach ideal gas behavior in actual practical situations. That's why the ideal gas model is of such widespread application and widespread value.

    Chet
     
  16. How does that work? I looked at a couple of refrigerant P-h graph and temperature trends downwards as pressure goes down at constant specific enthalpy?

    Also, what about for free expansion? I was reading this link: http://www.etomica.org/app/modules/sites/JouleThomson/Background2.html and they state that real gas always cool in free expansion?

    Thanks
     
  17. The reference you provided looks very solid. As far as your molecular interpretation is concerned, I must admit I'm not too adept at doping out that type of thing. I'm more of a continuum mechanics guy.

    Your reference indicates that, just beyond the ideal gas region, at constant enthalpy, the temperature decreases with pressure and, at constant internal energy, the temperature decreases with volume. That seems good enough for me. However, I don't rule out regions further beyond the ideal gas region where, for some gases, the reverse behavior might come into play. Have you seen any p-h diagrams where this is shown?

    Chet
     
  18. Hi

    From some searching I found this wiki article talking about inversion temperature (http://en.wikipedia.org/wiki/Inversion_temperature) found as 27/4T_c, which is above the vapor dome which I didn't pay any attention to and many diagrams cut off. I found this P-h diagram http://www.4shared.com/file/yU4FnV_U that extends further and it shows that indeed temperature does increase at constant h with decreasing P far above the vapor dome. It looks like I learned something today!

    The wiki link talks about attractive/repulsive intermolecular forces; what specific forces would these include? Also what happens when a supercritical fluid with has T > T_inv goes through a JT process?

    Thanks very much
     
  19. To find the answer to your JT question, just examine the p-H diagram that you were referring to. I guess that 6.75 Tc refers to the prediction of some EOS model (probably VanDer Waals).

    Chet
     
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