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Joule-thomson effect and linde process

  1. Oct 20, 2014 #1
    According to the j-t effect the decrease in the temperature is a result of the work done by molecule to expand thus increasing their PE and decreasing their KE, but in the linde process the next step is liquification, i am having trouble understanding how it happens, since the intermolecular distance between liquid molecules are shorter than gas, so the molecules have to get rid of the PE they just gained, but according to my professor the liquid reservoir doesn't allow heat exchange, so how do the molecules get rid of the PE.
    I think they get rid of it by doing work in expansion this is why not all the gas is liquified, but i am having trouble with this conclusion as how the gas expand ( increase molecular distance) and liquify (decrease molecular distance) at the same time, i am really confused
     
  2. jcsd
  3. Oct 21, 2014 #2

    DrDu

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    I don't see quite a problem here, small droplets will condense out of the gas on expansion. The same thing is happening also when clouds are forming in the rising air on a hot day: The gas cools off while rising and expanding until small droplets of water start to condense out.
    Ideally, during condensation, the temperature of the gas stays constant. So the energy the gas fraction needs to expand is taken from the heat of condensation of the part of the gas that condenses. It should be clear that the heat of condensation per mole is much larger than the heat absorbed on expansion by the gas fraction per mole, so only a tiny part of the gas really condenses out during each expansion step.
     
  4. Oct 21, 2014 #3
    so some of the gas is condensing and the heat needed to be absorbed from it, is absorbed by some gas expanding?
     
  5. Oct 22, 2014 #4

    DrDu

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    Exactly
     
  6. Oct 22, 2014 #5
    thanks
     
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