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Entropy in Partitioned tank

  1. Apr 28, 2012 #1
    Hi all,

    Sorry if this is posted in the wrong place, I'm new here.
    I have a thermodynamics question. Why is the temperature change 0 in the case of a partitioned tank containing an ideal gas when the partition is removed? This seems to run counter to the idea of an aerosol spray can getting cold when it is discharged.

    Thank you
     
  2. jcsd
  3. Apr 28, 2012 #2

    I like Serena

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    Welcome to PF, 74baja! :smile:

    It is called "free expansion" (look it up).

    The expanding gas does not do any work, since there is no counter pressure.
    Since there is also no exchange of heat with the environment, the change in internal energy is zero (dU=dQ+dW=0).
    This is conservation of energy.

    Since the internal energy of an ideal gas depends only on T (U=nCvT), T must have remained constant.
     
  4. Apr 28, 2012 #3
    I see, thank you. So, the change in temperature when a can is discharged into the atmosphere is due to the opposing pressure of the atmosphere, and the work that the can gas must do to overcome it?

    Thanks
     
  5. Apr 28, 2012 #4

    I like Serena

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    Yes.
    The expanding gas from the can has to push the opposing air away.
    As a consequence the molecules lose kinetic energy, which means that the temperature drops.
     
  6. Apr 28, 2012 #5

    rude man

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    No. Even if you sprayed into a vacuum, you would feel a colder can.

    By conservation of energy the kinetic energy acquired by the ejected propellant molecules has to come from somewhere, and that somewhere is the loss of heat energy (3kT/2 per molecule) of the remaining pressurized propellant.

    If the propellant is a liquid, the effect is much more pronounced since then the latent heat of vaporization is the major absorber of environmental heat. A can of freon (like you can't get any more) is a good example. So is butane lighter fluid.
     
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