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Free expansion of an ideal gas.

  1. Sep 30, 2009 #1
    I know that the work done on the system in any free expansion is 0 since the external pressure is 0. However.. is q necessarily 0? Does the temperature necessarily stay constant for an ideal gas?

    I've been trying to justify the reasons for which q is necessarily 0, but I can't find a reason... Does anybody know why this is so? I know that U varies only with T for an ideal gas, but that would require justifying that the change in temperature is also 0.. does anybody have a mathematical justification for why q is necessarily 0 for a free expansion of an ideal gas?

  2. jcsd
  3. Oct 1, 2009 #2


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    The problem is usually described as adiabatic free expansion of an ideal gas to address this issue.
  4. Oct 1, 2009 #3
    It's the definition of an adiabatic course
  5. Oct 1, 2009 #4

    Andrew Mason

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    Apply the first law. Does the gas do work? Does heat flow into or out of the gas from/to the surroundings? What does that tell you about internal energy of the gas?

  6. Oct 1, 2009 #5
    The gas does no work, but heat could, in theory, flow in or out of the gas. I don't see why it can't... The problem doesn't specifically say that the process is adiabatic, but... I guess it is.. Obviously, that would solve all of the problems.

    Thanks everybody!
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