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Internal Energy

  1. Nov 17, 2004 #1
    I have a question concerning internal energy and the First Law of Thermodynamics. If you plotted P vs V and the shape of the graph is an enclosed shape ie) square or triangle, would the total internal energy in one cycle be zero? (one cycle is for example-start at top left of square and finish at top left of square). I think this is right, but am a little uncertain being a Physics moron and all. I would appreciate anyone's input. Thanks
  2. jcsd
  3. Nov 17, 2004 #2
    If anyone has any idea on this I would love to hear from you. This thought is what I have based several questions on, and if I am incorrect I need to figure out something else. Thanks.
  4. Nov 18, 2004 #3


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    Internal Energy is a variable of state. Given some thermodynamic trajectory, for instance from 1 to 2, the change of internal energy is given by:

    [tex] \int_1^2 du=u_2-u_1[/tex]

    Due to the fact in ideal gases and perfect liquids the internal energy is only a function of the temperature:

    [tex] u=c_vT[/tex]

    then if 1 is the same thermodynamic state than 2 the change of internal energy is zero, no matter which trajectory it was.
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