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Thermodynamics basics

  1. Dec 18, 2005 #1
    Can anyone here help me to derive that during an adiabatic expansion, PV^gamma?is a constant, as well as other expressions similar to the above?
    I have found the following equation using definite integration and the basic formulae, nRdT/gamma - 1 = work done.
  2. jcsd
  3. Dec 18, 2005 #2


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    You can prove [tex] P V^{\gamma}[/tex] is constant for an ideal gas in an adiabatic process from the first law ( [itex] dU = dQ - pdV , dQ=0) [/itex] and the ideal gas law [itex] (pV=nRT) [/itex].
    How did you get your formula nRdT/gamma - 1 = work done.?
    Last edited: Dec 18, 2005
  4. Dec 19, 2005 #3

    Andrew Mason

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    Since heat flow (Q) is zero, use:

    [tex]nC_VdT = dU = PdV[/tex] and

    [tex]VdP + PdV = nRdT = n(C_P - C_V)dT[/tex]

    This will give two expressions for ndT. Integrate both expressions.
    This follows from the adiabatic condition. One can express the work as:

    [tex]W = \int_{V_i}^{V_f} PdV = \int_{V_i}^{V_f} \frac{PV^\gamma}{V^\gamma}dV = K\int_{V_i}^{V_f} \frac{dV}{V^\gamma}[/tex]
    Work that out to get the expression for Work.

    Last edited: Dec 19, 2005
  5. Dec 19, 2005 #4


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    Thanks for clearing the second part Andrew.
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