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Heat capaticy of isobaric process with real gas

  1. Oct 27, 2012 #1
    Hey,
    during last lecture of thermodynamics we did polytropic processes, and with no
    discuss teacher said, that isobaric process is in general polytropic. If we had ideal
    gas, then it is clear, bud what if the gas isn't ideal?
    I tried to proof it, but I stuck and don't know how to continue.
    If I write the definition of polytropic process
    [itex]c \mathrm{d}T = \mathrm{\delta}Q[/itex], where [itex]c[/itex] must be constant. For isobaric process we have
    [itex]c_P = \left(\frac{\mathrm{d}Q}{\mathrm{d}T}\right)_P[/itex], where index right down the derivation notes what state variable is constant. To define the state in thermodynamics, we
    need just two state variable, so
    [itex]c_P = \left(\frac{\mathrm{d}U(P,T)}{\mathrm{d}T}\right)_P + \left(\frac{P\mathrm{d}V
    (P,T)}{\mathrm{d}T}\right)_P[/itex]. And know I don't know how to continue. I tried said that
    [itex]U(T,V(P,T))[/itex] and chain rule, but it didn't look better . . .
    Thanks for any help . . .
     
    Last edited: Oct 27, 2012
  2. jcsd
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