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Stuck on the derivation of pV^gamma=c

  1. Dec 23, 2014 #1
    I have been tearing my hair out for a while over a step in the proof of the relation [itex]pV^{\gamma}=constant[/itex]. The textbook has assumed that we are dealing with an ideal gas undergoing an adiabatic process. Therefore [itex]dQ=0[/itex] and we get

    $$C_vdT + (c_p-c_V)\left(\frac{\partial T}{\partial V}\right)_pdV=0$$
    which gives
    $$dT=-(\gamma-1)\left(\frac{\partial T}{\partial V}\right)_pdV$$
    Where $$\gamma=\frac{C_p}{C_V}$$

    Now comes the part I don't get. They say that because we are dealing with an ideal gas, we have $$T=pV/nR$$ which gives $$\left(\frac{\partial T}{\partial V}\right)_p = \frac{T}{V}$$
    Why isn't [itex]\left(\frac{\partial T}{\partial V}\right)_p=p/nR[/itex]? Is there something obvious I'm missing? Would love to get this cleared up so I can get some sleep tonight.
     
  2. jcsd
  3. Dec 23, 2014 #2

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    What's dT? Remember, T is function of both V and P.
     
  4. Dec 23, 2014 #3
    Yes, but p is constant as indicated by the subscript in [itex]\frac{\partial T }{\partial V}_p[/itex].
     
  5. Dec 23, 2014 #4

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    Oops, dragged a "red herring" in front of you. Maybe it's too obvious.
    What's p/nR? Ideal gas. Rearrange things any way you wish, and p/nR is also equal to ____ ?
     
  6. Dec 23, 2014 #5
    T/V. You have my gratitude. I will now shed a tear for all the sleep this trivial thing has cost me.
     
  7. Dec 23, 2014 #6

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    Been there, done that.
     
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