Stuck on the derivation of pV^gamma=c

  • Thread starter Fosheimdet
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  • #1
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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.
 

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  • #2
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T=pV/nR
What's dT? Remember, T is function of both V and P.
 
  • #3
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Yes, but p is constant as indicated by the subscript in [itex]\frac{\partial T }{\partial V}_p[/itex].
 
  • #4
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Oops, dragged a "red herring" in front of you. Maybe it's too obvious.
p/nR \left(\frac{\partial T}{\partial V}\right)_p=p/nR? Is there something obvious I'm missing?
What's p/nR? Ideal gas. Rearrange things any way you wish, and p/nR is also equal to ____ ?
 
  • #5
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T/V. You have my gratitude. I will now shed a tear for all the sleep this trivial thing has cost me.
 
  • #6
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Been there, done that.
 

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