Stuck on the derivation of pV^gamma=c

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The discussion centers on the derivation of the equation pVγ = constant for an ideal gas undergoing an adiabatic process. The participants clarify that for an ideal gas, the relationship T = pV/nR leads to the conclusion that (∂T/∂V)p = T/V, rather than p/nR. This distinction is crucial for understanding the behavior of temperature and volume in adiabatic processes, where dQ = 0 and the specific heat capacities Cp and CV are involved in defining γ.

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  • Understanding of ideal gas laws, specifically T = pV/nR
  • Familiarity with adiabatic processes in thermodynamics
  • Knowledge of specific heat capacities Cp and CV
  • Basic calculus, particularly partial derivatives
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Fosheimdet
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I have been tearing my hair out for a while over a step in the proof of the relation pV^{\gamma}=constant. The textbook has assumed that we are dealing with an ideal gas undergoing an adiabatic process. Therefore dQ=0 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 \left(\frac{\partial T}{\partial V}\right)_p=p/nR? 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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Fosheimdet said:
T=pV/nR
What's dT? Remember, T is function of both V and P.
 
Yes, but p is constant as indicated by the subscript in \frac{\partial T }{\partial V}_p.
 
Oops, dragged a "red herring" in front of you. Maybe it's too obvious.
Fosheimdet said:
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 ____ ?
 
T/V. You have my gratitude. I will now shed a tear for all the sleep this trivial thing has cost me.
 
Been there, done that.
 

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