# Stuck on the derivation of pV^gamma=c

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.

Bystander
Homework Helper
Gold Member
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 $\frac{\partial T }{\partial V}_p$.

Bystander
Homework Helper
Gold Member
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 ____ ?

T/V. You have my gratitude. I will now shed a tear for all the sleep this trivial thing has cost me.

Bystander