# Derivation of the Adiabatic Relations

My professor offered this derivation of the adiabatic relation $$TV^{\gamma-1}=\text{const.}$$ :

Since $$\delta Q=0$$:

$$dU=\delta W,$$

using $$C_V=\left(\frac{\partial U}{\partial T}\right)_V\implies dU=C_V\, dT:$$

$$C_V\,dT=-PdV,\tag{1}$$

since for an ideal gas $$PV=nRT:$$

$$C_V\,dT=-\frac{nRT}{V}dV\implies C_V\frac{dT}{T}=-nR\frac{dV}{V},\tag{2}$$

integrating:

$$C_V\ln{T}=-nR\ln{V}+C_1,\quad C_1\in\mathbb{R}.\tag{3}$$

Defining $$\gamma=\frac{C_P}{C_V}$$ and using $$C_P-C_V=nR$$:

$$\ln{T}+(\gamma-1)\ln{V}=C_2\implies TV^{\gamma-1}=\text{const.}\quad\blacksquare$$

My question is why can we use the heat capacity at constant volume here? If we assume constant volume, doesn't equation (1) lead to a contradiction in that $$dV$$ should be zero? Conceptually, I would think that the $$C_V$$ relation should only be used when, in this case, the work done on the gas is converted directly into internal energy at constant volume.

The relation $$dU=C_VdT$$ holds for ideal gases, whether the process is at constant volume or not. This is because one can assume the internal energy to be a function of any two of the thermodynamic coordinates $$(T,V,P)$$, in particular, if $$U(T,V)$$ then
$$dU=\left(\frac{\partial U}{\partial T}\right)_V dT+\left(\frac{\partial U}{\partial V}\right)_T dV.$$
The first term on the right-hand side is $$C_VdT$$ because $$C_V$$ is defined as $$(\partial U/\partial T)_V$$. The second term is zero for ideal gases because the internal energy of an ideal gas is a function of temperature only $$U=U(T)$$, so the partial derivative $$(\partial U/\partial V)_T=0$$.
In the end, notice that to derive $$dU=C_VdT$$ for ideal gases one does not need to assume a process at constant volume $$(dV=0)$$.