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

Since $\delta Q=0$:

$$dU=\delta W,$$

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


since for an ideal gas $PV=nRT:$

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


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

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.


1 Answer 1


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 $(dV=0)$.


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