Internal energy

We can assume that internal energy of a homogeneous and pure phase is a function of temperature and volume, so $u=u(T, v) \\ du=\left(\frac{\partial u}{\partial T} \right)_v dT+\left(\frac{\partial u}{\partial v} \right)_T dv$ We know that heat capacity at constant volume is defined as $c_v=\left(\frac{\partial u}{\partial T} \right)_v$ , thus $du=c_v dT+\left(\frac{\partial u}{\partial v} \right)_T dv \ \ \ \ \ (1)$ Now remember from Thermodynamic square that $du=Tds-Pdv$ , where $s$ is entropy, and divide both sides of this differential relation by $dv$ in constant temperature to get $\left(\frac{\partial u}{\partial v} \right)_T=T \left(\frac{\partial s}{\partial v} \right)_T-P \ \ \ \ \ (2)$ From Maxwell relations we know that $\left(\frac{\partial s}{\partial v} \right)_T=\left(\frac{\partial P}{\partial T} \right)_v$ ; Put this back in $(2)$ : $\left(\frac{\partial u}{\partial v} \right)_T=T \left(\frac{\partial P}{\partial T} \right)_v-P$ And finally put the last relation in $(1)$ : $du=c_v dT+\left[ T \left(\frac{\partial P}{\partial T} \right)_v-P \right] dv$ Therefore, this relation is true for both ideal and real gases.

The above expression is valid for both ideal and real if you simplify it you will get dU=CvdT for ideal gases.