The internal energy of an ideal gas does not change in a reversible __________ process.

We know for an ideal gas the internal energy is given by:
Actually the internal energy (U) of any substance is a function of
$$\eqalign{ & dU = CvdT - \left[ {P + T\left( {\frac{{\left( {\frac{{\partial V}}{{\partial T}}} \right)p}}{{\left( {\frac{{\partial V}}{{\partial P}}} \right)T}}} \right)dV} \right] \cr & {\text{For an ideal gas, }}PV = RT \cr & {\text{So, }}\left( {\frac{{\partial V}}{{\partial T}}} \right)p = \frac{R}{P}\,\,{\text{and }}\left( {\frac{{\partial V}}{{\partial P}}} \right)T = - \frac{{RT}}{{{P^2}}} \cr & {\text{Hence, }}dU = CvdT \cr} $$
So, in case of isothermal process the internal energy remains constant.