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

Answer: Option A

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