For an isothermal process, the internal energy of a gas

Answer: Option C

The internal energy ($$U$$) is a function of

$$dU=CvdT-\left[P+T\left\{\frac{\left(\frac{\partial V}{\partial T}\right)p}{\left(\frac{\partial V}{\partial P}\right)T}\right\}dV\right]$$

For an ideal gas, $$PV=RT$$

So, $$\left(\frac{\partial V}{\partial T}\right)p=\frac{R}{P}\text{ and}\left(\frac{\partial V}{\partial T}\right)T=\frac{-RT}{P^2}$$

Hence, $$dU=CvdT$$
So, for an ideal gas if it undergoing isothermal change $$\left(dT=0\right)\Rightarrow dU=0$$
So, the questioned should be changed and should be mentioned for an ideal gas.