For an isothermal process, the internal energy of a gas

Solution(By Examveda Team)

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.