For the irreversible expansion of an ideal gas under isothermal conditions, the correct option is:

Solution & Explanation

\textbf{Answer:} (A) $\Delta U=0$, $\Delta S_{total}\ne 0$ \textbf{Solution:} For an isothermal process ($\Delta T=0$) of an ideal gas, $\Delta U=n{C}_v\Delta T=0$. Because the expansion is irreversible, the total entropy of the universe increases: $\Delta S_{total}=\Delta S_{sys}+\Delta S_{surr}>0$, i.e. $\Delta S_{total}\ne 0$. Hence $\Delta U=0$ and $\Delta S_{total}\ne 0$. Mapped: source lowercase answer 'a' → option A.