### Related Formula For any state function like Internal Energy (U$U$), the cyclic integral over a complete closed loop is identically zero: oint dU = 0 implies Delta U_textcyclic = 0$\oint dU = 0 \implies \Delta U_{\text{cyclic}} = 0$ ### Core Logic Internal energy (U$U$) depends only on the initial and final states of the thermodynamic system, not on the path followed. In all three listed cases, the ideal gas undergoes a complete cyclic path that returns to its original configuration state A$A$. ### Step 1: Final Evaluation Since every transformation begins and ends at point A$A$: Delta U_textCase-I = 0$\Delta U_{\text{Case-I}} = 0$ Delta U_textCase-II = 0$\Delta U_{\text{Case-II}} = 0$ Delta U_textCase-III = 0$\Delta U_{\text{Case-III}} = 0$ Therefore, Delta Utext (Case-I) = Delta Utext (Case-II) = Delta Utext (Case-III)$\Delta U\text{ (Case-I)} = \Delta U\text{ (Case-II)} = \Delta U\text{ (Case-III)}$. ### Pattern Recognition Do not waste time calculating path areas or values if the question asks for a state function change (Delta U$\Delta U$, Delta H$\Delta H$, Delta S$\Delta S$, Delta G$\Delta G$) over a cyclic loop. The answer is instantly zero for all cases! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics