An ideal gas undergoes a cyclic transformation starting from the point $A$ and coming back to the same point…


An ideal gas undergoes a cyclic transformation starting from the point $A$ and coming back to the same point by tracing the path $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{D} \rightarrow \mathrm{A}$ as shown in the three cases above.
Choose the correct option regarding $\Delta \mathrm{U}$ :
  1. $\Delta \mathrm{U}$ (Case-I) $=\Delta \mathrm{U}$ (Case-II) $=\Delta \mathrm{U}$ (Case-III)
  2. $\Delta \mathrm{U}$ (Case-I) $\gt \Delta \mathrm{U}$ (Case-III) $\gt \Delta \mathrm{U}$ (Case-II)
  3. $\Delta \mathrm{U}$ (Case-III) $\gt \Delta \mathrm{U}$ (Case-II) $\gt \Delta \mathrm{U}$ (Case-I)
  4. $\Delta \mathrm{U}$ (Case-I) $\gt \Delta \mathrm{U}$ (Case-II) $\gt \Delta \mathrm{U}$ (Case-III)

Solution

As internal energy ' $U$ ' is a state function, its cyclic integral must be zero in a cyclic process
$\therefore \Delta U \text { case }(I)=\Delta U \text { case }(\text { II })=\Delta U \text { case }(\text { III })$

Asked in: JEE Main 2025 (28 Jan Shift 2)

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