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 $\Delta \mathrm{U}$.

(D) $\Delta \mathrm{U}$ (Case-I) $=\Delta \mathrm{U}$ (Case-II) $=\Delta \mathrm{U}$ (Case-III)

(A) $\Delta \mathrm{U}($ Case-III $)>\Delta \mathrm{U}($ Case-II $)>\Delta \mathrm{U}($ Case-I)

(B) $\Delta \mathrm{U}($ Case-I $)>\Delta \mathrm{U}($ Case-II $)>\Delta \mathrm{U}($ Case-III $)$

(C) $\Delta \mathrm{U}($ Case-I $)>\Delta \mathrm{U}($ Case-III $)>\Delta \mathrm{U}($ Case-II $)$

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Chapter_pending

An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path

A \to B \to C \to D \to A

as shown in the three cases above.

Choose the correct option regarding

Δ U

(A) $Δ U ($ Case-III $) > Δ U ($ Case-II $) > Δ U ($ Case-I)
(B) $Δ U ($ Case-I $) > Δ U ($ Case-II $) > Δ U ($ Case-III $)$
(C) $Δ U ($ Case-I $) > Δ U ($ Case-III $) > Δ U ($ Case-II $)$
(D) $Δ U$ (Case-I) $= Δ U$ (Case-II) $= Δ U$ (Case-III)✓ Correct

Answer: (D)

Solution

Internal energy is a state function — it depends only on the state of the system, not on the path taken to reach it. In a cyclic process the system returns to its exact starting state, so the net change over the full cycle

A \to B \to C \to D \to A

must vanish:

\oint d U = 0

. This holds regardless of the shape of the p-V path, so it is equally true for Case-I, Case-II and Case-III:

Δ U (Case-I) = Δ U (Case-II) = Δ U (Case-III) = 0

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Chapter_pending

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