An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A rightarrow B rightarrow C rightarrow D rightarrow A as shown in the three cases above. Choose the correct option regarding Δ U .

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

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

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

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

Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 28 January, Evening Shift — Question 33

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 $\Delta \mathrm{U}$ .

Option A:
$\Delta \mathrm{U}($ Case-III) $>\Delta \mathrm{U}($ Case-II $)>\Delta \mathrm{U}($ Case-I)
Option B:
$\Delta \mathrm{U}($ Case-I) $>\Delta \mathrm{U}($ Case-II $)>\Delta \mathrm{U}($ Case-III $)$
Option C:
$\Delta \mathrm{U}($ Case-I $)>\Delta \mathrm{U}($ Case-III $)>\Delta \mathrm{U}($ Case-II $)$
Option D:
$\Delta \mathrm{U}($ Case-I $)=\Delta \mathrm{U}($ Case-II $)=\Delta \mathrm{U}($ Case-III $)$
Correct

Answer: D

Step-by-step solution

As internal energy ' $U$ ' is a state function, its cyclic integral must be zero in a cyclic process

$\therefore \Delta \mathrm{U}$ case (I) $=\Delta \mathrm{U}$ case (II) $=\Delta \mathrm{U}$ case (III)

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Chemistry
Chapter: Thermodynamics & Thermochemistry
Topic: Internal Energy and the First Law

Consider the following data : Heat of formation of CO 2( g)=-393.5 kJ mol^-1 Heat of formation of H 2 O( l)=-286.0 kJ mol^-1 Heat of…

Answer: D

Step-by-step solution

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