JEE Main 2025 Apr 7 Shift 2, Physics Q17: Thermodynamic Processes

JEE Main2025Apr 7, Shift 2Physics

Match List-I with List-II.

List-I	List-II
(A) Isothermal	(I) $Δ W$ (work done) = 0
(B) Adiabatic	(II) $Δ Q$ (supplied heat) = 0
(C) Isobaric	(III) $Δ U$ (change in internal energy) $\neq = 0$
(D) Isochoric	(IV) $Δ U = 0$

Choose the correct answer from the options given below :

A
(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
B
(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
C
(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
D
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Solution

An isothermal process holds $T$ constant, so for an ideal gas the internal energy (which depends only on temperature) is unchanged: $Δ U = 0$ , matching (IV).

An adiabatic process is defined by the absence of heat exchange with surroundings: $Δ Q = 0$ , matching (II).

An isobaric process happens at constant pressure with a temperature change, so $Δ U \neq = 0$ , matching (III).

An isochoric (constant volume) process involves no volume change, so no work is done: $Δ W = 0$ , matching (I).

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JEE Main 2025 Apr 7 Shift 2, Physics Q17: Thermodynamic Processes

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