Match List - I with List - II. Δ Q = Heat supplied Δ W = Work done by the system Δ U = Change in internal energy P = Pressure of the system Δ V = Change in volume of the system Choose the correct answer from the options given below: (a) (A)-(IV), (B)-(I), (C)-(III), (D)-(II) (b) (A)-(IV), (B)-(II), (C)-(III), (D)-(I) (c) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) (d) (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

Answer: (b). Isobaric: Δ W = PΔ V, so Δ Q = Δ U + PΔ V: (A)–(IV). Isochoric: Δ V = 0, Δ W = 0, so Δ Q = Δ U: (B)–(II). Adiabatic: Δ Q = 0: (C)–(III). Isothermal (ideal gas): Δ U = 0, so Δ Q = Δ W: (D)–(I). So the answer is (b).

Match List - I with List - II. Δ Q = Heat supplied Δ W = Work done by… | JEE Main 2025 (4 Apr, Shift 2)

Q65JEE Main 2025 · 4 APR · SHIFT 2Medium

Thermodynamic Processes

Match List - I with List - II.

$\Delta Q =$ Heat supplied

$\Delta W =$ Work done by the system

$\Delta U =$ Change in internal energy

$P =$ Pressure of the system

$\Delta V =$ Change in volume of the system

Choose the correct answer from the options given below:

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Answer: (b)

Isobaric: $\Delta W = P\Delta V$ , so $\Delta Q = \Delta U + P\Delta V$ : (A)–(IV).
Isochoric: $\Delta V = 0$ , $\Delta W = 0$ , so $\Delta Q = \Delta U$ : (B)–(II).
Adiabatic: $\Delta Q = 0$ : (C)–(III).
Isothermal (ideal gas): $\Delta U = 0$ , so $\Delta Q = \Delta W$ : (D)–(I).
So the answer is (b).

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Thermodynamic Processes question from JEE Main 2025 (4 Apr, Shift 2)

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