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Heat and Thermodynamics question

2025 · 7 Apr · Shift 2 · Q69
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Heat and Thermodynamics question

2025 · 7 Apr · Shift 2 · Q69

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1

Match List - I with List - II.

List - I List - II
(A) Isothermal (I) ΔW (work done) = 0
(B) Adiabatic (II) ΔQ (supplied heat) = 0
(C) Isobaric (III) ΔU (change in internal energy) ≠ 0
(D) Isochoric (IV) ΔU = 0

Choose the correct answer from the options given below :

  1. A
    (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  2. B
    (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  3. C
    (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  4. D
    (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
View written solutionFree

Correct answer: C

  1. Recall the defining properties of each thermodynamic process

    For an ideal gas:

    • Isothermal: temperature remains constant, so internal energy does not change. ΔU=0\Delta U = 0ΔU=0 Hence, (A)→(IV)\text{(A)} \to \text{(IV)}(A)→(IV)

    • Adiabatic: no heat is exchanged with surroundings. ΔQ=0\Delta Q = 0ΔQ=0 Hence, (B)→(II)\text{(B)} \to \text{(II)}(B)→(II)

    • Isochoric: volume remains constant, so no work is done because W=∫P dV=0W = \int P\,dV = 0W=∫PdV=0 Hence, (D)→(I)\text{(D)} \to \text{(I)}(D)→(I)

    • Isobaric: pressure remains constant. In general, temperature changes, so for an ideal gas internal energy changes. ΔU≠0\Delta U \ne 0ΔU=0 Hence, (C)→(III)\text{(C)} \to \text{(III)}(C)→(III)

  2. Final matching

    (A)−(IV), (B)−(II), (C)−(III), (D)−(I)\boxed{(A)-(IV),\ (B)-(II),\ (C)-(III),\ (D)-(I)}(A)−(IV), (B)−(II), (C)−(III), (D)−(I)​

  3. Compare with options

    This corresponds to Option C.

  4. Comparison with stored answer

    Stored correct answer: C

    Our derived answer: C

    So, they agree.

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