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Physics · Thermodynamics

JEE Main 2025 — 4 April, Evening Shift — Question 52

Match List - I with List - II.

List - IList - II
(A) Isobaric(I) Δ𝑄=Δ𝑊
(B) Isochoric(II) Δ𝑄=Δ𝑈
(C) Adiabatic(III) Δ𝑄= zero
(D) Isothermal(IV)   ⁣ ⁣Δ ⁣ ⁣ Q=  ⁣ ⁣Δ ⁣ ⁣ U+P  ⁣ ⁣Δ ⁣ ⁣ V\text{ }\!\!\Delta\!\!\text{ }Q=\text{ }\!\!\Delta\!\!\text{ }U+P\text{ }\!\!\Delta\!\!\text{ }V

ΔQ=\Delta Q= Heat supplied ΔW=\Delta W= Work done by the system ΔU=\Delta U= Change in internal energy P=P= Pressure of the system ΔV=\Delta V= Change in volume of the system

Choose the correct answer from the options given below:

  1. Option A:

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

  2. Option B:

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

  3. Option C:

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

    Correct
  4. Option D:

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

Answer: C

Step-by-step solution

Isobaric ⇒ΔQ=ΔU+∫PdV\Rightarrow \Delta Q=\Delta U+\int P d V

ΔQ‾=ΔU+PΔV\underline{\Delta Q}=\Delta U+P \Delta V

Isochoric ⇒ΔQ=ΔU\Rightarrow \Delta Q=\Delta U Adiabatic ⇒ΔQ=0\Rightarrow \Delta Q=0

Isothermal ⇒ΔQ=ΔW\Rightarrow \Delta Q=\Delta W

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Evening Shift
Subject
Physics
Chapter
Thermodynamics
Topic
Different Thermodynamic Processes

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Match List - I with List - II. List - I List - II --- --- (A)… | JEE Main 2025 PYQ with Solution · DhiX AI