JEE Main 2020PhysicsThermodynamicsEasyMCQ

JEE Main 2020 — Thermodynamics Question with Solution

JEE Main 2020 (04 Sep Shift 1)

Question

Match the CpCv ratio for ideal gases with different type of molecules:

Molecule Type Cp/Cv
(A) Monoatomic (I) 7/5
(B) Diatomic rigid molecules (II) 9/7
(C) Diatomic non-rigid molecules (III) 4/3
(D) Triatomic rigid molecules (IV) 5/3

Choose an option

Show full solutionCorrect option: C
Correct answer
C(A) – (IV), (B) – (I), (C) – (II), (D) – (III)

Step-by-step explanation

Relation between the ratio of specific heat and degree of freedom is given by,

γ=1+2f  ...(1)

For monoatomic molecule, A, f=3

 ⇒γ=1+23=53

For diatomic rigid molecule B, f=5

 ⇒γ=1+25=75

For diatomic non-rigid molecule C, f=7 ⇒γ=1+27=97

For Triatomic rigid molecule D, f=6 ⇒γ=1+26=43

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About this question

This is a previous-year question from JEE Main 2020, covering the Thermodynamics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.