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JEE Main 20204 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
4 September
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2020 · 4 September · Morning Shift · Physics · Thermodynamics

Match the C _ p C _ v ratio for ideal gases with different type of...

Match the CpCv\dfrac{{\text{C}}_{\text{p}}}{{\text{C}}_{\text{v}}} ratio for ideal gases with different type of molecules:

Molecule Type Cp/Cv{\text{C}}_{\text{p}} / {\text{C}}_{\text{v}}
(A) Monoatomic (I) 7/57 / 5
(B) Diatomic rigid molecules (II) 9/79 / 7
(C) Diatomic non-rigid molecules (III) 4/34 / 3
(D) Triatomic rigid molecules (IV) 5/35 / 3

Options

  1. A(A) – (IV), (B) – (II), (C) – (I), (D) – (III)
  2. B(A) – (III), (B) – (IV), (C) – (II), (D) – (I)
  3. C(A) – (IV), (B) – (I), (C) – (II), (D) – (III)
  4. D(A) – (II), (B) – (III), (C) – (I), (D) – (IV)

Correct answer

C. (A) – (IV), (B) – (I), (C) – (II), (D) – (III)

Step-by-step solution

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

γ=1+2f...(1)\gamma = 1 + \dfrac{2}{f} . . . ( 1 )

For monoatomic molecule, AA, f=3f = 3

⇒γ=1+23=53⇒ \gamma = 1 + \dfrac{2}{3} = \dfrac{5}{3}

For diatomic rigid molecule BB, f=5f = 5

⇒γ=1+25=75⇒ \gamma = 1 + \dfrac{2}{5} = \dfrac{7}{5}

For diatomic non-rigid molecule CC, f=7f = 7 ⇒γ=1+27=97⇒ \gamma = 1 + \dfrac{2}{7} = \dfrac{9}{7}

For Triatomic rigid molecule DD, f=6f = 6 ⇒γ=1+26=43⇒ \gamma = 1 + \dfrac{2}{6} = \dfrac{4}{3}

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