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JEE Main20204 Sep 2020Morning ShiftPhysicsThermodynamicsActual

Match the C p C v ratio for ideal gases with different type of molecules: Molecule Type C p / C v (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

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 + 2 f     . . . ( 1 ) For monoatomic molecule, A , f = 3 ⇒ γ = 1 + 2 3 = 5 3 For diatomic rigid molecule B , f = 5 ⇒ γ = 1 + 2 5 = 7 5 For diatomic non-rigid molecule C , f = 7 ⇒ γ = 1 + 2 7 = 9 7 For Triatomic rigid molecule D , f = 6 ⇒ γ = 1 + 2 6 = 4 3

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