Thermodynamics

JEE Main level, single correct, medium

11 of 190

Question 11 of 190

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

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

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\Rightarrow \gamma =1+\dfrac{2}{3}=\dfrac{5}{3}
For diatomic rigid molecule BB, f=5f=5
⇒γ=1+25=75\Rightarrow \gamma =1+\dfrac{2}{5}=\dfrac{7}{5}
For diatomic non-rigid molecule CC, f=7f=7 ⇒γ=1+27=97\Rightarrow \gamma =1+\dfrac{2}{7}=\dfrac{9}{7}
For Triatomic rigid molecule DD, f=6f=6 ⇒γ=1+26=43\Rightarrow \gamma =1+\dfrac{2}{6}=\dfrac{4}{3}

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