JEE PHYSICS MCQ: KINETIC THEORY OF GASES

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For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molecular specific heat

\left( {{C_V}} \right)

and gas constant

\left( R \right)

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Correct option: D

{C_V} = 3R

Option D (

{C_V} = 3R

) is correct because the molecular specific heat is calculated using the equipartition theorem, where each degree of freedom contributes R/2 to the specific heat.

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{V_e} =

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\rho

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