JEE PHYSICS MCQ: KINETIC THEORY OF GASES

1 Medium

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Application +4 | -1

Calculate the total number of degrees of freedom possessed by the molecules in

1 \mathrm{~cm}^3

\mathrm{H}_2

gas at N.T.P.

View answer and explanation

Correct option: C

{1.34375} \times 10^{20}

Option A (

{4.34375} \times 10^{20}

) is correct because it represents the total degrees of freedom calculated using Avogadro's number, the volume, and the degrees of freedom for a diatomic molecule.

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{C_{rms}} =

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{V_e} =

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