The average thermal energy for a mono-atomic gas is: (${{\rm{k}}_{\rm{B}}}$ is Boltzmann constant and ${\rm{T}}$, absolute temperature)

$\frac{3}{2}\,{{\rm{k}}_{\rm{B}}}{\rm{T}}$

$\frac{1}{2}\,{{\rm{k}}_{\rm{B}}}{\rm{T}}$

$\frac{5}{2}\,{{\rm{k}}_{\rm{B}}}{\rm{T}}$

$\frac{7}{2}\,{{\rm{k}}_{\rm{B}}}{\rm{T}}$

JEE PHYSICS MCQ: KINETIC THEORY OF GASES

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The average thermal energy for a mono-atomic gas is: (

{\mathrm{k}_\mathrm{B}}

is Boltzmann constant and

\mathrm{T}

, absolute temperature)

View answer and explanation

Correct option: B

\frac{3}{2}\,{\mathrm{k}_\mathrm{B}}\mathrm{T}

Option B is correct. A monoatomic gas has 3 degrees of freedom, so the average thermal energy is (3/2) * k_B * T.

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