A polyatomic gas with n degrees of freedom ha | vedclass

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Question

(C) According to the law of equipartition of energy,each degree of freedom contributes $\frac{1}{2}kT$ to the average kinetic energy of a molecule.For

Vedclass · Accepted Answer

$\frac{nkT}{2}$

Vedclass · Answer

(C) According to the law of equipartition of energy,each degree of freedom contributes $\frac{1}{2}kT$ to the average kinetic energy of a molecule.For a gas molecule with $n$ degrees of freedom,the total mean energy per molecule is the sum of the energies associated with each degree of freedom.Therefore,the mean energy per molecule $E = n 	imes (\frac{1}{2}kT) = \frac{nkT}{2}$.Thus,the correct option is $C$.

$A$ polyatomic gas with $n$ degrees of freedom has a mean energy per molecule given by (where $k$ is Boltzmann's constant and $T$ is temperature).

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