A polyatomic gas with n degrees of freedom ha | vedclass

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Question

(C) According to the law of equipartition of energy,the mean kinetic energy associated with each degree of freedom per molecule is $\frac{1}{2} K T$,w

Vedclass · Accepted Answer

$\frac{n K T}{2}$

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(C) According to the law of equipartition of energy,the mean kinetic energy associated with each degree of freedom per molecule is $\frac{1}{2} K T$,where $K$ is the Boltzmann constant and $T$ is the absolute temperature.Given that the gas has $n$ degrees of freedom,the total mean kinetic energy per molecule is the sum of the kinetic energy associated with each degree of freedom.Therefore,the mean kinetic energy per molecule $= n 	imes (\frac{1}{2} K T) = \frac{n K T}{2}$.

$A$ polyatomic gas with $n$ degrees of freedom has a mean kinetic energy per molecule given by (if $K$ is Boltzmann's constant):

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