The mean kinetic energy of 1 , mole of gas pe | vedclass

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Question

(D) According to the equipartition theorem,the average kinetic energy associated with each degree of freedom for a single molecule is $\frac{1}{2} kT$

Vedclass · Accepted Answer

$\frac{1}{2} RT$

Vedclass · Answer

(D) According to the equipartition theorem,the average kinetic energy associated with each degree of freedom for a single molecule is $\frac{1}{2} kT$,where $k$ is the Boltzmann constant and $T$ is the absolute temperature.For $1 \, mole$ of gas,the number of molecules is equal to the Avogadro number,$N_A$.Therefore,the mean kinetic energy for $1 \, mole$ per degree of freedom is $\frac{1}{2} kT 	imes N_A$.Since $R = k 	imes N_A$ (where $R$ is the universal gas constant),the expression becomes $\frac{1}{2} RT$.

The mean kinetic energy of $1 \, mole$ of gas per degree of freedom (on the basis of kinetic theory of gases) is

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