On the basis of the kinetic theory of gases,t | vedclass

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Question

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

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$\frac{1}{2}RT$

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(A) 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 $k 	imes N_A = R$ (the universal gas constant),the expression becomes $\frac{1}{2}RT$.

On the basis of the kinetic theory of gases,the mean kinetic energy $(K.E.)$ of $1\, mole$ per degree of freedom is:

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