Mean kinetic energy (or average energy) per g | vedclass

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(A) For a monoatomic gas,the degrees of freedom $(f)$ is $3$.According to the equipartition theorem,the average kinetic energy per molecule is $\frac{

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$\frac{3}{2}RT$

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(A) For a monoatomic gas,the degrees of freedom $(f)$ is $3$.According to the equipartition theorem,the average kinetic energy per molecule is $\frac{f}{2}kT = \frac{3}{2}kT$,where $k$ is the Boltzmann constant.For one gram mole of gas,the number of molecules is equal to the Avogadro number $(N_A)$.Therefore,the average kinetic energy per mole is $E = N_A 	imes \frac{3}{2}kT = \frac{3}{2}(N_A k)T$.Since $R = N_A k$,the expression becomes $E = \frac{3}{2}RT$.

Mean kinetic energy (or average energy) per gram molecule of a monoatomic gas is given by

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