The mean energy of a molecule of an ideal gas | vedclass

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(B) According to the kinetic theory of gases,the average kinetic energy of a molecule of an ideal gas is given by the equipartition theorem.For a mona

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$\frac{3}{2} KT$

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(B) According to the kinetic theory of gases,the average kinetic energy of a molecule of an ideal gas is given by the equipartition theorem.For a monatomic ideal gas,the molecule has $3$ degrees of freedom.The mean energy per molecule is given by $E = \frac{f}{2} k_{B} T$,where $f$ is the number of degrees of freedom.For a monatomic gas,$f = 3$,so the mean energy is $E = \frac{3}{2} k_{B} T$.Therefore,the correct option is $B$.

The mean energy of a molecule of an ideal gas is

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