According to the equipartition principle, each degree of freedom contribute following amount to the translational kinetic energy of molecule: (hence k = Boltzmann constant)

\frac{1}{2} k T

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k T

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\frac{3}{2} k T

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\frac{1}{3} k T

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Solution

The correct option is B $\frac{1}{2} k T$
According to the equipartition principle, each degree of freedom contributes an amount of $\frac{1}{2} k T$ to the kinetic energy of the molecule.

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Q. Assertion :The average translational kinetic energy per molecule of the gas per degree of freedom is

\frac{1}{2} K T

. Reason: For every molecule there are three rotational degree of freedom.

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For the theory of equipartition of energy the energy associated with the each molecule is f/ 2 KT if the gas has n number of moles , now how we can convert the moles into molecules and the energy associated with total molecule is found to be n×na×f/2KT. Here na×K=R(gas constant) how? And total energy associated with total molecule is n ×R×T×f/2.

Q. According to the Boltzmann's law of equipartition of energy, the energy per degree of freedom and at a temperature T K is :

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According to the equipartition principle, each degree of freedom contribute following amount to the translational kinetic energy of molecule: (hence k = Boltzmann constant)

The correct option is B 12kTAccording to the equipartition principle, each degree of freedom contributes an amount of 12kT to the kinetic energy of the molecule.

The correct option is B $\frac{1}{2} k T$
According to the equipartition principle, each degree of freedom contributes an amount of $\frac{1}{2} k T$ to the kinetic energy of the molecule.