The mean kinetic energy of a perfect mono atomic gas molecule at the temperature $T^{o} K$ is :

\frac{1}{2} k T

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k T

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\frac{3}{2} k T

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2 k T

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Solution

The correct option is C $\frac{3}{2} k T$
A monoatomic gas has 3 degrees of freedom, one each along for each axis (translation motion).
The kinetic energy per molecule per degree of freedom is $\frac{1}{2} k T$
i.e.
The kinetic energy per molecule for a monoatomic gas is $\frac{3}{2} k T$
Hence option C is correct.

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Similar questions

Q. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its electron in ground state, assuming average kinetic energy of hydrogen gas molecule =

\frac{3}{2} k T

Q. The average

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\frac{3}{2} k T

. Find the temperature at which the average KE of molecules equal to binding energy of its atoms.

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

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. 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.

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The mean kinetic energy of a perfect mono atomic gas molecule at the temperature ToK is :

The correct option is C 32kTA monoatomic gas has 3 degrees of freedom, one each along for each axis (translation motion).The kinetic energy per molecule per degree of freedom is 12kTi.e.The kinetic energy per molecule for a monoatomic gas is 32kTHence option C is correct.

The mean kinetic energy of a perfect mono atomic gas molecule at the temperature $T^{o} K$ is :