A polyatomic gas with n degrees of freedom has a mean energy per molecules given by :

\frac{n}{2} R T

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\frac{1}{2} R T

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\frac{n}{2} k T

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\frac{1}{2} k T

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Solution

The correct option is C $\frac{n}{2} k T$
Mean energy of gas per degree of freedom $= \frac{1}{2} k T$
where $k$ is the Boltzmann constant.
Thus mean energy of gas with $n$ degrees of freedom $= \frac{n}{2} k T$

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Similar questions

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.

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. If a gas has n degrees of freedom ratio of specific heats of gas is?

Q. A sample of a gas contains

N_{1}

molecules and the total kinetic energy at

- 123^{\circ}

C is

E_{1}

ergs. Another sample of a gas at

27^{\circ}

C has a total kinetic energy as

2 E_{1}

ergs. Assuming gases to be ideal, the number of gas molecules in the second sample will be:

The ratio of the specific heats $\frac{C_{p}}{C_{v}} = γ$ in terms of degrees of freedom (n) is given by :

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A polyatomic gas with n degrees of freedom has a mean energy per molecules given by :

The correct option is C n2kTMean energy of gas per degree of freedom =12kTwhere k is the Boltzmann constant.Thus mean energy of gas with n degrees of freedom =n2kT

The correct option is C $\frac{n}{2} k T$
Mean energy of gas per degree of freedom $= \frac{1}{2} k T$
where $k$ is the Boltzmann constant.
Thus mean energy of gas with $n$ degrees of freedom $= \frac{n}{2} k T$