Question

In a diatomic molecule, the rotational energy at a given temperature
A) Obeys Maxwell’s distribution.
B) Have the same value for all molecules.
C) Equals the translational kinetic energy for each molecule.
D) Is $\dfrac{2}{3}$ rd the translational kinetic energy for each molecule.

Answer 1

Kinetic energy per molecule of a gas does not depend upon the mass of the molecule but only depends upon the temperature of the gas. As for diatomic molecules like ${H_2}$, ${O_2}$, $He$ gases at the same temperature will have the same translational kinetic energy through their ${V_{rms}}$ are different.

Formula used:
(i) Kinetic energy of a gas,
$({E_{molecule}}) = \dfrac{1}{2}m{v_{rms}}^2$
$\Rightarrow \dfrac{1}{2}m(\dfrac{{3kT}}{m})$
$\Rightarrow \dfrac{3}{2}kT$, Since ${v_{rms}} = \sqrt {\dfrac{{3kT}}{m}}$
(ii) Kinetic energy of $1$($milligram$) gas,
$({E_{mole}}) = \dfrac{1}{2}M{v^2}_{rms}$
$\Rightarrow \dfrac{1}{2}M\dfrac{{3RT}}{M}$
$\Rightarrow \dfrac{3}{2}RT$, Since ${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}}$
(iii) Kinetic energy of $1gram$gas,
$({E_{gram}}) = \dfrac{3}{2}\dfrac{{RT}}{M}$
$\Rightarrow \dfrac{3}{2}\dfrac{{k{N_A}T}}{{m{N_A}}}$
$\Rightarrow \dfrac{3}{2}\dfrac{{kT}}{m}$
$\Rightarrow \dfrac{3}{2}{\text{rT}}$
Here, $m$=Mass of molecule, $K$=Boltzmann’s constant , $T$=Temperature,$M$=Gas’s molecular weight.
$R$= Universal gas constant=$8.314{\text{ }}{K^{ - 1}}mo{l^{ - 1}}$, ${V_{rms}}$=Root mean square velocity
${\text{Avogadro number = }}\dfrac{{{\text{Molecular mass}}}}{{{\text{Mass of 1 molecule}}}}$
$\Rightarrow 6.023 \times {10^{23}}$

Complete step by step solution:
Molecules of ideal gases possess only translational motion. So, they possess only translational energy. According to the question we have to find out the rotational energy of a diatomic molecule in terms of kinetic energy.
Translational kinetic energy of each molecule, ${{\text{K}}_{\text{T}}} = \dfrac{3}{2}{\text{kT}}$, where $k$ is constant………… (1)
Rotational kinetic energy of each molecule,${{\text{K}}_{\text{R}}} = 2\left( {\dfrac{1}{2}{\text{kT}}} \right)$
$\Rightarrow {\text{kT}}$, where $k$ is constant……….. (2)
Now dividing the equation (1) and (2) we get,
$\dfrac{{{{\text{K}}_{\text{R}}}}}{{{{\text{K}}_{\text{T}}}}} = \dfrac{{({\text{kT)}}}}{{\left( {\dfrac{3}{2}{\text{kT}}} \right)}}$
$\Rightarrow \dfrac{2}{3}$
Therefore, ${{\text{K}}_{\text{R}}} = \dfrac{2}{3}{{\text{K}}_{\text{T}}}$
Therefore, Rotational kinetic energy at a given temperature is $\dfrac{2}{3}$of the translational kinetic energy of a gas molecule. The translational kinetic energy and rotational kinetic energy both obey Maxwell’s law of distribution independent of each other. Different molecules have different rotational kinetic energy. Ideal gases only possess translational kinetic energy but non-ideal gases possess smaller rotational kinetic energy.

Hence the correct option is (A) and (D).

Note:
Kinetic theory of ideal gases explain that
All gases consist of molecules, molecules are rigid, spherical, and elastic in nature
The size of the molecule is negligible compared to the average distance between the molecules.
Molecules are in continuous random motion and in this phenomenon, the molecules collide with one another and also the walls of the container.
The collision is perfectly elastic and there is no force of attraction or repulsion between the molecules.
The collision is almost instantaneous i.e. the collisions last for very negligible time compared to the time of free path between the molecules.
Between two collisions a molecule moves in a straight line with uniform velocity.