# The oxygen molecule has a mass of 5.30 × 10–26 kg and a moment of inertia of 1.94×10–46 kg m2

Question:

The oxygen molecule has a mass of $5.30 \times 10^{-26} \mathrm{~kg}$ and a moment of inertia of $1.94 \times 10^{-46} \mathrm{~kg} \mathrm{~m}^{2}$ about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500 \mathrm{~m} / \mathrm{s}$ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

Solution:

Mass of an oxygen molecule, $m=5.30 \times 10^{-26} \mathrm{~kg}$

Moment of inertia, $I=1.94 \times 10^{-46} \mathrm{~kg} \mathrm{~m}^{2}$

Velocity of the oxygen molecule, $v=500 \mathrm{~m} / \mathrm{s}$

The separation between the two atoms of the oxygen molecule $=2 r$

Mass of each oxygen atom $=\frac{m}{2}$

Hence, moment of inertia $I$, is calculated as:

$\left(\frac{m}{2}\right) r^{2}+\left(\frac{m}{2}\right) r^{2}=m r^{2}$

$r=\sqrt{\frac{I}{m}}$

$\sqrt{\frac{1.94 \times 10^{-46}}{5.36 \times 10^{-26}}}=0.60 \times 10^{-10} \mathrm{~m}$

It is given that:

$\mathrm{KE}_{\mathrm{rot}}=\frac{2}{3} \mathrm{KE}_{\mathrm{trans}}$

$\frac{1}{2} I \omega^{2}=\frac{2}{3} \times \frac{1}{2} \times m v^{2}$

$m r^{2} \omega^{2}=\frac{2}{3} m v^{2}$

$\omega=\sqrt{\frac{2}{3} \frac{v}{r}}$

$=\sqrt{\frac{2}{3}} \times \frac{500}{0.6 \times 10^{-10}}$

$=6.80 \times 10^{12} \mathrm{rad} / \mathrm{s}$