Question

The oxygen molecule has a mass of $5.30\times {10}^{-26}kg$ and a moment of inertia of $1.94\times {10}^{-46}kg{m}^2$ 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 m/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.

Answer 1

Mass of an oxygen molecule, $m=5.30\times {10}^{-26}kg$
Moment of inertia, $I=1.94\times {10}^{-46}kg{m}^2$
Velocity of the oxygen molecule, v = 500 m/s
The separation between the two atoms of the oxygen molecule = 2r
Mass of each oxygen atom $=\frac{m}{2}$
Hence, moment of inertia I, is calculated as:
$\left(\frac{m}{2}\right)r^2+\left(\frac{m}{2}\right)r^2=m{r}^{2}r=\sqrt{\frac{I}{m}}=\sqrt{\frac{1.94\times {10}^{-46}}{5.36\times {10}^{-26}}}=0.60\times {10}^{-10}m$
It is given that:
$K{E}_{rot}=\frac{2}{3}K{E}_{trans}\frac{1}{2}I{\omega }^2=\frac{2}{3}\times \frac{1}{2}\times m{v}^{2}m{r}^2{\omega }^2=\frac{2}{3}m{v}^2\omega =\sqrt{\frac{2}{3}}\frac{v}{r}=\sqrt{\frac{2}{3}}\times \frac{500}{0.6\times {10}^{-10}}=6.80\times {10}^{12}rad/s$