Question

The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 ×10-46 kg m2 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 × 10 - 26 kg
Moment of inertia, I = 1.94 × 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 :

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

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

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

It is given that :

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

= $\frac{1 }{2}$ Iω2 = $\frac{2 }{3}$ × $\frac{1 }{2}$ × mv2

= mr2 ω2 = $\frac{2 }{ 3}$ mv2

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

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

= 6.80 × 10 12 rad / s