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