7.20 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.

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Vishal Baghel | Contributor-Level 10

4 months ago

Mass of the oxygen molecule, m = 5.30 × 10^-26 kg

Moment of inertia, I = 1.94×10^-46 kg m²

Velocity of the oxygen molecule, v = 500 m/s

the separation of atoms in oxygen molecule = 2r.

the mass of each oxygen atom = (m/2)

Moment of inertia I can be calculated as I = (m/2)r² + (m/2)r²

hence r = $\sqrt{I / m}$

r = sqrt (1.94×10^-46 / 5.30 × 10^-26 = 6.05 x 10^-11

It is given that KErotation = KEtranslation

(1/2)I $\frac{1}{2} m v^{2}$ = (2/3) (1/2)mv2

$m v^{2}$ = (v/r) $\sqrt{I / m}$

$ω^{2}$ = 6.8 x 1012 rad/s

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