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 m²

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 = `m/2`

Hence, moment of inertia I, is calculated as:

`(m/2)r^2 + (m/2)r^2 = mr^2`

r = `sqrt(I/m)`

`sqrt((1.94 xx 10^(-46))/(5.36 xx 10^(-26))) = 0.60 xx 10^(-10)m`

It is given that:

`KE_"rot" = KE_"trans`"`

`1/2 Iomega^2 =2/3 xx 1/2 "mv"^2`

`mr^2 omega^2 = 2/3 mv^2`

`omega = sqrt(2/3) v/r`

`= sqrt(2/3) xx 500/(0.6 xx 10^(-10))`

`= 6.80 xx 10^12 "rad/s"`

 

Answer 2

`m = 5.30 xx 10^(-26)`

`I = 1.94 xx 10^(-46) kg m^2`

`v = 500 "m/s"`

if m/2 si mass of each of oxygen and 2r is distnace between the two atoms as shown in figure then.

`I = m/2 r^2 + m/2r^2 = mr^2`

`r = sqrt(I/m) = sqrt((1.94 xx 10^(-46))/(5.30 xx 10^(-26)))`

`= 0.61 xx 10^(-10) m`

As K.E of rotation = 2/3 K.E of transalation

`:. 1/2 Iomega^2 = 2/3 xx 1/2 momega^2`

`1/2(mr^2)omega^2 = 1/2 mv^2`

`omega  = sqrt(2/3) v/r = sqrt(2/3) xx500 /(0.61xx10^(-10)) = 6.7 xx 10^(12) "rad/s"`