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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Solution

Given: The mass of the oxygen molecules is 5.30× 10 −26 kg, its moment of inertia about the axis passing through the centre of line joining the two atoms is 1.94× 10 −46 kg-m 2 and mean speed of the molecule is 500 m/s.

The rotational kinetic energy of the molecule is given as,

( KE ) rot = 1 2 I ω 2 (1)

Where, I is the moment of inertia of the molecule and ω is the angular velocity.

The translational kinetic energy of the molecule is given as,

( KE ) trans = 1 2 m v 2 (2)

Where, m is the mass of the molecule and v is the speed of the molecule.

According to the given condition,

( KE ) rot = 2 3 ( KE ) trans 1 2 I ω 2 = 2 3 ( 1 2 m v 2 ) ω=v 2m 3I

By substituting the values in the above equation, we get

ω=500× 2×5.30× 10 −26 3×1.94× 10 −46 =6.75× 10 12 rad/s

Thus, the average angular velocity of the molecule is 6.75× 10 12 rad/s .

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