Q. 203.9( 8 Votes )
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.
We need to find average angular velocity, ω of oxygen molecule.
We know that,
V = r ω
Mass of oxygen molecule, M = 5.30 × 10-26 kg
Moment of inertia of oxygen molecule, I = 1.94 × 10-46 kg m2
Mean speed of molecule, V = 500 m s-1
To find ω we need value of r (distance between axis of rotation and oxygen atom in the molecule)
The equation for moment of inertia of an oxygen molecule (system consisting of 2 equal masses (m) separated at a distance of r from axis),
I = mr2 + mr2
But m is mass of an oxygen atom,
∴ m = M/2
Thus, I = 2(M/2) r2
i.e. r = (I/M)2
= [ (1.94 × 10-46)/5.30 × 102] 1/2
= .606 × 10-10
KErot = 2KEtran/3
1/2 I ω2 = 2/3 × (m v2/2)
1/2 m r2 ω = 2/3 × mv2/2
ω = (2/3)1/2× v/r
v = 500 ms-1
r = .606 × 10-10 m
Thus, ω = (2/3)1/2 × 500/ (.606 × 10-10)
= 6.68 × 1012 rad s-1
Average angular velocity of molecule is 6.68 ×1012 rad s-1
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