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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Mass of oxygen molecule (Mo) = 5.3 × 10^-26 Kg
Moment of inertia (I) = 1.94 × 10^-46 kg.m²
Mean speed of molecule (v) = 500 m/s
Given,
K.E of rotation = (2/3)× KE of translation
(1/2)Iw² = (2/3)×(1/2)mv²
w = √{2mv²/I}
= √{2×5.3×10^-26×(500)²/1.94×10^-46}
= 6.75 × 10¹² rad/s
Moment of inertia (I) = 1.94 × 10^-46 kg.m²
Mean speed of molecule (v) = 500 m/s
Given,
K.E of rotation = (2/3)× KE of translation
(1/2)Iw² = (2/3)×(1/2)mv²
w = √{2mv²/I}
= √{2×5.3×10^-26×(500)²/1.94×10^-46}
= 6.75 × 10¹² rad/s
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13
Answer:
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)r2 + (m/2)r2 = mr2
r = ( I / m)1/2
(1.94 × 10-46 / 5.36 × 10-26 )1/2 = 0.60 × 10-10 m
It is given that:
KErot = (2/3)KEtrans
(1/2) I ω2 = (2/3) × (1/2) × mv2
mr2ω2 = (2/3)mv2
ω = (2/3)1/2 (v/r)
= (2/3)1/2 (500 / 0.6 × 10-10) = 6.80 × 1012 rad/s.
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