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

Asked by Abhisek | 1 year ago |  88

Solution :-

Given,

Mass of one oxygen molecule, m = 5.30 × 10–26 kg

Thus, the mass of each oxygen atom = $$\dfrac{m}{2}$$

Moment of inertia of it, I = 1.94 × 10–46 kg m2

Velocity of the molecule, v = 500 m/s

The distance between the two atoms in the molecule = 2r

Thus, moment of inertia I, is calculated as:

l =($$\dfrac{m}{2}$$)r2 + ($$\dfrac{m}{2}$$)r2 = mr2

r =  $$(\dfrac{I}{2})^ \dfrac{1}{2}$$

=$$\dfrac{(1.94 × 10^{-46)}}{(5.36 × 10^{-26} )}^\dfrac{1}{2}$$

=  0.60 × 10-10 m

Given,

K.Erot = ($$\dfrac{2}{3}$$)K.Etrans

($$\dfrac{1}{2}$$) I ω2 = ($$\dfrac{2}{3}$$) × ($$\dfrac{1}{2}$$) × mv2

mr2ω2 = ($$\dfrac{2}{3}$$)mv2

Therefore,ω = $$( \dfrac{2}{3})^\dfrac{1}{2}.\dfrac{v}{r}$$

= $$( \dfrac{2}{3})^\dfrac{1}{2}$$ (500 / 0.6 × 10-10) = 6.88 × 1012 rad/s.

Answered by Pragya Singh | 1 year ago

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