The oxygen molecule has a mass of 5.30xx10^(-26)kg and a moment of inertia of 1.94xx10^(-46 )kgm^(2) 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 (2)/(3) of its kinetic energy of translation. Find the average angular velocity of the molecules.

The oxygen molecule has a mass of 5.30xx10^(-26)kg and a moment of ine

1.	The oxygen molecule has a mass of 5.30xx10^(-26)kg and a moment of inertia of 1.94xx10^(-46 )kgm^(2) 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 (2)/(3) of its kinetic energy of translation. Find the average angular velocity of the molecules.
Answer» Solution :Rotational kinetic energy = `(2)/(3)` TRANSLATIONAL kinetic energy `(1)/(2)IOMEGA^(2)=(2)/(3)xx(1)/(2)MV^(2)` `omega^(2)=(2)/(3)xx(mv^(2))/(I)` `therefore omega=sqrt((2)/(3)(mv^(2))/(I))` `=sqrt((2)/(3)xx(5.3xx10^(-26)xx(500)^(2))/(1.94xx10^(-46)))` `=sqrt((265xx10^(-22))/(3xx1.94xx10^(-46)))` `therefore omega=sqrt(45.53xx10^(24))` `therefore omega=6.7477xx10^(12)" rad s"^(-1)~~6.75" rad s"^(-1)`