Suppose you have a room full of diatomic oxygen, where the molecular weight of a

Question

Suppose you have a room full of diatomic oxygen, where the molecular weight of an oxygen at is 16g/mol, and the separation of oxygen atoms in the molecule is 1.21*10-10 m.

a) Assuming we can neglect vibrational energy of the molecule, what is the total kinetic energy of an oxygen molecule if it has an angular velocity of 7.0*1012 rad/s (rotating about its COM) and a translational speed 550m/s?

b) Assuming this kinetic energy found in (a) represents the average KE of the molecules in the room, and that

KEav=(5/2)kT

for the diatomic molecule, what is the temperatire inside the room full of oxygen (k is the Boltzmann constant)

Please show your work! Thank you!

Explanation / Answer

KE(rot) = (1/2) I omega^2
I = M R^2
I = 16amu (1.12x10^-10m/2)^2
I = 3.33x10^-46 kg m^2
KE(rot) = (1/2) (3.33x10^-46 kg m^2) (7x10^12 rad/s)^2
KE(rot) = 8.17x10^-21 J
KE(rot,molecule) = KE(rot) * 2 = 1.63x10^-20 J

KE(trans) = (1/2) m v^2
KE(trans) = (1/2) (32amu) (550m/s)^2
KE(trans) = 8.03x10^-21 J

KE(tot) = 1.63x10^-20J + 0.803x10^-20J
KE(tot) = 2.43x10^-20J

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