Answer :
Answer: 6.65 rad/s
Explanation:
Firstly, we need to know that according the kinetic theory of gases, the average kinetic energy [tex]KE[/tex] of a molecule with [tex]i[/tex] degrees of freedom is:
[tex]KE=\frac{i}{2}kT[/tex] (1)
Where:
[tex]i=5[/tex] because oxigen is a diatomic molecule and has 5 degrees of freedom
[tex]k=1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K}[/tex] is the Boltzman constant
[tex]T=300 K[/tex] is the temperature
Then:
[tex]KE=\frac{5}{2}(1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K})(300 K)[/tex] (2)
[tex]KE=1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}[/tex] (3) With this value of average kinetic energy we can find the average angular velocity, with the following equation:
[tex]KE=\frac{1}{2}I \omega^{2}[/tex] (4)
Where:
[tex]I[/tex] is the moment of inertia
[tex]\omega[/tex] is the angular velocity
Now, [tex]I=m R^{2}[/tex] (5)
Being [tex]m=31.98 g/mol=0.03198 kg/mol[/tex] the molar mass of Oxigen molecule and [tex]R=0.121(10)^{-9}m[/tex] the distance between atoms
[tex]I=(0.03198 kg/mol)(0.121(10)^{-9}m)^{2}[/tex] (6)
[tex]I=4.682(10)^{-22} \frac{kg}{mol} m^{2}[/tex] (7)
Substituting (7) and (3) in (4):
[tex]1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}=\frac{1}{2} (4.682(10)^{-22} \frac{kg}{mol} m^{2}) \omega^{2}[/tex] (8)
Finding [tex]\omega[/tex]:
[tex]\omega=6.65 rad/s[/tex] (9) This is the average angular velocity for a molecule of O2