Question

For a molecule of O2 at room temperature (300 K), calculate the average angular velocity for rotations about the x or y axes. The distance between the O atoms in the molecule is 0.121 nm.

Answer 1

Answer: 6.65 rad/s

Explanation:

Firstly, we need to know that according the kinetic theory of gases, the average kinetic energy $KE$ of a molecule with $i$ degrees of freedom is:

$KE=\frac{i}{2}kT$ (1)

Where:

$i=5$ because oxigen is a diatomic molecule and has 5 degrees of freedom

$k=1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K}$ is the Boltzman constant

$T=300 K$ is the temperature

Then:

$KE=\frac{5}{2}(1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K})(300 K)$ (2)

$KE=1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}$ (3) With this value of average kinetic energy we can find the average angular velocity, with the following equation:

$KE=\frac{1}{2}I \omega^{2}$ (4)

Where:

$I$ is the moment of inertia

$\omega$ is the angular velocity

Now, $I=m R^{2}$ (5)

Being $m=31.98 g/mol=0.03198 kg/mol$ the molar mass of Oxigen molecule and $R=0.121(10)^{-9}m$ the distance between atoms

$I=(0.03198 kg/mol)(0.121(10)^{-9}m)^{2}$ (6)

$I=4.682(10)^{-22} \frac{kg}{mol} m^{2}$ (7)

Substituting (7) and (3) in (4):

$1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}=\frac{1}{2} (4.682(10)^{-22} \frac{kg}{mol} m^{2}) \omega^{2}$ (8)

Finding $\omega$:

$\omega=6.65 rad/s$ (9) This is the average angular velocity for a molecule of O2