# Question #1f4c4

Feb 25, 2016

You are right: at 280 K

#### Explanation:

In every system at thermal equilibrium, all molecules interact with each other by means of random collisions, thus transferring kinetic and potential energy among them. Eventually, each molecule reaches the same average kinetic energy (in time). The heavier molecule will have a lower quadratic speed to balance the mean kinetic energy of lighter molecules, which have higher (in the average) quadratic velocity:

$| {K}_{h} | = \frac{1}{2} {m}_{h} | {v}_{h}^{2} | = \frac{1}{2} {m}_{l} | {v}_{l}^{2} | = | {K}_{l} |$,
where $| K | = \frac{1}{2} m | {v}^{2} |$ is the average kinetic energy, and $| {v}^{2} |$ represents the average of square velocity.

Absolute temperature T is proportional to any quadratic term of kinetic energy (hydrogen and nitrogen have translational kinetic energy in the x, y, z directions, plus rotational kinetic energy around the two axes perpendicular to the molecular axis and (only at high temperatures) their molecules can oscillate by stretching and compressing along the internuclear distance.

Equipartition theorem states that any kind " i " of energies is transferred and interconverted through intermolecular random interactions until becoming evenly subdivided among all kinds of motion.

For each kind of motion, the relationship is $| {E}_{i} | = \frac{1}{2} {k}_{B} T$.
where ${k}_{B}$ is Boltzmann constant, given by the universal constant of gases, $R$, divided by Avogadro's constant.
Hence, if the temperature is the same, also every term of kinetic energies are the same, and vice versa.

For example, for the three directions x, y, z of translational motion:

$\frac{1}{2} m | {v}_{x}^{2} | = \frac{1}{2} {k}_{B} T$
$\frac{1}{2} m | {v}_{y}^{2} | = \frac{1}{2} {k}_{B} T$
$\frac{1}{2} m | {v}_{z}^{2} | = \frac{1}{2} {k}_{B} T$

Provided Pythagorean theorem states ${v}^{2} = {v}_{x}^{2} + {v}_{y}^{2} + {v}_{z}^{2}$, we have:
$| {K}_{\text{translational}} | = \frac{1}{2} m | {v}^{2} | = \frac{3}{2} {k}_{B} T$