# Do molecules of the ideal gas at a particular temperature have the same kinetic energy?

Jun 21, 2018

No. If that were true, then these Maxwell-Boltzmann distributions of speeds would be vertical lines: But since this speed distribution is just that---a distribution... there exist a myriad of speeds for a given temperature, and thus a myriad of different kinetic energies for a given temperature (but only a single average kinetic energy).

Molecules of an ideal gas at the same temperature have potentially different kinetic energies... but the same average kinetic energy.

As in the equipartition theorem , at high enough temperatures, the average molar kinetic energy is given by:

$\left\langle\kappa\right\rangle \equiv \frac{K}{n} = \frac{N}{2} R T$

in units of $\text{J/mol}$, where

• $N$ is the number of degrees of freedom, i.e. the number of coordinates for each type of motion, basically.

This number $N$ has contributions of:

${N}_{t r} = 3$ for translation (linear motion),

${N}_{r o t} = 2$ for rotational motion of linear molecules or ${N}_{r o t} = 3$ for rotational motion of nonlinear polyatomics, and

Up to ${N}_{v i b} = 1$ for vibration of polyatomics, but typically very small at room temperature. For simple molecules, like ${\text{N}}_{2}$ and ${\text{Cl}}_{2}$, the contribution to vibration is usually ignored due to negligibility.

• $n$ is the $\text{mols}$ of ideal gas.
• $R = \text{8.314472 J/mol"cdot"K}$ is the universal gas constant.
• $T$ is the temperature in $\text{K}$.

It is the average, in the sense that we take a sample of molecules that ALL have somehow DIFFERENT speeds AND traveling directions... and through observing all of them, the ensemble average leads to a distribution of speeds for a given temperature, which corresponds to a single observed (average) kinetic energy based on the temperature.

It is strictly NOT the same as observing a single molecule's velocity and using that to calculate the kinetic energy from $\frac{1}{2} m {v}^{2}$.