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

##### 1 Answer

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:

#<< kappa>> -= K/n = N/2RT# in units of

#"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# hascontributionsof:

#N_(tr) = 3# fortranslation(linear motion),

#N_(rot) = 2# forrotationalmotion oflinearmolecules or#N_(rot) = 3# forrotationalmotion ofnonlinearpolyatomics, and

Up to#N_(vib) = 1# forvibrationof polyatomics, but typically very small at room temperature. For simple molecules, like#"N"_2# and#"Cl"_2# , the contribution to vibration is usually ignored due to negligibility.

#n# is the#"mols"# of ideal gas.#R = "8.314472 J/mol"cdot"K"# is the universal gas constant.#T# is the temperature in#"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 #1/2mv^2#.*