Question

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

Answer 1

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).

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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` has contributions of:

`N_(tr) = 3` for translation (linear motion),

`N_(rot) = 2` for rotational motion of linear molecules or `N_(rot) = 3` for rotational motion of nonlinear polyatomics, and

Up to `N_(vib) = 1` for vibration of 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`.