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

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# 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#.