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