What is the average KE of N2 at 1 atm and 298K?

#K_(avg) ~~ 1.03 xx 10^20 "J/N"_2# #"molecule"#

The average kinetic energy at #"298 K"# and #"1 atm"# uses the #N/V# standard state (i.e. one molecule per unit volume), and for many molecules at #"298 K"#, we can assume the high-temperature limit to invoke the equipartition theorem:

#<< kappa >> = K_(avg)/N#

#= << kappa >>_(tr) + << kappa >>_(rot) + cancel(<< kappa >>_(vib))^"really small" + cancel(<< kappa >>_(el ec))^"practically zero"#

#~~ N_(tr)/2 k_B T + N_(rot)/2 k_B T#

where:

• #N_k# is the degrees of freedom for the #k#th type of motion (translation, rotation, vibration, electronic). To keep it simple I am ignoring vibrational and electronic energy states.
• The #N# in #K_(avg)/N# is a different #N# and is just the number of particles. Here we have #N = 1# for one #"N"_2# molecule.
• #k_B = 1.38065 xx 10^(-23) "J/K"# is the Boltzmann constant.
• #T# is the temperature in #"K"#.

For translational motion, #N_(tr) = 3#, because we are obviously in a three-dimensional world (#x,y,z#).

For rotational motion, a linear molecule has #N_(rot) = 2#, and a nonlinear molecule has #N_(rot) = 3#. Each degree of freedom corresponds to a rotation angle, either #theta#, #phi#, or some third angle #gamma#.

We ignore vibrational and electronic degrees of freedom, because that would wayyyyy overestimate #<< kappa >># at ordinary temperatures to use equipartition on them.

Therefore:

#color(blue)(<< kappa >>) = K_(avg)/1#

#~~ 3/2k_BT + 2/2 k_B T#

#~~ 5/2 k_BT#

#= 5/2 cdot 1.38065 xx 10^(-23) "J/K" cdot "298 K"#

#= color(blue)(1.03 xx 10^(-20) "J/N"_2)# #color(blue)("molecule")#