# If #He# (g) has an average kinetic energy of 5930 J/mol under certain conditions, what is the root mean square speed of #N_2# (g) molecules under the same conditions?

##### 1 Answer

Well, what's in common? Not their degrees of freedom, but their temperature.

#v_(RMS)("N"_2) = "650.7 m/s"#

The average kinetic energy of

Helium is an atom, which translates in 3 dimensions with zero rotational and vibrational degrees of freedom.

Hence, according to the **equipartition theorem**,

#<< kappa >> -= K/n = N/2RT# is the average kinetic energy, where we have that

#N = 3# for helium atom'slinear degrees of freedom.

What temperature is it at?

#T = 2/3 1/R << kappa >>#

#= 2/3 cdot 1/("8.314 J/mol"cdot"K") cdot "5930 J/mol"#

#=# #"475.5 K"#

Now, a pitfall would be to assume that

- Rotational degrees of freedom are non-negligible at room temperature.
- Vibrational degrees of freedom are negligible for diatomic molecules at room temperature.

So, what we find is that

#N = N_("trans") + N_("rot") + N_"vib" ~~ 3 + 2# because diatomic molecules rotate using two angles in spherical coordinates (

#theta,phi# ).

Fortunately, this does not matter because all we want is the **root-mean-square speed**, which depends only on molar mass and temperature.

#v_(RMS) = sqrt((3RT)/M)# where

#M# is the molar mass in#"kg/mol"# . Why is that necessary? Why not#"g/mol"# ? Well, what are the units of#R# ?

#color(blue)(v_(RMS)("N"_2)) = sqrt((3RT)/M)#

#= sqrt((3cdot"8.314 kg"cdot"m"^2"/s"^2//"mol"//"K" cdot "475.5 K")/"0.028014 kg/mol")#

#=# #color(blue)("650.7 m/s")#