# What is the ratio of average kinetic energy for hydrogen and helium molecules, if helium is twice as massive?

##### 1 Answer

Irrespective of their molar masses,

Well, I assume you mean only

- As an atom,
#"He"# can only translate, i.e. shoot forward and backwards via cartesian degrees of freedom, having only**translational average kinetic energy**#K_("trans")# . #"H"_2# can vibrate, at vibrational frequency#omega ~~ "4394.49 cm"^(-1)# .#"H"_2# can also rotate, with#2# degrees of freedom (#theta,phi# in spherical coordinates).

We honestly don't care about their molar masses, because they don't make a difference here. The **average molar kinetic energy** according to the

*equipartition theorem*is:

#<< kappa >> ~~ K_(avg)/n = N/2 RT# ,where:

#N# is thenumber of degrees of freedom(DOFs) for a particular type of motion (translational, rotational, and vibrational are the most important).#R# and#T# are known from the ideal gas law.

This theorem applies at the so-called **high temperature limit**, where the energy level spacings of each kind of motion (translational, rotational, vibrational) are much smaller than

- This (pretty much) always applies for translational DOFs.
- This usually applies for rotational DOFs, except for molecules with particularly low moments of inertia. This applies for
#"H"_2# with no problem (for#"H"_2# , it would be a problem below approximately#"88 K"# ). - For vibrational DOFs, given the high vibrational frequency of
#"H"_2# , we can safely say that ignoring the vibrational DOFs of#"H"_2# is more accurate than including them.

We can break up the kinetic energy into translational and rotational parts, then, to a good approximation:

#<< kappa >> ~~ << kappa >>_"trans" + << kappa >>_"rot"# , for#"H"_2#

For

So, the *ratio of the average kinetic energies* is:

#color(blue)((<< kappa >>_("H"_2))/(<< kappa >>_("He"))) = (<< kappa >>_("H"_2, "trans") + << kappa >>_("H"_2, "rot"))/(<< kappa >>_("He","trans"))#

#= (3/2 RT + 2/2 RT)/(3/2 RT)#

#= 5/2 xx 2/3 = color(blue)(5/3)#

So, on average, according to the equipartition theorem at high enough temperatures, an *ensemble* of *ensemble* of