Question

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

Answer 1

Irrespective of their molar masses, `"He"` ATOM has approximately `60%` the average molar kinetic energy that `"H"_2` MOLECULE does at `25^@ "C"` and `"1 atm"`.

[If you do not quite follow what it means to be at the high temperature limit, this answer may also help.](https://socratic.org/questions/at-absolute-temperature-molecules-have-which-type-of-energy)

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Well, I assume you mean only `"He"` is an atom, and not a molecule... `"He"` is an atom, and `"H"_2` is a molecule. Only then is one's molar mass twice that of the other.

• 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 the number 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 `k_B T` where `k_B` is the Boltzmann constant.

• 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 `"He"`, `<< kappa >>_"rot" ~~ 0`, because atoms cannot rotate and turn out to look any different under the hard sphere approximation.

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 `"H"_2` molecules will have `bb(1.67)` times the kinetic energy of an analogous ensemble of `"He"` atoms.