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

Jul 27, 2017

Irrespective of their molar masses, $\text{He}$ ATOM has approximately 60% the average molar kinetic energy that ${\text{H}}_{2}$ MOLECULE does at ${25}^{\circ} \text{C}$ and $\text{1 atm}$.

If you do not quite follow what it means to be at the high temperature limit, this answer may also help.

Well, I assume you mean only $\text{He}$ is an atom, and not a molecule... $\text{He}$ is an atom, and ${\text{H}}_{2}$ is a molecule. Only then is one's molar mass twice that of the other.

• As an atom, $\text{He}$ can only translate, i.e. shoot forward and backwards via cartesian degrees of freedom, having only translational average kinetic energy ${K}_{\text{trans}}$.
• ${\text{H}}_{2}$ can vibrate, at vibrational frequency $\omega \approx {\text{4394.49 cm}}^{- 1}$. ${\text{H}}_{2}$ 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:

$\left\langle\kappa\right\rangle \approx {K}_{a v g} / n = \frac{N}{2} R T$,

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 ${\text{H}}_{2}$ with no problem (for ${\text{H}}_{2}$, it would be a problem below approximately $\text{88 K}$).
• For vibrational DOFs, given the high vibrational frequency of ${\text{H}}_{2}$, we can safely say that ignoring the vibrational DOFs of ${\text{H}}_{2}$ is more accurate than including them.

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

$\left\langle\kappa\right\rangle \approx {\left\langle\kappa\right\rangle}_{\text{trans" + << kappa >>_"rot}}$, for ${\text{H}}_{2}$

For $\text{He}$, ${\left\langle\kappa\right\rangle}_{\text{rot}} \approx 0$, 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"))

$= \frac{\frac{3}{2} R T + \frac{2}{2} R T}{\frac{3}{2} R T}$

$= \frac{5}{2} \times \frac{2}{3} = \textcolor{b l u e}{\frac{5}{3}}$

So, on average, according to the equipartition theorem at high enough temperatures, an ensemble of ${\text{H}}_{2}$ molecules will have $\boldsymbol{1.67}$ times the kinetic energy of an analogous ensemble of $\text{He}$ atoms.