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

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

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Answer 1 · 2017-07-27T05:32:37

Irrespective of their molar masses, $He$ $"He"$ ATOM has approximately $60 %$ $60%$ the average molar kinetic energy that $H_{2}$ $"H"_2$ MOLECULE does at $25^{\circ} C$ $25^@ "C"$ and $1 atm$ $"1 atm"$ .

[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)

Well, I assume you mean only $He$ $"He"$ is an atom, and not a molecule... $He$ $"He"$ is an atom, and $H_{2}$ $"H"_2$ is a molecule. Only then is one's molar mass twice that of the other.

As an atom, $He$ $"He"$ can only translate, i.e. shoot forward and backwards via cartesian degrees of freedom, having only translational average kinetic energy $K_{trans}$ $K_("trans")$ .
$H_{2}$ $"H"_2$ can vibrate, at vibrational frequency $ω \approx {4394.49 cm}^{- 1}$ $omega ~~ "4394.49 cm"^(-1)$ . $H_{2}$ $"H"_2$ can also rotate, with $2$ $2$ degrees of freedom ( $θ, ϕ$ $theta,phi$ 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:

$⟨ κ ⟩ \approx \frac{K_{a v g}}{n} = \frac{N}{2} R T$ $<< kappa >> ~~ K_(avg)/n = N/2 RT$ ,

where:

$N$ $N$ is the number of degrees of freedom (DOFs) for a particular type of motion (translational, rotational, and vibrational are the most important).

$R$ $R$ and $T$ $T$ 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 $k_{B} T$ $k_B T$ where $k_{B}$ $k_B$ 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 $H_{2}$ $"H"_2$ with no problem (for $H_{2}$ $"H"_2$ , it would be a problem below approximately $88 K$ $"88 K"$ ).
For vibrational DOFs, given the high vibrational frequency of $H_{2}$ $"H"_2$ , we can safely say that ignoring the vibrational DOFs of $H_{2}$ $"H"_2$ is more accurate than including them.

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

$⟨ κ ⟩ \approx {⟨ κ ⟩}_{trans} + {⟨ κ ⟩}_{rot}$ $<< kappa >> ~~ << kappa >>_"trans" + << kappa >>_"rot"$ , for $H_{2}$ $"H"_2$

For $He$ $"He"$ , ${⟨ κ ⟩}_{rot} \approx 0$ $<< kappa >>_"rot" ~~ 0$ , 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:

$\frac{{⟨ κ ⟩}_{H_{2}}}{{⟨ κ ⟩}_{He}} = \frac{{⟨ κ ⟩}_{H_{2}, trans} + {⟨ κ ⟩}_{H_{2}, rot}}{{⟨ κ ⟩}_{He, trans}}$ $color(blue)((<< kappa >>_("H"_2))/(<< kappa >>_("He"))) = (<< kappa >>_("H"_2, "trans") + << kappa >>_("H"_2, "rot"))/(<< kappa >>_("He","trans"))$

$= \frac{\frac{3}{2} R T + \frac{2}{2} R T}{\frac{3}{2} R T}$ $= (3/2 RT + 2/2 RT)/(3/2 RT)$

$= \frac{5}{2} \times \frac{2}{3} = \frac{5}{3}$ $= 5/2 xx 2/3 = color(blue)(5/3)$

So, on average, according to the equipartition theorem at high enough temperatures, an ensemble of $H_{2}$ $"H"_2$ molecules will have $1.67$ $bb(1.67)$ times the kinetic energy of an analogous ensemble of $He$ $"He"$ atoms.

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

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