Question

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?

Answer 1

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 `"N"_2` would be higher, because it has more ways to move. But its RMS speed is lower due to its higher molar mass.

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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's linear 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 `N` is the same for `"N"_2`... it's not. `"N"_2` is a MOLECULE, which rotates and vibrates. As it turns out,

• 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")`