# 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?

Aug 6, 2018

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 ${\text{N}}_{2}$ would be higher, because it has more ways to move. But its RMS speed is lower due to its higher molar mass.

Helium is an atom, which translates in 3 dimensions with zero rotational and vibrational degrees of freedom.

Hence, according to the equipartition theorem,

$\left\langle\kappa\right\rangle \equiv \frac{K}{n} = \frac{N}{2} R T$

is the average kinetic energy, where we have that $N = 3$ for helium atom's linear degrees of freedom.

What temperature is it at?

$T = \frac{2}{3} \frac{1}{R} \left\langle\kappa\right\rangle$

= 2/3 cdot 1/("8.314 J/mol"cdot"K") cdot "5930 J/mol"

$=$ $\text{475.5 K}$

Now, a pitfall would be to assume that $N$ is the same for ${\text{N}}_{2}$... it's not. ${\text{N}}_{2}$ 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.

${v}_{R M S} = \sqrt{\frac{3 R T}{M}}$

where $M$ is the molar mass in $\text{kg/mol}$. Why is that necessary? Why not $\text{g/mol}$? Well, what are the units of $R$?

$\textcolor{b l u e}{{v}_{R M S} \left({\text{N}}_{2}\right)} = \sqrt{\frac{3 R T}{M}}$

= sqrt((3cdot"8.314 kg"cdot"m"^2"/s"^2//"mol"//"K" cdot "475.5 K")/"0.028014 kg/mol")

$=$ $\textcolor{b l u e}{\text{650.7 m/s}}$