# Finding root-mean-square speed of neon using kinetic energy

Find the root-mean-square speed of $$\ce{Ne}$$ atoms at the temperature at which their kinetic energy is $$\pu{6.24 kJ mol-1}.$$

I tried using the kinetic energy formula

$$\mathrm{KE} = \frac{mv^2}{2},$$

but I don't really understand how to achieve the necessary values.

I tried to do it by converting mass of one atom to mass of a mole since the given energy is per mole, but I'm still not getting the answer. I understand that I need to do $$\displaystyle\sqrt{\frac{3RT}{M}},$$ where $$R= \pu{8.3145 J mol^-1 K^-1}$$ and $$M = 20,$$ but how would I get the temperature?

• Hi Carol, did you convert mass of one atom to mass of a mole since your energy is per mole? You might find this helpful :) en.wikipedia.org/wiki/Root-mean-square_speed Oct 13, 2014 at 19:49
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• @Carol I'd probably use $E_k=\frac{3}{2}NKT$ to get the temperature and then the formula you quote :) Oct 13, 2014 at 20:17

As hinted in the comments by @AngusTheMan we use the equation

$$\mathrm{KE_{avg}} = \frac{3}{2}kT,$$

but ultimately we are looking for the $$v_\mathrm{rms}$$, so we'll also use

$$v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}.$$

You've been given your $$\mathrm{KE_{avg}},$$ $$R$$ and $$k$$ are constants, and $$M$$ is the mass of 1 mole of $$\ce{Ne}$$. We can set all of these equal and the solution will come from

$$v_\mathrm{rms} = \sqrt{\frac{3 \cdot R\cdot\mathrm{KE_{avg}}\cdot\frac{2}{3k}}{M}}.$$

• In the light of Hint answers revisited, it would be nice if the answer would actually contain an answer and wouldn't end abruptly. Jul 5, 2020 at 17:05