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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?

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    $\begingroup$ 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 $\endgroup$ Oct 13, 2014 at 19:49
  • $\begingroup$ Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question. $\endgroup$
    – Philipp
    Oct 13, 2014 at 19:51
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    $\begingroup$ @Carol I'd probably use $E_k=\frac{3}{2}NKT$ to get the temperature and then the formula you quote :) $\endgroup$ Oct 13, 2014 at 20:17

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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

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

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

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