R.M.S. Speed Relationships

For a sample of ideal gas, the r.m.s. speed of the molecules is 1900 m s⁻¹. The value of $kT$ is shown below. Find the mass of one molecule, in u. (9702/41/O/N/25 Q4(c))

1. Replace mean-square speed by r.m.s. speed squared.
   $$\langle c^2\rangle=c_{\mathrm{rms}}^2$$
   $$\tfrac{1}{2}mc_{\mathrm{rms}}^2=\tfrac{3}{2}kT$$
2. Make the molecular mass the subject.
   $$m=\dfrac{3kT}{c_{\mathrm{rms}}^2}$$
   $$m=\dfrac{2.4\times10^{-20}}{1900^2}$$
   $$m=6.65\times10^{-27}\ \mathrm{kg}$$
3. Change the mass to u.
   $$m=\dfrac{6.65\times10^{-27}}{1.66\times10^{-27}}$$
   $$m=4.0\ \mathrm{u}$$

A molecular mass of 4.0 u is consistent with helium. The same equation can find speed from temperature or mass from speed.