Answer
$\Delta V=1.2\times 10^2\frac{m}{s}$
Work Step by Step
We know that:
$V_p=\sqrt\frac{2RT}{M}$
and $V_{rms}=\sqrt\frac{3RT}{M}$
Now,
$\Delta V=V_{rms}-V_p=(\sqrt3 -\sqrt2)\times \sqrt\frac{RT}{M}$
We plug in the known values to obtain:
$\Delta V=(\sqrt3 -\sqrt2)\times \sqrt\frac{8.314(500)}{0.028}=1.2\times 10^2\frac{m}{s}$
