A spherical balloon of volume V contains Helium gas at a pressure P. How many  moles of Helium are there in the balloon if the average kinetic energy of the  Helium atoms is  $\stackrel{-}{K}$.

We know, ${\mathrm{v}}_{\mathrm{RMS}}=\sqrt{\frac{3\mathrm{kT}}{\mathrm{m}}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}}}=\sqrt{\frac{3\mathrm{PV}}{\mathrm{mass}\mathrm{of}\mathrm{gas}}}$ ,      (1)
where m, M are mass of each molecule, molar mass of gas respectively.
$\Rightarrow \frac{1}{2}m{v}_{RMS}^2=\frac{3kT}{2}=\stackrel{-}{K}$ 
Also, from (1), 
$\stackrel{-}{\mathrm{K}}=\frac{1}{2}{\mathrm{mv}}_{\mathrm{RMS}}^2=\frac{3\mathrm{mPV}}{2\left(\mathrm{mass}\mathrm{of}\mathrm{gas}\right)}$ 
But, $\frac{\left(\mathrm{mass}\mathrm{of}\mathrm{gas}\right)}{\mathrm{m}}=\mathrm{N}$ (number of molecules).
$\Rightarrow \stackrel{-}{\mathrm{K}}=\frac{3\mathrm{PV}}{2\mathrm{N}}\Rightarrow \mathrm{N}=\frac{3\mathrm{PV}}{2\stackrel{-}{\mathrm{K}}}$ 
Now, number of moles =  $\mathrm{n}=\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{A}}}=\frac{3\mathrm{PV}}{2\stackrel{-}{\mathrm{K}}{\mathrm{N}}_{\mathrm{A}}}$