Question

A spherical balloon of volume $4.00\times {10}^3{cm}^3$ contains helium at a pressure of $1.20\times {10}^{5}Pa$. How many moles of helium are in the balloon, if the average kinetic energy of the helium atoms is $3.60\times {10}^{-22}J$?

• A. $3.32mol$
• B. $4.12mol$
• C. $2.16mol$
• D. $2.82mol$

Answer 1

The correct option is A $3.32mol$

We know that the kinetic energy of a molecule is given by:

$\frac{1}{2}{m}_0{\overline{v}}^2=\frac{3}{2}{k}_{B}T\Rightarrow T=\frac{2}{3}\left(\frac{\frac{1}{2}{m}_0{\overline{v}}^2}{k_B}\right)$

So $T=\frac{2}{3}\left(\frac{3.60\times {10}^{-22}J}{1.38\times {10}^{-23}J/K}\right)=17.4K$

Now $PV=nRT\Rightarrow n=\frac{PV}{RT}$


$n=\frac{(1.20\times {10}^{5}N/m^2)(4.00\times {10}^{-3}{m}^3)}{(8.314J/mol.K)\left(17.4K\right)}=3.32mol$