Solved: A balloon 16 m in diameter is inflated with helium

Step 1 of 2

(a)Here we have calculated the mass of He in the balloon.

Given:

The diameter of a balloon = 16 m

Temperature \(=18^{\circ} \mathrm{C}+273=291 \mathrm{~K}\)

Pressure \(=98.7 \mathrm{KP}_{\mathrm{a}}=98.7 \mathrm{KP}_{\mathrm{a}} \times \frac{1 \text { atm }}{1.01325 \times 10^{2} \mathrm{KP}_{a}}=0.974 \mathrm{~atm}\)

In order to calculate the mass of He, 1st we have to calculate the volume of the balloon and then have to calculate the number of mole of He atoms by using the ideal gas equation.

Volume of balloon \(=4 / 3 \Pi r^{3}\)

\(\begin{aligned}&=4 / 3 \times 3.14 \times(8 \mathrm{~m})^{3} \\&=2.1 \times 10^{3} \mathrm{~m}^{3} \text { or } 2.1 \times 10^{6} \mathrm{~L}\end{aligned}\)

\(\begin{aligned} \text { Number of mole of He (n) } &=\frac{P V}{R T} \\ &=\frac{0.974 \mathrm{~atm} \times\left(2.1 \times 10^{6} \mathrm{~L}\right)}{0.0821 \mathrm{Latm} \mathrm{mol}^{-1} \mathrm{~K}^{-1} \times 291 \mathrm{~K}} \\ &=8.6 \times 10^{4} \mathrm{~mol} \mathrm{He} \end{aligned}\)

Mass of \(\mathrm{He}=8.6 \times 10^{4} \mathrm{~mol} \mathrm{He} \times 4.003 \mathrm{~g} / \mathrm{mol} \mathrm{He}=3.4 \times 10^{5} \mathrm{~g} \mathrm{He}\)

Thus the mass of He is found to be \(3.4 \times 10^{5} \mathrm{~g}\) He.