A spherical balloon of 21 cm diameter is to be filled up with hydrogen at N.T.P. from a cylinder containing the gas at 20 atmospheres at 27 ºC. If the cylinder can hold 2.82 liters of water, calculate the number of balloons that can be filled up.

No. of balloons that can be filled

$=\frac{\mathrm{V}\text{ of }{\mathrm{H}}_2\text{ available }}{\mathrm{V}\text{ of one balloon }}$

Calculation of total volume of hydrogen in the cylinder at N.T.P.

$\frac{{\mathrm{P}}_1{ \mathrm{V}}_1}{{ \mathrm{T}}_1}=\frac{{\mathrm{P}}_2{ \mathrm{V}}_2}{{ \mathrm{T}}_2}$

$$\begin{gathered} {\mathrm{P}}_1 = 1 \mathrm{atm} {\mathrm{P}}_2 = 20 \mathrm{atm} \\ {\mathrm{V}}_1 = ? {\mathrm{V}}_2 = 2.82 \mathcal{l} \\ {\mathrm{T}}_1 = 273 \mathrm{K} {\mathrm{T}}_2 = 273 + 27 = 300 \end{gathered}$$

$\therefore {\mathrm{V}}_1=\frac{20\times 2.82\times 273}{300\times 1}=51.324 \mathrm{lit}=51324 \mathrm{mL}$

Actual volume to be transferred into balloons = 51324 – 2820 mL = 48504 mL [$\because$ 2820 mL of ${\mathrm{H}}_2$ will remain in cylinder]

$\text{ Volume of one balloon }=\frac{4}{3}\pi r^3=\frac{4}{3}\times \frac{22}{7}\times {\left(\frac{21}{2}\right)}^3$

$\left[\because \mathrm{r}=\frac{\text{ diameter }}{2}\right]$

= 4851 mL = 4.851L

No. of balloons that can be filled up

$=\frac{48504}{4851}=9.999\approx 10$