Volume of one balloon (to be filled)
$\left(\frac{4}{3}\right)\mathrm{\pi }{\mathrm{r}}^3=\left(\frac{4}{3}\right)\times \left(\frac{22}{7}\right)\times {\left(\frac{21}{2}\right)}^3$
$\text{=}\text{4851}\text{mL}\text{=}\text{4}\text{.851}\text{litre}$
Let $\mathrm{‘}\mathrm{n}\mathrm{’}$ balloons are filled, then total volume of ${\mathrm{H}}_2$ used in filling balloons $=4.851\times \mathrm{n}\mathrm{}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{e}$
The cylinder of ${\mathrm{H}}_2$ used in filling balloons will also have ${\mathrm{H}}_2$ in it, after n balloons are filled. Also volume of cylinder $\text{=}\text{2}\text{.82}\text{litre}$
Hence Total volume of ${\mathrm{H}}_2$ at NTP = volume of $\mathrm{\text{'}}\mathrm{n}\mathrm{\text{'}}$ balloons + volume of cylinder
$=4.851\times \mathrm{n}+2.82\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$ ….(i)
The volume of available ${\mathrm{H}}_2$ at NTP can be derived by the data
Given $\text{P}\text{=}\text{1}\text{atm,}\text{V}\text{=}\text{?,}\text{T}\text{=}\text{273}\text{K}$
and $\text{P}\text{=}\text{20}\text{atm,}\text{V}\text{=}\text{2}\text{.82}\text{litre,}\text{T}\text{=}\text{300}\text{K}$
$\therefore \frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{{\mathrm{P}}_2{\mathrm{V}}_2}{{\mathrm{T}}_2}$
or $\frac{1\times \mathrm{V}}{273}=\frac{20\times 2.82}{300}$
$\therefore \text{V=}\frac{20\times 2.82\times 273}{300}=51.32\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$ ...(ii)
From equation $\left(1\right)$ and $\left(\mathrm{I}\mathrm{I}\right)$
$4.851\times \mathrm{n}+2.82=57.32\mathrm{}$
$\mathrm{n}=10$