A spherical balloon of 21 cm diameter is to be filled up with H _ 2...

Volume of one balloon (to be filled)
$\left(\dfrac{4}{3}\right)\pir^{3} = \left(\dfrac{4}{3}\right) \times \left(\dfrac{22}{7}\right) \times {\left(\dfrac{21}{2}\right)}^{3}$
$\text{=}\text{4851}\text{mL}\text{=}\text{4}\text{.851}\text{litre}$
Let $‘n’$ balloons are filled, then total volume of ${H}_{2}$ used in filling balloons $= 4.851 \times nlitre$
The cylinder of ${H}_{2}$ used in filling balloons will also have ${H}_{2}$ in it, after n balloons are filled. Also volume of cylinder $\text{=}\text{2}\text{.82}\text{litre}$
Hence Total volume of ${H}_{2}$ at NTP = volume of $'n'$ balloons + volume of cylinder
$= 4.851 \times n + 2.82$ ….(i)
The volume of available ${H}_{2}$ 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 $\text{P}\text{=}\text{20}\text{atm,}\text{V}\text{=}\text{2}\text{.82}\text{litre,}\text{T}\text{=}\text{300}\text{K}$
$∴ \dfrac{{P}_{1}{V}_{1}}{{T}_{1}} = \dfrac{{P}_{2}{V}_{2}}{{T}_{2}}$
or $\dfrac{1 \times V}{273} = \dfrac{20 \times 2.82}{300}$
$∴ \text{V=}\dfrac{20 \times 2.82 \times 273}{300} = 51.32$ ...(ii)
From equation $( 1 ) and ( II )$
$4.851 \times n + 2.82 = 57.32$
$n = 10$