Question

A spherical balloon of $21 \,cm$ diameter is to be filled up with $H _{2}$ at NTP from cylinder containing the gas at $20\, atm$ at $27^{\circ} C$. The cylinder can hold $2.82$ litre of water a NTP. Calculate the number of balloons that can be filled up.

Answer 1

Volume of one balloon (to be filled)
$\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7} \times \frac{21}{2}^{3}$
$=4851 mL =4.851$ litre
Let ' $n$ ' balloons are filled, then total volume of
$H _{2}$ used in filling balloons $=4.851 \times n$ litre
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 $=2.82$ litre
Hence, Total volume of $H _{2}$ at NTP = volume
of ' $n$ ' balloons $+$ volume of cylinder
$=4.851 \times n +2.82 \ldots$...(i)
The volume of available $H _{2}$ at NTP can be
derived by the data
Given $P=1\, atm , V=?, T=273 \,K$
and $P=20\, atm, V=2.82$ litre, $T=300\, K$
$\therefore \frac{ P _{1} V _{1}}{ T _{1}}=\frac{ P _{2} V _{2}}{ T _{2}}$
or $\frac{1 \times V }{273}=\frac{20 \times 2.82}{300}$
$\therefore V =\frac{20 \times 2.82 \times 273}{300}=51.32$...(ii)
From equation (1) and (II)
$4.851 \times n+2.82=51.32$
$n =10$