Question

A spherical balloon of 21 cm diameter is to be filled up with $H_2$ at NTP from a cylinder containing the gas at 20 atm at ${27}^{o}C$. The cylinder can hold 2.82 litre of water at NTP. The number of balloons that can be filled up is___________.

Answer 1

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

$\Rightarrow$ 4851 ml $=$ 4.851 litre

Let 'n' balloons are filled, then the 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 litres

$\therefore$ Total volume of $H_2$ at NTP $=$ volume of 'n' balloons + volume of cylinder = $4.851\times n+2.82$ ................(i)

The volume of $H_2$ available at NTP can be derived from the data.

Given P$=$1atm ,V$=$ ?, T$=$273K and P $=$ 20 atm, V $=$2.82 litres, T $=$300K

$\therefore$ $\frac{P_1{V}_1}{T_1}$ $=$ $\frac{P_2{V}_2}{T_2}$

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)

By equations (i) and (ii), both are same.

$4.851\times n+2.82$ = $51.32$

$⟹n=10$.