Question

A balloon of diameter 20 metre weighs 100 kg. Its pay-load is $x\times {10}^{4}g$, if it is filled with He at 1.0 atm and ${27}^{o}C$. Density of air is $1.2kg{m}^{-3}(R=0.082d{m}^{3}atm{K}^{-1}mo{l}^{-1}$).
Then value of x is...........(as nearest integer)

Answer 1

Mass of balloon $=100kg=10\times {10}^{4}g$

Volume of balloon$=\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times \frac{22}{7}\times {(\frac{20}{2}\times 100)}^3$

$=4190\times {10}^{6}c{m}^3=4190\times {10}^{3}litre$

Mass of gas (He) in balloon $=\frac{PVm}{RT}(\because Pv=\frac{w}{m}RT)$
$=\frac{1\times 4190\times {10}^3\times 4}{0.082\times 300}$

$=68.13\times {10}^{4}g$

$\therefore$ Total mass of gas and balloon

$=68.13\times {10}^4+10\times {10}^4=78.13\times {10}^{4}g$

Mass of air displaced $=1.2\times 4190=5028kg$

$=502.8\times {10}^{4}g$

Pay load $=$ mass of air displaced - mass of (balloon + gas)

pay load $=502.8\times {10}^4-78.13\times {10}^4=424.67\times {10}^{4}g$

So, value of x is 4.