Q. When a balloon of radius $10m$ having of mass $100kg$ is filled with helium, at $1.66$ bar in which given density of air is $1.2kgm^{- 3}$ and $R=0.083$ bar $dm^{3 }K/mol\left.$ , the pay load will be:

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$1138.12kg$

$3811.12kg$

$1381.12kg$

$3112.38kg$

Solution:

Volume of displaced air = Volume of balloon
$=\frac{4}{3}\pi r^{3}=\frac{4}{3}\times \frac{22}{7}\times \left(\right.10m\left(\left.\right)^{3}=4190.5m^{3}$
Mass of air displaced $=$ volume $\times $ density
$=4190.5\times 1.2=$
$5028.6kg$
Mass of $He=4n$ moles
$=4\times \frac{P V}{R T}=\frac{4 \times 1 . 66 bar \times \left(4190 . 5 \times \left(10\right)^{3} \left(dm\right)^{3}\right)}{0 . 083 bar d m^{3} K^{- 1} \left(mol\right)^{- 1} \times 300 K}$
Pay-load $=$ Mass of air displaced $-$ Mass of filled balloon
$=5028.6-\left(\right.1117.48+100\left.\right)=3811.12kg$

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