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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 $$27^oC$$. Density of air is $$1.2 kg m^{-3} (R=0.082 dm^3 atm K^{-1} mol^{-1}$$).
Then value of x is...........(as nearest integer)


Solution

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

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

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

$$=4190\times 10^6 cm^3=4190\times 10^3 litre$$

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

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

$$\therefore$$ Total mass of gas and balloon

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

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

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

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.

Chemistry

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