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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