Question

Payload is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the payload, when a balloon of radius 10 m of mass 100 kg is filled with helium at 1.66 bar at 27 $$\displaystyle ^{\circ}C$$. (Density of air = 1.2 kg m$$\displaystyle ^{3}$$ and R = 0.083 bar dm $$\displaystyle {3}$$ K$$\displaystyle ^{-1}$$ mol$$\displaystyle ^{-1}$$)

Solution

The volume of the balloon is $$\displaystyle V = \frac {4}{3} \pi r^3$$.The radius of balloon is 10 m.Hence, the volume of the balloon is $$\displaystyle V= \frac {4}{3} \times 3.1416 \times (10)^3 = 4186.7 m^3$$.The mass of displaced air is obtained from the product of volume and density. It is $$\displaystyle 4186.7 \times 1.2 = 5024.04 kg$$.The number of moles of gas present are $$\displaystyle n = \frac {PV}{RT}$$ $$\displaystyle =\frac {1.666 \times 4186.7 \times 10^3}{0.083 \times 300} =279.11 \times 10^3$$.Note: Here, the unit of volume is changed from $$m^3$$ to $$dm^3$$. $$1 m^3 = 1000 dm^3$$.Mass of helium present is obtained by multiplying the number of moles with molar mass. It is $$\displaystyle 279.11 \times 10^3 \times 4 = 1116.44 \times 10^3 g = 1116.4$$ kg.The mass of filled balloon is the sum of the mass of the empty ballon and the mass of He. It is  $$\displaystyle 100+1116.4=1216.4$$ kg.Pay load $$\displaystyle =$$ mass of displaced air $$\displaystyle -$$ mass of balloon $$\displaystyle = 5024.04-1216.44=3807.6$$ kgChemistryNCERTStandard XI

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