Class 11 Chemistry Gases & Solutions States of Matter

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 $_{∘}C$. (Density of air = 1.2 kg m$_{3}$ and R = 0.083 bar dm $3$ K$_{−1}$ mol$_{−1}$)

Solution: The volume of the balloon is $V=34 πr_{3}$.

The radius of balloon is 10 m.

Hence, the volume of the balloon is $V=34 ×3.1416×(10)_{3}=4186.7m_{3}$.

The mass of displaced air is obtained from the product of volume and density. It is $4186.7×1.2=5024.04kg$.

The number of moles of gas present are $n=RTPV $ $=0.083×3001.666×4186.7×10_{3} =279.11×10_{3}$.

Note: Here, the unit of volume is changed from $m_{3}$ to $dm_{3}$.

Mass of helium present is obtained by multiplying the number of moles with molar mass. It is $279.11×10_{3}×4=1116.44×10_{3}g=1116.4$ kg.

The mass of filled balloon is the sum of the mass of the empty ballon and the mass of He. It is $100+1116.4=1216.4$ kg.

Pay load $=$ mass of displaced air $−$ mass of balloon $=5024.04−1216.44=3807.6$ kg

The radius of balloon is 10 m.

Hence, the volume of the balloon is $V=34 ×3.1416×(10)_{3}=4186.7m_{3}$.

The mass of displaced air is obtained from the product of volume and density. It is $4186.7×1.2=5024.04kg$.

The number of moles of gas present are $n=RTPV $ $=0.083×3001.666×4186.7×10_{3} =279.11×10_{3}$.

Note: Here, the unit of volume is changed from $m_{3}$ to $dm_{3}$.

$1m_{3} =1000dm_{3}$.

Mass of helium present is obtained by multiplying the number of moles with molar mass. It is $279.11×10_{3}×4=1116.44×10_{3}g=1116.4$ kg.

The mass of filled balloon is the sum of the mass of the empty ballon and the mass of He. It is $100+1116.4=1216.4$ kg.

Pay load $=$ mass of displaced air $−$ mass of balloon $=5024.04−1216.44=3807.6$ kg

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