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What will be the passenger carrying capacity of a balloon of diameter 20 m and mass 100 kg filled with helium at 1.0 atm at 27^(@)C. Density of air is 1.2 "kg"//m^(3) and average weight of a passenger is 65 kg. |
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Answer» 55 = `(4)/(3) pi r^(3)= (4)/(3) xx (22)/(7) xx 10^(3)=4190.47 m^(3) or 4190.47 xx 10^(3) L` `"Mass of air displaced by balloon" = 4190.47 xx 1.2 = 5028.56 g`. Number of moles of He in balloon, `n = (PV)/(RT) = (1 xx4190.47 xx 10^(3))/( 0.082 xx 300)= 170344` `"Mass of filled balloon" = 681.376 + 100 = 781.376 g`. `"Pay load of balloon" = "mass of air displaced" – "mass of filled balloon" = 5028.56 - 781.376 = 4247.184 g` `"Passenger carrying capacity" = ("Payload")/("Average wt.of Passenger")= 4247.184 65 = 65.3 = 65 "Passengers"`. |
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