A recreational (open) hot air balloon (i.e., Pinside is approximately Poutside)
ID: 1974714 • Letter: A
Question
A recreational (open) hot air balloon (i.e., Pinside is approximately Poutside) has a volume of 2139 m3 when fully inflated. The total weight of the balloon, basket, ballast and pilot is 1852.2 N (417 lbs). By how much must the density of the air in the balloon be smaller than that of the surrounding atmosphere in order to keep the balloon floating level near the ground?Explanation / Answer
The balloon will float when the weight of the displaced air is equal to the weight of the inflated balloon. --> the weight of the ballon (without air) is equal to the weight of the displaced air - weight of the hot air in the balloon. Let x = density of the air outside in kg/m^3, and y = the amount of kg/m^3 that the inner hot air is less dense than the outside air. 1852.2N = 2139 (m^3)*x (kg/m^3)*9,81 - 2139(m^3)* (x - y) (kg/m^3)*9,81 1852.2 = 2139*9,81*x - 2139*9,81*x + 2139*9,81*y 1852.2 = 2139*9,81*y y = 0,088286312 kgm^3. The hot air inside must be less dense by 0,088286312 kg/m^3 compared to the outside air.