Question

What is the minimal mass of helium (density 0.18 kg/m3) needed to lift a balloon carrying two people in a basket, if the total mass of people, basket, and balloon (but not gas) is 300 kg ? the density of the air is 1.2 kg/m3?

Answer 1

Answer:   
 M[min] = M[basket+people+ balloon, not gas] * ΔR/R[b] 
 Î”R is the difference in density between the gas inside and surrounding the balloon. 
 R[b] is the density of gas inside the baloon.   
 ==================================== 
 Let V be the volume of helium required. 
 Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V   
 U = 1.225gV newtons 
 ---- 
 Weight of Helium = Volume of Helium * Density of Helium * g 
 W[h] = 0.18gV N   
 Net Upward force produced by helium, F = Upthrust - Weight = (1.225-0.18) gV = 1.045gV N  -----

 
 Weight of 260kg = 2549.7 N 
 Then to lift the whole thing, F > 2549.7 
 So minimal F would be 2549.7 
 ---- 
 1.045gV = 2549.7 
 V = 248.8 m^3   
 Mass of helium required = V * Density of Helium = 248.8 * 0.18 = 44.8kg (3sf)   
 =====   
 Let the density of the surroundings be R 
 Then U-W = (1-0.9)RgV = 0.1RgV   
 So 0.1RgV = 2549.7 N 
 V = 2549.7 / 0.1Rg   
 Assuming that R is again 1.255, V = 2071.7 m^3 
 Then mass of hot air required = 230.2 * 0.9R = 2340 kg   
 Notice from this that M = 2549.7/0.9Rg * 0.1R so   
 M[min] = Weight of basket * (difference in density between balloon's gas and surroundings / density of gas in balloon)   
 M[min] = M[basket] * ΔR/R[b]