total weight = (251+350)(9.81)
+ rhogas 9.81(4/3) pi (9.5)^3
where rhogas is the mass per unit volume of the gas in the balloon
weight must = weight of air displaced = rhoout g (4/3)pi(9.5)^3
rho gas = kg/mol * n = .029 n
where n is mols in PV =nRT
T = 273 + T that we want
rho air = same but T = 273+20 = 293
assume P = 1 atm inside and out
1)We consider a hot-air balloon of mass 251 kg (basket and envelope). The spherical envelope of the balloon has a diameter of 19 m when fully inflated. To what temperature (in °C) must the enclosed air be heated for the balloon to carry five standard men? Assume the surrounding air is at 20°C and is treated as an ideal gas. Use 29 g/mol for the molar mass of air. (Assume the mass of each standard man is 70 kg.)
so ive been looking through my notes and such for a long time and i cant find out how to do this!! i dont need an answer persay, just maybe a little bit of help on how to start this question off right. I know for sure it has something to do with bouyancy. its just our prof hasnt explained it very well lately. thanks!
1 answer
1 answer