A typical volume of a modern hot air balloon is 2500 cubic metres, and a typical maximum temperature of the hot air is 120 degrees Celsius. Given these figures, and an outside air temperature and density of 15 ∘C and 1.225 kg/m3 respectively, compute the maximum mass (in kilograms) of the balloon, basket and payload.

Answer 1

Archimedes says:
volume of fluid displaced * density of fluid * g = upward buoyant force = weight is neutrally buoyant
mass = weight/g = volume*density

2500 * 1.225 = 3063 kg up

now subtract the mass of air in the balloon
2500 * rho
what is rho, the density of air inside at 120 C ?
approximate as perfect gas ?
P V = n R T
P is close to the same in and out or balloon would burst
V the same
so
n T is about constant
density proportional to mass in the balloon volume
n hot T hot = n cold Tcold
n hot /n cold = Tcold/Thot
Tcold = 273 + 15 = 288
T hot = 273 + 120 = 393
so density hot/density cold = 288/393 = .733
.733 * 1.225 = .8977 kg/m^3
so mass of air in balloon = .8977*2500 = 2244 kg
then finally
mass of other stuff = mass of cold air displaced - mass of hot air = 3063 -2244 = 819 kg

Answer 2

The only parameters we need to convert are the temperature difference (ΔT=120−15=105) and the outside air temperature in Kelvin (T = 288.15 K).

Now the lift of this hot air balloon is found using:

L=ρVg(ΔTT+ΔT)
L=1.225⋅2500⋅9.81(105288.15+105)=mg
m⋅9.81=8023.727
m=818kg

Answer 3

818.225