Need help calculating density of hot air..?

Question

I need to calculate the density of the hot air inside a hot air balloon that we made in class out of a garbige bag.....

Measurments recorded-

Weight of Bag - 18.7grams

Weight of rocks added to bag - 8.8grams

Diameter of bad - 64.5cm

hsight of bag - 74cm

ρA - 1.2kg/m^3

Let-

FB – Buoyant Force

ρA – Density of Normal Air

V – Volume

g – Gravity

ρHA - Density of hot air

Ecuation worked out to use-

ρA x V x g = Weight of bag x g

NEEDED ASAP!!!!

any relevent sites would be useful!!

Answer 1

Since the height of the bag is different from the diameter, assume it has a shape of a cone capped with a hemisphere (traditional hot-air balloon shape):

V = (1/3)πr^2(h - r) + (1/2)(4/3)πr^3

V = (1/3)π[hr^2 - r^3 + 2r^3]

V = (1/3)πr^2(h + r)

For the balloon to provide lift,

ρA*V*g - ρHA*V*g ≥ m*g

(ρA - ρHA)V ≥ m

ρA - ρHA ≥ m/V

ρA - m/V ≥ ρHA

ρHA ≤ ρA - m/[(1/3)πr^2(h + r)]

ρHA ≤ 1.2 kg/m^3 - (18.7 g + 8.8 g)/[(1/3)π(32.25 cm)^2(74 cm + 32.25 cm)]

ρHA ≤ 1.2 kg/m^3 - (3)(0.0275 kg)/[π(0.3225 m)^2(1.0625 m)]

ρHA ≤ 1.2 kg/m^3 - (0.0825 kg)/(0.3471669 m^3)

ρHA ≤ 1.2 kg/m^3 - 0.2376380 kg/m^3

ρHA ≤ 0.962 kg/m^3

If you instead assume a cylindrical shape,

ρHA ≤ 1.2 kg/m^3 - (0.0275 kg)/[π(0.3225 m)^2(0.74 m)]

ρHA ≤ 1.09 kg/m^3