Question

A hot air balloon is a sphere of radius 8 m. the air inside is at a temperature of 60°C. how large a mass can the balloon lift when the outside temperature is 20°C? (Assume air is an ideal gas, R = 8.314 J mol^-1 K^-1 , 1 atm. = 1.013 × 10³ Pa; the membrane tension is 5 Nm^-1.)

Answer 1

Let the pressure inside the balloon be and the outside pressure be P₀

P_i − P₀ = \(\frac{2T}{r}\)

[T = Surface Tension, r = radius balloon]

Considering the air to be an ideal gas

P_iV = n_iRT_i

Where, V is the volume of the air inside the balloon, n_i is the number of moles inside and T_i is the temperature inside, and P₀V = n₀RT₀ where V is the volume of the air displaced and n₀ is the number of moles displaced and T₀ is the temperature outside.

So, n_i = \(\frac{P_{i}V}{RT_i}=\frac{M_{i}}{M_A}\)

Where M_i is the mass of the air inside and M_A is the molar mass of the air and n₀ = \(\frac{P_0V}{RT_0}=\frac{M_0}{M_A}\) where M₀ is the mass of air outside that has been displaced. If W is the load it can raise, then

W + M_ig = M₀g

Or W = M₀g − M_ig

In atmospheric air, 21% O₂ and 79% N₂ is present.

∴ Molar mass of air

M_A = 0.21 × 32 + 0.79 × 28 = 28.84 g

∴ Weight raised by balloon

or W = \(\frac{M_AV}{R}(\frac{P_0}{T_0}-\frac{P_i}{T_i})\)g        here, P₀ = P_i

= \(\frac{0.02884× \frac{4}{3}π×8^3}{8.314}\) × 1.013 × 10⁵ (\(\frac{1}{293} − \frac{1} {333}\)) × 9.8 N

= 3044.2 N

Therefore, Mass lifted by balloon = \(\frac{W}{g}\)

= \(\frac{3044.2\, N}{9.8\frac{m}{s^2}}\)

= 310.6 Kg