Let the pressure inside the balloon be and the outside pressure be P0
Pi − P0 = \(\frac{2T}{r}\)
[T = Surface Tension, r = radius balloon]
Considering the air to be an ideal gas
PiV = niRTi
Where, V is the volume of the air inside the balloon, ni is the number of moles inside and Ti 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, ni = \(\frac{P_{i}V}{RT_i}=\frac{M_{i}}{M_A}\)
Where Mi is the mass of the air inside and MA 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 + Mig = M0g
Or W = M0g − Mig
In atmospheric air, 21% O2 and 79% N2 is present.
∴ Molar mass of air
MA = 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 = Pi
= \(\frac{0.02884× \frac{4}{3}π×8^3}{8.314}\) × 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