$A$ hot air balloon is a sphere of radius $8 \ m$. The air inside is at a temperature of $60^{\circ} \ C$. How large a mass can the balloon lift when the outside temperature is $20^{\circ} \ C$? Assume air is an ideal gas,$R = 8.314 \ J \ mol^{-1} \ K^{-1}$,$1 \ atm = 1.013 \times 10^5 \ Pa$,and the membrane tension is $S = 5 \ N/m$.

(N/A) The pressure inside the balloon is $P_i$ and the outside pressure is $P_0$. The excess pressure is given by $P_i - P_0 = \frac{2S}{r}$,where $S = 5 \ N/m$ and $r = 8 \ m$.
$P_i - P_0 = \frac{2 \times 5}{8} = 1.25 \ Pa$.
Given $P_0 = 1.013 \times 10^5 \ Pa$,so $P_i = P_0 + 1.25 \ Pa \approx 1.013 \times 10^5 \ Pa$.
The volume of the balloon is $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (8)^3 \approx 2144.66 \ m^3$.
Using the ideal gas law $PV = nRT = \frac{m}{M} RT$,the mass of air is $m = \frac{PVM}{RT}$.
The molar mass of air is $M \approx 29 \times 10^{-3} \ kg/mol$.
Mass of displaced air $M_0 = \frac{P_0 V M}{R T_0} = \frac{1.013 \times 10^5 \times 2144.66 \times 29 \times 10^{-3}}{8.314 \times (273 + 20)} \approx 2577 \ kg$.
Mass of air inside $M_i = \frac{P_i V M}{R T_i} = \frac{1.013 \times 10^5 \times 2144.66 \times 29 \times 10^{-3}}{8.314 \times (273 + 60)} \approx 2345 \ kg$.
The lifting capacity is $M_{lift} = M_0 - M_i = 2577 - 2345 = 232 \ kg$.