Question

The mass of a hot-air balloon and its cargo (not including the air inside) is 220 kg. The air outside is at 10.0°C and 101 kPa. The volume of the balloon is 550 m3. To what temperature must the air in the balloon be warmed before the balloon will lift off? (Air density at 10.0°C is 1.244 kg/m3.)

Answer 1

Total mass of balloon:
m = m₀ + m_hotair

Required buoyant force:
B = m_net*g

Origin of buoyant force based upon volume and background fluid density:
B = rho_bg*g*V

Thus:
rho_bg*g*V = m_net*g
rho_bg*V = m_net

Mass of hot air:
m_hotair = rho_hot*V
rho_bg*V = m₀ + rho_hot*V
rho_hot = (rho_bg*V - m0)/V

From ideal gas law:
rho_hot = P*M/(R*T_hot)
rho_bg = P*M/(R*T_cold)

Thus:
P*M/(R*T_hot) = (P*M/(R*T_cold)*V - m₀)/V
R*T_hot/(P*M) = V/(P*M/(R*T_cold)*V - m₀)
T_hot = V*P*M/(R*(P*M/(R*T_cold)*V - m₀))
550*101000*28.97/(8.314*(*101000*28.97/(8.314*283.15)*550 - 220))

Data:
P = 101 kPa
M = 28.97 kg/kmol
R = 8.314 kPa-m^3/kmol-K
T_cold = 283.15 K
m₀ = 220 kg
V = 550 m³

T_hot = 550*101000*28.97/(8.314*(*101000*28.97/(8.314*283.15)*550 - 220))
T_hot = 417.5 K = 144.5 C