Question

The mass of a hot-air balloon and its cargo (not including the air inside) is 140 kg. The air outside is at 10.0°C and 101 kPa. The volume of the balloon is 490 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

To lift the balloon, buoyancy force of outside air must be equal to weight of balloon +air

Using Force balance

Fb = W(inside air) - W(balloon)

density of outside air*V*g = density of inside air*V*g + mass of ballon*g

rho_out*V = rho_in*V + Mb

(rho_out - rho_in)*V = Mb

rho_out - rho_in = Mb/V ......... eq 1

Using given values:

Mb = mass of balloon = 140 kg

rho_out = 1.244 kg/m^3

V = Volume of balloon = 490 m^3

Now Using ideal gas law:

PV = nRT

n = mass/Molecular weight = m/Mw

PV = m*R*T/Mw

m/V = density = P*Mw/(RT)

Since P*Mw/R is constant, So density is inversely proportional to the temperature

rho_in/rho_out = T_out/T_in

T_out = 10 C = 283 K

rho_in = rho_out*T_out/T_in

Using above expression in eq 1.

rho_out - rho_out*T_out/T_in = Mb/V

1 - T_out/T_in = Mb/(V*rho_out)

T_in = T_out/(1 - Mb/(V*rho_out))

Using known values:

T_in = 283/(1 - 140/(490*1.244))

T_in = 367.4 K = 367.4 - 273 = 94.4 C

So temperature must be increased to 94.4 C before lift off

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