Question

A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is 500.0 m^3 and the surrounding air is at 15.0°C. What must the temperature of the air in the balloon be for it to lift a total load of 290 kg (in addition to the mass of the hot air)? The density of air at 15.0°C and atmospheric pressure is 1.23kg/m^3.

Answer 1

Answer:

272° C

Explanation:

Given :

Volume of the balloon, V = 500 $m^3$

The temperature of the surrounding air, $T_{air} = 15^\circ C$

Total load, $m_{T}$ = 290 kg

Density of the air, $\rho_{air} = 1.23 \ kg/m^3$

We known buoyant force,

$F_B = \rho_{air} V$

For a 290 kg lift,  $m_{hot} = \frac{F_B}{g} = 290 \ kg$

$m=\rho V$

∴ $m_{hot}=\rho_{hot} V ; \ \ \ \ \ \frac{F_B}{g}-m_{hot} = 290 \ kg$

  $(\rho_{air} - \rho_{hot}) V= 290 \ kg$

  $\rho_{hot} = \rho_{air}- \frac{290}{V} \ kg = 1.23 \ kg/m^3 - \frac{290 \ kg}{500 \cm^3}$

  $\rho_{hot}= 0.65 \ kg/m^3 =\frac{\rho M}{R T_{hot}}$

 ∴ $\rho_{hot} T_{hot}= \rho_{air} T_{air}$

   $T_{hot}= T_{air}\left[\frac{\rho_{air}}{\rho_{hot}}\right]$

          $=288 \ K \times \frac{1.23 \ kg/m^3}{0.65 \ kg/m^3}$

         = 545 K

          $=272^\circ C$

Therefore, temperature of the air in the balloon is 272 degree Celsius.

Answer 2

To lift a load more than the weight of the balloon, the temperature of the air in the balloon has to be higher than the air in the surrounding.

• The temperature of the air in the balloon to lift a total load of 290 kg is approximately 272.12°C.

Reasons:

Given information are;

Volume of the balloon = 500.0 m³

Temperature of the surrounding air = 15.0°C

Density of air at 15.0°C = 1.23 kg/m³

Required:

The temperature required to lift 290kg.

Solution:

Let, $\rho _{air , b}$, represent the density of the air in the balloon, we have;

$\rho _{air , b}$ × 500.0 + 290 = 1.23 × 500

Therefore;

$\displaystyle \rho _{air , b} = \frac{1.23 \times 500- 290}{500} = 0.65$

According to the Ideal Gas Law, we have;

ρ₁ × R × T₁ = ρ₂ × R × T₂

Therefore;

$\displaystyle T_2 = \mathbf{\frac{\rho_1 \times T_1}{\rho_2}}$

Therefore;

$\displaystyle T_2 = \frac{1.23\times288.15}{0.65} \approx 545.27$

The temperature of the balloon, T₂ ≈ 545.27 -  273.15 = 272.12

The temperature of the air in the balloon, T₂ ≈ 272.12 °C

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