# Mass of air inside air balloon

I have an exercise here with a solution.

The basic outline of the problem is:

Calculate the mass of air inside a hot air balloon, given the mass of the air balloon at rest is $m_b = 500~\mathrm{kg}$.

The solution goes like this:

The buoyant force of the air and the gravitational pull on the balloon are in equilibrium:

$$F_{B,a} = F_{g,a}$$

From which follows:

$$\rho_a g V = m_b g$$

Rewriting as an expression for volume:

$$V = \frac{m_b}{\rho_a}$$

Using the ideal gas law in terms of density:

$$V = \frac{m_b R T_o}{M_a P}$$

$T_o$ is the temperature outside the air balloon, $M_a$ is the molar mass of dry air.

Rewriting the ideal gas law, in this case in terms of the amount of moles of air inside the balloon:

$$n = \frac{PV}{RT_i} = \frac{m_b}{M_a} \frac{T_o}{T_i}$$

Here $T_i$ is the temperature inside the balloon.

Finally, the mass of the air inside is given by the relation:

$$m_a = m_b \frac{T_o}{T_i}$$

There's one thing I'm missing here: why does the mass of the air inside the balloon pulling it down not play a part ?

• Do you mean the "mass of the air balloon" (the balloon and the envelope) is mb, or the mass of the air inside the balloon is mb? Jul 20, 2014 at 23:45

From Wikipedia: (1) The most-used size is about $2,800 m^3$ (99,000 cu ft), and can carry 3 to 5 people.
(2) At sea level and at 15 °C, air has a density of approximately $1.225 kg/m^3$ ($0.001225 g/cm^3$, $0.0023769 slugs/ft^3$).
So, the air in a $2,800 m^3$ balloon filled by a fan (without heating) would have a mass of nearly 3500 kg.