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? – BowlOfRed Jul 20 '14 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.