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 ?
