One indication whether the 'inner' balloons are filled with helium or something heavier is whether the 'outer' balloon still floats in air or not.
Whether it floats or not depends on whether its overall density is lower than air's density or not. Its overall density $d$ is simply given by:
$$d=\frac{\Sigma m}{V}$$
Where $\Sigma m$ is the sum of all the masses that make up the balloon and $V$ its total volume.
The sum of all masses means the sum of:
- Mass of outer balloon material
- Mass of helium
- Masses of all inner balloons material
- Masses of all gas contained in the inner balloons
If the inner balloons are filled with air and there are too many of them the overall density will exceed that of air and the outer balloon will not float.
But even if the inner balloons are all filled with helium, the outer balloon may still not float. That's because a larger number of inner balloons making up the same volume of a smaller number of inner balloons with create more mass because of the increased surface area and the inherent 'cost' in mass of larger number of inner balloons. For that reason:
Obviously, if the inner balloons were filled with something moderately lighter than helium, they would float, but they're huddled at the bottom, so it's either helium or plain air.
... is not necessarily correct.
If the mass of the outer balloon material, the total volume of the balloon, the mass of the inner balloon material and the volume of the inner balloons was known, then depending on whether the outer balloon still floats in air or not, the lower or lower limit of the density of the inner balloon gas could be calculated.
Call $m$, $d$ and $V$, with resp. suffixes $o$ and $i$ for outer and inner balloons, the mass of balloon material, the density of the filling gas and volume of the balloons. Let there be $n$ identical inner balloons. The overall density of the balloon is then:
$$d=\frac{m_o+d_o(V_o-nV_i)+nm_i+nd_iV_i}{V_o}$$
If the balloon just about floats in air, then:
$$d=d_{air}$$
From which we can deduce:
$$d_i=\frac{d_{air}V_o-m_o-d_o(V_o-nV_i)-nm_i}{nV_i}$$
Best Answer
Start from the ideal gas law: PV = NkT where V is volume, N is number, k is Boltzmann's constant, T is temperature, and P is pressure. Nowhere does the mass density or mass of each atom appear.
If the Helium in the balloon is at the same temperature and roughly the same number density as the air outside, it will be at the same pressure as the air outside. In practice, the Helium will be slightly overpressured because the pressure of the Helium must balance the pressure of the air outside + the surface tension of the balloon.
Although both the air and the Helium will have roughly the same number density and pressure, the Helium will be less mass dense because each Helium atom is lighter.
The more dense air falls down around the less dense balloon, pushing it up.