MR. P said:
...three senerios...three balloons of equal volume balloon
1 is filled with He2 balloon
2 is filled with H2 balloon
3 is unique and retains the same volume as 1 &2 but holds a perfect vacuum
all these balloons are at STP what would be the difference in their 'bouancy' or rather their lifting capability?
This all boils down to the ideal gas law:
PV=N_mRT
where N
m is the number of molecules in the balloon. I hate that form of it, however, so I'm going to define:
n=\frac{N_mN_A}{V}
where n is the number density and N
A is Avogadro's number. So we have:
P=nkT
where k is Boltzmann's constant. Anyway, let's look at this. In the hydrogen and helium balloons, we know that the internal pressure must equal the outside pressure. If it didn't, then the molecules would push on the balloon, making it expand or contract until the pressures were equal. Likewise, we can treat the temperatures as being equal because we'll assume that the gaseous contents were exposed to the external environment and reached thermal equilibrium with it. This means:
n=\frac{P}{kT}
is a constant as well.
However, any number of hydrogen or helium molecules has a smaller total mass than the same number of air molecules (mainly diatomic nitrogen). To see this, just look on the periodic table. Atomic hydrogen and helium have masses of about 1 amu and 4 amu, respectively, while nitrogen has a mass of about 14 amu. By Archimedes principle, the buoyancy force is given by:
F=-m_{air}g
where m
air is the mass of air that
would be in the balloon if it weren't filled with hydrogen or helium. The total weight of the balloon is given by:
F=m_{balloon}g
If we neglect the weight of the elastic balloon material, this gives a total force of
F=(m_x-m_{air})g=-m_{air}(1-\frac{m_x}{m_{air}})g
For the cases you listed:
\frac{m_{He_2}}{m_{air}}\sim\frac{4}{14},\ \frac{m_{H_2}}{m_{air}}\sim\frac{1}{14},\ \frac{m_{vac}}{m_{air}}=0
Thus, the ratios of the upward forces are:
\frac{F_{He_2}}{F_{H_2}}\sim\frac{10}{13},\ \frac{F_{H_2}}{F_{vac}}\sim\frac{13}{14}
This was done on-the-fly and I have no means of checking it, so please let me know if I made any errors.
NOTE: If the balloon were made of an elastic material, it couldn't maintain a vacuum at the same volume (the external pressure would squash it), but the above calculation assumes that it does anyway. For this case, one can instead imagine that the "balloon" is made of a light rigid material.