Hoping some physics experts can help me with a simple, but confusing question.

Helium baloons float because they are less dense than the surrounding air.
If one wanted to do so, one could asily calculate the amount of helium required to lift a 200kg person in a baloon.

Now, suppose one is designing an emergency evac system for climbers on Mt Everest, which has, at the top, a pressure of only 300mm Hg.

Would one require more or less Helium to lift a man at this altitude?

Assuming a Helium tank contains 291cf of Helium at 2600psi, how many tanks would one need and what size baloon, to create neutral boyancy of said 200kg person at that alititude?

The mass of helium required (surprisingly) remains the same.

Let's look at the buoyancy problem from a force equilibrium point of view. The buoyant force = air density * balloon volume * g = m_air*g, which must balance helium weight + balloon weight, + man's weight = m_He*g + (m_b + m_m)g. The "g" factor cancels out, and you can solve for m_b+m_m = m_air - m_He.

Both air and helium being nearly ideal gases, both will expand according to the ideal gas law as the pressure drops. Therefore, the ratio of their densities remains constant. The constant is the ratio of their molecular weights, 29g/mol for air and 4g/mol for He, so the ratio is R = 29/4. Now look at the m_air - m_He term in the equation above. Since the air and helium masses are both calculated using the volume of the balloon, you get m_air / m_He = (d_air * V) / (d_He *V) = d?air / d_He = 29/4. Therefore, m_air - m_He = 29/4 * m_He - m_He = (29/4 - 1)*m_He = 25/4*m_He.

Putting this into the equation gives m_b + m_m = 25/4*m_He, so m_He = 4/25*(m_b + m_m). Since 4/25 is 16%, the helium required to loft a specified mass is 16% of that mass, regardless of height. Of course, at greater heights, this same mass of helium will expand to a larger volume, so you'll need a larger balloon. The balloon must be pretty light for this to be feasible in the first place, so this will modestly increase the total weight that must be lofted. For a nominal 200kg, this would be 32kg of helium (although I doubt very much that a 200kg person could make it to the top of Everest!).

As to the number of tanks, 291cf is 8.24m^3, which at 22.4m^3/kmol is 0.3679kmol. Since helium has a molecular weight of 4g/mol, this gives 1.4715kg of helium in each tank. To get 32kg of helium, you would therefore need 21.75, which rounds up to 22 tanks of helium, regardless of altitude.

As to the balloon volume, 32kg of helium is 8kmol, which would occupy 22.4m^3/kmol * 8kmol = 179.2m^3 at standard conditions (sea level atmospheric pressure, 0 degrees Celsius). Assuming the temperature at the top of Everest is not too cold, the main effect on the balloon volume will be pressure. the pressure ratio is 760mmHg/300mmHg = 2.533, so the balloon will be larger by this factor. The volume of the balloon is therefore 179.2m^3 * 2.533 = 454m^3. The volume of course does depend on altitude via pressure. The diameter of such a balloon would be 9.5355m, i.e. 31.28ft.

Hope that helps!

--Stuart Anderson