How do you calculate the lift generated by hydrogen, helium, or  (theoretically) a vacuum in a balloon (probably mylar)?  How high could such balloons rise, given an ideal situation and a standard  mylar or similar balloon material?  What difference does a hot gas  make over a cold gas?  What other gases could provide lift in our atmosphere (i.e. are lighter than air)?

The question you asked has to do with physics as a basic science, and with one of its specialized branches, which is fluid mechanics. The example you mentioned, the balloon, is an example for buoyancy and buoyant force. Perhaps you remember Archimedes’ principle about floatation and buoyant force. That theory says that there is a force exerted on any body immersed or half-immersed  inside a fluid. That force is termed the buoyant force, and it is equal to the weight of the displaced volume from the fluid. The second point to discuss is some mechanics that will not exceed your secondary school information.

You are aware of Newton’s laws, I am sure. Thus for a body moving with variable velocity (acceleration) like the balloon, we should be able to establish an equation of motion. There are forces that pull that balloon down. These forces are the weight of the people on board and the equipment (motor and fuel tanks). This is in addition to the weight of the balloon ( I mean the place where the gas is inside, whatever the gas name was. However, the balloon is moving upwards. Thus there should be forces that are pushing the balloon upwards.

This force is the one called the buoyant force. The buoyant force is equal to:

F (b) = (displaced fluid density)x(gravity acceleration)x(displaced volume)

Consider this fact that a lot do not consider, the balloon is a body immersed in a fluid, which is air. The displaced volume is the volume that the balloon replaces of air to occupy. Then the buoyant force for the balloon equals:
F (b) = (air density)x(9.81 m/sec^2)x(volume of the gas filled balloon)

In an equation
Net force = Upward pushing forces - Downward pulling forces
F = ma = F (up) - F (down)
Thus the complete equation

F = ma = F (b) - Weight of equipment- Weight of the gas container
(This equation ignores some effects; such as: wind direction, surrounding temperature and pressure)
Consider with me these facts; the weight of the balloon equipment is perfectly constant, and so is the buoyant force (in fact it changes somehow because as we ascend upward the density of air decreases, but this is not so significant because air density itself is 1.2 Kg/m^3 and variations will be in limits of 0.01 to 0.1 ). Thus if you are the driver of a balloon, how can you change the speed of the balloon? The only answer is to change the weight of the gas container. That container is full of a gas, probably hydrogen. By increasing the heat for hydrogen gas, its density decreases.
There is one more equation that I would like to add:

Weight = (density)x(gravity acceleration)x(volume)

Thus if  return back to the equation
ma = F (b) - Weight of equipment- Weight of the gas container

m = mass of the balloon equipment, and the mass of the gas (total mass)
F (b) = (density of air)x(gravity acceleration)x( volume of the balloon) = nearly a constant value
Weight of the equipment = constant
Weight of the gas container = (gas density)x(gravity acceleration)x(volume)

For other gases, here are some with their densities in Kilogram per cubic meters:

air  = 1.29
Oxygen = 1.43
Hydrogen  = 0.0899
Helium = 0.179

You had a question about if a free vacuum. Okay, just create a vacuum, and we will discuss it later. But till that time, you should guess that the contribution of the gas weight will vanish.
You asked about the maximum height. I am not sure about the value nore the figure, because this all depends on the fuel you have. If you have enough containers to reach the moon, and you are able to resist variation in temperature, pressure, and oxygen % in air, you can make it I suppose. However, atmosphere does not extend to the moon, as you definitely know.

That is all for now. Waiting for a reply or further explanation if you want. Thanks for your question, John. You really refreshed my mind.

Moataz Attallah
The American University in Cairo-Egypt
Mechanical Engineering undergraduate student
Email: mizoa@aucegypt.edu

Abbott, A.F. Ordinary Level Physics. Heinemann Educational Books, London. Fourth edition 1984.

Serway, Raymond. Physics for Scientists and Engineers. Saunders Collage publishing, USA. Third updated edition 1990.