Helium lift (part 1)?

Question

What is the lifting power of Helium? For example how many cubic feet of Helium would it take to lift a 5 pound weight?

Edward...are you going to answer the question or not? / Can someone please give me a straight answer?

Answer 1

The weight is
W=g(p(air) - p(He))V

Where the required volume V is
V= W/[g(p(air) - p(He))]
g - acceleration due to gravity
p(air) - density of air
p(He)) - density of helium

Some additional explanation.
_________
Let density p be mass m per unit volume v
p=m/v
also weight W is a product of mass m times acceleration due to gravity g.
W=mg
A buoyant force is cased by a fluid attempting to displace an intruding object. In this case Helium is 'intruding' on air space. We can express it as a difference in densities of fluid a certain material is submerged. That is
p(air) - p(He) Why minus? Well The gravity of earth is pushing our Helium downward while air is pushing it in the apposite direction. The weight is equal to the product of acceleration due to gravity g
W= g p(He) V where V is the volume of the object.
the force that air exerts on the Helium balloon is
F=g p(air) V

The lifting force is the difference between these forces
Fb=F-W=g(p(air) - p(He))V (equation up above)

If it is 5 pound of mass then it is equal to 5/2.205=2.27 kg
p(air)=1.292 kg/m^3 at 20 deg C
p(He)=0.1786 kg/m^3 at 0 C
V= W/[g(p(air) - p(He))]
V=mg/[g(p(air) - p(He))]=
V=m/[(p(air) - p(He))]=
V=2.27 /[1.292 - 0.1786]=
V=2.04 m^3 or if insist on British units where 1 feet cube = 0.0283 meter cubed
V=72.0 ft^3

Answer 2

Just search it on google, but I imagine it is a lot.