Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Helium Lift Power
Name: Paul
Status: other
Age: 30s
Location: N/A
Country: N/A
Date: 1999 - Revised May 2008

I would like to know the lifting power of Helium eg: How much helium to lift 1kg Any information would we great.

That depends on the density of air and the density of the helium. For simplicity's sake, let's say the air and the helium are at the same temperature and pressure. In that case, four grams of helium will occupy about the same volume as 28.5 gm/mole of air. To lift an object, you need for the mass of the object + the mass of the helium to be less than the mass of the air it displaces. Since any object you want to lift will probably have a much greater density than air or helium, let's neglect its volume for simplicity's sake (in other words, we'll neglect the mass of the air displaced by the object, because it will probably be only a small part of the mass of the air displaced by the helium and the object together). Then we can say that you need the mass of the object + helium to be less than the mass of the air displaced by the helium.

        M + Mh < Ma,

where M is the mass of the object, Mh is the mass of the helium, and Ma is the mass of the air displaced. Our shortcut of neglecting the volume of the object just says that

        Ma = (28.5/4)Mh,

that is, a given mass of of helium will displace (28.5/4) times its mass of air.

Knowing that 4 grams of helium occupies that same volume as 28.5 grams of air, we can see that four grams of helium will lift almost 25 grams:
        M + Mh < Ma
        M + Mh < (28.5/4)Mh
        M < (28.5/4)Mh - Mh
        M < Mh [(28.5/4) - 1]
        M < Mh (7.125 - 1)
        M < Mh (6.125)
        M < 4g (6.125)
        M < 24.5 g.
An easier way to look at this is to note that since four grams of helium displaces 28.5 grams of air, the "payload" for the helium is 28.5 g - 4 g = 24.5 g.

On a per gram basis, this means that one gram of helium will lift a payload (including the mass of the balloon) of (24.5 g payload)/(4 g helium) = 6.125 g payload /g helium. You can also see this in the above derivation if you just figure the mass of helium to be 1 g instead of 4 g.

So, to lift 1000 grams (1 kg), you need about 163 grams (~0.16 kg) of helium:
        M < Mh (6.125)
        M/(6.125) < Mh
        Mh > (1000g)/ (6.125)
        Mh > 163.3 g.

How much volume of helium is this? It depends on the temperature and pressure. Neglecting the volume of the object to be lifted becomes a more serious error as it becomes a larger fraction of the total volume. Under normal conditions (ambient temperature and pressure), this is a small error.

Richard Barrans Jr., Ph.D.

Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (, or at Argonne's Educational Programs

Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory