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Ok, this would be a problem of essentially displacement as the lift is caused by displacing "air".

Ok, so one needs the densities of air and helium at some given temperature and pressure... "STP" is common, although one really would need to do it at the ambient temperature... or to do the final conversion using the simple formula PV=nRT.

According to answers.Yahoo.com, the densities of air and helium at STP are:

Density of helium = 0.0001785 g/cm³
Density of air = 0.001293 g/cm³,

Ok, so your questions is how many cm³ of helium are needed to lift 1 gm.... let's try to make an equation.

So, if we displace X cm³ of air with Helium we have:

(X cm³)*(Density of Air g/cm³) - (X cm³)*(Density of Helium g/cm³) = (mass displaced in grams).

Set the amount of mass being displaced to 1 gram, and putting in the densities we have:

(X cm³)*(0.001293 g/cm³) - (X cm³)*(0.0001785 g/cm³) = 1 g
(X cm³)*(0.001293 g/cm³ - 0.0001785 g/cm³) = 1 g
(X cm³) = 1g/(0.001293 g/cm³ - 0.0001785 g/cm³)

(X cm³) = 897 cm³

You had asked in liters... with 1000 cm³/liter, the answer would be:

0.897 liters of helium would displace 1 gram of oxygen without taking into account the weight of the container, or any pressure imposed by inflating container such as a rubber balloon.

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15y ago

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