Physics
posted by Gabe on .
The boiling point of helium at 1 atm is 4.2 K.
What is the volume occupied by helium gas
due to evaporation of 70 g of liquid helium at
1 atm pressure and a temperature of 6.2 K?

This is not a trivial question. At those temperatures, the ideal gas law does not apply. Additionally, there are intermolecular forces that "compress" the volume.
Here is an estimate. The density of liquid helium is about .125g/ml, so 70 grams occupies about .560Liters (appx) as a liquid.
So this volume is the starting point, as is 4.2K as the "absolute" zero for the gas.
Now we fiddle with the Gas law:
PV=nRT
P(V+.560)=nR(6.24.2)
1(V+.560)=70/4.003 * 0.082057*2
V=2.31 liters
check this . Remember, this calculation is a stretch at low temps. 
There are various nonideal forms of the gas law that take into account molecular sizes and intermolecular attraction.
One of them is called the van der Waals equation. You have not been provided with the two constant paramters (a and b) that are needed to use that equation. The other is called the law of corresponding states, if I recall correctly. Temperatures are referred to the boiling point are part of the equation. I suspect that is what you need to take a look at to answer the question. Your reading assignment should provide a clue of wh8ch equation to use. 
This may help. It describes a way of using critical point properties to apply the Van der Waals equation of state:
http://www.phys.uri.edu/~gerhard/PHY525/tln30.pdf