A13.0 l sample of helium gas has a pressure of 29.0 atm. what volume would this gas occupy at 3.50 atm? assume ideal behavior.

From the equation;
P1V1=P2V2
V2=P1V1÷P2
V2=29.0×13.0÷3.50
V2=107.71L
therefore V2=107.7L

This uses an equation known as Boyle's law. Boyle's law states that at a constant temperature, the product of the pressure that a gas exerts and the volume it occupies is constant, meaning they are inversely proportional. This means the equation is:
P1V1 = P2V2
We have 3 knowns:
P1 = 29.0 atm
V1 = 13.0 L
P2 = 3.50 atm
V2 = unknown
Plugging into our equation:
(29.0 atm)*(13.0 L) = (3.50 atm)*(V2)
V2 = (29.0 atm * 13.0 L)/(3.50 atm)
V2 = 108 L = Volume occupied at 3.50 atm
This makes sense, as when the pressure decreased, Boyle's Law predicts that the volume will increase to compensate for the lost pressure, and it did.