A 13.0 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.