A helium tank has a volume of 13.0 L at a pressure of 26.5 atm. What is the volume of the same amount of helium at a pressure of 1.25 atm?

275.6 L

Explanation: We use Boyle's Law to solve this.

The initial pressure (P1) is 26.5, the initial volume (P1) is 13.0 L, and the final pressure (P2) is 1.25. Using Boyle's Law, we plug in the values into the equation.

P1V1=P2V2

(26.5 atm)(13.0 L) = (1.25 atm) V2

As a regular equation, we solve for V2.

344.5 atm/L= (1.25 atm) V2

344.5 atm/L / 1.25 atm = V2

275.6 = V2

The atm value cancelled out and we are just left with the L value.

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.