# A volume of helium occupies 11.0 L at a pressure of 98.0 kPa. What is the new volume if the pressure drops to 86.2 kPa?

Dec 18, 2016

#### Answer:

The new volume is $= 12.5 L$

#### Explanation:

We use Boyle's Law

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$

Pressure, ${P}_{1} = 98.0 k P a$

Volume, ${V}_{1} = 11.0 L$

Pressure, ${P}_{2} = 86.2 k P a$

Volume, ${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2$

$= 98.0 \cdot \frac{11.0}{86.2} = 12.5 L$