# If 18 L of a gas at room temperature exerts a pressure of 5 kPa on its container, what pressure will the gas exert if the container's volume changes to 12 L?

This a pretty straight forward application of Boyle's Law, expressed by the equation: ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$
$18 \cdot 5 = 12 \cdot {V}_{2} \text{ solve for } {V}_{2}$
${v}_{2} = 5 \frac{18}{12} = 5 \frac{3}{2} = \frac{15}{2} = 7.5 k P a$