# If 32 L of a gas at room temperature exerts a pressure of 64 kPa on its container, what pressure will the gas exert if the container's volume changes to 24 L?

Mar 19, 2016

The decrease in volume causes the pressure to increase to 85 kPa.

#### Explanation:

This is an example of Boyle's law, which states that the pressure and volume of a gas kept at constant temperature vary inversely, meaning that if the volume increases, the pressure will decrease and vice versa. The equation to use is ${V}_{1} {P}_{1} = {V}_{2} {P}_{2}$.

Given
${V}_{1} = \text{32 L}$
${P}_{1} = \text{64 kPa}$
${V}_{2} = \text{24 L}$

Unknown
P_2"

Solution
Rearrange the equation to isolate ${P}_{2}$, substitute the given values into the equation and solve.

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

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

P_2=((32cancel"L")xx(64"kPa"))/(24cancel"L")

${P}_{2} = \text{85 kPa}$ rounded to two significant figures.

As expected from Boyle's law, when the volume decreased from 32 L to 24 L, the pressure increased from 64 kPa to 85 kPa.