# If 8 L of a gas at room temperature exerts a pressure of 28 kPa on its container, what pressure will the gas exert if the container's volume changes to 9 L?

Jun 19, 2016

The final pressure will be 25 kPa.

#### Explanation:

This is an example of Boyle's law, which states that the volume of a gas kept at constant temperature varies inversely with the volume. This means that when volume increases, pressure decreases. The equation required to solve this problem is ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, where ${P}_{1}$ and ${P}_{2}$ are the initial and final pressure, while ${V}_{1}$ and ${V}_{2}$ are the initial and final volume.

Given
${V}_{1} = \text{8 L}$
${P}_{1} = \text{28 kPa}$
${V}_{2} = \text{9 L}$

Unknown
${P}_{2}$

Solution
Rearrange the equation to isolate ${P}_{2}$. Substitute the variables with their values and solve.

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

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

P_2=(8cancel"L"*28"kPa")/(9cancel"L")="25 kPa" rounded to two significant figures