# If 2/5 L of a gas at room temperature exerts a pressure of 15 kPa on its container, what pressure will the gas exert if the container's volume changes to 7/2 L?

Oct 14, 2017

When the volume increases to $\text{7/2 L}$, the pressure decreases to $\text{1.7 kPa}$.

#### Explanation:

This problem involves Boyle's law, which states that the volume of a gas is inversely proportional to its pressure as long as temperature and mass remain constant. This means that if volume increases, the pressure decreases, and vice versa.

The equation for Boyle's law is:

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

where:

$P$ is pressure, and $V$ is volume.

Organize the data:

Known

${P}_{1} = \text{2/5 L}$

${V}_{1} = \text{15 kPa}$

${V}_{2} = \text{7/2 L}$

Unknown

${P}_{2}$

Solution

Rearrange the equation to isolate ${P}_{2}$. Plug in the known values and solve.

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

P_2=(15"kPa"xx2/5color(red)cancel(color(black)("L")))/(7/2color(red)cancel(color(black)("L")))="1.7 kPa"