# To what pressure would you have to compress 48.0 L of oxygen gas at 99.3 kPa in order to reduce its volume to 16.0 L?

Aug 11, 2016

$298 k P a$

#### Explanation:

The answer can be determined by using Boyle's Law:

Let's identify our known and unknown variables.

color(red)("Knowns:"
${P}_{1}$= 99.3 kpa
${V}_{1}$= 48.0 L
${V}_{2}$= 16.0 L

color(maroon)("Unknowns:"
${P}_{2}$

Rearrange the equation to solve for the final pressure by dividing both sides by ${V}_{2}$ to get ${P}_{2}$ by itself like this:

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

Plug in your given values to obtain the final pressure:

${P}_{2} = \left(99.3 \setminus k P a \times 48.0 \setminus \cancel{\text{L")/(16.0\cancel"L}}\right)$ = $298 k P a$