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?

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Answer 1 · 2016-08-11T02:05:18

$298 kPa$

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

$P_2=(P_1xxV_1)/V_2$

Plug in your given values to obtain the final pressure:

$P_2=(99.3\kPa xx 48.0\ cancel"L")/(16.0\cancel"L")$ = $298kPa$