If 3 L3L of a gas at room temperature exerts a pressure of 15 kPa15kPa on its container, what pressure will the gas exert if the container's volume changes to 5 L5L?

1 Answer
Aug 3, 2016

The gas will exert a pressure of 9 kPa9kPa

Explanation:

Let's begin by identifying our known and unknown variables.

The first volume we have is 3 L3L , the first pressure is 15kPa15kPa and the second volume is 5 L5L. Our only unknown is the second pressure.

The answer can be determined by using Boyle's Law:
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Rearrange the equation to solve for the final pressure by dividing both sides by V_2V2 to get P_2P2 by itself like this:

P_2=(P_1xxV_1)/V_2P2=P1×V1V2

Plug in your given values to obtain the final pressure:

P_2=(15\kPa xx 3\ cancel"L")/(5\cancel"L") = 9kPa