If 13/4 L134L of a gas at room temperature exerts a pressure of 16 kPa16kPa on its container, what pressure will the gas exert if the container's volume changes to 3/8 L38L?

1 Answer
Jan 2, 2017

The gas will exert a pressure of 138.7 kPa138.7kPa

Explanation:

We can obtain the final pressure via Boyle's Law:

http://slideplayer.com/slide/698879/http://slideplayer.com/slide/698879/

Let's identify the known and unknown variables:

color(blue)("Knowns:")Knowns:
- Initial Volume
- Final Volume
- Initial Pressure

color(magenta)("Unknowns:")Unknowns:
- Final Pressure

All we have to do is rearrange the equation to solve for the final pressure. We do this by dividing both sides by V_2V2 in order to get P_2P2 by itself like this:
P_2=(P_1xxV_1)/V_2P2=P1×V1V2

Insert the given values into the equation to solve for P_2P2:

P_2= (16kPa xx 13/4cancel"L")/(3/8\cancel"L") = 138.7 kPa