If #17/3 L# of a gas at room temperature exerts a pressure of #24 kPa# on its container, what pressure will the gas exert if the container's volume changes to #19/4 L#?

1 Answer
Aug 19, 2016

Answer:

The gas will exert a pressure of #544/19# kPa.

Explanation:

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

The first volume we have is #17/3#L, the first pressure is #24kPa# and the second volume is #19/4L#. Our only unknown is the second pressure.

We can obtain the answer using Boyle's Law:
www.physbot.co.uk

The letters i and f represent the initial and final conditions. All we have to do is rearrange the equation to solve for the final pressure.

We do this by dividing both sides by #V_f# in order to get #P_f# by itself like so:
#P_f=(P_ixxV_i)/V_f#

Now all we do is plug in the values and we're done!

#P_f=(24\kPa xx 17/3\ cancel"L")/(19/4\cancel"L")# = #544/19kPa#