If #7/5 L# of a gas at room temperature exerts a pressure of #6 kPa# on its container, what pressure will the gas exert if the container's volume changes to #2/3 L#?

1 Answer
Aug 1, 2016

Answer:

The gas will exert a pressure of #63/5# kPa.

Explanation:

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

The first volume we have is #7/5# L, the first pressure is #6kPa# and the second volume is #2/3L#. 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=(6\kPa xx 7/5\ cancel"L")/(2/3\cancel"L")# = #63/5kPa#