If #18 L# of a gas at room temperature exerts a pressure of #8 kPa# on its container, what pressure will the gas exert if the container's volume changes to #15 L#?

1 Answer
Jun 4, 2016

Answer:

It will be #9600# Pa.

Explanation:

First of all it is always useful to try to "guess" the answer in order to have an idea of what is going on.
We have a box with a certain volume (18 L) and we are going to shrink this volume. So we expect that the pressure will increases because we are compressing the same gas in a smaller volume.

Then the equation that regulates this is

#P_iV_i=P_fV_f#

The initial pressure multiplied by the initial volume is equal to the final pressure multiplied by the final volume. If the volumes on the right decreases, the pressure has to increase in order to maintain the same product. Now with the numbers #P_i=8000# Pa #V_i=18# L #P_f=?# Pa #V_f=15# L.
The equation is then

#8000 * 18 = P_f*15#

then

#P_f=8000*18/15=9600# Pa.

This pressure is bigger than the initial one, as expected.