If #2 L# of a gas at room temperature exerts a pressure of #36 kPa# on its container, what pressure will the gas exert if the container's volume changes to #12 L#?

Question

Olivia Askew · Accepted Answer

The new pressure is #6kPa#.

Let's start off with identifying our known and unknown variables.
The first volume we have is #2# L, the first pressure #36kPa# and the second volume is #12L#. Our only unknown is the second pressure.

We can obtain the answer using Boyle's Law which shows that there is an inverse relationship between pressure and volume as long as the temperature and number of moles remain constant.

The equation we use is

where the numbers 1 and 2 represent the first and second conditions. All we have to do is rearrange the equation to solve for the pressure.

We do this by dividing both sides by #V_2# in order to get #P_2# by itself:
#P_2=(P_1xxV_1)/V_2#

Now all we have to do is plug in the given values:
#P_2=(36\kPa xx 2\ cancel"L")/(12\cancel"L")# = #6 kPa#

Nicholas Agrusa · Answer

undefined Using Boyle's Law, we can solve for the new pressure of the gas:

$P_1V_1 = P_2V_2$

Where:
$P_1 = 36 \, kPa$ (initial pressure)
$V_1 = 2 \, L$ (initial volume)
$V_2 = 12 \, L$ (final volume, after change)

$P_2$ is the unknown pressure we need to find.

Substituting the values into the equation:

$36 \, kPa \times 2 \, L = P_2 \times 12 \, L$

$72 \, kPa = 12 \, L \times P_2$

$P_2 = \frac{72 \, kPa}{12 \, L} = 6 \, kPa$

Therefore, if the container's volume changes to $12 \, L$, the gas will exert a pressure of $6 \, kPa$.