If #32 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 #8 L#?

Question

Nathan Ashcroft · Accepted Answer

The pressure is 32kPa.

Let's identify our known and unknown variables.

The first volume we have is #32# L, the first pressure is #8kPa#, and the second volume is #8L#. 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.
We will use this equation:

All we have to do is rearrange the equation to solve for #P_2#

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=(32\kPa xx 8\ cancel"L")/(8\cancel"L")# = #32 kPa#

Xavier Adame · Answer

undefined According to Boyle's Law, the pressure of a gas is inversely proportional to its volume when the temperature is constant. Using the formula $P_1V_1 = P_2V_2$, where $P_1$ is the initial pressure, $V_1$ is the initial volume, $P_2$ is the final pressure, and $V_2$ is the final volume:

$P_1 = 8 \, \text{kPa}$
$V_1 = 32 \, \text{L}$
$V_2 = 8 \, \text{L}$

$P_2 = \frac{P_1 \cdot V_1}{V_2}$
$P_2 = \frac{8 \, \text{kPa} \times 32 \, \text{L}}{8 \, \text{L}}$
$P_2 = 32 \, \text{kPa}$

Therefore, the pressure the gas will exert if the container's volume changes to 8 L will be 32 kPa.