If #3 L# of a gas at room temperature exerts a pressure of #15 kPa# on its container, what pressure will the gas exert if the container's volume changes to #5 L#?

Question

Amelia Adams · Accepted Answer

The gas will exert a pressure of # 9 kPa#

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

The first volume we have is #3 L# , the first pressure is #15kPa# and the second volume is #5 L#. Our only unknown is the second pressure.

The answer can be determined by using Boyle's Law:

Rearrange the equation to solve for the final pressure by dividing both sides by #V_2# to get #P_2# by itself like this:

#P_2=(P_1xxV_1)/V_2#

Plug in your given values to obtain the final pressure:

#P_2=(15\kPa xx 3\ cancel"L")/(5\cancel"L")# = #9kPa#

Olivia Askew · Answer

undefined Using the ideal gas law, $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles of gas, $R$ is the ideal gas constant, and $T$ is temperature in Kelvin, we can solve for the final pressure ($P_2$) when the volume changes.

Given $P_1 = 15 \, \text{kPa}$, $V_1 = 3 \, \text{L}$, and $V_2 = 5 \, \text{L}$, we rearrange the equation to solve for $P_2$:

$P_1V_1 = P_2V_2$

$15 \, \text{kPa} \times 3 \, \text{L} = P_2 \times 5 \, \text{L}$

$45 \, \text{kPa}\, \text{L} = 5P_2$

Divide both sides by 5 L:

$P_2 = \frac{45 \, \text{kPa}\, \text{L}}{5 \, \text{L}} = 9 \, \text{kPa}$

So, the gas will exert a pressure of $9 \, \text{kPa}$ when the container's volume changes to $5 \, \text{L}$.