# A gas with a volume of 4.0 L at 90.0 kPa expands until the pressure drops to 20.0 kPa. What is its new volume if the temperature doesn't change?

Feb 27, 2018

$\text{18 L}$

#### Explanation:

You know that the temperature of the gas is being held constant, so you should be able to identify that you're dealing with a situation covered by Boyle's Law here.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{P}_{1} {V}_{1} = {P}_{2} {V}_{2}}}}$

Here

• ${P}_{1}$, ${V}_{1}$ are the pressure and the volume of the gas inside the balloon at an initial depth
• ${P}_{2}$, ${V}_{2}$ are the pressure and the volume of the gas inside the balloon at a final depth

Basically, Boyle's Law states that when the temperature and the number of moles of gas are being kept constant, increasing the pressure of a gas will cause its volume to decrease and decreasing the pressure of a gas will cause its volume to increase.

In other words, pressure and volume have an inverse relationship when temperature and number of moles of gas are constant.

In your case, the pressure drops from

$\text{90.0 kPa " -> " 20.0 kPa}$

so you should expect the volume of the gas to increase.

Rearrange the equation to solve for ${V}_{2}$, the new volume of the gas.

${V}_{2} = {P}_{1} / {P}_{2} \cdot {V}_{1}$

Plug in your values to find

V_2 = (90.0 color(red)(cancel(color(black)("kPa"))))/(20.0color(red)(cancel(color(black)("kPa")))) * "4.0 L" = color(darkgreen)(ul(color(black)("18 L")))

The answer is rounded to two sig figs, the number of sig figs you have for the initial volume of the gas.