# 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?

##### 1 Answer

#### 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.

#color(blue)(ul(color(black)(P_1V_1 = P_2V_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

#"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 = P_1/P_2 * 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.