# 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

#### Answer:

#### Explanation:

Before doing any calculations, try to predict what you *expect* to see happen to the volume of the sample.

The problem tells you that the *temperature* of the gas **remains unchanged**. No mention is made about the *number of moles* of gas present in the sample, which means that you can assume it to be **constant** as well.

As you know, temperature is actually a measure of the **average kinetic energy** of the gas molecules. Gas pressure is a result of the *frequency* and *intensity* of the collisions that take place between the gas molecules and the walls of the container.

If a pressure of *temperature*, then **decreasing** would result in an **increase** in volume.

Think about it like this - the average kinetic energy of the gas molecules remains **unchanged**, which means that in order to produce a **lower pressure** they must hit the walls of the container **less often**.

The molecules are moving with the same average kinetic energy, which is why they must hit the walls of the container *less often*.

This can only be achieved by an **increase in volume**.

So, when temperature and number of moles are kept constant, pressure and volume have an **inverse relationship** - this is known as **Boyle's Law**.

When pressure **increases**, volume **decreases**, and when pressure **decreases**, volume **increases**.

This means that you can expect the volume of the gas to **increase**.

Mathematically, you can write this as

#color(blue)(P_1V_1 = P_2V_2)" "# , where

Rearrange and solve for

#V_2 = P_1/P_2 * V_1#

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

As predicted, the volume of the gas **increased** as a result of the *decrease* in pressure.