# Boyles law in use

We know in Boyles law

`P_1\ \ V_1=P_2\ \ V_2`

**NOTE:**

Temperature remains constant

Also when you sit on an exercise ball it will get smaller

This is Boyles law.

`P_1` and `V_1` refer to the pressure and volume before you sit on the exercise ball **and** `P_2` and `V_2` refer to the pressure and volume after you sit on the exercise ball.

Reducing the volume of a gas increases the pressure of the gas.

Push the gas into a smaller volume and the gas particles will collide more frequently with the walls. They will exert more pressure.

**NOTE 1:**

The SI unit used to measure pressure is `Pa` (Pascals)

The SI unit used to measure volume is `m^3` (metre cubed)

**NOTE 2:**

You can use other units for volume and pressure such as `ml`, pounds per square inch, `mm` of mercury or atm (atmospheres) as long as you use the same units either side of the equation.

**Examples**

1. A sample of gas is at atmospheric pressure `(100,000 Pa)` and has volume `50cm^3`

Determine the new pressure if the volume is decreased to `35cm^3`

`P_1V_1=P_2V_2`

Therefore `100,000xx50=P_2xx35`

`5,000,000=P_2xx35`

Therefore `P_2=(5,000,000)/35`

`P_2=142,857\ \Pa`

2. A sample of gas has a volume of `12.0\ \l` at a pressure of 1.00 atm. If the pressure of gas is increased to `1000\ \mm\ \ Hg`, what is the new volume of the gas?

But we must first get the two pressures in the same units of measurement

`1\ \atm=760\ \mm\ \Hg`

Therefore:

`P_1=760`

`V_1=12`

`P_2=1000`

`V_2=?`

`760xx12=1000V_2`

`9120=1000V_2`

`9120/1000=V_2`

`V_2=9.12l`

3. A carbon dioxide gas cylinder contains `180cm^3` of gas at a pressure of `6xx10^6 Pa`. Calculate the volume of the gas at atmospheric pressure, `1.0xx10^5 Pa`

`6.3xx10^6xx180=1.0xx10^5xxV_2`

`V_2=(6.3xx10^6xx180)/(1.0xx10^5)`

Therefore `V_2=11340\ \cm^3`