# Gay-Lussac's Law In Use

We know in Gay-Lussac's law

`P_1/T_1=P_2/T_2`

**NOTE:**

Volume remains constant

Also you know that if you put a sealed container in a fire it will make the pressure inside the canister increase.

This is Gay-Lussac's law

`P_1` and `T_1` refer to the pressure and temperature before you throw the container in the fire **and** `P_2` and `T_2` refer to the pressure and temperature after you throw the container in the fire.

Increasing the temperature of the gas increases the pressure in the container.

Increasing the temperature of a gas speeds up the molecules and they have more energy (kinetic energy) to hit and push the side of the container. The molecules will exert more pressure.

**NOTE 1:**

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

The SI unit used to measure temperature is `K` (Kelvin)

**NOTE 2:**

You can use other units for pressure such as pounds per square inch or `mm` of mercury or atm (atmospheres) as long as you use the same units either side of the equation but you must use Kelvin for temperature.

**NOTE 3:**

Never forget to convert temperature to Kelvin

Temperature in Kelvin `=` Temperature in degrees Celsius `(\ ^(circ)C)+273`

You must always use temperature measured in Kelvin in any gas law equation.

**Examples**

1. A `30\ \L` sample of nitrogen inside a rigid, metal container at `20^circC` is placed inside an oven whose temperature is `50^circC`. The pressure inside the container at `20^circC` was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased to `50^circC` ?

`P_1/T_1=P_2/T_2`

`P_1=3atm`

`P_2=?`

`T_1=20^circC+273=293K`

`T_2=50^circC+273=323K`

Therefore `3/293=P_2/323`

`P_2=(3xx323)/293`

`P_2=3.31atm` (to three significant figures)

2. If a gas in a closed container is pressurized from `15` atmospheres to `16` atmospheres and its original temperature was `25^circC`, what would the final temperature of the gas be in degrees Celsius?

`P_1/T_1=P_2/T_2`

`P_1=15atm`

`P_2=16atm`

`T_1=25^circC+273=298K`

`T_2=?`

Therefore `15/298=16/T_2`

`T_2=(16xx298)/15`

`T_2=317.9K`

But don't forget to convert back from Kelvin to degrees Celsius

Therefore `T_2=317.9-273=44.9^circC`

3. The temperature of a sample of gas in a steel container at `30\ \kPa` is increased from `-100^circC` to `1.00xx10^3\ \^circC`. What is the final pressure inside the tank.

`P_1/T_1=P_2/T_2`

`P_1=30\ \kPa`

`P_2=?`

`T_1=-100^circC+273=173K`

`T_2=1.00xx10^3\ \^circC=1000^circC+273=1273K`

Therefore `P_2=(P_1xxT_2)/T_1`

`P_2=(30xx1273)/173`

`P_2=221\ \kPa` to three significant figures