# Gay-Lussac's Law Formula

Gay-Lussac's Law shows the relationship between the Temperature and Pressure of a gas. At a fixed volume, the temperature and pressure of a gas are directly proportional to each other. Since temperature and pressure have a direct relationship, if the pressure goes up then the temperature goes up and if the temperature goes down then the pressure goes down (and vice versa)

The equation for Gay-Lussac's Law is:

*T _{1}* = Initial Temperature (

*Kelvin - K*)

*P _{1}* = Initial Pressure (

*atm or mmHg*)

*T _{2}* = Final Temperature (

*Kelvin - K*)

*P _{2}* = Final Pressure (

*atm or mmHg*)

*Note:* Temperature must be in Kelvin for the equation to work. You calculate Kelvin temperature by adding 273 to the Celsius temperature.

Gay-Lussac's Law Formula Questions:

1.) A gas has a pressure of 699.0 mm Hg at 40.0 °C. What is the temperature at a pressure of 760.0 mm Hg?

Answer: For this problem the Initial Pressure is *P _{1}* = 699.0

*mmHg*. The Initial Temperature is

*T*= 40.0 + 273 = 313

_{1}*K*. The Final Pressure is

*P*= 760.0

_{2}*mmHg*and the Final Temperature (

*T*) is what you are asked to solve for.

_{2}Plug into the Gay-Lussac's Law Equation

*T _{2}* = 344.75 K

The temperature increased to 344.75 K.

2.) Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C.

Answer: The Initial Pressure is *P _{1}* = 1.00

*atm*. The Initial Temperature is

*T*= 20.00 + 273 = 293

_{1}*K*. The Final Temperature is

*T*= 30.0 + 273 = 303

_{2}*K*. The Final Pressure (

*P*) is what we are trying to find in the problem.

_{2}Plug into the Gay-Lussac's Law Equation

*P _{2}* = 1.03

*atm*

The pressure of the container increased to would be 1.03 *atm*.