1. Gay-Lussac’s Law: Pressure-Temperature relationship
Formula:
P1 = P Kelvin (K) = oC
T T2
P1 and T1 are the beginning pressure and Kelvin temperature
P2 and T2 are the ending pressure and Kelvin temperature

2. Gay-Lussac’s Law: Pressure-Temperature relationship
P1 = P2
T T2
The gas in a container is at a pressure of 3.00 atm at 25oC. What would the gas pressure in the container be at 52oC?
So in order to answer this question you have to first: convert the 25oC and 52oC to Kelvin temp.
25oC = 298 K and 52oC = 325 K
Then you must set up the equation correctly.

3. Gay-Lussac’s Law: Pressure-Temperature relationship
P1 = P2
T T2
P1 = 3 atm P2 = ?
T1 = 298K T2 = 325 K
3atm = P2
298K K
P2 must be on one side of the equation by itself. To do this you have to cross multiply.
3atm = P atm x 325 K = P2
298K K K
P2 = atm

4. Gay-Lussac’s Law: Pressure-Temperature relationship
1. at 120oC, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure be at 205oC?
1.30 atm
2. at 122oC the pressure of a sample of nitrogen gas is atm. What will the pressure be at 205oC?
1.295 atm
3. a sample of helium gas has a pressure of 1.2 atm at 22oC. At what degree Celsius temperature will the helium reach a pressure of 2.0 atm?