This law is incorrectly named after Joseph Gay-Lussac and was actually
discovered by French inventor and physicist Guillaume Amontons (1663-1705).

Absolute temperature is ALWAYS required with Gay-Lussac’s Law.
(either Kelvin or the less common Rankine scale can be used).
For a Temperature Converter, click here.

P1               P2
——     =     ——
T1               T2

The above formula is Gay-Lussac’s Law named after the French chemist and physicist Joseph Louis Gay-Lussac (1778 – 1850). The law states that the pressure of a fixed mass of gas at a constant volume is directly proportional to its absolute temperature.

In other words, when temperature increases, pressure increases.
When pressure decreases, temperature decreases.

Do you want to solve for:

Similar to the calculations with Boyle’s Law or Charles’ Law, every word problem involving Gay-Lussac’s Law will always give us 3 of the 4 variables. Of those 3 variables, we have to determine which two “pair up” (or which two were measured at the same time). To pair these correctly, these get designated as “P1” and “T1” or “P2” and “T2” but never as “P1” and “T2” or “P2” and “T1”.

1) The temperature of a gas is 30 degrees Celsius and its pressure is 760 torr. If the temperature originally was 40°C, what was the original pressure?

The two variables that were measured at the same time (and can get “paired up”) are 30°C (T₂) and 760 torr (P₂).
Yes, we could have classified these as T₁ and P₁ but either way, they are paired properly.
We are told that the original temperature was 40°C (T₁) and we must solve for the original pressure (P₁).
As it says at the top of this page, we always must use absolute temperature with this formula so we’ll have to convert both temperatures to Kelvin.
T₁ = 40°C + 273.15 = 313.15 Kelvin     T₂ = 30°C + 273.15 = 303.15 Kelvin

Solving Gay-Lussac’s Law for P₁ we get:
P₁ = P₂ • T₁ ÷ T₂     P₁ = 760 torr • 313.15 K ÷ 303.15 K     P₁ = 785.07 torr

Using the calculator, we click on the P1 button.
We then enter the 3 numbers 760   313.15 and 303.15 into the correct boxes then click “CALCULATE” and get our answer of 785.07 torr.

2) A gas has a temperature of 300 Kelvin and a pressure of 1.5 atmospheres. If the previous pressure was 1.2 atmospheres, what was the original temperature?

Pairing up and classifying the data we have T₂ = 300K     P₂ = 1.5 atm     P₁ = 1.2 atm and we must solve for T₁.
Solving Gay-Lussac’s Law for T₁ we get:
T₁ = P₁ • T₂ ÷ P₂     T₁ = 1.2 atm • 300K ÷ 1.5 atm     T₁ = 240K

Using the calculator, we click on the T1 button.
We then enter the 3 numbers in the correct boxes, then click “CALCULATE” and get our answer of 240K.

3) A gas with 2 atmospheres of pressure has a temperature of 250 Kelvin.
Calculate the pressure if the temperature increases to 300K.

The two items of data which “pair up” are 2 atmospheres (P₁) and 250 K (T₁).
The remaining data is the new temperature of 300K (T₂) and we must now solve for the fourth data item (P₂).
Solving Gay-Lussac’s Law for (P₂) we get:
P₂ = P₁ • T₂ ÷ T₁     P₂ = 2 atm • 300 K ÷ 250 K     P₂ = 2.4 atmospheres

Using the calculator, we click on the P2 button.
We then enter the 3 numbers into the correct boxes then click “CALCULATE” and get our answer of 2.4 atmospheres.

4) A gas at a temperature of 60° Fahrenheit has a pressure of .9 atmospheres. What is the new temperature if the pressure increases to 1.1 atm?

The two variables that were measured at the same time and which can get “paired up” are 60°F (T₁) and .9 atm (P₁).
Before we continue, we must convert that temperature to an absolute scale.
To change Fahrenheit to Rankine, we add 459.67 and so we get:
T₁ = 60°F + 459.67 = 519.67 Rankine.
P₂ is 1.1 atm and so we are ready to calculate T₂.
Solving Gay-Lussac’s Law for T₂ we get:
T₂ = P₂ • T₁ ÷ P₁     T₂ = 1.1 atm • 519.67R ÷ .9 atm     T₂ = 635.15R

Using the calculator, we click on the T2 button.
We then enter the 3 numbers into the correct boxes then click “CALCULATE” and get our answer of 635.15 Rankine.

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