# Gay-Lussac’s Law

Gay-Lussac’s Law says that gas pressure is directly proportional to gas temperature when volume is held constant. The equation is:

Gay-Lussac’s law says that, as you increase the pressure of a gas, its temperature also increases, if volume is held constant. As you increase the temperature of a gas, its pressure also increases.

The following is a graph of gas pressure as a function of temperature. As you can see, pressure and temperature are directly related. As pressure is increased, so is temperature.

**Sample problem:**

A cylinder contains a gas with a pressure of 125 kPa at a temperature of 200. K. Find the temperature of the gas which has a pressure of 100. kPa.

The original pressure of the gas is 125 kPa so we’ll call that P_{1}. The temperature at this pressure is 200. K, which we’ll call T_{1}. The pressure on the gas is then changed to 100. kPa, which is called P_{2}. We want to find the temperature at this new pressure, which is T_{2}.

P_{1} = 125 kPa

T_{1} = 200. K

P_{2} = 100. KPa

T_{2}= ?

Gay-Lussac’s Law is:

To solve for T_{2}, we first cross-multiply:

Which becomes:

Finally, to isolate T_{2}, divide each side by P_{1}:

To get:

Plugging in our values for T_{1}, P_{2} and P_{1}:

*The temperature of the gas at this new pressure is 160. K.*

**Sample Problem**

A container, designed to hold a pressure of 2.5 atm, is filled with 20.0 mL of air at room temperature (20 °C) and standard pressure (1 atm). Will it be safe to throw this container into a fire where temperatures of 600°C will be reached?

To determine whether it will be safe to throw this container into a fire of 600 °C, we need to determine whether the pressure will exceed 2.5 atm at 600 °C.

Therefore, we need to calculate the final pressure P_{2} and see if it is greater or less lthan 2.5 atm. This final pressure is the pressure when the temperature reaches 600 °C, which is T_{2}. The initial temperature T_{1} is 20 °C and the initial pressure P_{1} is 1 atm.

T_{1} = 20 °C

P_{1} = 1 atm

T_{2} = 600 °C

P_{2} = ?

Remember that, for all gas law problems, we need to convert temperatures to K.

T_{1} = 20 °C = 293.15 K

P_{1} = 1 atm

T_{2} = 600 °C = 873.15 K

P_{2} = ?

Gay-Lussac’s Law is:

To isolate the variable P_{2}, we need to multiply each side of the equation by T_{2}:

We get:

Plugging in our values for P_{1}, T_{2} and T_{1}:

*Since P2 is 3 atm, which exceeds 2.5 atm, the container will explode under this pressure.*