Gay Lussac’s Law

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-lussacs-law

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.

gay-lussac-graph

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 P1. The temperature at this pressure is 200. K, which we’ll call T1. The pressure on the gas is then changed to 100. kPa, which is called P2. We want to find the temperature at this new pressure, which is T2.

P1 = 125 kPa

T1 = 200. K

P2 = 100. KPa

T2= ?

Gay-Lussac’s Law is:

gay-lussacs-law

To solve for T2, we first cross-multiply:

gay-lussacs-cross-multiply

Which becomes:

gay-lussac-p1t2

Finally, to isolate T2, divide each side by P1:

gay-lussac-t2

To get:

gay-lussac-t2equals

Plugging in our values for T1, P2 and P1:

gay-lussac-160K

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 P2 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 T2. The initial temperature T1 is 20 °C and the initial pressure P1 is 1 atm.

T1 = 20 °C

P1 = 1 atm

T2 = 600 °C

P2 = ?

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

gay-lussacs-293

T1 = 20 °C = 293.15 K

P1 = 1 atm

T2 = 600 °C = 873.15 K

P2 = ?

Gay-Lussac’s Law is:

gay-lussac-p1t2

To isolate the variable P2, we need to multiply each side of the equation by T2:

gay-lussacs-isolate

We get:

gay-lussacs-p2

Plugging in our values for P1, T2 and T1:

gay-lussacs-3atm

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