Gay-Lussac's Gas Law Examples

Ideal Gas Law Example Problems

Gay-Lussac's gas law is a special case of the ideal gas law where the gas is held at constant volume.
Gay-Lussac's gas law is a special case of the ideal gas law where the gas is held at constant volume. Patrick Foto / Getty Images

Gay-Lussac's gas law is a special case of the ideal gas law where the volume of the gas is held constant. When the volume is held constant, the pressure exerted by a gas is directly proportional to the absolute temperature of the gas. These example problems use Gay-Lussac's law to find the pressure of gas in a heated container as well as the temperature you would need to change the pressure of gas in a container.

Gay-Lussac's Law Example

A 20-liter cylinder contains 6 atmospheres (atm) of gas at 27 C. What would the pressure of the gas be if the gas was heated to 77 C?

To solve the problem, just work through the following steps:

The cylinder's volume remains unchanged while the gas is heated so Gay-Lussac's gas law applies. Gay-Lussac's gas law can be expressed as:

Pi/Ti = Pf/Tf

where
Pi and Ti are the initial pressure and absolute temperatures
Pf and Tf are the final pressure and absolute temperature

First, convert the temperatures to absolute temperatures.

Ti = 27 C = 27 + 273 K = 300 K
Tf = 77 C = 77 + 273 K = 350 K

Use these values in Gay-Lussac's equation and solve for Pf.

Pf = PiTf/Ti
Pf = (6 atm)(350K)/(300 K)
Pf = 7 atm

The answer you derive would be:

The pressure will increase to 7 atm after heating the gas from 27 C to 77 C.

Another Example

See if you understand the concept by solving another problem: Find the temperature in Celsius needed to change the pressure of 10.0 liters of a gas that has a pressure of 97.0 kPa at 25 C to standard pressure.

Standard pressure is 101.325 kPa.

First, convert 25 C to Kelvin (298K).  Remember that the Kelvin temperature scale is an absolute temperature scale based on the definition that the volume of a gas at constant (low) pressure is directly proportional to the temperature and that 100 degrees separate the freezing and boiling points of water.

Insert the numbers into the equation to get:

97.0 kPa / 298 K = 101.325 kPa / x

solving for x:

x = (101.325 kPa)(298 K)/(97.0 kPa)

x = 311.3 K

Subtract 273 to get the answer in Celsius.

x = 38.3 C

Tips and Warnings

Keep these points in mind when solving a Gay-Lussac's law problem:

  • The volume and quantity of gas are held constant.
  • If the temperature of the gas increases, pressure increases.
  • If temperature decreases, pressure decreases.

Temperature is a measure of the kinetic energy of gas molecules. At a low temperature, the molecules are moving more slowly and will hit the wall of a container less frequently. As temperature increases, so does the motion of the molecules. They strike the walls of the container more often, which is seen as an increase in pressure. 

The direct relationship only applies if the temperature is given in Kelvin. The most common mistakes students make working this type of problem is forgetting to convert to Kelvin or else doing the conversion incorrectly. The other error is neglecting significant figures in the answer. Use the smallest number of significant figures given in the problem.

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Helmenstine, Todd. "Gay-Lussac's Gas Law Examples." ThoughtCo, Jan. 4, 2018, thoughtco.com/guy-lussacs-gas-law-example-607555. Helmenstine, Todd. (2018, January 4). Gay-Lussac's Gas Law Examples. Retrieved from https://www.thoughtco.com/guy-lussacs-gas-law-example-607555 Helmenstine, Todd. "Gay-Lussac's Gas Law Examples." ThoughtCo. https://www.thoughtco.com/guy-lussacs-gas-law-example-607555 (accessed January 23, 2018).