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## Chemistry: Gay-Lussac’s Law (Gas Laws) with 2 examples

## Gay Lussac’s Law Practice Problems

## Gases: Gay-Lussac’s Law

## What is Gay-Lussac’s Law?

Gay-Lussac’s law is a gas law which states that the pressure exerted by a gas (of a given mass and kept at a constant volume) varies directly with the absolute temperature of the gas. In other words, the pressure exerted by a gas is proportional to the temperature of the gas when the mass is fixed and the volume is constant.

This law was formulated by the French chemist Joseph Gay-Lussac in the year 1808. The mathematical expression of Gay-Lussac’s law can be written as follows:

**P ∝ T ; P/T = k**

Where:

- P is the pressure exerted by the gas
- T is the absolute temperature of the gas
- k is a constant.

The relationship between the pressure and absolute temperature of a given mass of gas (at constant volume) can be illustrated graphically as follows.

From the graph, it can be understood that the pressure of a gas (kept at constant volume) reduces constantly as it is cooled until the gas eventually undergoes condensation and becomes a liquid.

## Formula and Derivation

Gay-Lussac’s law implies that the ratio of the initial pressure and temperature is equal to the ratio of the final pressure and temperature for a gas of a fixed mass kept at a constant volume. This formula can be expressed as follows:

**(P _{1}/T_{1}) = (P_{2}/T_{2})**

Where:

- P
_{1}is the initial pressure - T
_{1}is the initial temperature - P
_{2}is the final pressure - T
_{2}is the final temperature

This expression can be derived from the pressure-temperature proportionality for gas. Since P ∝ T for gases of fixed mass kept at constant volume:

P_{1}/T_{1} = k (initial pressure/ initial temperature = constant)

P_{2}/T_{2} = k (final pressure/ final temperature = constant)

Therefore, **P _{1}/T_{1} = P_{2}/T_{2}** = k

Or, **P _{1}T_{2} = P_{2}T_{1}**

## Examples of Gay-Lussac’s Law

When a pressurized aerosol can (such as a deodorant can or a spray-paint can) is heated, the resulting increase in the pressure exerted by the gases on the container (owing to Gay-Lussac’s law) can result in an explosion. This is the reason why many pressurized containers have warning labels stating that the container must be kept away from fire and stored in a cool environment.

## Solved Exercises on Gay-Lussac’s Law

**Exercise 1**

**The pressure of a gas in a cylinder when it is heated to a temperature of 250K is 1.5 atm. What was the initial temperature of the gas if its initial pressure was 1 atm.**

Given,

Initial pressure, P_{1} = 1 atm

Final pressure, P_{2} = 1.5 atm

Final temperature, T_{2} = 250 K

As per Gay-Lussac’s Law, P_{1}T_{2} = P_{2}T_{1}

Therefore, T_{1} = (P_{1}T_{2})/P_{2} = (1*250)/(1.5) = 166.66 Kelvin.

**Exercise 2**

**At a temperature of 300 K, the pressure of the gas in a deodorant can is 3 atm. Calculate the pressure of the gas when it is heated to 900 K.**

Initial pressure, P_{1} = 3 atm

Initial temperature, T_{1} = 300K

Final temperature, T_{2} = 900 K

Therefore, final pressure (P_{2}) = (P_{1}T_{2})/T_{1} = (3 atm*900K)/300K = 9 atm.

## Frequently Asked Questions – FAQs

### What is Gay Lussac’s law formula?

The law of Gay-Lussac is a variant of the ideal gas law where the volume of gas is held constant. The pressure of a gas is directly proportional to its temperature while the volume is kept constant. P / T = constant or Pi / Ti = Pf / Tf are the standard calculations for Gay-Lussac ‘s law.

### What does Charles law state?

The physical theory known as Charles’ law states that a gas’s volume equals a fixed value as determined on the Kelvin scale compounded by its temperature.

### What is the importance of Gay Lussac’s law?

The meaning of this gas law is that it illustrates that rising a gas’s temperature induces a relative increase in its pressure (assuming that the volume does not change). Likewise, reducing the temperature allows the strain to decrease proportionally.

### How does Avogadro’s law apply to everyday life?

The Law of Avogadro states that the amount of the gas is directly proportional to the number of gas moles. You are driving more molecules of gas into it when you blow up a football.

### What are the applications of Avogadro’s law?

The relationship between a gas’s relative vapour density and its relative molecular mass is defined. Establishes the relationship between the volume of a gas at STP and gram molecular weight.

**Example**

The gas in an aerosol can is under a pressure of 3.00atm at a temperature of 25oC. It is dangerous to dispose of an aerosol can by incineration. What would the pressure in the aerosol can be at a temperature of 845oC?

**Solution**

## Gay-Lussac’s Law Example

If 10.0 L of oxygen exerts 97.0 kPa at 25 degrees Celsius, what temperature (in Celsius) is needed to change its pressure to standard pressure?

To solve this, you first need to know (or look up) standard pressure. It’s 101.325 kPa. Next, remember that gas laws apply to absolute temperature, which means Celsius (or Fahrenheit) must be converted to Kelvin. The formula to convert Celsius to Kelvin is:

K = degrees Celsius + 273.15

K = 25.0 + 273.15

K = 298.15

Featured Video How To Balance Equations In Chemistry

Now you can plug the values into the formula to solve for the temperature:

TTTAll that’s left is to convert the temperature back to Celsius:

C = K – 273.15

C = 311.44 – 273.15

C = 38.29 degrees Celsius

Using the correct number of significant figures, the temperature is 38.3 degrees Celsius.

**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? Featured Video What Is a Gas?

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:

P_{i}/T_{i} = P_{f}/T_{f}

where

P_{i} and T_{i} are the initial pressure and absolute temperatures

P_{f} and T_{f} are the final pressure and absolute temperature

First, convert the temperatures to absolute temperatures.

T_{i} = 27 C = 27 + 273 K = 300 K

T_{f} = 77 C = 77 + 273 K = 350 K

Use these values in Gay-Lussac’s equation and solve for P_{f}.

P_{f} = P_{i}T_{f}/T_{i}

P_{f} = (6 atm)(350K)/(300 K)

P_{f} = 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 containerless frequently. As temperature increases so do 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.

Gases have various properties that we can observe with our senses, including the gas pressure, temperature (T), mass, and the volume (V) that contains the gas. Careful, scientific observation has determined that these variables are related to one another and that the values of these properties determine the state of the gas.

The relationship between temperature and volume, at a constant number of moles and pressure, is called** Charles and Gay-Lussac’s Law** in honor of the two French scientists who first investigated this relationship. Charles did the original work, which was verified by Gay-Lussac. They observed that if the pressure is held constant, the volume **V** is equal to a constant times the temperature **T**:

V = constant * T

For example, suppose we have a theoretical gas confined in a jar with a piston at the top. The initial state of the gas has a volume qual to 4.0 cubic meters, and the temperature is 300 Kelvin. With the pressure and number of moles held constant, the burner has been turned off and the gas is allowed to cool to 225 Kelvin. (In an actual experiment, a cryogenic ice-bath would be required to obtain these temperatures.) As the gas cools, the volume decreases to 3.0 cubic meters. The volume divided by the temperature remains a constant (4/300 = 3/225 ). Here is a computer animation of this process:

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