Lesson Explainer: Gay-Lussac’s Law Physics • 9th Grade

In this explainer, we will learn how to use the formula 𝑃𝑇=constant (Gay-Lussac’s law) to calculate the pressure or temperature of a gas that is heated or cooled at a constant volume.

Gay-Lussac’s law relates the pressure and temperature of an ideal gas when all other factors remain constant.

First, let us understand what an β€œideal gas” is. A gas is made up of very small particles that move around, occasionally colliding with each other. In an ideal gas, we assume that these particles are so small that they take up no individual volume and that there are no interactions between these particles.

Definition: Ideal Gas

An ideal gas is made up of particles that occupy negligible space and do not interact with each other.

Now, let us understand the concept of the pressure of an ideal gas.

If we consider a small section of the wall of the container, we will see particles moving in random directions, and some of them collide with the wall.

Each collision exerts a small force on the wall. Over the whole surface of the container at any instant, there is a constant force pushing on the walls.

When we divide this force by the total area it is being acted on, we get the pressure.

Pressure has units of newtons per square metre; notice that these units are force divided by area.

Now that we understand that the pressure of a gas is caused by collisions of the gas particles on the walls of our container, we can look at the effects of temperature on these particles.

The particles in our container have kinetic energy; they move in random directions with some speed.

If we heat the gas, the energy of the gas is increased. The thermal energy added to the gas is equal to the sum of the kinetic energy added to each particle of gas. This increase in kinetic energy means that the particles are moving faster.

Similarly, if we cool the gas, energy in the gas is lost. As with heating, the thermal energy lost from the gas is equal to the sum of the kinetic energy lost from the particles of the gas.

We can visualize this in a diagram, where the red particle has been given energy in the form of heat, which has become kinetic energy. This causes the red particle to move with greater speed. The blue particle, on the other hand, has less kinetic energy and so is moving with a lower speed.

We may be familiar with temperature expressed in units of degrees Fahrenheit or degrees Celsius, but the SI unit of temperature is actually kelvins. Recall that the conversion between kelvins and degrees Celsius is simply kelvinCelsius=+273.15.

The conversion between degrees Fahrenheit and degrees Celsius is CelsiusFahrenheit=59Γ—(βˆ’32).

This is shown in the following diagram.

If you cool a gas to 0 K, the particles will stop moving entirely. It should be noted that this is not physically possible, although scientists can get quite close. The lowest temperatures ever recorded are around 500Γ—10 K!

The effect of temperature on the pressure of a gas can be understood by thinking about the collisions taking place in our container at any point in time.

Let us keep our container at a fixed volume and increase the temperature. This causes the particles to increase in speed. The following diagram shows an example of this:

If the particles are moving faster, they will collide with the walls of our container more often. The force exerted on the wall per collision will also increase, because more momentum will be transferred each time a particle collides with the wall. Because collisions occur more frequently, and each collision exerts a higher amount of force on the walls, the pressure of the gas increases.

So, an increase in temperature causes an increase in pressure.

Let us look at an example question about this relationship.

Example 1: Understanding Gay-Lussac’s Law

For a gas at a constant volume, if the temperature is , then the pressure .

  1. increased, stays the same
  2. decreased, stays the same
  3. increased, decreases
  4. increased, increases
  5. decreased, increases

Answer

To answer this question, we should imagine what happens to the particles colliding with the walls of a container filled with the gas.

Recall that there is a relationship between the temperature and pressure of an ideal gas at constant volume. If temperature increases or decreases, pressure will also change. This immediately rules out options A and B.

If the temperature is decreased, as in options B and E, the particles of gas will be traveling slower, so less collisions will take place.

As we have learned, this means there is a lower amount of force being exerted on the walls per unit area, meaning that pressure has decreased.

This rules out option E and again rules out option B.

Now let us consider what happens when the temperature of the gas is increased, as in options A, C, and D. Particles have more energy and are therefore moving faster. At any moment, there will be more collisions between the particles and the walls of the container. Each collision will also exert a greater force on the wall, as there will be a higher change in momentum of the particle in each collision.

This means that the pressure of the gas increases. This rules out options A and C and corresponds to option D.

The correct answer is, therefore, option D: for a gas at a constant volume, if the volume is increased, then the pressure increases.

This relationship between pressure and temperature was discovered in the 19th century, and the exact relationship is known as β€œGay-Lussac’s law.”

Gay-Lussac’s law states that the pressure exerted by an ideal gas is directly proportional to its absolute temperature, if the volume and amount of gas remain constant.

Definition: Gay-Lussac’s Law

The pressure of an ideal gas of fixed mass and fixed volume is directly proportional to the gas’s absolute temperature.

The term directly proportional means that if pressure, 𝑃, increases by some factor, temperature, 𝑇, increases by the same factor. This can be written as π‘ƒβˆπ‘‡.

Another way to write this relationship is to include some constant, π‘˜: 𝑃=π‘˜π‘‡.

Dividing both sides by 𝑇, we get 𝑃𝑇=π‘˜.

So, the pressure divided by the absolute temperature of the gas is constant, provided we also keep the volume and the amount of gas constant.

This has some interesting implications. If the absolute temperature of the gas is decreased all the way to zero, the particles stop moving entirely. This means that there are no collisions taking place with the walls of the container, so the pressure of the gas must also be zero. However, as mentioned previously, it is physically impossible to cool a gas to absolute zero.

Or, going in the other direction, if we increase the temperature of the container hugely, there will be many collisions taking place, each exerting a greater force on the wall, so the pressure of the gas will become very large.

A possible graph of this relationship is shown here:

The constant in our equation, π‘˜, depends on many other factors, such as the gas we are considering and the volume.

Let us take a look at a series of temperature changes at constant volume. The following diagram shows our container of gas at three different temperatures:

As we have learned, the pressure divided by the temperature at each of these points is constant. That means that 𝑃𝑇=𝑃𝑇=𝑃𝑇.

Plotting these points on a graph of pressure and volume, we can see that all of these points lie on the same line: 𝑃=π‘˜π‘‰.

Using this relationship, we can calculate the pressure of a gas after a temperature change at constant volume.

If we know the pressure, π‘ƒοŠ§, and temperature, π‘‡οŠ§, of a gas before a temperature change and the temperature afterward, π‘‡οŠ¨, then we can calculate the pressure afterward, π‘ƒοŠ¨.

Starting with 𝑃𝑇=𝑃𝑇, we can multiply both sides by π‘‡οŠ¨ to give an expression for the pressure after the temperature change: 𝑃=𝑃𝑇𝑇.

Let us work through an example question looking at the pressure change when a gas experiences a temperature change at constant volume.

Example 2: Using Gay-Lussac’s Law to Find the Pressure of a Gas

A gas with a volume of 2 m3 is initially at a temperature of 300 K and a pressure of 500 Pa. It is heated to 375 K while being kept at a constant volume. What is the pressure of the gas after it has been heated?

Answer

Gay-Lussac’s law states the pressure of an ideal gas of fixed mass and fixed volume is directly proportional to the gas’s absolute temperature.

This can be written as 𝑃𝑇=π‘˜.

In this question, we are asked to consider two points in time: before and after heating.

At point 1, we are given 𝑃=500Pa and 𝑇=300K. At point 2, we are given 𝑇=375K. And we are asked to calculate π‘ƒοŠ¨.

Because 𝑃𝑇 is constant, we can write 𝑃𝑇=𝑃𝑇.

If we multiply both sides by π‘‡οŠ¨, we are left with an expression for π‘ƒοŠ¨: 𝑃=𝑃𝑇𝑇.

We can now substitute the values given to us into the equation: 𝑃=500Γ—375300𝑃=625.PaKKK

Similar to calculating the pressure of a gas after a temperature change at constant volume, Gay-Lussac’s law can also be used to calculate the temperature of a gas after a pressure change at constant volume.

If we know the pressure, π‘ƒοŠ§, and temperature, π‘‡οŠ§, of a gas before a pressure change and the pressure afterward, π‘ƒοŠ¨, then we can calculate the temperature afterward, π‘‡οŠ¨.

Starting with 𝑃𝑇=𝑃𝑇, we can divide both sides by π‘ƒοŠ¨, 𝑃𝑃𝑇=1𝑇, then take the reciprocal to get an expression for π‘‡οŠ¨: 𝑇=𝑇𝑃𝑃.

Let us work through an example question looking at the temperature change when a gas experiences a pressure change at constant volume.

Example 3: Using Gay-Lussac’s Law to Find the Temperature of a Gas

A gas with a volume of 0.5 m3 is initially at a temperature of 240 K and a pressure of 800 Pa. The gas is heated while being kept at a constant volume until its pressure is 1β€Žβ€‰β€Ž000 Pa. What is the temperature of the gas after it has been heated?

Answer

Gay-Lussac’s law states the pressure of an ideal gas of fixed mass and fixed volume is directly proportional to the gas’s absolute temperature.

This can be written as 𝑃𝑇=π‘˜.

In this question, we are asked to consider two points in time: before and after a pressure change.

At point 1, we are given 𝑃=800Pa and 𝑇=240K. At point 2, we are given 𝑃=1000Pa. And we are asked to calculate π‘‡οŠ¨.

Because 𝑃𝑇 is constant, we can write 𝑃𝑇=𝑃𝑇.

If we divide both sides by π‘ƒοŠ¨, then take the reciprocal, we are left with an expression for π‘‡οŠ¨: 𝑇=𝑇𝑃𝑃.

We can now substitute the values given to us into the equation: 𝑇=240Γ—1000800𝑇=300.KPaPaK

Finally, we will practice performing calculations based on pressure–temperature graphs.

Example 4: Using Gay-Lussac’s Law to Find the Pressure of a Gas

The graph shows how the pressure of a gas in a container of a fixed volume changes with the temperature of the gas.

  1. What is the pressure of the gas at 200 K?
  2. What is the pressure of the gas at 400 K?
  3. If the gas was heated to 800 K, what would the pressure of the gas be?

Answer

Part 1

To find the pressure of the gas at 200 K, we can read directly from the graph.

The pressure of the gas at 200 K is 90 kPa, or 90β€Žβ€‰β€Ž000 Pa.

Part 2

To find the pressure of the gas at 400 K, we can again read this value directly from the graph.

The pressure of the gas at 400 K is 180 kPa, or 180β€Žβ€‰β€Ž000 Pa.

Part 3

Recall that Gay-Lussac’s law states that the pressure, 𝑃, of an ideal gas at fixed volume and mass is directly proportional to the temperature (in kelvins), 𝑇, of the gas.

With a constant, π‘˜, this can be expressed as: 𝑃=π‘˜π‘‡.

Dividing both sides of this equation by 𝑇 shows us that the pressure divided by the temperature of the gas is constant: 𝑃𝑇=π‘˜.

The value of the constant, π‘˜, can be found by calculating the gradient of the graph.

Starting with 𝑃=90kPa and 𝑇=200K, we can calculate that π‘˜=90200π‘˜=0.45/.kPaKkPaK

We can use this to calculate the pressure at the final temperature point, 𝑇=800K: 𝑃=π‘˜π‘‡π‘ƒ=0.45/Γ—800.kPaKK

Giving us our final pressure, in kilopascals, 𝑃=360.kPa

We can summarize what we have learned in this explainer in the following key points.

Key Points

  • Gay-Lussac’s law relates the pressure and temperature of a constant amount of an ideal gas at constant volume.
  • Gay-Lussac’s law states that the pressure of a constant amount of ideal gas at constant volume is directly proportional to the temperature of the gas: π‘ƒβˆπ‘‡. Another way to write this is to include a constant, π‘˜: 𝑃=π‘˜π‘‡.
  • This can be used to relate the pressure and temperature of an ideal gas at different stages in heating and cooling: 𝑃𝑇=𝑃𝑇=β‹―=𝑃𝑇.

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