Lesson Explainer: Charles’ Law | Nagwa Lesson Explainer: Charles’ Law | Nagwa

Lesson Explainer: Charles’ Law Physics

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

Charles’ law relates the volume 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.

Recall that volume is a measure of how much space something occupies. When we deal with gases, it can be hard to imagine the volume that lots of tiny particles take up, so it is often useful to think of the gas being placed inside a container.

Now, we are comfortable with the volume of a gas. Next, we can look at the effects of temperature on the gas.

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 is thus 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 and is sometimes referred to as absolute temperature. 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 relationship between temperature and volume of a gas was discovered from experimental findings.

Let us consider an experiment where a fixed amount of an ideal gas is heated in a container that has one wall that can move and can thus expand and contract. We can keep the pressure of the gas constant by placing a mass on top of the moving wall so that no matter how the volume of the gas changes, the force pushing down on it is the same.

Now, we can heat the gas. We will see that the gas expands as its temperature increases.

Let us work through an example question on the effects of temperature on the volume of a gas at constant pressure.

Example 1: The Effects of Temperature Changes on the Volume of a Gas

On a hot day, a party balloon is filled with helium. The following day is much colder, but the air pressure is the same. Assuming that no helium escapes from the balloon in that time, will the balloon have a larger volume than, a smaller volume than, or the same volume as that it had on the previous day?

  1. A smaller volume
  2. A larger volume
  3. The same volume

Answer

Recall that when a gas is heated at constant pressure, the volume of the gas also increases. This relationship can be seen in an experiment where a container with one wall free to move is heatedβ€”that is, kept at constant pressure by a mass placed on top of itβ€”as shown in the following diagram:

This question is very similar to that experiment: we have a container of gas (a balloon) that is able to expand and contract. The temperature of the gas decreases from the day it is filled to the next day, but the pressure of the gas remains constant.

As we have learned, at constant pressure, a lower temperature corresponds to a lower volume, so the volume of the balloon decreases.

The correct answer is A: the balloon will have a smaller volume than it had on the previous day.

The exact relationship between the volume and temperature of a gas was discovered in the 18th century, and is known as Charles’ law.

Charles’ law states that the volume of a fixed amount of an ideal gas at constant pressure is directly proportional to its absolute temperature.

Definition: Charles’ Law

The volume of a fixed amount of an ideal gas at constant pressure is directly proportional to its absolute temperature.

The term directly proportional means that if absolute temperature, 𝑇, increases by some factor, volume, 𝑉, will increase by the same factor. This can be written as π‘‰βˆπ‘‡ or, introducing a constant π‘˜, 𝑉=π‘˜π‘‡.

This is a linear relationship, and an example is shown in the following graph.

This has some interesting features. Notice that if the temperature of the gas decreases to absolute zero, 0 K, the volume of the gas will also decrease to zero. In reality, gases are not ideal; the gas will become a solid or liquid before it gets close to absolute zero. On the other hand, heating the gas by a huge amount will cause its volume to increase massively.

We can relate the temperature and volume of a gas at different points during heating by first dividing the equation by absolute temperature, 𝑇: 𝑉𝑇=π‘˜.

This shows us that, at any point during heating, the volume of the gas divided by its temperature in kelvins is constant.

For example, consider the temperature and volume of a gas before heating, π‘‡οŠ§ and π‘‰οŠ§, and after heating, π‘‡οŠ¨ and π‘‰οŠ¨: 𝑉𝑇=𝑉𝑇.

We can use this relationship to calculate the volume of a gas after a temperature change. Starting with 𝑉𝑇=𝑉𝑇, then multiplying both sides by π‘‡οŠ¨ gives us an expression for the volume after a temperature change, π‘‰οŠ¨: 𝑉=𝑉𝑇𝑇.

Let us practice using this relationship in an example question.

Example 2: Calculating the Volume of a Gas after Heating Using Charles’ Law

The temperature of 14 m3 of gas is initially 350 K. It is heated to 450 K while being kept at a constant pressure. What is the volume of the gas after it is heated?

Answer

Charles’ law gives us an equation relating the temperature, 𝑇, and volume, 𝑉, of a gas at constant pressure: 𝑉𝑇=π‘˜.

This is true for any point during heating or cooling, as long as the pressure of the gas remains constant. In this question, we consider two points in time, which we will write as point 1 and point 2, where the gas has temperatures and volumes of π‘‡οŠ§ and π‘‰οŠ§ and π‘‡οŠ¨ and π‘‰οŠ¨ respectively: 𝑉𝑇=𝑉𝑇.

Multiplying both sides of this equation by π‘‡οŠ¨, we get an expression for the volume after heating: 𝑉=𝑉𝑇𝑇.

Initially, the gas has volume 𝑉=14m and temperature 𝑇=350K. After heating, the gas has volume π‘‰οŠ¨ and temperature 𝑇=450K. Inserting these values into the equation gives 𝑉=14Γ—450350,mKK which gives a final volume of 𝑉=18.m

We can also use Charles’ law to calculate the temperature of a gas after a temperature change, where we know the volume and temperature before the change and the volume afterward.

Starting with 𝑉𝑇=𝑉𝑇, then dividing both sides of this equation by π‘‰οŠ¨, 1𝑇=𝑉𝑉𝑇, we can then obtain an expression for π‘‡οŠ¨ by taking the reciprocal: 𝑇=𝑇𝑉𝑉.

Let us practice using this relationship in an example question.

Example 3: Calculating the Temperature of a Gas before Cooling Using Charles’ Law

A hot gas that initially has a volume of 20 m3 is allowed to cool under constant pressure. When the gas reaches a temperature of 320 K, it has a volume of 16 m3. What is the initial temperature of the gas?

Answer

Charles’ law gives us an equation relating the temperature, 𝑇, and volume, 𝑉, of a gas at constant pressure: 𝑉𝑇=π‘˜.

This is true for any point during heating or cooling, as long as the pressure of the gas remains constant. In this question, we consider two points in time, which we will write as point 1 and point 2, where the gas has temperatures and volumes of π‘‡οŠ§ and π‘‰οŠ§ and π‘‡οŠ¨ and π‘‰οŠ¨ respectively: 𝑉𝑇=𝑉𝑇.

Dividing both sides of this equation by π‘‰οŠ§ gives us 1𝑇=𝑉𝑉𝑇.

We can then obtain an expression for π‘‡οŠ§ by taking the reciprocal 𝑇=𝑇𝑉𝑉.

Initially, the gas has volume 𝑉=20m and temperature π‘‡οŠ§. After cooling, the gas has volume 𝑉=16m and temperature 𝑇=320K. Inserting these values into the equation gives 𝑇=320Γ—2016,Kmm which gives an initial temperature of 𝑇=400.K

We can extend the relationship between two stages in heating or cooling to multiple stages of heating and cooling: 𝑇𝑉=𝑇𝑉=β‹―=𝑇𝑉.

The following graph shows three points during the heating and cooling of a gas at constant pressure.

Sometimes, we must consider scenarios where the volumes and temperatures are only expressed as ratios related to one another. To handle this, we can start with our equation relating the temperature and volume at two points in a temperature change, 𝑉𝑇=𝑉𝑇, then dividing both sides by π‘‰οŠ¨: 𝑉𝑇𝑉=1𝑇.

Now, multiplying both sides by π‘‡οŠ§, we obtain an expression relating the ratio of the volumes before and after to the ratio of the temperatures before and after: 𝑉𝑉=𝑇𝑇.

Let us work through an example question where we must work in terms of these ratios.

Example 4: Calculating the Ratio of Gas Volume before and after Heating Using Charles’ Law

A gas is cooled while being kept at a constant pressure. If its temperature in kelvins changes by a factor of 0.25, by what factor does the volume of the gas change?

Answer

Charles’ law gives us an equation relating the temperature, 𝑇, and volume, 𝑉, of a gas at constant pressure: 𝑉𝑇=π‘˜.

This is true for any point during heating or cooling, as long as the pressure of the gas remains constant. In this question, we consider two points in time, which we will write as point 1 and point 2, where the gas has temperatures and volumes of π‘‡οŠ§ and π‘‰οŠ§ and π‘‡οŠ¨ and π‘‰οŠ¨ respectively: 𝑉𝑇=𝑉𝑇.

Then, dividing both sides by π‘‰οŠ§ gives us 𝑉𝑇𝑉=1𝑇.

Finally, multiplying both sides by π‘‡οŠ¨, we obtain an expression relating the ratio of the volume before and after to the temperature before and after: 𝑉𝑉=𝑇𝑇.

The question tells us that the temperature changes by a factor of 0.25, meaning 𝑇𝑇=0.25. Inserting this value into the equation gives 𝑉𝑉=0.25.

From this, we can see that the volume of the gas has changed by a factor of 0.25.

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

Key Points

  • Charles’ law relates the volume and temperature of a constant amount of an ideal gas at constant pressure.
  • Charles’ law states that the volume of a constant amount of ideal gas at constant pressure is directly proportional to the temperature of the gas: π‘‰βˆπ‘‡. Another way to write this is to include a constant, π‘˜: 𝑉=π‘˜π‘‡.
  • This can be used to relate the volume and temperature at different stages in heating and cooling: 𝑉𝑇=𝑉𝑇=β‹―=𝑉𝑇.
  • Charles’ law can also be used to relate the ratio of the volume of the gas before, π‘‰οŠ§, and after, π‘‰οŠ¨, heating or cooling to the temperature before, π‘‡οŠ§, and after, π‘‡οŠ¨, heating or cooling: 𝑉𝑉=𝑇𝑇.

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