# Video: Gay-Lussac’s Law

In this video, we will learn how to use the formula 𝑃/𝑇 = constant (Gay–Lussac’s Law) to calculate the pressure or temperature of a gas that is heated or cooled at a constant volume.

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### Video Transcript

In this video, we’re learning about Gay-Lussac’s law. This law helps us to understand what happens to a particular type of gas when its temperature changes while it’s held inside a fixed volume container.

To start getting a sense for this law, let’s imagine that we have a container of fixed volume and inside the container is some gas. We know that this gas because it is a gas is made up of particles that are in motion in every sort of direction. We also know that as these gas particles move around within this chamber. They move with some average or overall speed. And it’s this speed which indicates the temperature of this gas: the faster the average speed of the particles, the higher the temperature of the gas. Because these particles are moving around every which way, they collide with one another and especially they collide with the walls of this container.

For example, if we were to take a zoomed-in view of a section of the container wall and watch it for some amount of time, then we would see particles of this gas approach the wall, run into it, and bounce off. And as we kept watching, this will continue to happen with other particles in the gas. Each one of those collisions between a gas particle and the wall of the container exerts a force on that container wall. In each case, there’s some component of that force which points straight out perpendicular to the wall surface. If we add up all the forces pushing out on the wall over some amount of time and over some particular area of the wall surface, then what we’re getting a measure of is the pressure that this gas exerts.

We can recall that the units of pressure where pressure is represented using a capital 𝑃 are equal to newtons, unit of force, divided by square metres, the unit of area. And notice something about this pressure on the container wall. When we consider the different gas particles running into the wall, the faster they’re moving, the more force they exert pushing outward. And as we said, the faster the average speed of these particles, the higher the temperature of the gas. This tells us that there is a relationship between the pressure a gas exerts and the temperature of that gas. These two quantities are connected.

To see how, let’s imagine we do an experiment with this gas. Let’s say that we start to heat it. We start to put heat energy into this gas in the chamber. When that happens, the average speed of the particles picks up. Their velocity vectors are longer. And that means when we look again at an up-close view of our container wall, that the particles now running into the wall are coming in with a higher speed and therefore they exert a stronger force on the wall. In other words, the total outward push on the container wall is increasing. Looking over here at our units for pressure, we see that if the overall force is growing, but our area is staying the same, then the pressure itself must be increasing.

When it comes to pressure and temperature then of this gas in this fixed container, what we’re seeing is that when temperature goes up, so does pressure. A higher temperature means faster moving gas particles. And those faster moving particles push harder on the container when they run into it, increasing the pressure. And if we continue this experiment, say we increase the temperature of our gas even more by adding in more heat, then in that case our particles will move even faster and therefore hit against the wall of the container even harder. And so once again, as we increased temperature, we would see an increase in pressure.

We can describe this relationship between pressure and temperature this way. We can say that pressure is proportional to temperature. That means that as temperature changes, say we increase it by doubling it, then pressure will change the same way. It will also double. Another way we can write this expression, 𝑃 is directly proportional to 𝑇, is to say that pressure is equal to some constant value — we’ll call it capital 𝐶 — multiplied by 𝑇. This constant capital 𝐶 is sometimes called the constant of proportionality because it’s what makes 𝑃 proportional to 𝑇.

Now notice what happens if we do a bit of rearrangement with this equation. Let’s say we divide both sides of the equation by the temperature 𝑇. In that case, that term cancels on the right-hand side. Then, what we find is that 𝑃 divided by 𝑇 is equal to some constant. An important thing to realise is that for this relationship to be true, the volume of the gas we’re working with needs to be held constant. It has to be fixed. In other words, the size of the container the gas is in must stay the same. But if it does, then we can say that as temperature goes up, pressure will go up by the same amount and vice versa, as temperature goes down, pressure will also decrease by that amount.

When we write the relationship this way using an equal sign, then this relationship goes by the name of the person who discovered it. It’s called Gay-Lussac’s law, named after Joseph Louis Gay-Lussac. We talked a bit earlier about the units of one of the terms in this equation, pressure. In order to understand the law more fully, it’s very helpful to also consider the other term temperature. Here’s a question: in the SI system, what are the units of temperature? We might be most familiar with temperature on the Fahrenheit scale, abbreviated degrees F. Or perhaps, we’ve seen how to convert between degrees Fahrenheit and another temperature scale degrees Celsius. In fact, neither one of these scales represents the SI temperature scale.

The proper scale to use for SI calculations involving temperature is the Kelvin scale, abbreviated capital K. And interestingly, while temperatures in the Fahrenheit and Celsius scales are called degrees Fahrenheit and degrees Celsius, in the Kelvin scale, we don’t use the word degrees. We just say kelvin. Because these three temperature scales, Fahrenheit, Celsius, and Kelvin, are different, they report temperatures differently. For example, on the Fahrenheit scale, we know that water boils at 212 degrees. 212 degrees Fahrenheit is equal to 100 degrees Celsius and that’s equal to approximately 373 kelvin.

Now, let’s say we we’re doing a calculation involving Gay-Lussac’s law. And we wanted to calculate the ratio of a certain gas’s pressure to its temperature. If that gas was at 100 degrees Celsius or 373 kelvin, we can see that when we go to actually substitute in the numerical value of the temperature, it matters very much which temperature scale we use. Depending on the scale we pick, we get a very different answer. Whenever using Gay-Lussac’s law or any equation involving temperature in the SI system, we always want to use the Kelvin temperature scale rather than Celsius or Fahrenheit.

Before we look at an example to get some practice with this idea, let’s look at a second way of writing out Gay-Lussac’s law. And to do it, let’s go back to considering this fixed-volume chamber of gas that we’ve been heating up. Let’s say that back towards the beginning of the experiment when our heating flame was just this big and the temperature of our gas wasn’t so high, let’s say the pressure of the gas at that time was 𝑃 one and the temperature of the gas at that time was 𝑇 one. As we know though, we increased the heating of this gas so that the average temperature went up and the pressure on the wall of the container went up. We can say that at that latter moment, our gas now has a pressure 𝑃 two and a temperature 𝑇 two.

Gay-lussac’s law in telling us that pressure divided by temperature is a constant for a gas at fixed volume indicates that if we take the ratio 𝑃 one divided by 𝑇 one, then that’s equal to the ratio 𝑃 two divided by 𝑇 two. In other words, if we have a gas which goes through a change so that its pressure and therefore its temperature are different, we can write that the ratio of pressure to temperature before the change is equal to that same ratio after the change. This is what it means to say that at any one instant, the pressure of the gas divided by its temperature is a constant. Now, let’s look at an example of Gay-Lussac’s law.

A gas is heated from 19 degrees Celsius to 80 degrees Celsius, while the volume is kept constant. If the initial pressure was 2000 pascals, what is the pressure of the gas after it has been heated? Give your answer to three significant figures.

Alright, so in this example, we have a gas which starts out at 19 degrees Celsius, but then it’s heated and its temperature goes up to 80 degrees Celsius. We’re told when the temperature of the gas is 19 degrees Celsius, it has a pressure of 2000 pascals. Then as the temperature of the gas changes, we want to see what impact — if any — this has on the pressure of the gas. As we consider what kind of method would help us figure out this final gas pressure, we find a clue in the problem statement: this gas was heated and the volume was kept constant.

When the volume of a gas is held constant over some process, that means that Gay-Lussac’s law applies to that process. This law tells us that for a gas held at constant volume, the pressure of that gas divided by its temperature is equal to a constant value. In other words, even as the pressure or the temperature of the gas will change, so long as the volume of the gas is constant, that ratio 𝑃 divided by 𝑇 is always the same.

Another way of writing this is to consider some change that happens in a gas and to say that the pressure divided by the temperature before the change 𝑃 one divided by 𝑇 one is equal to the pressure divided by temperature after the change. This second way of writing the law is especially helpful for us because we indeed have a gas that has gone through a change. It’s been heated. So we can say the initial pressure of our gas divided by its initial temperature is equal to its final pressure — that’s what we want to solve for — divided by its final temperature.

Since it’s 𝑃 two, the final gas pressure that we want to solve for, let’s rearrange this equation to let us do that. If we multiply both sides of the equation by 𝑇 two, the temperature of the gas after it’s heated, then that term cancels on the right-hand side and we have 𝑃 two all by itself. We can write the resulting equation this way. We can say that 𝑃 two, the final pressure of the gas, equals 𝑃 one times the temperature ratio 𝑇 two divided by 𝑇 one.

Now at this point in our solution, it’s important to notice something. See the temperatures we’ve been given in the problem statement, 19 degrees Celsius; 80 degrees Celsius. That is these temperatures are given as values on the Celsius scale. But that’s not the SI unit of temperature. The SI temperature unit is kelvin. In other words, before we plug in 𝑇 two and 𝑇 one for our calculation, we want to convert these given temperatures in to temperatures on the Kelvin scale. If we forgot to do that, our answer would be off.

So we want to know how to convert a temperature from degrees Celsius to kelvin. And roughly speaking, a temperature in kelvin is equal to that same temperature in degrees Celsius plus 273. So for example, if we had a temperature in Celsius of zero degrees, that is freezing, then the equivalent temperature in kelvin would be zero plus 273 or 273 kelvin. Using this relationship, let’s write out the initial and final temperatures of our gas in kelvin. 𝑇 one, the initial temperature of the gas, will be equal to 19 plus 273 and all this will be in kelvin. All that together is 292 kelvin. This value in kelvin is the value we’ll use to substitute in for 𝑇 one in our equation.

Next, what about 𝑇 two, the temperature of our gas after it’s been heated? That’s equal to 80 plus 273 all in kelvin. And this equals 353 kelvin. That’s the temperature we’ll use for 𝑇 two in our equation. With these values plugged in, notice that the units kelvin cancel out. But be careful. This doesn’t mean that if we had used degrees Celsius instead of kelvin, the units would also have cancelled and therefore wouldn’t have made a difference. It’s true that if we had forgotten to convert our initial temperatures in degrees Celsius into kelvin and simply plug them into the equation, then those units would have also cancelled.

But notice what our fraction would have been. Instead of 353 divided by 292, we would have had a fraction of 80 divided by 19. These two fractions are not the same thing. All that to say whenever we’re working within the SI system as we usually are, then whenever an equation involves temperature, we’ll want to make sure to use the Kelvin temperature scale in our calculation. Okay, we’re almost ready to solve for 𝑃 two, the final pressure of this gas.

In order to do so, let’s plug in for 𝑃 one, the initial pressure. And that’s given as 2000 pascals. This unit, pascals, is the base unit for pressure in the SI system. As the unit of pressure, a pascal is equal to a newton of force divided by a square metre of area. So when we see 2000 pascals, that’s another way of saying 2000 newtons of force per square metre of area. And that value can now go in for 𝑃 one in our equation. We’re now ready to calculate 𝑃 two, the pressure of the gas after it’s been heated. We see the units of this result will be pascals. And when we calculate this expression and round it to three significant figures, we find a result of 2420 pascals. This is the pressure of the gas after it’s been heated.

Let’s summarise now what we’ve learned in this lesson on Gay-Lussac’s law. Earlier on, we saw that for a gas at constant volume, the gas pressure is proportional to the gas’s temperature. We can write that out this way: 𝑃 is proportional to 𝑇. And this means that for example if the temperature doubled, then the pressure also would double.

Moreover, we saw that this relationship implies another one that gas pressure is equal to a constant multiplied by the gas temperature. And if we divide both sides of this equation by the temperature 𝑇, we arrive at this formulation known as Gay-Lussac’s law that the pressure of a gas divided by its temperature when that gas is held at fixed volume is a constant. This implies that if we have a system again held at constant volume with an initial and a final state, then we can say that the pressure divided by the temperature of that initial system state is equal to the pressure divided by the temperature at the final system state.

Along with that, we saw that because we were working within the SI system, whenever we use temperatures in this law, those temperatures must be in kelvin. If we’re given them in degrees Fahrenheit or degrees Celsius, they’ll need to be converted to the Kelvin scale. And lastly, as we saw that the SI unit of temperature is kelvin, we also learned that the unit of pressure is the pascal, abbreviated Pa and that one pascal represents a newton of force spread over one square metre of area.