Video: Using Boyle’s Law and Gay-Lussac’s Law to Find the Pressure of a Gas

A gas has an initial pressure of 500 Pa, a volume of 0.1 m³, and a temperature of 300 K. It is allowed to expand without changing temperature until its volume is 1 m³. Then, the temperature of the gas is increased, while the volume is kept constant, to 360 K. What is the final pressure of the gas?

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Video Transcript

A gas has an initial pressure of 500 pascals, a volume of 0.1 meters cubed, and a temperature of 300 kelvin. It is allowed to expand without changing temperature until its volume is one meter cubed. Then the temperature of the gas is increased, while the volume is kept constant, to 360 kelvin. What is the final pressure of the gas?

Okay, so in this question, we’ve got a gas which is going through two stages of change. Initially, it starts out with the pressure of 500 pascals, a volume of 0.1 meters cubed, and a temperature of 300 kelvin.

The first stage of change it goes through is an expansion. And this occurs without changing temperature. So the temperature stays constant. Now this expansion occurs until the volume is one meter cubed. Then it undergoes the second stage of expansion, which is when the temperature of the gas is increased while the volume is kept constant this time. And the temperature is increased to 360 kelvin. What we need to do is to find the final pressure of the gas.

Let’s start out then by drawing diagrams to represent the gas before any changes occurred, as well as after stage one of change and stage two of change as well. So here’s our gas initially before any changes occurred.

We’ve been told that the pressure of the gas initially is 500 pascals. So let’s call this 𝑃 sub naught for the initial pressure. And that’s 500 pascals. As well as this, we’re told that the volume of the gas initially — we’ll call this 𝑉 sub naught for the initial volume — is 0.1 meters cubed.

Thirdly, we’re told about the initial temperature, which we’ll call 𝑇 sub naught. And it happens to be 300 kelvin. Now it’s at this point that the gas undergoes stage one of change. So what happens to the gas now is that the volume increases to one meter cubed. So our diagram is going to look slightly larger.

Now if the volume is going to increase and the temperature, remember, is kept constant, then the pressure must change as well. Let’s call the pressure here 𝑃 sub one, because this is after stage one of the change. And we don’t yet know what this pressure is. So the easiest thing to do is to assume that it changes, which we’ll see later that it actually does. As well as this, we’re told that the volume increases to one meter cubed. And we’re calling this 𝑉 sub one.

Now interestingly, the temperature is kept constant. So 𝑇 sub one, the temperature after stage one, is the same as 𝑇 sub naught, because the temperature was kept constant throughout this process. So this is equal to 300 kelvin. And that’s all the information we have after stage one of the procedure.

So let’s move on to stage two then. This time, the temperature is increased, while the volume is kept constant. So the size of the box representing our gas is gonna stay the same because the volume is staying constant. When it comes to the pressure, however, once again, we don’t know what it is. So we’ll say 𝑃 two is equal to question mark, where 𝑃 two represents the pressure after stage two of the change.

Secondly, we’re told that this stage involves a constant volume. So the volume is the same from here to here. Therefore, 𝑉 two is equal to 𝑉 one, the volume after stage one. And this volume was equal to one meter cubed, which is what we’ve been told in the question.

Now as for the temperature, we’re told that the temperature increases to 360 kelvin. The question actually asks us to find out the final pressure of the gas. In other words, what’s the pressure after the two stages of change have occurred? And that’s this pressure here.

We’re trying to find the value of 𝑃 two. Now there are two main ways of doing this. There may be more. But we’ll discuss two of them in this video. One of them involves a bit more memorization, while the other one involves a bit more analysis. And it’s up to you whichever method you prefer because, in the end, they’re both doing the same thing, just in different ways.

Now at this point, we’ve got all the information we can from the question. So let’s erase it and go through both the methods. Let’s start with the method that requires a bit more memorization in terms of what we know rather than what we can work out. In this method, we’ll be able to work out what the value of 𝑃 one is and then what the value of 𝑃 two is.

We’ll do this in two steps, firstly by studying the change of the gas from here to here and then from here to here. So let’s consider stage one first of all. Stage one is where the volume of the gas is allowed to increase at a constant temperature. We keep the temperature the same. It’s 300 kelvin throughout.

Now because the temperature is kept constant and we’re looking at the change in volume and a potential change in pressure as well, we can recall what’s known as Boyle’s law. Boyle’s law tells us that the product of pressure and volume is constant if the temperature is constant. And we’ve already satisfied the “if” statement.

In this situation, in stage one, the temperature is constant. So Boyle’s law tells us that, before any changes occurred, the product of pressure and volume is the same as after the changes occurred, which is 𝑃 one multiplied by 𝑉 one, because the product of pressure and volume has to be constant. So it has to be the same before the change and after the change.

Now this equation will allow us to work out the value of 𝑃 one, because we know the value of 𝑃 naught, we know the value of 𝑉 naught, and we know the value of 𝑉 one. So we can rearrange the equation by dividing both sides of the equation by 𝑉 one. And that gives us an expression for 𝑃 one.

At this point, we can substitute in the values. So we say that 𝑃 one is equal to 𝑃 naught multiplied by 𝑉 naught divided by 𝑉 one. And when we evaluate the fraction on the right-hand side, we find that 𝑃 one is equal to 50 pascals, at which one we can say that the application of Boyle’s law to the first stage of change of the gas has given us that 𝑃 one is equal to 50 pascals.

Now we know all of the important quantities of the gas after stage one of change. We already knew 𝑉 one and 𝑇 one. And we’ve just worked out 𝑃 one. So we can look at stage two of the change that the gas undergoes, from here to here. But before we do that, let’s write down that 𝑃 one is equal to 50 pascals.

Okay, now in the second stage of transformation, the volume is kept constant and the temperature is increasing. And the pressure is probably changing as well. So to work out what’s going on here, we need to recall another law. We need to recall Gay-Lussac’s law. This law tells us that pressure divided by the temperature is constant if the volume is kept constant.

Now we’ve already satisfied this condition because volume is kept constant. It stays at one meters cubed. So the pressure divided by the temperature must also stay constant in stage two. In other words, we can say that the pressure before stage two occurs divided by the temperature before stage two occurs is equal to the pressure after stage two has occurred divided by the temperature after stage two has occurred.

Now in this equation, we already know 𝑃 one because we worked it out here. We already know 𝑇 one because we’ve been given it in the question. And we already know 𝑇 two because we’ve also been given that in the question. And we need to work out what 𝑃 two is. So we rearrange the equation by multiplying both sides by 𝑇 two. Doing this gives us 𝑃 one multiplied by 𝑇 two divided by 𝑇 one is equal to 𝑃 two.

At this point, we can plug in our values. We say that 𝑃 two is equal to 𝑃 one multiplied by 𝑇 two divided by 𝑇 one. Evaluating the right-hand side of the equation then, we find that the value of 𝑃 two is 60 pascals. And this becomes the final answer to our question because, remember, we were trying to find the final pressure of the gas.

Now this method that we’ve used requires a bit of memorization on our part because we need to remember Boyle’s law and Gay-Lussac’s law. And we need to apply them to the information that we’ve been given. So we need to ensure that we know the two laws in the correct order. For example, it has to be 𝑃𝑉 is equal to constant. But it’s very easy to misremember and say something like 𝑃 divided by 𝑉 is constant or any similar mistake.

So if you’re confident with your gas laws and you remember which one is which and you remember them correctly, then use this method. If you’re not confident with the gas laws, then there is another way of doing it. And luckily, we only need to remember one thing. But before we look at method two, let’s reset the diagrams to have only the information we knew beforehand. Remember, we only worked out the value of 𝑃 one in method one. We didn’t know this at the start. And of course, the same is true for 𝑃 two.

Now let’s look at method two then. For method two, there’s only one thing we need to remember as we’ve said already. We need to remember the ideal gas equation. One form of this equation is the following. The pressure of an ideal gas multiplied by its volume is equal to the number of moles of gas that we have multiplied by the molar gas constant, which is just a number — it’s just a constant — multiplied by the temperature of the gas.

Now the clever thing about this equation is that if we rearrange it by dividing both sides by the temperature, then on the left-hand side, we’ve got all the things that are changing throughout the various processes that occur to the gas. But on the right-hand side, we’ve got quantities that are constant throughout, because first of all, we’ve got 𝑅, which is just a constant — it’s the molar gas constant. And secondly, we’ve got 𝑛, the number of moles of gas.

Now even though the gas is expanding and changing its temperature, there’s nothing been told to us about the number of moles of gas changing. In other words, the amount of gas or the quantity of gas that we have is the same throughout the two stages.

Therefore, we can apply the logic that whatever the pressure, the volume, or temperature may be at any stage in the process, the value of 𝑃 multiplied by 𝑉 divided by 𝑇 must be exactly the same. It must be a constant. This is because if 𝑛 is a constant and 𝑅 is a constant, then 𝑛𝑅 is a constant. And the right-hand side of the equation is a constant. So this is the logic that we’ll apply to the two stages in order to work out the value of 𝑃 two.

Once again, to make life easier for ourselves, we can break it down into the two stages. Now because 𝑃 multiplied by 𝑉 divided by 𝑇 is constant throughout, let’s look at the initial pressure, volume, and temperature and the pressure, volume, and temperature after stage one.

Now in this equation, we know the value of 𝑃 naught, 𝑉 naught, and 𝑇 naught because we’ve been given them in the question. And as well as this, we know 𝑉 one and 𝑇 one. So the only thing we don’t know is 𝑃 one. And once again, we’re calculating 𝑃 one. However, this time, we’ve not had to remember Boyle’s law. All we’ve had to remember is the ideal gas equation, which is a really important equation to know anyway. And we’ve rearranged it to have a constant on one side. So let’s use this method to work out 𝑃 one then.

First thing we need to do is to rearrange the equation. We multiply both sides by 𝑇 one and divide by 𝑉 one. Doing this leaves us with 𝑇 one divided by 𝑉 one multiplied by 𝑃 naught times 𝑉 naught over 𝑇 naught is equal to 𝑃 one, at which one we can substitute in all our values.

So what we have here is 𝑇 one divided by 𝑉 one multiplied by 𝑃 naught times 𝑉 naught over 𝑇 naught. And interestingly, because the temperature is kept constant in stage one, 𝑇 naught is equal to 𝑇 one. And so we have the same fraction as earlier. We’ve got 500 times 0.1 divided by one, except as we’ve already said we’ve come to that conclusion without needing to remember any gas laws. And so once again, we find that 𝑃 one is equal to 50 pascals.

Now we can apply the same logic to the properties of the gas before stage two has occurred and the properties after stage two has occurred. See, now that we’ve worked out 𝑃 one and we know 𝑉 one and 𝑇 one from the question already, as well as knowing 𝑉 two and 𝑇 two from the question, we can work out the value of 𝑃 two.

Once again, we rearrange in the same way to say that 𝑇 two divided by 𝑉 two multiplied by 𝑃 one times 𝑉 one over 𝑇 one is equal to 𝑃 two. Now at this point, we can plug in our values, which looks like the following. 𝑇 two divided by 𝑉 two is equal to 𝑃 one multiplied by 𝑉 one divided by 𝑇 one.

But remember, 𝑉 one is the same as 𝑉 two because the volume is kept constant in stage two. And that’s exactly what we see here. One meter cubed and one meter cubed for the volumes of 𝑉 one and 𝑉 two cancel out. And then when we evaluate the fraction on the left-hand side, we find that 𝑃 two is equal to 60 pascals. So both of our methods have arrived at the same conclusion. The final pressure of the gas is 60 pascals. So pick whichever method you’re more comfortable with.

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