### Video Transcript

A bike tyre is pumped up using a foot pump. The pressure of the gas within the tyre increases. But the volume remains the same. What happens to the temperature of the gas?

In order to answer this question, let’s start by underlining all the important bits first. So we know that we’ve got a bike tyre. And it’s pumped up using a foot pump. Luckily, none of this is majorly important stuff. It’s just nice to be able to visualise. But we do know that the pressure of the gas within the tyre increases. We also know that the volume remains the same.

What we’re asked to do is to find out what happens to the temperature of the gas. So we’ve got this bike tyre with some gas inside it. And initially, the pressure is, let’s say, 𝑃 one. After we pump it up, we know that we’ve got a pressure 𝑃 two, because the pressure changes. In fact, the pressure increases. Therefore, we know that 𝑃 two is larger than 𝑃 one.

However, there’s something else that we know. We know that the volume of the gas, 𝑉 one, which is what we’ll call the initial volume, is exactly the same as 𝑉 two, which is what we’ll call the final volume. This is because we’re told in the question that the volume remains the same. What we’re asked to do is to find out what happens to the temperature, in other words, how did the values 𝑇 one and 𝑇 two linked together.

To do this, we need to use something known as Gay-Lussac’s law. This is an experimental law. In other words, it was discovered just by experimenting rather than by calculating it theoretically. And it tells us the following. The pressure 𝑃 of a gas is directly proportional to the temperature 𝑇 if the volume remains constant.

Now the final bit, i.e., the bit that says if the volume remains constant, is very important. Luckily for us, however, that’s exactly what we’ve got here. We’ve said that 𝑉 one, the initial volume, is equal to 𝑉 two, the final volume. If they’re equal, then the volume must have stayed constant.

So we’ve got a constant volume, which means we can apply Gay-Lussac’s law. We’ve said that the pressure is directly proportional to the temperature. Now we’ve been told that, after pumping up the tyre, the pressure increases. That is, 𝑃 two is greater than 𝑃 one.

So if the pressure and temperature are directly proportional, then if the pressure has increased, then the temperature must have also increased. In other words, 𝑇 two is greater than 𝑇 one. Or the other way to read it is 𝑇 one is less than 𝑇 two. And a good way to think about this is to think of a graph that shows direct proportionality. And this is an example of a graph that shows exactly that.

We’ve got pressure on the horizontal axis and temperature on the vertical axis. And the relationship between these two is that they’re directly proportional to each other. In other words, let’s say this pressure was 𝑃 one and this pressure is 𝑃 two, because we know that 𝑃 two is larger than 𝑃 one. Well then, this temperature here must be 𝑇 one and this temperature here must be 𝑇 two. So we can clearly see that 𝑃 two being larger than 𝑃 one means that 𝑇 two is also larger than 𝑇 one. So our final answer is that the temperature of the gas increases.