# Question Video: Using Gay-Lussac’s Law to Find the Pressure of a Gas Physics • 9th Grade

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?

05:04

### Video Transcript

A gas with a volume of two cubic meters is initially at a temperature of 300 kelvin and a pressure of 500 pascals. It is heated to 375 kelvin while being kept at a constant volume. What is the pressure of the gas after it has been heated?

This question is about a gas that’s kept at a constant volume. So, we could imagine it as a gas inside a box like this with a fixed volume. A gas consists of particles that are free to move around. And so we’ve got all these particles, that’s the pink circles in our diagram, flying around inside the box. Here, we’ve indicated the direction of each particle’s motion with an orange arrow. Those particles can collide with each other, and also importantly they can collide with the walls of the box.

Let’s consider a particle colliding with one of the walls, for example, like this one right here is about to do. Remember that we’re told that the volume is constant, which means that the walls of the box are fixed. This means that when the particle collides with the wall, then it will simply bounce off. This process exerts a force on the wall. This force is always going to have a component that’s acting outward perpendicular to the wall.

Since in reality there’s going to be a much, much larger number of particles flying around than we’ve drawn in this sketch, then that means that there’s continually going to be a large number of particles colliding with and bouncing off the walls of the box. So, that’s a large number of collisions happening across the area of the walls. Since each of these collisions will exert a force on the wall with an outward component, then that means that there’s a force acting all over the area of the walls of the box. A force acting across a surface with a particular area results in a pressure. And so there’s a pressure being exerted on the walls of the box.

In this question, we’re told that this pressure is initially equal to 500 pascals and that this is the case when the temperature of the gas is 300 kelvin. Let’s label the initial pressure as 𝑃 one and the initial temperature as 𝑇 one. We’re told that the gas is then heated to a temperature of 375 kelvin. We’ll label this new temperature as 𝑇 two. Let’s also label this new pressure at the temperature 𝑇 two as 𝑃 two. And this pressure is what we’re being asked to find.

To understand why the pressure of this gas will change when the temperature changes, we can recall that temperature is related to the average speed of particles in a gas. And in particular the higher this temperature, the greater the average speed of the particles. Now, if we have a box of gas like this with a constant volume and we imagine a particle colliding with one of the walls, then the faster that particle is moving, the greater the force it will exert on the wall. If we have many particles moving with a greater average speed, then on average each particle will exert a greater force on the wall when it collides with it. This means that a greater average speed results in a greater value of pressure.

Since we know that the average speed indicates the temperature of the gas, then this means that a higher temperature results in a greater pressure. This is described by our law known as Gay-Lussac’s law, which says that pressure 𝑃 is directly proportional to temperature 𝑇. This can also be written as 𝑃 is equal to a constant multiplied by 𝑇. It’s important to note that Gay-Lussac’s law applies for gas held at a constant volume. Since the question tells us that this gas is kept at a constant volume while it’s heated, then we know that this formula will apply.

If we divide both sides of the equation by the temperature 𝑇, then on the right-hand side, the 𝑇 in the numerator will cancel with the 𝑇 in the denominator. This leaves us with an equation that says 𝑃 divided by 𝑇 is equal to a constant. So that means that for a given gas kept at a fixed volume, the pressure of that gas divided by its temperature will always have the same constant value. In this question, we know that at some instant initially the gas has a pressure of 𝑃 one equal to 500 pascals and a temperature of 𝑇 one equal to 300 kelvin. Then, after it’s heated, it has a temperature 𝑇 two of 375 kelvin and a pressure that we’ve labeled as 𝑃 two.

Since Gay-Lussac’s law tells us that the pressure divided by the temperature is constant, then this means that 𝑃 one divided by 𝑇 one must be equal to 𝑃 two divided by 𝑇 two. The quantity that we’re trying to find is this pressure 𝑃 two. And we can make 𝑃 two the subject of the equation if we multiply both sides by the temperature 𝑇 two. On the right, the 𝑇 two in the numerator and denominator cancel out. Then, writing the equation the other way around, we have that 𝑃 two is equal to 𝑃 one times 𝑇 two divided by 𝑇 one.

Let’s now substitute our values for the pressure 𝑃 one and the temperatures 𝑇 one and 𝑇 two into this equation. We find that the pressure 𝑃 two is equal to 500 pascals multiplied by 375 kelvin divided by 300 kelvin. The units of kelvin cancel from the numerator and denominator, which leaves us with units of pascals. Evaluating the expression then gives a result for 𝑃 two of 625 pascals. So then we have found that the pressure of the gas after it has been heated is equal to 625 pascals.