Video Transcript
10 cubic meters of helium gas is allowed to cool from a temperature of 400 kelvin to a temperature of 290 kelvin. The helium is kept at a constant volume. If the final pressure of the gas is 1200 pascals, what was its pressure before cooling? Give your answer to the nearest pascal.
This question is asking us about a gas that’s being cooled at a constant volume. We could imagine that as a gas inside a box like this one, where the volume of this box remains fixed. Let’s recall that a gas is made up of particles that are free to move around. In our sketch, we’ve drawn some of these particles in pink. And the orange arrows show their individual velocities at some particular instant in time.
Now all these particles are all flying around in all different directions. This means that they can collide with each other. And importantly, they can also collide with the walls of the box. When a particle collides with one of the walls, like we can see this one here is about to, then since the volume of the box is fixed and so the walls can’t move, the particle must simply bounce off the wall, changing its direction and exerting a force on the wall with an outward component. The faster that the particle was moving, the greater the force that it will exert on the wall when it collides with it.
In our sketch here, we just drew a few particles to give an idea of what’s going on. In reality, though, there will be far more particles than this flying around and colliding with all the walls of the container. Each one of these collisions will exert a force on the wall with an outward component. A force acting all across the area of the walls means that there’s a pressure exerted on the walls of the container. Since a faster moving particle exerts a greater force when it collides with a wall, then the greater the average speed of the particles in the gas, the greater the pressure that the gas will exert on the container walls.
Let’s recall that the average speed of the particles in a gas indicates the temperature of that gas. The higher the temperature, the greater the average speed. We can say then, for a gas held at a constant volume, that a greater temperature means a greater pressure. This relationship can be expressed mathematically by Gay-Lussac’s law. This says that pressure, 𝑝, is directly proportional to temperature, 𝑇. It’s important to keep in mind that this law only applies when a gas is kept at a constant volume.
We know that in this case the volume is indeed kept constant while the helium is cooled. That means that Gay-Lussac’s law does apply to this situation. Now we could equally choose to write the law as saying that the pressure 𝑝 is equal to a constant multiplied by the temperature 𝑇. If we then divide both sides of this equation by 𝑇, we can cancel the factors of 𝑇 from the numerator and denominator on the right, which leaves us with an equation that says 𝑝 divided by 𝑇 is equal to a constant. What this means is that for a gas that’s kept at a constant volume, the pressure of that gas divided by its temperature will always have the same value. This fact is the key to answering this question.
We’re told that the helium gas starts out with a temperature of 400 kelvin. We’ll label this as 𝑇 one. Let’s also label the pressure of the gas at this initial temperature as 𝑝 one. This is the value that the question is asking us to find. We’re told that the helium gas is then cooled to a temperature of 290 kelvin. We’ll label this final temperature as 𝑇 two. At this temperature, the gas has a pressure of 1200 pascals. We’ll label this final pressure as 𝑝 two.
Now from Gay-Lussac’s law, we know that the pressure divided by the temperature will have the same value at all points in time. This means that we know that the initial pressure, 𝑝 one, divided by the initial temperature, 𝑇 one, must be equal to the final pressure, 𝑝 two, divided by the final temperature, 𝑇 two.
Since the quantity that we want to calculate is the pressure 𝑝 one, we’ll need to rearrange this equation to make 𝑝 one the subject. To do this, we can multiply both sides by 𝑇 one. The 𝑇 one terms then cancel from the numerator and denominator on the left. And we have that 𝑝 one is equal to 𝑝 two multiplied by 𝑇 one divided by 𝑇 two. Since we know the values of all three quantities on the right-hand side of this expression, we can go ahead and substitute those values in.
We find that the pressure 𝑝 one is equal to 1200 pascals multiplied by 400 kelvin divided by 290 kelvin. The units of kelvin cancel from the numerator and denominator, leaving overall units of pascals. Then evaluating this expression, we get a result for 𝑝 one of 1655.172 etcetera pascals.
Let’s notice, though, that the question asks for our answer to the nearest pascal. Rounding our result to the nearest pascal, we find that the pressure 𝑝 one is equal to 1655 pascals. Our final answer, then, is that to the nearest pascal, the initial pressure of the helium gas is 1655 pascals.
