Video Transcript
A gas with a volume of 0.5 cubic meters is initially at a temperature of 240 kelvin and a pressure of 800 pascals. The gas is heated while being kept at a constant volume until its pressure is 1000 pascals. What is the temperature of the gas after it has been heated?
This question is about a gas that’s being kept at a constant volume, and we could imagine that as a gas inside a box like this one where the volume of that box remains fixed. A gas is made up of particles that are free to move around in all directions. This means that the particles inside this box are all flying around so 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 — and, for example, we could consider this particle here — then since the volume of the box is fixed and so the wall can’t move, the particle must just bounce off, changing the direction of its velocity and exerting a force on the wall with an outward component. The faster that the particle was moving, the greater the force that it’ll exert on the wall when it collides with it.
Now, in this sketch, we’ve just drawn a few particles, but in reality there’s going to be absolutely loads of them flying around and colliding with the walls. Each of these collisions will exert a force on the wall with an outward component. A force acting all over the area of the walls means that there’s a pressure exerted on the walls of the box. And since a faster moving particle will exert a greater force when it collides with a wall, then if the particles in the gas have a greater average speed, the gas will exert a greater pressure on the walls of the box.
We can 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 our gas held at a constant volume, that a greater temperature means a greater pressure. This idea can be expressed mathematically by Gay-Lussac’s law, which says that pressure 𝑃 is directly proportional to temperature 𝑇.
It’s important to remember that this law only applies when the gas is held at a constant volume. Since we are told in the question that the volume of this gas is indeed kept constant, then we know that this equation will apply. Instead of writing 𝑃 is proportional to 𝑇, we could instead write that pressure 𝑃 is equal to a constant multiplied by temperature 𝑇. These two statements are exactly equivalent.
If we now take this equation and divide both sides of it by the temperature 𝑇, then on the right-hand side, the 𝑇 in the numerator cancels with the one in the denominator. Then, we end up with an equation that says 𝑃 divided by 𝑇 is equal to a constant. That is, for a gas held at a constant volume, the pressure of that gas divided by its temperature will always remain the same.
We are told in the question that this gas has an initial pressure, which we’ve labelled as 𝑃 one, that’s equal to 800 pascals and an initial temperature that we’ve called 𝑇 one equal to 240 kelvin. We are also told that the gas is then heated until its pressure reaches a value of 1000 pascals. We’ve labelled this new pressure as 𝑃 two and the temperature at which the gas reaches this pressure as 𝑇 two.
So, 𝑇 two is the temperature of the gas after it has been heated, and that’s exactly what the question is asking us to find. We know that at a temperature of 𝑇 one the gas has a pressure of 𝑃 one and that at this new temperature of 𝑇 two the pressure is 𝑃 two. Since Gay-Lussac’s law tells us that the pressure divided by the temperature will always be constant, then we can say that 𝑃 one divided by 𝑇 one must be equal to 𝑃 two divided by 𝑇 two. We’re going to want to make 𝑇 two the subject of this equation.
If we multiply both sides by 𝑇 two, then we can cancel the 𝑇 two in numerator and denominator on the right-hand side. This gives us that 𝑇 two multiplied by 𝑃 one divided by 𝑇 one is equal to 𝑃 two. If we then multiply both sides by 𝑇 one divided by 𝑃 one, then on the left we can cancel the 𝑃 one in the numerator and denominator and also the 𝑇 one in the numerator and denominator. We end up with an equation that says 𝑇 two is equal to 𝑃 two times 𝑇 one divided by 𝑃 one.
We can now substitute in our values for the pressures 𝑃 one and 𝑃 two and the temperature 𝑇 one. We find that 𝑇 two is equal to 1000 pascals multiplied by 240 kelvin divided by 800 pascals. Notice that the units of pascals cancel out, and this leaves us with just units of kelvin. Evaluating the expression gives a result for 𝑇 two of 300 kelvin. So, we have found that the temperature of the gas after it has been heated is equal to 300 kelvin.
