A hot gas that initially has a volume of 20 cubic metres is allowed to cool under constant pressure. When the gas reaches a temperature of 320 Kelvin, it has a volume of 16 cubic metres. What is the initial temperature of the gas?
Okay, so here’s our scenario. Say that we have a container that’s filled with a gas. And here this gas is represented by these pink dots. Now our container has three sides to it. And the top side is a freely moving section that has a weight on top of it. And the reason we’ve put that weight there is because our problem statement tells us that this gas is cooling at a constant pressure. In other words, the pressure pushing in on the gas is the same all throughout this process. And so to indicate that, we have this constant weight, which is always pressing down on the gas.
We’re told that, at first, the gas has a volume of 20 cubic metres. And let’s call that volume 𝑉 sub 𝑖 for the initial volume. So 𝑉 sub 𝑖 is 20 cubic metres. And the gas then cools down until it reaches a temperature of 320 Kelvin. Let’s call that temperature, 320 Kelvin, 𝑇 sub 𝑓, 𝑓 because this is a final temperature of the gas. And we’re told further that when the gas temperature has decreased to 320 Kelvin, the gas then has a volume of 16 cubic metres. And we’ll call that volume 𝑉 sub 𝑓.
Knowing all this, we want to solve for the initial temperature of the gas. We can call that temperature 𝑇 sub 𝑖. We can see then that, in this exercise, we’re working with gas temperature and gas volume under the condition that gas pressure is held constant. There’s a particular gas law that describes the relationship between volume and temperature under constant pressure. It’s called Charles’s law. And it simply states that, at a constant pressure, the volume of a gas is proportional to its temperature.
A mathematically equivalent way of writing this is to say that gas volume is equal to a constant — we can call it 𝑘 — times temperature. Now this statement of Charles’s law is useful for our case because if we divide both sides of the equation by the gas temperature 𝑇. Then that term cancels out from the right-hand side. And we see that gas volume divided by gas temperature is equal to a constant. And this is quite helpful because it tells us that we can choose any volume that this particular gas at constant pressure reaches. And so long as we use the corresponding temperature at that volume, that volume divided by that temperature is equal to the same value. As we’ve written it, it’s equal to the constant 𝑘.
Here’s how we can connect this with our particular scenario. We can take the initial volume of our gas. We called it 𝑉 sub 𝑖. And if we divide that volume by the initial temperature of the gas, 𝑇 sub 𝑖, then Charles’s law says this is equal to some constant. We can call it 𝑘.
Now we don’t know what that constant is. But we don’t need to know because 𝑘 is also equal to the ratio of the final volume of our gas, 𝑉 sub 𝑓, divided by its final temperature. Remember, according to Charles’s law, we can choose any volume of the gas, whether it’s 𝑉 sub 𝑖 or 𝑉 sub 𝑓 or somewhere in between. And as long as we use the corresponding gas temperature, 𝑉 divided by 𝑇 is the same value. So this means we can leave off the constant 𝑘 entirely.
And we now have this relationship. And we can recall that we know three of these four variables. We know 𝑉 sub 𝑖, 𝑉 sub 𝑓, and 𝑇 sub 𝑓. And we want to solve for the fourth, 𝑇 sub 𝑖. To do that, let’s rearrange this expression so 𝑇 sub 𝑖 is on one side all by itself. If we multiply both sides of the equation by 𝑇 sub 𝑖, it cancels out on the left. And then if we multiply both sides by 𝑇 sub 𝑓 divided by 𝑉 sub 𝑓, then over on the right, our final volume as well as our final temperature cancel out. And we see that our initial gas temperature is equal to our final gas temperature multiplied by 𝑉 sub 𝑖 divided by 𝑉 sub 𝑓.
And since we know all three of these values, we can substitute them in now. 𝑇 sub 𝑓 is 320 Kelvin, 𝑉 sub 𝑖 is 20 cubic metres, and 𝑉 sub 𝑓 is 16 cubic metres. Notice that the units of cubic metres cancel out. And we’ll be left simply with units of Kelvin in our final answer. When we calculate that answer, we find a result of 400 Kelvin. That’s the initial temperature of this gas.