### Video Transcript

A gas in a container has a pressure of 400 kilopascals, a temperature of 360 kelvin, and a volume of 1.2 cubic meters. The gas is moved into a new container, which has a volume of 0.8 cubic meters. No gas is lost in the process. If the pressure of the gas in the new container is 500 kilopascals, what is the temperature of the gas in the new container?

In this example, we have a gas that starts out in one container, we’ll call it container one, which is then transferred with no loss of the gas to a second container; we’ll call it container two. We want to solve for the temperature of the gas once it’s in the second container; we’ll call that 𝑇 two. To help us do that, we’re going to assume that this gas is an ideal gas. This means it’s comprised of many identical particles that are negligibly small. It also means that we can describe this gas using the ideal gas law.

This law says that the pressure of an ideal gas multiplied by that gas’s volume is equal to the number of moles of the gas times a constant multiplied by the gas’s temperature. We can apply this ideal gas law to our gas when it’s in container two and also separately when it’s in container one. When the gas is in container two, the ideal gas law says that the gas’s pressure, 𝑃 two, times its volume, 𝑉 two, is equal to the number of moles of gas in container two, 𝑛 two, times the molar gas constant, 𝑅, multiplied by the gas’s temperature, 𝑇 two. We can likewise write a similar equation for when the gas was in container one. That expression will read 𝑃 one, the pressure of the gas in container one, times 𝑉 one, the gas’s volume in this container, equals 𝑛 one times 𝑅 times 𝑇 one.

Note that in both of these equations the quantity 𝑅 has no subscript. This is because 𝑅 is a constant called the molar gas constant, so its value is always the same. For that reason, we can leave out subscripts when writing 𝑅. Recalling that when the gas is transferred from container one to container two, no gas is lost in the process, we understand this to mean that the number of moles of gas that was in container one equals the number of moles of gas in container two. In other words, 𝑛 one equals 𝑛 two. Since these quantities are the same, let’s call them by the same name. We’ll say that we have 𝑛 moles of gas total. That leaves us then with these two applications of the ideal gas law. And as we’ve seen, it’s 𝑇 two, the temperature of the gas in the second container, that we want to solve for.

To begin doing that, we can divide both sides of this first equation by 𝑛 times 𝑅. Doing so will make 𝑛 and 𝑅 both cancel on the right-hand side. We find then that 𝑇 two equals 𝑃 two times 𝑉 two divided by 𝑛 times 𝑅. In our problem statement, we’re told the pressure of the gas in the new container, that’s 𝑃 two, as well as the volume of this container. Along with this, we can recall the value for the molar gas constant 𝑅. However, even if we know all of these quantities, we still don’t know the number of moles of gas 𝑛. Without that value, we can’t solve for 𝑇 two.

Let’s note though that the same quantity of moles, 𝑛, appears in our other ideal gas law equation. We can isolate 𝑛 by dividing both sides by 𝑅 times 𝑇 one. This will cause both of those factors to cancel out on the right-hand side of our equation. We find then that the number of moles of our gas is equal to 𝑃 one times 𝑉 one divided by 𝑅 times 𝑇 one. That is, it’s the gas’s pressure in the first container times the volume of that container divided by the gas’s temperature in this container multiplied by the gas constant. We can now take this fraction and substitute it in for 𝑛 in our equation for the temperature 𝑇 two.

When we make this substitution, notice that the molar gas constant 𝑅 will cancel out from our denominator. If we were to then multiply both the numerator and denominator of our right-hand side by the temperature 𝑇 one, that temperature would cancel in our denominator. And with just a little bit of regrouping, we would have this expression for the temperature 𝑇 two of our gas in the second container. It’s equal to the temperature of the gas in the first container, 𝑇 one, times the ratio 𝑃 two divided by 𝑃 one times the ratio 𝑉 two divided by 𝑉 one.

All five of these quantities are given to us in our problem statement. 𝑇 one is 360 kelvin. 𝑃 one, the pressure of the gas in the first container, is 400 kilopascals. 𝑃 two is given as 500 kilopascals, while the volume 𝑉 one is 1.2 cubic meters and 𝑉 two is 0.8 cubic meters. This entire expression is equal to the temperature 𝑇 two. And notice that the units of kilopascals cancel from numerator and denominator, as do the units of cubic meters. When we compute this expression, we find a value of exactly 300 kelvin. This is the temperature of the gas in the new container.