### Video Transcript

A cylinder with a movable lid contains gas with a volume of 0.022 cubic meters. The gas has an initial temperature of 348 kelvin. The temperature is increased to 512 kelvin and the lid rises to equalize the pressure to the value it had before heating. What volume does the gas in the cylinder occupy after it is heated? Give your answer to three decimal places.

Let’s say that this is our cylinder containing gas, and we’ve drawn it to show that this lid is movable. It can slide up or down to expand or decrease the volume. We’re told that the temperature of the gas in this cylinder is increased from its original value. This results in an increase in pressure in that volume, which then raises the movable lid so that the final pressure of the gas in this cylinder is equal to the pressure it had initially. We’ve seen that this happens through the volume of the gas expanding, and it’s that final volume that we want to solve for.

If we treat the gas in this cylinder as an ideal gas, then we can describe it using what’s called the ideal gas law. In this law, 𝑃 is pressure; 𝑉 is volume; 𝑛 is the number of moles of a gas, that is, its quantity; 𝑅 is a constant value called the gas constant; and 𝑇 is the gas’s temperature. It’s fairly common for the quantities of pressure, volume, and temperature to vary in a given scenario. It’s less common though for the amount of gas, that is, the number of moles of the gas, to change. And 𝑅 we see can’t change because it’s a constant. In our situation, the amount of gas, the number of moles of the gas, is a constant. That is, the number of particles of this gas is the same, even though the gas expands as it’s heated.

If we divide both sides of the ideal gas law by the temperature 𝑇, then that factor cancels out on the right. What we have then is an equation where on the left there are a set of factors which are fairly likely to change being equal to a set of factors which are not. And indeed, we’ve seen that, in our scenario, 𝑛 times 𝑅 is a constant value. That has an important implication for us. It means if we were to have an initial pressure of our gas, we’ll call it 𝑃 one; an initial gas volume, we’ll call it 𝑉 one; and an initial gas temperature, 𝑇 one, then 𝑃 one times 𝑉 one divided by 𝑇 one would equal a final gas pressure, we’ll call it 𝑃 two, times a final gas volume, we’ll call it 𝑉 two, divided by a final gas temperature, 𝑇 two.

Because 𝑃 times 𝑉 divided by 𝑇 is equal to a constant, we can say that our gas’s pressure times its volume divided by its temperature at any instant is equal to its pressure times its volume divided by its temperature at any other moment. In our problem statement, we’re given some of what we could call initial and final conditions for the pressure, volume, and temperature of our gas. For example, at first, the volume of the gas is 0.022 cubic meters. We can label that quantity 𝑉 one. Likewise, the gas initially has a temperature of 348 kelvin. We’ll call that 𝑇 one. Then, the gas temperature is increased to 512 kelvin, and we’ll call that 𝑇 two. We want to solve for the volume of the gas after it’s been heated, which we’ll call 𝑉 two.

In this equation, we see that 𝑉 two is present, but it’s not yet isolated. We can make 𝑉 two the subject of this equation though by multiplying both sides by 𝑇 two divided by 𝑃 two. That way, on the right-hand side, 𝑃 two cancels out as does 𝑇 two. That gives us an equation where 𝑉 two is the subject. It’s equal to 𝑉 one times the ratio of pressures, 𝑃 one to 𝑃 two, times the ratio of temperatures, 𝑇 two to 𝑇 one.

Looking back at the information we’ve been given, we see that we’ve been told the two temperatures as well as the initial volume 𝑉 one, but we’re not told anything about the specific values of 𝑃 one and 𝑃 two. However, we are told that the lid in our cylinder did rise enough so that the pressure in the cylinder at the end was the same as it was at the beginning. That is, the pressures were equalized so that 𝑃 one is equal to 𝑃 two. So, whatever is the particular value of these pressures, they’re the same, and therefore their ratio is equal to one. Our equation for 𝑉 two then simplifies to 𝑉 one times 𝑇 two divided by 𝑇 one.

To carry out this final calculation, let’s clear some space at the top of our screen and then substitute in our values for 𝑉 one, 𝑇 two, and 𝑇 one. Notice that both of our temperatures are expressed in kelvin. Therefore, these units will cancel out, and we’ll be left with final units of meters cubed. Rounding this result to three decimal places, we get 0.032 cubic meters. This is the final volume of our gas after it’s expanded to equalize pressures. And note that this volume is indeed larger than 𝑉 one.