What is Charles’s Law?

Charles’s law shows the relationship between volume and temperature, but in order for Charles’s law to be correct the pressure and the amount of gas must remain the same, they must remain constant. Charles’s law can be expressed in the equation \(V=k\) times \(T\), where \(V=Volume\), k[/latex] is a constant, (meaning it’s always the same) and \(T\) is the absolute temperature expressed in kelvins.

Say the constant was 2, (I’m just making that number up, all right) and the temperature is 5, and the volume is 10. That equation makes sense. This is the volume, this is the constant, and this is the temperature. Now say the volume is 6, well, the constant is going to have to be the same because it’s always the same, times \(T\)—trying to figure out what temperature is going to be.

Using algebra, we divide both sides by 2, and \(T\) equals 3. Now (I’m just going to erase all this real quick) \(T\) equals 3, so volume went down and so did temperature, so we can see here that volume and temperature are directly related. In other words, what happens to one variable will also happen to the other.

If volume decreases temperature will decrease; if temperature decreases volume will also decrease; and if volume increases temperature will increase; and if temperature increases volume will also increase. Now, I want to look at a more practical application of Charles’s law.

We can also express Charles’s law like this: \(V_1\) divided by \(T_1\), equals \(V_2\), divided by \(T_2\). Now look at this example problem: A 2.0 liter balloon is at a temperature of 223 kelvin. When the balloon is heated to 323 kelvin, what is the volume? This is the first volume right here, so that’s \(V_1\), it’s at a temperature of 223, so that’s \(T_1\).

Then the balloon is heated to 323, so that’s \(T_2\), then what is the volume? \(V_2\) is what we’re looking for here, so we need to get \(V_2\) by itself on the side of the equation. Now notice here that the temperature changed, it went from 223 to 323. (Since volume changed temperature also has to change here, or excuse me).

Since temperature changed, volume also has to change here, so that’s why we’re looking for \(V_2\), and notice that temperature went from 223 to 323, so it increased. We can also know that volume is going to increase, and we’ll see that in a moment once we work this equation. We’re going to multiply both sides by \(T_2\) to get \(V_2\) by itself on the side of the equation. Then we’re going to do \(V_2\) equals (all right) \(T_2\) is 323k, times \(V_1\), which is 2.0 liters.

Then divide by \(T_1\), which is 223k. Now I’ll save you having to do all the math, so the answer is 2.90. Now we have to put the correct units on here, we can cross out the \(k\)s, so the only thing left is the \(L\), and that should be correct because volume is always expressed in liters, it’s never going to express in a temperature unit.

All right, so notice here the volume was originally 2 and now it’s 2 and 9 tenths, and the temperature was originally 223 and now it’s 323, so temperature increased, and since temperature and volume are directly related volume also increased. Now, Charles’s law makes sense because imagine a balloon filled up with molecules of a certain gas.

Well if that gas is heated up, then the molecules start moving faster. When the molecule start moving faster they start bumping up against the sides of the balloon more often, and they start pushing the walls of the balloon out increasing the volume of the balloon.

Now if the balloon is cooled off, then the molecules inside start moving slower, and they don’t start hitting the sides of the wall as often. Since they’re not hitting the sides of the wall as often, the walls start to come in and the volume decreases. That’s a practical application of Charles’s law. The important thing to remember is that volume and temperature are directly related.