 So, now that we have a collection of different thermodynamic relationships, what is it exactly that we're going to do with them? Suppose we know that the thermal expansion coefficient is defined this way, 1 over V dV dT at constant P. How is that useful to us? To explain that, we'll work a numerical example. To use the thermal expansion coefficient, I'll ask a question related to how much the volume changes as the temperature changes. So let's say we have a sample of water. Let's say just to keep the numbers as a real world example, let's say you have a bottle of water right next to you that contains 20 ounces of water. And, right, initially the temperature is 20 degrees Celsius, close to room temperature. And so you walk outside on a hot summer day and the temperature increases to 30 degrees Celsius. The volume of that water is going to expand as the temperature increases by 10 degrees Celsius. The question is how much can you expect the volume of the water to expand in response to that 10 degrees Celsius change? If we know, let's say we go and look up the fact that someone has measured the thermal expansion coefficient for water at room temperature and that value is 2.1 times 10 to the minus 4 per Kelvin. So that's the value of this thermal expansion coefficient. To answer the question of how much the volume changes, we remind ourselves again what the definition of alpha is. Alpha is 1 over v, dv, dt at constant pressure. If I rearrange that equation, let's go ahead and put the v and the alpha and the dt on this side. So v alpha dt is equal to dv, as long as I do that process at constant pressure. So if I integrate dv to get delta v, the change in volume is going to be the integral of v alpha dt. So from t1 to t2, integrate v times alpha times dt. So how do I do that integral? I can do it the easy way or the hard way. If I'm extra careful, I know that the volume is changing as the temperature changes. In fact, the thermal expansion coefficient is a little bit different at 30 degrees Celsius than at 20 degrees Celsius. But over a 10 degree temperature change, I can expect that those two numbers won't change a whole lot from everyday experience. The water may expand, but it's not going to expand a whole lot. So that's going to be approximately equal to, if I pull the v and the alpha out of the integral, assuming that they're not going to change very much, assuming they're approximately constant. I've got v alpha times the integral of dt or v times alpha times the change in temperature. So that's the simplest way to approach this problem if we can assume the volume's not changing much, because the temperature changes relatively slight. So then the change in volume is going to be the initial volume times the thermal expansion coefficient, 2.1 times 10 to the minus 4 per degree. And then the temperature change, 10 degrees Celsius or 10 degrees Kelvin. And if I multiply those numbers together, 20 times 10 times this small number. Units wise, Kelvin cancels as it should. And my change in volume just has units of the same volume that I started with. So that 20 ounce container of water is going to expand in volume by 0.042 ounces as I heat it up by 10 degrees. So the volume is going to change from 20 to 20.042 ounces. And at least to the number of sig figs that I cared about when I wrote the problem down, that's a negligible change in volume. That change of volume of 0.04 ounces, an ounce is about 30 milliliters. So that's in the vicinity of a little larger than 1 milliliter. So the volume is going to only change by 0.04 ounces, a little larger than 1 milliliter. That's not significant if we're talking about the volume of a bottle of water. If we were doing an analytical chemistry experiment where we had filled a volumetric flask with exactly 20.000 ounces or 500 milliliters exactly, then that volume change of a milliliter might be significant. So it depends on the context of the problem, whether that's significant. But this gives us an example of if we know the numerical value for one of these thermodynamic derivatives. How quickly v changes with respect to t, then we can use that in a problem to quantify how much the volume changes in response to a particular temperature change. So that's very often how we use these derivatives. If we know dx dy at constant z is equal to some numerical value, then by rearranging the equation in essentially the same way we did over here, then the change in x is going to be w times dy. Or and for small changes, that's going to tell us roughly speaking, change in x is w times the change in y for a quantitatively finite change in delta x and delta y. So often we'll use these thermodynamic relationships in the way we have here. If we're given a thermodynamic relationship, we can determine change in volume when the temperature changes. Probably more often, you don't know when you're given a problem when you run into a situation in the real world, which thermo relationship you need. So for example, let me phrase another example. More like you would run across in an everyday situation. Let's take a very similar problem. Let's say I have my 20 ounce bottle of water, 20 degrees Celsius, one atmosphere pressure. But now let's imagine instead of walking outside in the summer and the water heating up at constant pressure. Now let's say I squeeze the bottle of water. If you grab the bottle of water next to you, squeeze it until you've exerted two atmospheres pressure rather than the one atmosphere provided by the atmosphere. If we do that without changing the temperature, we can ask how much can you expect the volume of the water to have decreased in response to that change in pressure? So notice I haven't given you any more information than that. That's typically the problem we find ourselves faced with. We know how much some variables are changing. We would like to know how much other variables are changing and it's up to us to figure out which thermodynamic relationship we're interested in. So if we want to know how much V is changing in response to a change in P, we would like to know that derivative. How quickly does V change as P changes? We're doing that isothermally, I didn't say that. But if we add the restriction that we do that isothermally, then what we're interested in the thermodynamic relationship we're interested in is dV dP at constant T. So now we figured out from the statement of the problem, which thermodynamic relationship we want. Now it's up to us to find what dV dP at constant T is equal to. And sometimes, as in this case, we can say, okay, that's one I recognize, that's one I've seen before. We've seen the definition of the isothermal compressibility is minus one over V dV dP at constant T. Other times, as we'll run across soon, we might find a derivative that we don't recognize and we have to do some calculus or some arithmetic or some algebra to figure out what that derivative is equal to. So sometimes, if we're given a thermodynamic relationship, we can use it. Sometimes the statement of the problem provides a hint to us of what thermodynamic relationship we need. But either way, we need to be able to find out what those thermodynamic relationships are.