 So one of the nice things about the fundamental equation for the Gibbs energy in this case or in any of the energies is that it immediately tells us several or a thermodynamic derivatives. So for example, reading directly off of this equation, derivative of g with respect to p at constant t is going to be this coefficient v. So we know the rate at which the Gibbs free energy changes when I change the pressure of an object is the volume of the object. So let's see if we can understand what that expression means a little bit. Typically, the way we'll use that expression would be if we exert pressure on something, change the pressure on an object. And we want to know how much is Gibbs free energy changes. This is the equation we can use to determine that. So we'll rearrange this equation, for example, to say dg is equal to vdp. That, of course, is exactly like this equation if I ignore the temperature term. So if we're at constant temperature, either because I'm ignoring this term or because I know this derivative, Gibbs free energy change is volume times change in pressure. If I want to know the finite change in pressure, changing Gibbs energy rather than the infinitesimally small changing Gibbs energy, I just need to integrate this expression integral of vdp. So what do we do with this? There's a couple of things we could do. One, if we can assume that the thing I'm changing the pressure of is relatively incompressible. So for example, let's suppose I have a bottle of water. I can exert pressure. I can squeeze on this water bottle and I can't change its volume very much. So water doesn't shrink. Its volume doesn't change very much as the pressure changes. So if I squeeze the water, it's not going to change its volume very much. We say that water is not quite perfectly incompressible, but close to an incompressible fluid, in which case I could say since the volume is not changing as I change the pressure on the bottle of water, I can pull the v out of the integral. v doesn't depend on the pressure. And then integral of dp is just delta p. So for example, let's work a numerical example. Let's say I have this bottle of water, which the label tells me is half a liter, 500 milliliters. If that bottle of water is at room temperature, if it has a volume of half a liter. And if I squeeze the water bottle, so right now it's at a pressure of one atmosphere. And I'm going to exert pressure. I'm going to squeeze on the bottle, exert a pressure. Let's say I squeeze hard enough to be exerting a pressure of two atmospheres. The question is, how much have I changed the Gibbs free energy of the water in a bottle? So what is delta G? Well, we have an expression that can answer that question quite simply. It's just volume times the change in pressure if I can assume the water is incompressible. So that's going to be half a liter multiplied by the change in pressure from one atmosphere to two atmospheres, so that's one atmosphere. That will give me the Gibbs free energy change in liters times atmospheres, which is not a unit of energy, but it's not the most convenient unit of energy. To convert that to a more convenient unit of energy, I need to convert liter atmospheres to joules, for example. And we could look up a conversion factor. There's 101.325 joules in a liter atmosphere. What's convenient, often, is to remember the gas constant 8.314 joules per mole Kelvin is the same as the gas constant 0.08206.08206 liter atmospheres per mole Kelvin. So this many joules is equal to that many liter atmospheres. That ratio is 101.325. So 0.5 liter atmospheres times 101. That works out to a couple of sig figs to be 51 joules. So what we've been able to figure out is that I've raised the free energy of the water in this bottle by 51 joules when I squeeze on it and exert a pressure of two atmospheres rather than just the one atmosphere that the room air is pressing on the bottle with right now. So we can calculate the Gibbs free energy change when I compress, when I increase the pressure on an object, if it's incompressible. Not everything can be treated as incompressible, however. If this bottle contained not a liquid, not an incompressible liquid like water, but if it contained a gas, maybe an ideal gas, then we would have to make a different assumption. The good news is we can still do that. If we back up to before we made our assumption that the liquid was incompressible, this result is always true. Free energy change is volume times dp integrated from some initial pressure to some final pressure. If we don't pull the v out of the integral, if we acknowledge that the volume might in fact be changing as I change the pressure, as I increase the pressure, the volume is going to decrease for something like your gas, the volume is a function of pressure, so I have to leave it inside this integral. If it's an ideal gas, so here's our second set of conditions, for an ideal gas, we know exactly what the volume of an ideal gas is. It's nRT over P, but that's an integral we can do. If we've done this at constant temperature, the n and the R and the T can come out of the integral. Integral of 1 over P is just natural log of P. And if I evaluate that from P1 to P2, lower limit P1, upper limit P2, ln of P2 minus ln of P1 is the same as ln of P2 over P1. So this expression, delta G is positive nRT, log P2 over P1, that's an expression we can use for the free energy of compression of an ideal gas. This expression would be the free energy of compression of an incompressible object. Let's go ahead and work an example comparable to this one. Let's say that we had half a liter of an ideal gas, still at 298 Kelvin. And again, I want to raise the pressure from one atmosphere to two atmospheres. Same problem as before, but now I'm squeezing a gas. If I imagine that this water bottle is now filled with air instead of with water, then when I squeeze it, exert two atmospheres of pressure on it, the volume will shrink quite a bit. So this is the equation we would need to use. In order to use this expression, I'm going to need to know how many moles of gas are in the container. So I know V and T and P, so I can calculate the number of moles. PV equals nRT, so n is equal to PV over RT. One atmosphere times 298. Sorry, one atmosphere times half a liter divided by the gas constant divided by 298. We don't need to spend time doing ideal gas law problems, so I'll just tell you what the answer. That gives me 0.02 moles of gas in this half liter bottle at room temperature. So that's now enough information to calculate the free energy change when I compress that gas. It's nRT log P2 over P1. So putting numbers in just to make sure the units work out. So I've got 0.02 moles of the gas. The gas constant I'll use in this case is because I want the answer to come out in joules. I'll use the gas constant in joules, 8.314 joules per mole Kelvin. I'm multiplying that by the temperature of 298 Kelvin and also the natural log of this ratio of pressures, P2 of two atmospheres divided by P1 of one atmosphere. So let's double check in this natural log. Atmospheres cancel atmospheres. The natural log is of a unitless quantity, as it should be. In the nRT product, moles cancel top and bottom. Kelvin cancel top and bottom. And I'm left with an answer in just joules. And if we do the arithmetic, it turns out to be the case that the free energy when I compress a gas inside this container changes by 35 joules rather than the 51 joules when I compress the liquid. So those are the numerical answers. One interesting question we can ask is why does the free energy change by less when I compress the gas than when I compress the liquid? One way to think about that answer is the rate at which the free energy changing, the free energy goes up. Volume is positive. So the free energy is always going to go up when I increase the pressure. The rate at which the free energy is going up is the volume of the substance. For the liquid, the volume was always the same. I could squeeze harder and harder and harder. Every additional atmosphere of pressure I add increases the free energy by the same amount. But for the gas, when I increase the pressure, the volume keeps shrinking. So the fact that the gas has shrunk, that its volume has decreased as I began squeezing on it, means the free energy starts rising less and less because the volume is smaller. So the reason the free energy goes up less for the gas than for the liquid is because the gas actually does shrink, can be compressed as I exert pressure on it. So we have two different expressions now. We have, I guess, three different expressions. This is the most general expression. If we want to know how the free energy changes as we change the pressure, this is always correct. If we make assumptions that the substance is incompressible or an ideal gas, we can arrive at two different results. In other circumstances, if the pressure dependence of the volume had a different form other than the ideal gas behavior, as it will for other substances, we could obtain other expressions still.