 So since we know how the Gibbs free energy changes with pressure, in particular dGDP at constant T is equal to V, that allows us to calculate the Gibbs free energy change when I change the pressure from P1 to P2. Most commonly what we'll do with that expression is use it to calculate the Gibbs free energy at one pressure relative to the Gibbs free energy at another pressure, where all I've done now is I've written delta G as G at P2 minus G at P1, so the final pressure is equal to the initial pressure plus this change. In general, if the volume is changing as the pressure changes, if the object is not incompressible, if it can be compressed, then we have to do an integral here. For the particular case of an ideal gas, as we've seen, inserting nRT over P here gives us an expression which has the form nRT and then the integral of 1 over P became log P evaluated between P1 and P2. So Gibbs free energy at a final pressure is equal to Gibbs free energy at an initial pressure plus nRT log P2 over P1. So far so good. That's for an ideal gas. I can write that expression one more way. These are all extensive quantities, extensive Gibbs free energy, and then including the moles here makes this the change in the extensive free energy. If I divide through by n to convert the extensive free energies into intensive, I can say that the Gibbs free energy per mole at P2 is equal to the Gibbs free energy per mole at P1 and then RT log P2 over P1. Those equations are equivalent to one another just multiply or divide by n to convert between them. Now, what I can do with this expression, what we very often do with this expression is take as our P1 a standard starting point. In fact, we can say we can define a standard state for an ideal gas and I'll define the standard pressure to be P with this superscript circle on top. Most commonly, standard state is defined to be one bar. So if I say an ideal gas at one bar pressure is in the standard state, then I can use this expression to say the molar Gibbs free energy at whatever pressure I want, this is my P2, is equal to the molar Gibbs free energy at standard pressure plus RT natural log of that ratio of the pressure I'm interested in relative to the standard pressure. So I've just rewritten this equation with P1 equal to standard pressure P2 equal to P. So this is an expression that tells us what the free energy is at any pressure I'm interested in relative to the free energy at the standard state conditions. And we often write standard state conditions G at P0, G at standard pressure. We can write this as standard Gibbs free energy. So the molar Gibbs free energy can be written relative to the standard Gibbs free energy. So if, for example, we tabulate the Gibbs standard free energies under standard conditions for a variety of materials, then we can very easily calculate the molar Gibbs free energy at a range of different pressures with this correction or addition onto the standard free energy. So that equation's worth noting because it will be the starting point for several other things we'll do in the future. One comment I'll make at this point, it can be very tempting to think of this equation as, I'll write this in a different color to indicate that it's not the correct thing to do. It can be tempting to memorize or remember this equation as G being G0 plus RT log P. In fact, occasionally you might see it written down that way. That's at best misleading and at worst just plain wrong because a couple of reasons. First of all, we've dropped the P0. The reason it might be tempting to drop the P0 is P0 is just equal to 1. So P over P0, P divided by 1, that's just the same thing as P, right? But because this quantity has units on it, that's not quite a safe thing to do. Pressure divided by 1 bar, if we're working in bar, you're dividing by 1. But if you're working in atmospheres, if you're working in tor, it's not the same thing. So the telltale sign that we've made a mistake here is the natural log is being taken of some quantity with units on it. If you ever find yourself taking the natural log of a quantity with units, then something has gone wrong. Possibly something as benign as forgetting to divide by 1 or forgetting to write down that you're dividing by 1. But in other circumstances, if you're using this pressure in units other than bar, this would be a mistake that would affect your answer. So this is not the right way to remember that equation. Inside this natural log, always remember to take the pressure divided by the standard pressure. But in any case, we have this equation for calculating free energies as a function of pressure, specifically for an ideal gas. And that's an equation we'll use several times in the future.