 So we've seen that the pressure for a gas that obeys the Van der Waals model can be written with this equation, an equation that we call the Van der Waals equation of state. So equation of state, sometimes we abbreviate it as EOS for equation of state. The reason we call this an equation of state is because it describes how to calculate the pressure as a function of the volume and the temperature. So if I know V and T, if I also know the number of molecules, I can calculate the pressure. So the macro state of the system turns out can be described by V and T. If I know the volume and the temperature, and in this case the number of molecules of the gas, that describes the state of the system well enough that that's all I need to know to describe the pressure for the macro state of the system. And as we've seen, there are other equations I could also calculate, for example the internal energy, and all I need to do that is the temperature as well. So knowing just a few thermodynamic properties like the volume and the temperature is enough to tell me the state of the system, and this equation of state tells me how to get the pressure from the state of the system. So this equation is useful and convenient. It's sometimes more useful to think not in terms of the number of molecules of the gas, but the number of moles of the gas. So as we've already seen, if we replace Boltzmann's constant with units of joules per Kelvin, for example, with gas constant and units of joules per Kelvin per mole, then if I multiply by the number of moles rather than the number of molecules, molecules times K is the same as moles times R. So I'm just going to rewrite this equation now, writing it in terms of moles rather than molecules. Everywhere I have an N, a big N. In the first equation I'll write a little N in the second equation. And just like Boltzmann's constant turning into the gas constant, which are really the same number just expressed in different units, joules per Kelvin or joules per Kelvin per mole. Same thing here. I can think of the molecular volume as being a particular number per molecule, or I can think of it as a volume per mole. And so when I change from big N number of molecules to little N number of moles, I need to be thinking about the molecular volume here in terms of volume per mole rather than volume per molecule. Likewise, in the second term, the A here might have units. We'll write down in a second what these units should be. But the A is going to have an extra units of per moles twice to reflect the fact that I'm counting my molecules in units of moles in this case. But other than that, these are exactly the same expression. And I can write the pressure now in terms and calculate it from the number of moles of molecules as well as the temperature and the volume. I can make the equation one step more convenient still by noticing that all these properties, the temperature and the volume and the number of moles, those are all extensive properties. They scale, I'm sorry, only the volume and the number of moles are extensive properties. Temperature is intensive, of course, and pressure is intensive. This equation will get a little more convenient if I write everything in terms of intensive properties. So to turn the volume into an intensive property, I need to think about the molar volume, the volume divided by number of moles. So if I rewrite each of these V's, if I divide it by N to write it as a V bar, then I can write pressure as maybe the easiest one. Well, let's deal with the first term first. So I divide on top and bottom by the same factor, if I divide by N. So N or T divided by N and V minus NB divided by N. Then that just gets rid of the N in the numerator. V divided by N becomes a V bar in the denominator. NB divided by N becomes just a B in the denominator. Likewise, I have a V underneath an N twice here. I have a V squared underneath an N squared. So that looks like a V bar squared in the denominator and I still have an A on top. So I've rewritten this equation in terms of molar volumes rather than extensive volumes. And not only did the equation get a little bit simpler, it's also a little more convenient because now all the properties that I'm talking about, the temperature, the molar volume and the pressure, those are all intensive properties. So this is still an equation of state. It's still called the Van der Waals equation of state. But what I'm doing now is I'm calculating the pressure as a function of the molar volume and the temperature rather than the extensive volume of the temperature. And as a bonus, I don't any longer have to know how many molecules or how many moles of the gas that I'm talking about. So let's do an example and make sure we understand how to use these expressions and see what they tell us about the pressure of a Van der Waals gas. So let's take a fairly ordinary gas, nitrogen gas, rather than room temperature, let's use 273 Kelvin standard temperature because we know a little bit about ideal gases anyway at standard temperature and pressure. If I say the molar volume of this gas is 2 liters for every mole. So I've given you the V bar and the T that we need to be able to calculate the pressure. The question is going to be what is the pressure of this gas under these conditions? To do that, of course, I'm also going to have to tell you the values of the A and B constants. So the value of A for nitrogen gas is always going to have units of volume squared times the pressure divided by mole squared because that means to when I divided by moles per liter twice, it needs to result in units of pressure. So I could give it to you in units of liter squared per mole squared times bar or times atmosphere or Pascal or any other pressure unit. The B constant, that's the molecular volume. So it has the same units as molar volume and it tells us how large each of the individual molecules is or how large one mole of those molecules are. And that value for nitrogen has been measured to be 0.0387 liters per mole. So a mole of nitrogen molecules, not when you fill up a balloon, but if you pack them next to each other taking up all available space, they would take up 0.0387 liters for a mole of those molecules. So now that's enough information for us to calculate the pressure of this gas. And we can do that two ways. First of all, let's calculate the pressure if it were an ideal gas just to get our bearings. So we can already look at this and say if it's an ideal gas, we know at standard temperature and pressure, 273 kelvin and 1 atmosphere, the molar volume would be 22.4 liters for a mole. That's the number you may remember from doing ideal gas calculations multiple times. This volume, 2 liters per mole, that's roughly 10 times smaller, a little smaller than 10 times smaller than the gas should occupy at standard temperature and pressure. So we should expect that this pressure, both the Van der Waals and the ideal gas pressure, are going to turn out to be relatively high pressures, about 10 times higher than standard pressure. But of course we don't have to just guess. We can say the ideal gas law tells us pressure is nRT over V or RT over V bar. So we might need to stop and think about what value of R that we need to use, what value of the gas constant to use. So since my volume is going to be in units of liters per mole, I'm going to want moles in the gas constant. I'm going to want a gas constant that involves liters. And since I've seen that the Van der Waals problem is going to involve pressures of bar, let's use the value of R that's in units of liter bar per mole kelvin. So 0.08314 liter bars per mole kelvin is the value of the gas constant in those units. If I multiply that by 273 kelvin, divide by the molar volume, I find, let's double check that the units work correctly, but liters over moles on top, liters over moles on bottom cancel, kelvin in the numerator and the denominator up top also both cancel. So I'm left with units of just bar and we find that if this nitrogen was behaving as an ideal gas, its pressure would be 11.3 bar if I compress it down to a molar volume of only 2 liters per mole. The more interesting question is what is the Van der Waals pressure? If the gas behaves not as an ideal gas, but as a Van der Waals gas, what's its pressure? And we certainly expect the Van der Waals prediction to be more accurate than the ideal gas prediction because nitrogen does in fact have some finite volume, does in fact have some attraction between the molecules. So here we need to use the Van der Waals equation of state. So we just plug these various numbers into that expression. Again if we use 0.08314 for the gas constant in units of liter bar per mole kelvin, 273 kelvin, instead of dividing by 2 liters per mole, we're dividing by 2 minus the molecular volume, 0.0387 liters per mole. So already we can see roughly what size correction the finite molecular volume is having rather than dividing by 2, we're dividing by 1.96 or so. So there's a couple percent difference in the volume occupied by the molecules. And then for this last term we're subtracting 1.37 liters squared bar per mole squared divided by the molar volume itself squared. So if I take that calculation one step at a time, this first term that looks almost like the ideal gas expression, RT over V, but it's a volume that's been corrected by this finite volume. If we do just that portion, we obtain 11.57 bar, and then the second term what we subtract from that is about a third of a bar, a third of an atmosphere. So altogether, if I subtract those from one another, I get a value of 11.2 after rounding off to three sig figs, which is not very different than the value I started with. So there's a couple interesting things to point out here. First, the finite molecular volume, as we expected, had some effect to increase the pressure. That was partially canceled by the interactions between the molecules themselves decreasing the pressure. Those two contributions didn't exactly cancel each other, they somewhat canceled each other, and the net result is the Van der Waals gas has a pressure of 11.2 bar, about 1% lower than the ideal gas law predicted. And that's not too surprising. We wouldn't have felt too badly before we learned about the Van der Waals equation of state going ahead and using the ideal gas law to predict the pressure of a gas. Once we get up to pressures of about 10 or 11 atmospheres or so, we might begin to be worried that the ideal gas law is not giving us a good result anymore. We're not yet any more under conditions where the ideal gas law can be expected to be valid. And we see that a more accurate estimate of the pressure indeed does give an answer that's different than the ideal gas law. But it's only beginning to differ by about 1% or so at this point. So this is one example of doing a calculation with the ideal gas equation of state. And things are relatively simple. If I give you the molar volume and I give you the temperature and I ask you to calculate the pressure, it gets a little more complicated on the other hand if what we know is the temperature and pressure and what we're interested in and solving for is the molar volume. So that'll take a separate video lecture for us to explain how to work with that sort of problem.