Let's take a look now at another equation of state. The virial equation of state. So, what is a virial expansion that plays a role in these equation of state? The virial expansion expresses the compressibility. And you'll remember that the compressibility is P V bar, divided by RT. As an infinite series expansion in either the density and remember that the density is the inverse of the molar volume. So I could either write this as an expansion in density or an expansion in inverse molar volume, or finally we can use the pressure. So expressed in inverse molar volume the expansion is as shown here. It's 1 plus B2VT over V bar. Plus B3VT over V bar squared and a term over V bar cubed, and so on. Infinite series. If we express, instead, in the pressure, 1 plus B2PT times the first power of the pressure, plus B3PT times pressure squared, pressure cubed, and so on, an infinite series expansion. The constant terms that are appearing, the B terms are known as Virial coefficients. They are indexed by two subscripts, the first subscript tells us at what term in the expansion do they appear, the second, the third, the fourth. The second subscript is simply are the volume term or a pressure term. And so, for instance, we would call B2V a second virial coefficient. They are functions of temperature, and that's emphasized by including temperature in parentheses afterwards. So if and when we determine them, we'll do them at certain temperatures. And we'll get to that momentarily. Do, take a moment to look at the equations and think about sort of the sensibility at certain limits, and in particular, let's consider the limit as pressure is going to 0. So as the pressure is going to 0, and as my molar volume as a result of there being essentially no pressure is going to infinity. And if the molar volume is going to infinity, then, of course, the density is going to zero because they're inversely related. Well, what would happen to these equations? If pressure is going to zero, pressure cubed goes to zero really quickly. All the terms are going to zero, including even this one. Although most slowly, this one. And ultimately, I'll get compressibility equal to one. Which is what I expect. I should have ideal gas behavior as my pressure goes to zero. Similarly, if molar volume is going to infinity, I keep dividing things by infinity or powers of infinity, they all go away, and I'm left with compressibility equals one. So the virial equation of state has the correct form to show ideal gas behavior at that limit of infinitely low pressure. Well, where do these virial coefficients come from? So, let me show you a plot of compressibility against pressure. Very near no pressure at all, so these are [COUGH] very low pressures, somewhere between zero and a tenth of a bar. [COUGH] And remember that a tenth of a bar is about a tenth of an atmosphere. And what we see for these isotherms. And this is for ammonia gas. Is zero degrees C, 100 degrees C, 200 degrees C. If I extrapolate, they're all going to ideal gas behavior at the lowest pressure. More over they're all linear. And let me think about what happens at very low pressure. So as the pressure is going to 0, as numbers go to 0 if you square or cube or take their fourth power they go to 0 much faster obviously. So the only term that will survive that contains P within it. In the virial expansion, it will be Z is equal to 1 plus B2P, function of temperature, divided, sorry, multiplied times P. So that is a linear equation, in P. And sure enough, these are lines, as a function of P. And so the slope, of this experimental measurement, gives you, B2P. And each of these isotherms, has a different slope. And that illustrates to you that yes the coefficient does depend on temperature. But, by doing a series of measurements, you can infact determine B2PT, for a given gas at given tempuratures. You can also manipulate the two virial expansions. It takes about maybe 15 lines of equations to do this. It's not awful, but it's a little much to put on a slide, so I'll just give you the result. You can show that B2V(T), so that's the coefficient that appears in the molar volume, virial expansion, Is equal to something quite simple. It's just R times T times B2PT. So by doing this experiment, you simultaneously get both of these two virial coefficients, second virial coefficients. Let's just look at, how the various terms in the virial expansion. What sort of orders of magnitude they take on,at, under different sets of conditions. So here are some data for argon at room tempature, 298 Kelvin. I'll show you for one bar, so that's atmospheric pressure, 10 bar, and 100 bar. And these all are the B 2V component of that stage, as in units of volume per mole. 'Cuz remember compressibility is unitless. 1 is certainly unitless. So, if we're going to divide by Vbar, B2VT has to have the same units as Vbar, in order to be unitless. What are the units on Vbar? Volume per mole. In any case, at one bar, argon, it really behaves pretty ideally, so, the one comes in. And the second term in the virial expansion is 0 point 00064. So the deviation from the ideal behavior is 6 times 10 to the minus 4th. Pretty small. And the third and fourth terms don't contribute at all. On the other hand as I go up to 10 bars, so I increase the pressure by a factor of 10. And it turns out each of the contribute, well, okay, the one contribution that wasn't zero before, also goes up by about an order of magnitude. So it went from 6.4 times 10 to the minus 4th, to 6.5 times 10 to the minus 3rd. And the 3rd virial term begins to creep up a bit. It's actually non-negligible. There's a little bit a contribution from all the other terms, but it's still pretty far out past the decimal place. If I increase the pressure yet again by an order of magnitude, now the second virial term is up by about a factor of ten. The third virial term is up by about a factor of ten. And there's more a contribution from the final terms. So, you can sort of see that the, the length of the expansion you need to consider increases as you move further and further away from ideal gas length conditions. So, okay it's an equation of state. It's got some constants appearing in it. Can we gain any physical insight from these constants? That is what does B2V, what does it mean? Well it actually is possible to assign a pretty straight forward and simple meaning to B2v. So let me just write again the low pressure virial expansion in pressure. It says that the compressibility is 1 plus B2P, times pressure. And the other terms drop out because the pressure is low enough they don't contribute. Well, when we multiply both sides by RT over P. So that'll leave only the molar volume on the left-hand side, one times RT over P gives RT over P. And the Ps will cancel out over here. I get RT times B2P. But, remember, RT over P. What is that? That's Vbar for an ideal gas. RTB2P, what is that? That's B2V. We, showed that on a prior slide, or I told you that you could actually derive that relationship. So, I'll just replace those in there. It says Vbar for my real gas is Vbar for an ideal gas plus B2V. And all I have to do is rearrange that ever so slightly. It says B2V is Vbar for the real gas minus Vbar for an ideal gas. That is, B2V is the difference between the observed molar volume and the ideal gas molar volume for a given set of pressure and temperature conditions. And that's a pretty intuitive phenomenon, and intuitive quantity, if you will. So, let me take a moment and, and see if that's that is intuitive for you. I'll let you answer a question that will address that in a little bit more detail. Well, now that we have this intuitive feeling for what B2V means. Let's actually take a look at some B2V data as a function of temperature for a variety of gases. So I've got nitrogen gas, methane, carbon dioxide on this slide. And I am plotting b2v versus temperature. And remember that the physical interpretation of b2v is it's the deviation of the real volume from the ideal gas volume. So here, this line tracking across the data at, at zero. That would be if the observed volume was indeed the ideal gas volume. Right? And, in that case, B2V would be zero. And so you see, as for helium for example. As we start very, very cold, the B2V is a negative number, so the actual molar volume is smaller than the ideal gas volume. The gas molecules are clumping together would be a way to think about that. They're occupying less volume than an ideal gas would. And then as I raise the temperature, I go up, up, up, I pass through the ideal gas volume, and I go a bit above, and then I kind of level out after that. And then helium hovers pretty close to the ideal gas volume at higher temperatures. Nitrogen, methane, carbon dioxide, they all show the same behavior. That at lower temperature, they are occupying less volume then an ideal gas. They are sticking together for lack of a better way to think about it. As you warm them, they all increase, and they all do ultimately pass through on this scale. We don't see Co2 pass through here, but clearly they're all approaching that ideal line ultimately. And so again what we're seeing is that attractive forces are dominating in the low temperature region. That all of these gases are occupying less molar volume, and we're seeing that expressed as B2V. And ultimately we get repulsive at high temperature, they all pass through, once you're on this side of the line. Your molar volume, because you've got a positive V to V, is larger than the ideal gas volume. So they must be repelling each other and be forced to occupy more space. All right, well, we've invoked, a physical phenomenon, or really two. We've invoked attraction and repulsion to explain changes in molar volume. So it's really appropriate at this point that we spend a little time thinking about the physics behind attraction and repulsion. And so in the next video, we will take a look at molecular interactions.