 All right, so we're on our way to understanding the BET model, how that predicts multi-layer adsorption onto surfaces. We can predict how much adsorption there is of molecules on the first layer or into the nth layer in terms of the amount of the surfaces uncovered, this data knot quantity. We'd really rather know how the volume of a gas adsorbs depend on the pressure in the gas above the surface so that we can do some experimental studies of adsorption. So that's the next thing to consider is what do these surface coverage levels in molecules in different layers tell us about how much volume of a gas would be absorbed. So let's take just a few quick examples to get our bearings. So suppose I tell you that theta sub 1 equal to 1, 100% of the surface is covered at the first level, none is covered at the second or third or fourth or or zeroth level. So basically I've got one monolayer covering the surface, so that means the total amount of gas adsorbed is whatever it would have taken to cover a monolayer because that's what I've done. I've covered the surface by one monolayer. That would be for example if I get rid of these molecules and I cover the surface by exactly one monolayer. On the other hand if I say I've got a full bi-layer covered, I've got a surface, I've got layer number one adsorbed, I've got a full second layer absorbed, 100% of the surface is covered by two molecules. Clearly in this case the adsorbed volume would be twice what it would take to cover one monolayer. But let's do a slightly trickier case. What if half of the surface is covered with one molecule, half of the surface is covered with two molecules? So let's go ahead and put these molecules back, half my surface, I need an even number of molecules, half my surface is covered by only one molecule, half covered by two. Then from the picture you can see that it's one and a half monolayers worth of surface. One way to think about that is half the surface is covered by one molecule, half the surface is covered by twice as many molecules. So half times two gives me one, half times one gives me a half, all together that adds up to one and a half monolayers worth of coverage. So what we're building our way to understanding with these examples is the total amount of gas adsorbed onto the surface is the monolayer coverage multiplied by some number. And what that number is is each layer of coverage gets the fraction of the surface covered by that many molecules. Go ahead and multiply by that many molecules and add those numbers up. So all the way from no coverage up to infinite layers of coverage. So for example, one molecule times the fraction that's covered by one, two times the fraction that's covered by two molecules, three times the fraction of coverage that's covered by three molecules and so on. So this is a general equation to understand how the adsorbed volume depends on the amount that's adsorbed into each one of these layers. So that's now in a form we can do something with because we have a result for what the surface coverage in the nth layer is. So this, I can rewrite this as monolayer times the sum of n times theta n. Theta n itself is c times k to the n, p to the n, and theta not, the amount of uncovered surface, bare surface. So three of these terms in here depend on n. The other two that don't, c and theta not, I'll pull those out of the sum. And in fact, let's go ahead and write v adsorbed relative to vm. So I'll go ahead and pull vm over to the other side. So outside of the summation, I've got a c, theta not, inside the sum, I've got n, k to the n, p to the n. That sum we can do something with. Let me go ahead and write that just for short hand. I'll write that as c theta sum of nx to the n, where this x is k times p. This quantity k times p is raised to the nth power. That makes it look a little more like something we know how to manipulate with summation notation. So let's see what we're going to do to get to our next result. We're interested in trying to understand what this infinite sum is. The sum of 0x to the 0 plus 1x to the 1 plus 2x to the 2. So that's basically x plus 2x squared plus 3x cubed plus 4x to the fourth and so on. The fact that I've got an n out front as well as an n up in the exponent makes me think I should do something like take a derivative. So if I just had the sum of x to the n, that would be one thing. If I take the derivative of that sum, the derivative is going to pull down the exponent in each one of those terms. So that derivative itself is going to give me pull down the exponent to give me an n. The exponent decreases from n down to n minus 1. And that's what I have left. We can go ahead and include all of those terms. So that looks a little bit like the thing I'm interested in. It differs because the term I'm interested in has an x to the n. The one I've just been playing with down here has an x to the n minus 1. So instead of taking the derivative of this sum, what I actually want to do is multiply the whole thing by x. So if I multiply by x over here, multiply by x over here, then that will have the effect of x times x to the minus 1. That'll give me x to the n, which is what I'm after. All right. So what I've just figured out is this infinite sum that I'm interested in, sum of n x to the n, that's all equal to x times the derivative of the summations of just the x to the n's. 1 plus x plus x squared plus x cubed and so on without any coefficients in front. The reason that's useful, that transformation that I've just done is useful, is because this summation is a power series that we may be familiar with. 1 plus x plus x squared plus x cubed and so on. That is 1 over 1 minus x. So if you write down the Taylor series for 1 over 1 minus x, that is exactly what the Taylor series is. So what we have figured out now is, again, this summation that is the one that I'm actually interested in is equal to x times the derivative of this thing, which is equal to 1 over 1 minus x. So I can take, let's go ahead and take the derivative of 1 over 1 minus x. That's 1 minus x squared in the denominator with a negative sign. So 1 minus x to the negative 1 pulls down a negative 1 and I get 1 minus x quantity squared. I also need to use chain rule to take the derivative of the 1 minus x itself. So that brings in another negative sign. The x out front sticks around. So this net result of x times this derivative is x over 1 minus x quantity squared. All right. So all this work on the side has convinced us that the summation of nx to the n is equal to x over 1 minus x quantity squared. So I'll use that to rewrite this result. My adsorbed volume over the monolayer volume is equal to c theta naught times the sum. The sum is equal to x over 1 minus x quantity squared. I'll go ahead and write these x's now as kp's. So the x in the numerator looks like k times p. The x in the denominator looks like 1 minus kp, the 1 minus x in the denominator looks like 1 minus kp, and that gets squared. All right. So now I am almost there. I've got the adsorbed volume written as a function of pressure on the right hand side. That's good. But it's also a function of theta naught, which is something I don't know. So I still need to do a little more work to figure out what theta naught is equal to. So the key here is to remember that every one of these surface sites is going to be occupied either by zero molecules or one or two or three or four all the way up to infinity. So the sum of theta naught and theta 1 and theta 2 and theta 3 and so on has to add up to 1. So I can say theta naught is going to be 1 minus the sum of all these. I'll go ahead and put parentheses around that. So it's going to be 1 minus the sum of all the theta n's. This time the sum isn't over all the n's, just over n equals 1 up to infinity. So that's what theta naught is. The theta n's, I know what those are. Theta n is equal to this expression. So I can write 1 minus the sum of c, k to the n, p to the n, theta naught, which again if I pull out the c theta, all that's left inside is k to the n, p to the n. If I resurrect my notation here, that k, p to the quantity to the n is, I'll just write that as x to the n. That is the same summation, summation of x to the n is equal to 1 minus x if I sum from zero to infinity. So again, 1 plus x plus x squared plus x cubed and so on adds up to 1 over 1 minus x. Here I'm only starting from the n equals 1 term. So I don't have 1 plus x plus x squared plus x cubed. I just have x plus x squared plus x cubed plus x to the fourth and so on. So I can rewrite this infinite sum using this result, but I have to write 1 over 1 minus x. That includes an extra term, the n equals zero term. I have to remember that I didn't include that term and subtract it back out. So again, that minus 1 is because I'm not including the n equals zero term in this sum. So and I've forgotten the subscript zero, so that theta is a theta naught. So is this one. Let's rewrite this 1 minus c theta naught inside the parentheses here. If I put this all over one denominator, 1 minus x, 1 over 1 minus x, when I subtract 1 minus x over 1 minus x, it looks like that. This one and this one cancel and so I just have an x over 1 minus x. All this is equal to theta naught. So my goal was to solve for theta naught. I've got a theta naught on the right hand side as well, but I can isolate all these theta knots. If I do a little bit of algebra now, I'm going to multiply on both sides by 1 minus x to get rid of this fraction. So on the left side, I've got theta naught times 1 and theta naught times minus x. So I've multiplied on the left by 1 minus x. On the right, I have this 1 that gets multiplied by 1 minus x. The c theta times x, the denominator has gone away. C theta times x, that's got a theta naught in it. So I'm going to move this over to the left hand side. So I'll move that over here as a plus c theta times x. So all of that stuff is equal to 1 minus x. Rearranging on the left side, I've got theta times 1 minus x plus cx. So I'm going to bring that over to the right hand side that looks like 1 minus x plus cx that's now showing up in the denominator. And that's what I've obtained for my value of theta naught. So that's just doing a fair amount of algebra to figure out the value of theta naught. The reason I wanted theta naught is to plug it into this expression so I know how much volume of gas would adsorb under the surface. So if I take this result for theta naught and plug it into this expression, we're in the home stretch now. What I get is c times theta naught, which looks like when I write these x's now, I'll remember that x is equal to kp. So I'll write 1 minus kp on the top. In the denominator, I've got 1 minus kp plus c times kp. That's writing this theta naught. I need to multiply that by kp over 1 minus kp quantity squared. So this 1 minus kp cancels one of the two in the denominator down here and nothing else will cancel. So I'm ready to write a final looking result now. Total amount of volume of gas that's going to adsorb under the surface relative to the volume of a monolayer is going to be c times kp in the numerator. In the denominator, let's put this one first. So let me write 1 minus kp and also this longer term, 1 minus kp plus ckp. So we'll leave that expression there. We've obtained now exactly what we're looking for. I have an expression for how the volume of gas that will adsorb under the surface depends on the pressure. All I need to know to do that is what the pressure is, the equilibrium constant for that second third fourth layer adsorption reaction, as well as that BET constant c that tells us the ratio of the first layer adsorption constant relative to those higher level equilibrium constants. So there's an expression that we can use. What we'll do next is take a look at what that expression tells us and what it means about how gases actually do adsorb under surfaces and compare that to the Langmuir model.