 Okay, so we have a model for adsorption. This lattice model where molecules can either be adsorbed in one of these lattice sites on the surface or up in the gas phase. We've written down a partition function for what the partition function looks like when n of the molecules are stuck to the surface with an energy of epsilon for each one of them out of the total m different surface sites in total. So if we want to understand how many molecules are adsorbed to the surface or how that depends on the experimental parameters like pressure and temperature the easiest way to proceed would be to say since we're going to have an equilibrium based on what we know about equilibrium we can say the chemical potential of the molecules in the gas phase must be equal to the chemical potential of the molecules adsorbed to the surface when the system is at equilibrium. We can understand what each one of these is. In the gas phase we know the chemical potential of a gas standard state chemical potential plus kT log of the pressure of that gas over the standard pressure the pressure at which this is the chemical potential. In our case mainly to make things simpler qualitatively simpler we're going to assume we're working under conditions where the standard state chemical potential is zero so we can ignore that term. Basically we've chosen a zero of energy such that the chemical potential is zero for the gas in a standard state. For the adsorb species we can also write down a chemical potential. That's going to turn out to involve a little bit more work. The chemical potential of anything is derivative of one of the energies with respect to the number of molecules. Remember we're talking about the number of molecules adsorbed on the surfaces capital N so we want to take the derivative of an energy maybe energy or enthalpy or Gibbs free energy or in this case Helmholtz free energy with respect to number of molecules. We can take the derivative with any energy we want as long as we're careful to hold the right things constant. A, the Helmholtz free energy is going to be easiest to write from our partition function so I'll take DADN we just have to make sure and do that. Normally we would say DADN at constant temperature and volume if we were talking about a three-dimensional system but in this case since the system we're talking about is the molecules stuck to this surface, this two-dimensional surface the two-dimensional surface doesn't have a volume it has an area and the way we're characterizing the area of the surface is in the total number of sites that the surface has. If I double the surface area of this surface to which the molecules are absorbing I'm doubling the total number of sites to which the molecules can absorb so I'm going to want to do this derivative at constant T and constant M this is the equivalent of the volume for the system that we're talking about. So like I said that's going to take a little bit of work to figure out what DADN is equal to so let's start working on that. If we know a partition function we can calculate a Helmholtz free energy it's minus kT log Q since we know what Q is I can write the Helmholtz free energy as minus kT times the log of this term this partition function I've got various terms being multiplied and divided so I'll break apart that product of terms the log of this product is going to be the sum of a bunch of logarithms so I've got log of M factorial in the denominator so with a negative sign I've got log of N factorial and I've got a M minus N factorial also in the denominator log of M minus N factorial and lastly I've got log of this E to the minus N epsilon over kT so log of E to something is just the exponent and epsilon over kT alright so that's what I've got for the Helmholtz free energy seeing a log of something factorial makes you think hopefully that we should be using Sterling's approximation and that's what we'll do next log of M factorial using Sterling's approximation is M log M minus M log of N factorial is N log N minus N the negative sign turns that into negative N log N minus a negative N likewise minus log of M minus N factorial is minus M minus N log M minus N minus a negative M minus N and I still have this minus N epsilon over kT alright so now that's simple enough that it's expanded enough that we can start to simplify it a little bit looks like I have a minus M term here and a positive M term there that are going to cancel I've got a positive N and a negative N they're going to cancel and that's all I've got that's going to cancel for the moment so I'll just leave that that's my expression for the Helmholtz free energy can't simplify that too much more Helmholtz free energy is not really what I was after in order to do this equilibrium problem I need to take the derivative of the Helmholtz energy with respect to N so let's go ahead and do that if I take DADN at constant T because there's some temperatures that show up here and at constant M which is our equivalent of volume that's going to be kT so the M terms, M's being held constant so these derivative of this term goes away because M is constant derivative of minus N log N I've got using the product rule minus log N if I take the derivative of this N and if I take the derivative of the log term minus N leaving that one alone derivative of log is 1 over N similarly I can take the derivative of this term minus M minus N log M minus N with respect to N so taking the derivative of the first term minus minus N becomes positive 1 times log M minus N I'm going to leave the M minus N alone so I've got minus M minus N derivative of log M minus N is 1 over M minus N but chain rule with the negative sign in front of that N means I need to bring out another negative sign and lastly derivative of minus N epsilon over kT with respect to N gives me minus epsilon over kT and again we've got some cancellation that's going to happen N over N is 1 with a negative sign M minus N over M minus N is 1 with a positive sign so this entire term cancels this entire term so what I've got now is minus kT times so I've got two different log terms negative log N and positive log M minus N so I'll combine those the sum of these two logs or in this case the difference of these two logs becomes the log of a quotient so on top with the positive sign I've got M minus N and then the bottom with the negative sign because of this negative sign I've got the N so I've combined those two log terms to look like this and I'll just leave the minus epsilon over kT alone alright one more simplification to make that's you might think we can distribute this kT inside the brackets here we certainly could do that and it might make it look a little bit more simple but I'm going to save this kT because I know it's going to end up cancelling that kT the one other simplification that we can make is if we take a look at these variables M and N those are both extensive quantities in our problem M is the total size of the surface the number of lattice sites on this surface N is the total amount of molecules that are stuck onto that surface so what's more useful experimentally is not the total number of molecules on the surface but what fraction of the molecules on the surface what fraction of the sites on the surface are occupied by molecules so if a total of N molecules occupy M sites that's the fraction of sites that are occupied on the surface so we'll call this the surface coverage that's the variable we're going to often prefer to work with if every lattice site is occupied by a molecule then the surface coverage would be one or a hundred percent if there's no molecules on the surface N is zero so that's zero percent so somewhere between zero percent and a hundred percent coverage is this ratio so since N over M is a useful thing to calculate let's rewrite this fraction M minus N over N let's take each one of the terms in that fraction and divide it by M so on the top I've divided by M on the bottom I've divided by M and that's all I've done in this rewriting and then now that I've got N over M showing up I can rename that to be theta my surface coverage M over M is of course just one so this fraction becomes one minus theta on the top theta on the bottom so that's what I've got for DADN which is the chemical potential of the adsorbed species so now I can go back to after all this work on the side I can go back to this statement that the chemical potential of the gas this term involving the pressure must be at equilibrium to the chemical potential of the adsorbed species the term that I've just obtained over here so that tells us chemical potential of gas KT log P over P naught must be equal to this term minus KT log one minus theta over theta minus epsilon over KT alright so far so good as promised a KT over on this side I'll cancel a KT over on this side and now I'll bring this negative sign in and distribute it inside these brackets and I think this will probably be the last writing of this equation I'll get log of P over P naught on the left negative log instead of writing negative log I'll flip the argument of the logarithm so it's positive log of the inverse of this fraction theta over one minus theta and the negative sign applied to this epsilon over KT term turns it into a positive epsilon over KT alright so that is probably a good place to stop for the moment what we've obtained at this point is a useful equation this tells us a relationship between the pressure this gas of molecules that are not adsorbs onto the surface has some pressure we're doing this experiment at some temperature the molecules when they bind have some binding energy epsilon and a certain fraction of the surface will be covered by molecules so depending on which variable we want to solve we can understand how the surface coverage depends on the pressure and the temperature or we can understand how the pressure depends on the surface coverage and the temperature here's a nice thermodynamic relationship between the pressure the temperature and what fraction of the surface is occupied by molecules so we what to do with that will explore what that equation means that tells us about this adsorption process a little more in the next video