 Hi, well I'm Stephen Asherba, and I want to help you out a little bit with calculating the work associated with moving some molecules up gradient in terms of concentration. In other words, the work against a concentration gradient. So the physical situation here is, I just imagine I've got a box that's got a bunch of molecules. They saw you, molecules. And there's low concentration here, high concentration here, and the way we're going to do this is to calculate the Gibbs energy change associated with taking those molecules and kind of compressing a given number of molecules into a smaller and smaller space until we get up to that new concentration. So I've just imagined that I've gotten a certain number of molecules there at C1 and the very first step is going to be taking those molecules, putting them into a smaller space. Now, because I know that the Gibbs energy is H minus TS, and I'm saying that this is an isothermal process, and we're also going to say that the enthalpy, that it's an ideal solution. And what that means is that the change in Gibbs energy going from here to here would include a change in the enthalpy because it's ideal solutions. We're going to say that that goes away. And because it was an isothermal process, I just factor out that T. So I have the change in Gibbs energy and that little step is equal to minus the temperature times the change in the entropy. The next thing that we're going to do is going to use the differential equation of state for a change in entropy. And I've written it here as we know it. That's a Cv over T times the change in temperature. But of course, that's going to go away because it's isothermal. Second term in that differential equation of state should be alpha over kappa times dv over v. But we're assuming ideal solutes and we're sort of using the analogy of an ideal gas. So that would be alpha over kappa just became nr over v. So I've written it that way. Now another step in here is that since we're sort of thinking about these solutes, so it's kind of more convenient to think of that in terms of concentrations. So if I have a concentration here and it changes a little bit there, turns out since the concentration is just moles per volume, then the change, the relative change in volume dv over v turns out to be equal to the negative of the relative change in concentrations. That is to say, since volume here is getting smaller, the concentrations will get bigger in a relative sense. So if we put that all together, dg is, that was taken away because it was an ideal solute that just leaves minus t. That's that times the change in entropy which turned into that quantity there times nr. So that's that part there. So I have to change and give energy now and going from that tiny little step. And the minuses cancel out there and so I have nrt at the top and concentration at the bottom. So if I want to calculate the change in Gibbs energy resulting from that tiny little concentration going in that first step, I just take nrt divided by that concentration multiplied by the change in concentration. And of course, if we do that many steps, then if I want to calculate the accumulated change in the Gibbs energy, I just have to, you know, I'm going to integrate that and nrt will come out. I have that dg integrates out to a delta g. That is to say from here to here and the 1 over cdc integrates out to a log. So we end up with nrt log c2 over c1. So that's quite important there. The change in Gibbs energy just depends on nrt times the ratio of, the log of the ratio of the concentrations. One more note here. We've established previously that as long as that sequence of concentrations takes place reversibly and isothermally and isobarically, then we can say that that change in the Gibbs energy is the work that you would have to do in order to effect that concentrating process. So I just said that that's equal to w prime, the non-PV work. So this is basically our answer. If I want to get the work done by associated with, let's say, moving a molecule from this concentration to that concentration, I just say it's nrt times the log of c2 over c1 and that's the work that's done. Thanks.