 Hi, well, I'm Stephen Nashiba, and I want to tell you a little bit about how to calculate numerically the path dependence of work done against the concentration gradient. And it may well turn out to be that there is no path dependence, that it's actually path independent. That's the whole goal of the exercise is to see whether that's true. So I have a, I'm imagining a physical situation where I have a solute and it's a molecule or a mole of molecules or find themselves at this low concentration side of this box and they're going to be moved up into this high concentration side of the box. So we know that that's upgrading and that's gonna, that's gonna cost some non-PV work. W prime will be, will be positive. But we're trying to decide now what path we should take them along. I mean, for example, one could imagine that I'm just gonna mark the path with the distance parameter called epsilon, okay? Where epsilon will be zero here and one when we arrive at our destination, which is the second concentration. So on a graph here of epsilon looking at from zero to one, this would be, I could just say, well, it's just, you know, along that path, the concentration just keeps going up, you know, as a straight line. That would be given by this algebraic formula. The concentration as a function of epsilon would be C1 plus that difference times epsilon and you can just kind of see how that works. If epsilon were zero, meaning worth the start, then the concentration is C1 and when epsilon is one, then the C1s go away and I just get C2. So it satisfies the boundary conditions and it's just like, just like that. Now, one could imagine other paths. For example, I could square or cube that epsilon and we'd have different paths, but still the same endpoints. So, given that, how do we calculate the accumulated work that, that, that happens along the path? Well, we've previously established that the change in Gibbs energy is actually, if done reversibly, would be the, the, the little bit of work that would be done along every step. And we've also established that the change in that little bit of, that increase in Gibbs energy is given by nRT over C times DC. That is, the nR times the temperature divided by the concentration and multiplied by the increase in concentration as, as we go along that path. Now, if I want to think about this in terms of a force to a distance, then what I'm going to do here is instead of saying there's an increase in the concentration, I'm going to say, well, there's a gradient of concentration, DC over epsilon. So that's the concentration gradient, still multiplied by that factor times the distance. So this looks like a force times a distance. And the force again is just proportional to the concentration gradient multiplied by that factor, okay? So, how do you accumulate the work done as you, as you, uh, uh, uh, mark through a force? Well, you just integrate the little bit of work that's done every step of the way. So I've just indicated that with this integral. That is to say, integrating all the DW primes gives me a total work done. And if my independent variable is epsilon, and I'm going from epsilon, going from zero out to one, according to that, uh, idea, then I just integrate the integrand here, the force, which is this quantity right here, you know, through the, uh, through that distance, which, you know, a Python could be done with something like trap Z. The only, uh, the only sort of extra analytical work that one has to do here is, of course, you have to construct that integrand. Well, that's nRT divided by C of epsilon. That's easy. You just lay out epsilon going from zero to one, and we got that. Here you'd have to do an actual derivative. You'd have to take the, you know, how the concentration changes with respect to epsilon. In other words, you'd have to take the derivative of that quantity, which in this case is pretty simple. dC d epsilon, there's no epsilon there, looks just like C2 minus C1. So that's what would go right there. But of course, if this, if our path were different, then it would be a different derivative. And then what, when you're done, you plug that into the integrand and use trap Z again to integrate that. Okay.