 All right, so we can work now if we choose either in terms of molecules or in terms of moles when we work an equilibrium problem because we can now write the equilibrium constant either in terms of molecules or in terms of moles and we know how to convert back and forth between them. But notice that all the equilibrium problems we've discussed so far have involved gas phase species, H2, Br2, HBr, bromine atoms, those are all gas phase species. If we're talking about an equilibrium problem involving gases, working in terms of moles is actually not the most convenient quantity to work in terms of. If I told you there was some reaction involving the nitrogen and oxygen molecules in the air in this room or the room where you are, you're probably more likely to be able to tell me what the partial pressures of those gases are than to tell me how many moles of those gases are present in the container you're studying or in the room that you're in. So we'd really rather work in terms of pressures rather than in terms of moles, but luckily if the gases are ideal, the number of moles and the pressure are related to each other. So let's go back to our expression for the equilibrium condition written in terms of Kn. Yeah, we'll use K big n. Equilibrium condition tells us there's this relationship between the number of molecules of each species raised to the stoichiometric coefficients. Combine those in the right way. They should equal the equilibrium constant written in terms of molecules. But that relationship between molecules and pressures is the ideal gas law. The partial pressure of any gas is equal to, times the volume is equal to the number of molecules of that gas times Boltzmann's constant times temperature. So if I rewrite this product of molecules raised to stoichiometric coefficients as Pv over Kt, where P is the partial pressure of that particular gas, then that's exactly the same case of big n. If I pull out of that sum, or out of that product, the terms that don't depend on the species, so V over Kt, I can pull V over Kt out of that product. All that's left inside is pressures raised to some stoichiometric coefficient. I pulled V over Kt out some number of times, new sub i times for each species, all together, new sub i for all the species added up. Or the way we prefer to write that, the sum of all those stoichiometric coefficients is just the total change of the stoichiometric coefficient during the process. So I've got some number of factors of V over Kt raised to this change in stoichiometry, multiplying this quantity which is a product of partial pressures raised to stoichiometric coefficients. This term looks a lot like an equilibrium constant. We know how to calculate equilibrium constant as molecules raised to stoichiometric coefficients, or this one would be moles raised to stoichiometric coefficients. This one is pressures raised to stoichiometric coefficients. This thing we'll call Kp, equilibrium constant, written in terms of pressures. The result we've got is that the molecule-based equilibrium constant is equal to V over Kt times the pressure-based equilibrium constant, or, sorry, that V over Kt is raised to this delta-new power. Because usually, if we've calculated the value of Kn, we'd like to get a value of Kp. I'll flip that equation around, and I'll write instead Kp is equal to Kn, the K big n. I'll need to multiply by Kt over V. I've flipped this upside down, raised to the delta-new. So if I have a value for Kn, that maybe I've gotten by obtaining partition functions or looking it up, I can convert that into a Kp by multiplying by some appropriate number of ratios of Kt divided by V. So that tells us now how to work equilibrium problems either in terms of molecules, if we like, in terms of moles, if we would rather, or perhaps most convenient in terms of pressures. So that'll make more sense with an example, so that's what we'll do next.