Hi, well I'm Professor Steven Nashuban and I wanted to tell you a little bit about changes in the Gibbs energy and its relationships with the Clausius inequality and for this little chat we're going to be assuming that the temperature and pressure are fixed. So first idea that I want to just get a cross here is that by definition the Gibbs energy is equal to the enthalpy minus temperature times entropy, that's fine. Then what that means is that if I have a system that undergoes a change, let's say from x to y or y to x, then the change in the Gibbs energy will be equal to the change in the enthalpy and since temperature is going to be held constant but that I can write the change in that second term as the temperature which factors out times the change in the entropy of the system and what we've previously seen is that delta g for at least for a reversible process at constant temperature and pressure is equal to the non-pv work so that would be equals w, w prime there. Let's just take a moment to think about how g changes in which one of these two states x or y has the higher Gibbs energy. Well I can tell by looking at this that if I were to, let's say we've reversibly, move some of these molecules off to the right as I sort of crowded them up on the right here, that's going to take some work to do that because it's clear that this is the thermodynamically stable state. So what that means is going from here to here obviously w prime is positive and of course what that means is that delta g is positive and going from y to x which must mean that this is the high Gibbs energy state and that would be the low g state. So going from low Gibbs energy to high Gibbs energy. The second part of this that I want to argue for you is that the transition from a high Gibbs state to a low Gibbs state which would correspond to, well if we carried it out reversibly then I would say w prime is negative that is to say the system has the capacity of doing work on the surroundings or if we just back off from that a second it's, even if it didn't do any work it's clear that the transition from x to y would be spontaneous and since that's a high g state and that's a low g state then what we're really arguing is that spontaneity is associated with delta g being negative that is going from high g to low. I just want to argue that out through the clousiest inequality here. So clousiest says that for the system any change in the system has an entropy change and we know that that's greater than or equal to q that's the heat going in and out divided by temperature of the surrounding. I've added this equal sign because in the inequality holds for spontaneous irreversible processes the equality holds for reversible processes. So I've just done a couple things and getting over here I've said well that the temperature of the system is the same as the temperature of the surrounding, it's all isothermal so I can replace that temperature of the surrounding for the temperature. I've also said that since we're talking about an isobaric isothermal process then I can replace q with delta h and so I haven't really invoked reversibility only in the sense that there's mechanical equal to every which allows me to replace q with delta h. So I still have that inequality here and not just rearranged that t delta s being greater than or equal to delta h move that over to the all onto one side and so since we already know that delta g is delta h minus t delta s that condition translates into this inequality. So the idea here is that a spontaneous process is associated with negative delta g and a reversible non-spontaneous process is associated with delta g equals zero. Okay.