 Let's start right now by recalling what we know about reversible versus irreversible processes and in particular when we were first thinking about reversible and irreversible processes we concluded that the heat involved in any reversible process is always going to be larger than the heat for an irreversible process. So that it turns out is going to be key to understanding whether processes will be spontaneous or not because we know that irreversible processes are spontaneous and reversible processes are equilibrium that the nice edge between spontaneous and not spontaneous. The other thing we know is that the Clausius theorem tells us something about reversible heat. Reversible heat divided by temperature is equal to ds so if reversible heat is always bigger than irreversible heat dq over t for reversible process is always going to be larger than dq over t over that same temperature for an irreversible process. So that means that in general either ds is equal to dq over t if the process is reversible or it's greater than dq over t ds is greater than dq over t if we have an irreversible process so I can say ds is always either bigger than or equal to dq over t where the greater than sign would apply to any spontaneous process in other words any irreversible process and the equal sign would apply to any equilibrium process. That may seem like a relatively minor change on either the irreversible heat inequality that we've seen before or the Clausius theorem it's relatively close to that we've just now got an inequality instead but notice what this statement says this is a statement about whether a process is going to be spontaneous or not if we know whether ds is larger than or smaller than dq over t we can tell whether the process is going to be spontaneous or not and each of these properties the entropy the heat the temperature those are all statements about the system itself or the heat flowing into the system so those are all statements about the system they don't involve the surrounding so we can measure properties of just what's happening in the system and make a conclusion about whether a process is going to be spontaneous or in equilibrium or non spontaneous so that's significant but it's not all that convenient because now we're back to using a path function heat is a path function we'd much rather have this defined in terms of state functions so in an attempt to get rid of the path function dq we can go back to the first law energy is heat plus work that'll let us rewrite the heat in terms of some other functions and in order to get rid of the dw here let's make an additional assumption let's assume we're doing something at constant volume so that there's no pv work if there's no change in volume no pv work then the then work is going to be equal to 0 and under those conditions du will be equal to dq so then I can rewrite this expression ds greater than or equal to dq over t if we're at constant volume and dq is the same thing as du ds will be greater than or equal to du over t and now I've got this expression written purely in terms of state variables entropy internal energy temperature or all state variables the price I've had to pay is I'm restricting our attention now only to processes at constant volume so if I rearrange this equation a little bit bring the t out of the denominator so I'll say tds is bigger than or equal to du by moving the t over to the left hand side I can subtract a du tds minus du is bigger than or equal to 0 or actually the form that will usually use this if I reverse these if I take du minus tds instead of the other way around I've changed the sign of the thing on the left hand side so that's going to be instead of positive number it's going to be a negative number so this is the most significant expression we've got so far on the board I can make a prediction about whether a process will be spontaneous or not spontaneous spontaneous in equilibrium or not spontaneous based on whether the sign of this quantity du minus tds is less than zero for spontaneous process equal to zero for an equilibrium process or if that quantity du minus tds is positive then it won't happen it'll be a non spontaneous process so to give you an idea of how this works in practice let's take an example of a process where we can predict without doing any math whether the process is going to be spontaneous or not and we'll see what this equation says about the same process what and the process we'll consider is the melting of ice solid water turning into liquid water and we know whether that's spontaneous or not if we know one additional piece of information solid ice will melt and form liquid water if the temperature is above the boiling point it will stay frozen it will not be spontaneous to melt if the temperature is below the boiling point so we can ask that question so we can ask is this spontaneous or not at a variety of different temperatures we can do it at negative 10 Celsius we'll do it also at zero degrees Celsius and we'll do it also at positive 10 degrees Celsius so we ought to expect in fact I think we'll do these in a different order we ought to expect it to be spontaneous above the melting point in equilibrium at the melting point and non spontaneous below the melting point the other information I have to give you an order for us to solve this problem is the values of du and ds so the change for a mole of water when I melt a mole of water that entropy of fusion for water is a value we can look up that's 22.1 joules per mole kelvin likewise there's a change in the energy when water melts and becomes liquid that change in the energy is 6.04 kilojoules per mole or to get the units the same 6040 joules per mole so there's all the information now we can begin answering the question let's start actually with this third case where we know it's going to be spontaneous if I want to know whether water is going to spontaneously melt at this temperature all I have to ask is what is the sign of the change in the energy minus t times the change in the entropy I've got this written here in terms of du's differentials but if I as long as the temperature is remaining constant while that ice is melting I can turn those ds into deltas essentially I just integrated du to become delta u ds to become delta s so at 10 degrees C delta u minus t delta s if I've got so I've given you the entropy of fusion and the energy of fusion per mole the molar entropy the molar energy of fusion let's imagine we have one mole of water and we want to know whether it will melt or not the energy change will be the molar energy change multiplied by the number of moles so 6040 joules per mole times one mole that would be 6040 joules I want to subtract from that the temperature so in this case above the melting point 10 degrees Celsius or 283 kelvin so I'm going to subtract 283 kelvin times the entropy of this process the entropy of melting the entropy of fusion is 22.1 joules per mole kelvin times one mole so 22.1 joules per kelvin kelvin cancels kelvin joules and joules are the same as they should be so this process 6040 joules this product is 6250 joules so I've written out that intermediate step out to show you that the t delta s product that's a larger number than the delta u I'm subtracting this large number from a smaller number the result is negative negative 210 joules so what have we learned from that if delta u minus t delta s is a negative number that means the process is spontaneous as we expected we know that water melts above 273 kelvin above its boiling it's melting point of zero degrees Celsius so we expected that that should turn out to be a spontaneous process if we consider one of the different cases let's do the one below the melting point very similar calculation at negative 10 Celsius or 263 kelvin I need the energy of fusion minus the temperature of 263 times the entropy of fusion now since the temperature has changed the second term has changed also 263 times 22.1 that number is 5810 so now I have a larger number 6000 odd joules minus a smaller number 5800 and some joules not kelvin should have been more explicit with my units kelvin's cancel and I'm left with just joules so that difference is positive 200 and some joules so again what does that mean the fact that energy minus t delta s is a positive number if it were negative it would be spontaneous process if it were equal to zero that's an equilibrium it's neither of those cases so that so what this equation says is that process won't happen it's a non-spontaneous process so what we've learned although we knew it already is for case a at negative 10 degrees Celsius below the melting point water does not spontaneously melt and we can predict that by knowing the energy and the entropy of fusion and just to complete the third case which was case B the details of which won't surprise you at this point so if I take 60 40 joules minus and here in order to get them to cancel I need to do this at exactly 0 degrees Celsius so 273.15 kelvin that product 273.15 kelvin multiplying 22.1 joules that'll give me 60 40 so if I ask whether ice will melt and form liquid water at a temperature of zero degrees Celsius at its melting point that difference delta u minus t delta s that comes out to exactly zero which means the process is in equilibrium so that means solid water liquid water are in equilibrium with each other I can have both of them existing converting back and forth between each other at this temperature of zero kelvin so these numerical results are not a surprise they match what we already know about water the significant result is now that we have a way of predicting even for cases where we don't know the answer already we have quantities that we can evaluate just for the system I didn't have to ask questions about the surroundings what kind of container the water was in whether it was isolated or not all I had to know was that I was doing that process at constant volume and I use the energy temperature and entropy of that process in order to make a prediction about whether the process is spontaneous or not so clearly that's a useful thing to be able to do so this quantity delta u minus t delta s is going to turn out to be fairly important so our next step is going to be to explore that quantity a little more