 We are discussing desalination and told you that the osmotic pressure is p-p this is your system as I said the membrane units here the reverse osmotic units here do not you see water they will use brackish water here you get pure water that is very soft typically below 100 ppm Tds and this pressure we found was equal to RT log of let me call this is here this is water this is component 1 plus salt this component 2 this is component 1 so this is mu 1 liquid pure minus this minus this is v into pi RT by v liquid we calculated the osmotic pressure this is typically you are working in cc atmoshers you are working in atmoshers this is typically suppose I should use 80 to temperature of 300 degrees by 18 into log x1 and log x1 is 1 minus x2 or the log of 1 minus x2 which is minus x2 is a minus sign somewhere yes this is a minus sign which is approximately 18 into 300 by 18 is about 1000 to 50 10% 1300 if you like into x2 so we said typically this is 70 atmoshers for seawater and about 25 atmoshers for brackish water this is only an estimate simply brackish water from the ground and your membranes here are designed to withstand this 25 they are not meant for seawater we also have one unit for seawater desalination if you want to take a look at it you can in cc that is under that project that I told you about Surya gel and if you contact some of you are interested you can ask Dr. Kumaravel you fix an appointment with him he will show you the unit he has got a PV photovoltaic for electricity generation that power is used to drive the reverse osmosis in reverse osmosis the major this thing is to pump the the sink to P prime pump your seawater to C prime P prime that is from one atmosphere it will go to 70 atmoshers and he works at 500 liters per day so he actually buys the seawater in tanks so it is filled and it is pumped and the pure water is taken out of course you will also have to take out the concentrated brain and disposal of this is a problem and the many problem first is the membrane cost as I said most important thing about the membrane is in India I think the most important thing is biodegradability and the fact that we should manufacture it otherwise it is the Japanese who manufacture it so they will sell you the whole desalination unit for 5 lakhs or something you will be thrilled and then the membrane will be 10 lakhs every replacement then you will be stuck with it I mean it is just a manner of speaking what I mean is that components will be very expensive so you have to manufacture the membrane yourself to manufacture the membrane economically it requires a very large unit so you need a large number of desalination units so it is all coupled all the states have to come together you must all agree on using the same membrane so there is lots of coupled non-technical issues but the crux of the matter is that we have seawater and so on and the sun in plenty energy from the sun those are only two things we have in plenty nobody can take it away from us so we may as well use that the thing I have pointed out was that Ws dot the rated which work is done per unit produced would be pi into V for example Ws dot by m dot so just be pi into V which would be for seawater osmosis it will be 70 atm into 18 cc per gram mole doing this in joules it is about 80 cc atm 8 joules so divided by 10 if you like roughly this is joules per gram more so this is about 130 or if you are doing it per gram actually it is even into 1 and just put 1 per gram so this is about 7 joules per gram okay you compare this with 540 calories in fact that is joules is even more we are talking of 210 about 2200 joules per gram for evaporation and for freezing if you want to take out water by from salt water by freezing out the ice you can do that it is called the Omanov process introduced in Japan way back more than 60 70 years ago that will be about 80 calories per gram or about what 320 joules per gram this is freezing actually I put down figures symbolically the 2100 has to be modified we have what is called multiple effective operation you will read about it so there you can recover most of the latent heat therefore it will come to about 210 you can get a steam what is called a steam economy of 10 that means for every mole of water evaporated in the boiler you can actually by decreasing the pressure you can evaporate 10 moles of water at various stages because you evaporate one mole that condenses when it condenses use the latent heat to heat another gram of water so this is done very effectively so you can get economies of about 10 so 210 and here you can get an economy about 3 because there there is in freezing you have to first produce a cooling plant and there you have a Carnot efficiency limitation therefore eventually the source of energy is the sun so really here you can produce an economy of about 3 here you can produce an economy about 10 so you are talking of 210 joules per gram in practice and here you are talking about 100 joules per gram in spite of that of course there are many other complications here because the freezing unit has to be maintained compare that with 7 and even if you have an efficiency of 3 you are still talking of 20 so in principle the reverse osmosis process is very very economical but it is sort of deceptive in all these cases when people give you arguments like this it is porous because this implies that you have a separate membrane manufacturing unit and there is work done in creating the membrane a lot of work done in making the membrane because these membranes should not have pinholes so you are talking of membranes that will withstand a mechanical pressure of 70 atmospheres on one side and are thin enough to allow water to pass through and not the salt and have no pinholes in them so it is a huge process of membrane manufacturing well understood technology will is well understood so what you really have to do is come if you want to compare this with this or the evaporation process you must take the whole plant and calculate the energy requirements you will find they are all comparable except reverse osmosis probably more efficient in principle by about a factor of about 20 to 30% especially on large scale it can be much more efficient so this is as far as these are nation is concerned then two other things I wanted to discuss was a very similar I do not know if I already discussed this they discussed freezing point depression I am talking about water and ice you want to lower the freezing point of ice this is not of direct day-to-day interest for us but in very cold countries they have this ice formation and they want to know if this is water component 1 plus salt is pure component 1 I simply equate like all phase equilibrium problems I have only water in both phases so I have to equate µ1 solid pure is equal to µ1 liquid pure at the same pnp plus RTLn, 1 x 1 well I do not know µ1 I can relate them but I cannot relate the I cannot get across the phase boundary directly because the saturation pressure is where the equilibrium is so one possibility is simply make note that at saturation pressure the two are equal and apply corrections in the the pointing correction but I told you this log ? 1 x 1 what I am looking at is the freezing point is equal to µ1 liquid solid pure – µ1 liquid pure by RT it is only at the triple point that these the pure solid and the pure liquid will be at equilibrium with one another because water does not have it has a melting range so I can go to the saturation pressure at which it melts that is one possibility but if I am not sure that I am in the correct region of phase space the easier way to handle it and also because I am interested in this case in the temperature of the freezing point so I am really looking at T so I take this and differentiate this with respect to T I will get H1 liquid pure – H1 solid because it will be change in sign this will be the latent heat of melting of one by RT square this is H1 liquid pure – H1 solid pure now when x1 equal to 1 which is pure water ? 1 is 1 and T should be equal to the melting point of the solid Tm1 or the normal freezing point we will call it Tf1 this is the normal normal means at one atmosphere freezing point it is 0 degrees C typically for water so it is a differential equation for x1 in T ? 1 you have to know a model so you are solving it with this condition if you assume that this is approximately constant you get ? 1 x1 is equal to Lm1 by R into this is x1 equal to 1 to x1 equal to sorry x1 at x1 equal to 1 we have seen this is 1 by Tm1 you can say approximately if you like because basically the latent heat of melting is also a function of temperature actually you can do this rigorously also because we want to do it rigorously you just have to know what is this as a function of temperature we will do this incidentally Lm1 is strictly a function of temperature in this particular case it is an negligible change because you will find the freezing point depression is a few degrees lower those few degrees ? h does not change I will come back to this let me look at this here this is the expression here since at x1 equal to 1 ? 1 is 1 this is simply log ? 1 x1 T is the point at which the solution freezes solution containing salt freezes the solid phase that produces pure ice so this is the actual freezing point and this is the I called it Tm1 Tf1 keep changing sorry Tm same as Tf1 this is the freezing point of water in the presence of salt this is the freezing point of pure water so if you have we call this ? T freezing point depression this is freezing point depression physical chemists call it this is defined as T- Tf1- T and this is actually much smaller than you can verify this ? T will be typically the order of a degree where as Tf1 is doing degrees k so this is 273 so this is typically very small so I can write this as Lf1 sorry by R Tf1 Tf1 x Tf1- T so there will be a minus sign sorry is a plus sign Tf1- T is ? T freezing point depression by T Tf1 so T Tf1 is actually approximately Tf1 square typically in these applications you try to lower the freezing point by adding a small amount of salt so x1 is close to 1 x2 is small so when x1 is close to 1 ? 1 is 1 so we have approximately log ? 1 x1 dilute solutions which is what we have is the same as log x1 which is approximately – x2 log of 1- x2 and therefore just – x2 ? T freezing point depression I am missing a sign somewhere here Lf T Tf1 you get Tf1- T that is what I defined as this is all right or the previous one okay this is Lf again sorry log ? 1 x1 T – there is a minus sign here in the integration thanks I needed it – so define this way ? T freezing point is simply equal to R Tf1 square by Lf1 x x2 actually solving a phase equilibrium problem but the information you get from the phase equilibrium problem is the depression in the freezing point upon addition of solute actually it has a most physical chemistry use it to calculate molecular weights of polymers what they do is weigh a certain amount of polymer M2 will be the weight of the polymer they add it to known amount of solvent so this is the expression they measure this quantity right if you make a polymeric substance you can measure this R T and L are known these are for water or for the solvent in question so if you dissolve a polymer for example in a solvent and for the solvent you know it is freezing point and it is latent heat then this coefficient is known and you know x2 from this calculation you know the amount of polymer you added therefore you can calculate its molecular weight if M2 is the polymer molecular weight is the total weight of polymer added to 1000 grams of solution you get M2 by molecular weight of polymer divided by M2 by molecular weight of polymer plus moles of water solvent say 1000 by 18 that is known therefore you can calculate the molecular weight of polymer but coming back here you can calculate the freezing point depression that you need in order to get x2 so yeah what what log x1 is log of 1- x2 when x2 is small it is approximately minus this one is log of 1- x2 log of 1- x is x- x it is a Taylor expansion when x is small so we will calculate this RTF1 so RTF1 for example for water by LF1 square this is for water which is component 1 this will be about I will do it in calories because I know LF1 in calories calories per gram I know RTF1 squared by LF1 TF1 squared is 273 squared by LF1 would be about 80 calories per gram this is by 80 if you have the calculate so for water plus solute delta T freezing point depression is approximately given by this must be this must come out in degrees k because RTF as the same units as LF what does this come to as anybody have a calculator is something like 90 or 100 so 104 let us say 104 x2 approximately you can see what will happen in the case of boiling point elevation do the same thing I think I will do it on this board because I want to copy the boiling point elevation results there very similar so let me write down first you have liquid and vapor this one is water component 1 plus solute which is 2 here this is water so for the solution you will still get µ1 liquid plus RTLn ?1 x1 on the other side will be µ1 vapor pure instead of solid pure so it will become vapor pure so you do the same differentiation instead of liquid latent heat of melting you will get latent heat of evaporation call it L1 B it is called boiling point elevation so I will call it L1 B but L1 B will be H1 liquid sorry vapor pure minus liquid pure but notice this is vapor minus liquid so you pick up a minus sign here because when you differentiate you will get liquid minus vapor µ by T will give you – H by T2 this is still valid T is not equal to Tf1 it is equal to Tb1 which is normal boiling point so this is Tb1 again we will integrate assuming that this is Lb1 L1 B if you like since the latent heat of evaporation of water so you will still get this there is a minus sign that has changed now because of this minus sign this minus sign will go so this minus sign will disappear we will get Tb1 we will pick up a minus sign here this is called boiling point elevation defined as T minus T boiling point of 1 because you know the boil it will turn out that the boiling point increases so T will be greater than and you want this quantity to be a positive quantity ? T boiling point elevation it is still true that this is much greater than the boiling point of 1 so this will change to Lb1 L1 B by T1 B so here the minus sign is restored at this point so ? T boiling point elevation will be RT boiling point of 1 the whole squared by L boiling point of 1 x 2 it is boiling point elevation so it is defined as T minus Tb1 the other one is defined as Tf1 minus T so this is now called boiling point elevation so again I get I have to evaluate RT B1 squared by Lb1 for water this will be 2 x 373 whole squared boiling point this ET is now 540 x ET you do that for me 373 squared this 540 calories per gram but 29 approximately 29 x 2 x 2 of course oh this is sorry this is just 29 then I have for ? T boiling point elevation is equal to 29 x 2 ? T freezing point depression by ? T boiling point elevation is that correct then this by that is 104 by 29 3.6 for the same amount of addition you get a larger depression and a boiling point elevation actually trying to think of okay anyway you can see that these two are connected and sometimes this is used for consistency of measurements if you are measuring x 2 for example in solution if you are measuring for the same solution you can measure your ? T freezing point you can measure your boiling point and you can verify that the ratio ratio is independent of x 2 in dilute solutions this is independent of x 2 in fact it is independent of two the value next to no matter what the solute is for a given solvent the ratio of boiling point elevation to the freezing point depression is a constant function only of the solute so actually the number of such applications is any number you just have to in every situation you start with the chemical potential model for the phase if it is solid liquid or vapor and then work through it you cannot go wrong.