 So, welcome to the 18th lecture of the NPTEL course on cryogenic engineering. We have done various topics till now and the broad topics we can classify them as first the introduction to cryogenic engineering. We studied the properties of different cryogens and therefore, properties of cryogenic fluids. Then we studied the properties of materials at cryogenic temperature and the last topic which we covered was gas liquefaction and refrigeration system. Having done all these four topics, now I will be changing some gears and we will be going to now gas separation which is a very important topic of cryogenic engineering. Having liquefied gas, having liquefied gases together and air being a mixture of various gases, whenever we liquefy air, we got lot of gases in liquefied form in air and now we have to separate oxygen, nitrogen and other gases may be argon from liquid air and therefore, we have to use a gas separation technique and this is the most important thing because gases are needed by every industry nowadays and therefore, this topic has got a very special relevance from users point of view from gas user point of view. So, topic is now gas separation. What we are going to study under these are different sub topics which are basics of gas separation, then ideal gas separation system just as we got a ideal system for every system to compare with, we got a ideal gas separation system also. Then we will study properties of mixture and the governing laws. Now, as you possibly know that we got several laws which govern the mixtures and they help us basically in understanding the partial pressure business and the total pressure and their relationship and they are very important to analyze these mixtures, so as to get the properties of mixtures at different temperatures and pressures. Then we will study principles of gas separation in broad and then separation process itself, the rectification and plate calculations. So, plate calculations actually essentially mean that there is a column which has got lot of plates and we have to calculate number of plates involved in that. So, this is a very important topic and which involves actually it covers a bit of chemical engineering also because it has got some mass transfer aspects associated with it. We will keep it of course for mechanical engineers however and therefore from this point of view I have basically going to cover this topic from mechanical engineering point of view. The current topic will be covered in around 7 to 10 lectures just as we had liquefaction in 7 to 10 lectures. Similarly, as we proceed ahead the number of lectures possibly will be around 7 to 10 and then tutorials and assignments are also included at the end of each lecture. So, in this topic what we going to learn is the outline of today's lecture will be the basics of gas separation in which we will study some gas separation methods and then we will find what is the ideal gas separation system and we will find what is the ideal work requirement for ideal gas separation system. These two sub topics will be considered will be taught in this lecture of first lecture of gas separation. As mentioned earlier the cryogenic industry is huge going to the various application of the cryogens both in liquid and gaseous states. For example, the use of inert gases like argon in chemical and welding industries has increased in the recent past. Basically I am highlighting the utility of these gases in various industries. Liquid nitrogen is used as pre coolant in most of the cryogenic systems also cryogens like liquid oxygen, liquid hydrogen are used in rocket propulsion. In the recent past liquid hydrogen is being considered as a fuel in an automobile. The production of ammonia in RCF industry for example, the fertilizer industry requires separation of purge gases like nitrogen argon and other inert gases. This production has to be done at cryogenic temperature and therefore, separation of gases play a very important role in fertilizer industry also. For most practical purposes it is considered as a mixture of 78 percent nitrogen, 21 percent oxygen and 1 percent argon. So, if you want to use nitrogen, oxygen and argon we have to basically separate them out from air. The other ingredients are helium, neon, krypton which occur in negligible quantities. So, normally as far as helium, neon, krypton are considered the you know take them out from air becomes rather difficult. So, normally air will be used to obtain nitrogen, oxygen and argon which are abundantly used in abundantly required in the industries. So, air serves as a raw material. Air is a raw material for the production of most of the gases and the process of separation of the gas mixture into its individual component is called gas separation. That is what normally we will define as. In other words, this topic of gas separation deals with separation of various gas mixtures and their purification. Now, how do we separate gas mixture? There are various techniques available. We just touch upon those techniques and then come down to the technique of cryogenic gas separation. So, what are these different techniques to separate various constituents of a gas mixture? One is using synthetic membrane. We will see each of these techniques in brief to support to understand how the separation occurs using synthetic membranes. The second technique is adsorption which is a physical phenomena as possibly most of you know. The absorption are the technique of separation of a gas mixture and then the cryogenic distillation. So, broadly there are four methodologies of which the first three are normally at room temperature while the third last one is cryogenic distillation or separation at cryogenic temperature or at low temperature. So, we will first study what these three techniques are and then we would go to cryogenic distillation in detail. So, let us see what synthetic membrane mean and how does synthetic membrane separate a gas mixture? Synthetic membranes are the porous media which allows only a certain gas molecules to pass through. So, very simple. So, you can see from this figure. This figure shows that the membrane in some cylinder and this membrane allows to separate a mixture of A and B molecules or a mixture of A and B components of gases under the piston which you see from the left side. Synthetic membranes are the porous media which allows only a certain gas molecules to pass through. The membrane in the figure allows only gas A to pass and hence separation occurs. So, in this particular figure what you say this membrane allows only A to pass through it while B cannot pass through it. So, what will happen? If I go on pushing this piston on the right side, the A will get A will pass through the membrane and on the right side of A as you can see right side of membrane what you can see is only the component A. That means, only the gas molecule with gas A gas will come through this membrane while the B gas would be always on the left side because B cannot pass through the synthetic membrane. For example, thin sheet of palladium allows only hydrogen gas to pass through. So, if I use palladium material as membrane material and then have a mixture of various gases in which one of the components is hydrogen and then if I use such kind of arrangement only hydrogen will be in this particular location where A is right now because this membrane will allow only hydrogen to pass through it while other gas members will not be able to pass through it alright. So, this is a way the membranes can be used. They are being used various membranes they got a preferential treatment preferential passage for particular gas and according to which they could be used. The second technique is adsorption. So, adsorption is a physical process in which only a certain kind of gas molecules are adhered to the adsorbing surface. Most of you know this adsorbent that you got a surface phenomena and if you got a A and B gases then only one of the gases can adhere to the mixture or it can get adsorbed on this adsorbent while other gas cannot. So, adsorbate in the figure adheres only gas A. So, this is adsorbate and it the gas only can get adsorbed on the surface of this adsorbate. So, it allows only A to adhere to the surface while B cannot adhere to the surface in a way therefore, a mixture of A and B will get separated. A will adhere to the gas adsorbate surface while B will be freely lying over here and then in a way this is separation of gases A and B. So, the adsorbate in the figure adheres only gas A to the surface hence the separation occurs. So, what you can see here the layer here will have only A components adhere to it while all the B components eventually will be all the A's would be getting adsorbed while B would remain there meaning which A and B have got separated. For example, finely divided nickel adsorbs hydrogen on its surface. So, if I want to separate hydrogen I can use nickel as adsorbate and hydrogen will get adsorbed on the surface of this nickel. Second one is absorption which we know that we got a sponge basically. So, we know that absorption is a chemical process in which substance in one physical state is taken into other substance at a different physical state. So, it is basically solid can take liquid or solid can take gas and one physical state is taken into other substance at a different physical state. For example, liquid being absorbed by a solid or gases being absorbed by liquid. So, what is the example? When an incoming stream of containing carbon dioxide is passed through a solution of sodium hydroxide the latter absorbs the gas and hence decreases the CO to contain in the outgoing stream. So, if I got a sodium hydroxide, sodium hydroxide will absorb CO2 the actually chemical reaction will happen between sodium hydroxide and carbon dioxide meaning which if an incoming stream has CO2 the quantity of CO2 can be lessened from this mixture and in a way separation occurs. So, if I got a sufficient quantity of sodium hydroxide all the CO2 from the incoming gases can be removed because sodium hydroxide will absorb CO2 there will be chemical reaction and may be some other compounds get formed during this chemical reaction. In a way this is a method of separation and therefore, this absorption is used abundantly in the chemical reaction. Hence, this chemical process helps in the separation of the mixture. So, in this case I am removing CO2 from the mixture by absorbing CO2 by sodium hydroxide. So, having done membrane adsorption and absorption having understood the basics of this now let us come to the distillation. Now, distillation is a process of separation based on differences in volatilities or the boiling points. So, I have got different gases and all these different gases have got their boiling points and they are different depending on the characteristics of these gases the boiling points will be decided and then the unique boiling point for each gas. Choosing these volatilities or the differences in the boiling point we have got a process that is called as distillation. So, here you can see that there is something called a schematically I have shown here something called a distillation column to which mixture of A and B gas A has got a different boiling point gas B has got a different boiling point and if this mixture enters this distillation column what you can see is the mixture A is at the bottom at the end and mixture B at the top. So, one which has got a high boiling point of around 90 Kelvin which is oxygen boiling point you will have it at the bottom and the liquid you will get for gas A at the bottom and one can convert this liquid to gas if one wants gas or one can get this liquid one wants liquid A also. Similarly, on the top what you get is a lower temperature here the Q is removed here the Q is added. So, you can see that Q in and Q out over here the low temperature boiling point gas for example, nitrogen which has got a boiling point of 77 Kelvin will be obtained at top of this distillation column this is just a schematic. So, a mixture of A and B enters the column and at the end this is now a cryogenic separation because the boiling points of these two gases A and B are different they can get separated A is coming at the bottom B is going at the top of this column. So, if the process of distillation occurs at cryogenic temperature this process is called as cryogenic distillation. The commercial production of gases like oxygen, nitrogen, argon, neon, krypton and xenon are obtained by cryogenic distillation of liquid air. So, liquid air actually is a source of all these gases and therefore, if you liquefy air that means all these gases are in the liquefied form in the mixture and this mixture can be fed to the column and according to the difference in the boiling points of these gases they can be separated and this is what we are going to see in the coming slides and coming lectures. The separation of a gas separation of a mixture can be done at both room temperature and at cryogenic temperature as we just saw. For example, in the case of air the following processes are possible. So, I just showed an example where the cryogenic distillation is done, but there is possibility that one can do this separation at room temperature also. So, if I got air at 300 Kelvin I can liquefy this air at 78 Kelvin which is the boiling point of air and from where I can get liquid oxygen and liquid nitrogen which have got boiling point of 90 Kelvin and 77 Kelvin and therefore, because this separation is occurring at lower temperature. First I am liquefying air and then I am separating the two gases from liquid air I call this process as cryogenic separation. The other process is now I have got air first at 300 Kelvin and first at room temperature itself at 300 Kelvin itself I am dividing or I am separating this mixture assuming that air is a mixture of nitrogen and oxygen only I got oxygen at 300 Kelvin and I got nitrogen at 300 Kelvin. That means, I am doing this separation of air at room temperature itself and then I get oxygen at 90 Kelvin and nitrogen at 77 Kelvin if I want liquid nitrogen and liquid oxygen. But you can see from here that the separation has occurred at room temperature for which I used special techniques called pressure swing adsorption, temperature swing adsorption and they are basically adsorption techniques of separating oxygen and nitrogen. These are more widely used these days it has got its plus points and negative points similarly cryogenic separation has got its plus point and negative points. Some of the advantages of cryogenic separation over room temperature separation are the separation at lower temperature is most economical. We will find later that as the separation temperature decreases the power input requirement or the work of separation requirement lessens. So, here we can understand that there is an increased difference in the boiling point of the ingredients. So, we can do that separation is effective because of the difference in volatilities. A large quantity of the gases can be separated in the cryogenic distillation and high purity of the gas can be obtained. If you do the separation at lower temperature we can get high purity of the gas we can obtain high purity of the gas at cryogenic temperature. These are the certain advantages of cryogenic separation. Let us understand is gas mixing is a reversible process what do I mean by that? That means if I mix the gas can I unmix it and can I mix them together. So, meaning which is this process a reversible process. So, consider a closed chamber filled with gas A and gas B as shown here and initially the gases are separated by impervious wall. That means A is here and B is here and there is the impervious wall which will not allow A and B to pass through. They are separate entities in this chamber if the wall is removed now the gases would mix. So, suppose I remove this wall you will find that A and B are together. Now, if I go back and put this wall again can I get A and B separated is it possible? This is not possible and therefore, I will say that the replacement of the wall would not result in separation of gas and therefore, I can claim that this process is not reversible. It is clear that the mixing of two different gases is an irreversible process because unmixing or separation of the mixture requires work input. I have to do some work on this then only I can unmix the gas. The system in which all the processes are reversible is called ideal system as you know this. So, mixing and unmixing if it is possible if it is reversible process then I will call this ideal separation system. Although in reality such a system does not exist a system can be conceived as you know that every time normally we deal with ideal system of separation and therefore, we can conceive such a system wherein if I calculate work of separation or work of mixing and unmixing I have to first conceive an ideal system. I am going to conceive a system constitute to serve the required purpose as explained in the next slide. So, I am basically having a ideal system of gas separation now and how does this work ideal gas separation system. So, consider a closed chamber filled with a mixture of gas A and gas B as shown over here and there are two pistons one is red and one is black and again I got a mixture of gas A and gas B. This two pistons as sure by arrows can be moved inside and can be taken outside also. The temperature of this gas mixture is T m the pressure of this gas mixture is P m the temperature and the mixture pressure are T m and P m respectively the partial pressure of gas A and gas B are P 1 A and P 1 B. Depending on the amount of gas A and gas B we got a partial pressure of gas A and gas B termed as P 1 A and P 1 B. The chamber has two frictionless opposing pistons made from semi permeable membrane as shown in the figure. So, these are not only pistons they are also membranes and they are semi permeable membrane meaning which they allow only one type of gas A or B to pass through it as we have seen earlier. So, we got a piston which will compress this gas at the same time it will separate one of the components of this mixture. As seen earlier a semi permeable membrane is a film which allows only one kind of gas to pass through, but not the other alright. So, we will this membrane will allow only one type of gas to pass through it. The left piston which is a red it will allow only gas A to pass through it and it will not allow the gas B to pass through it alright. So, basically when I start pressing this left piston to the right it will let A go through it, but it will not allow B to go through it. Similarly, when I press B on the left side it will allow only B to pass through it and it will not allow A to pass through it. Similarly, the right piston allows only gas B to pass through, but not gas A. This is what we call as a semi permeable membrane working like a piston in this close chamber. So, as if I go on pressing this two membranes or pistons you can find the separation has started occurring. So, I am getting on the left side of the membrane only A and I am getting right side of this green piston or green membrane only the component B. So, when I basically reach both the pistons reach together A and B would get separated. When the both the pistons are moved inward the mixture is separated and it would look like this understood. So, I got a mixture A I got a gas A and gas B separated from mixture of gases A and B as seen here. Since the processes are reversible we are claiming this particular process to be reversible what should happen the system interacts with the surroundings to maintain a constant temperature. So, all these processes are happening at temperature being constant T m. So, what I am doing basically I am pushing this piston B and piston A inwards. So, I have to do some work W A and W B respectively and at the same time because the compression process occurring some q r is being released some heat generated is being released. So, that the temperature T m is maintained I am saying that the temperature T m is all through constant and because of this q r temperature T m is being maintained because of this interaction with the surrounding the constant temperature is maintained. The work of separation is the work required to compress the compress each gas from its partial pressure P 1 A to P m and P 1 B to P m. The piston A is compressing the gas B while the piston B is compressing the gas A alright. So, W B is the work done on the piston B while W A is the work done on piston A and what is this compressing? Compressing gas B from partial pressure P 1 B to the mixture pressure P m and the whole process is carried out at temperature T m. So, here you can see that the gases A and B are being separated have been separated here. Since the left piston is permeable to gas A the gas A exerts no pressure on the left piston alright. When this piston moves the pressure is exerted on this piston by gas B and not by gas A. Similarly, the gas B exerts no pressure on the right piston it will be gas A which will exert pressure gas B will just pass through this gas A just passes through this. When both the pistons are moved inward the mixture is separated at constant temperature T m because we are calling this the temperature remaining constant in this case alright. So, this is a separated gases A and B the pressure here is P m and temperature being T m the pressure being P m and temperature being T m the entire processes are assumed to be reversible. So, now I find that the mixture has been separated, but because this process is reversible now this process should also reach back to the earlier point where it started from. The process is reversed due to the difference in the concentration of gas A and gas B. So, slowly gas A will start diffusing through piston A and gas B will start diffusing through the piston B because they are semi permeable membrane and slowly the concentration balance will get created slowly there will be difference in concentration on the left side of the piston and right of the piston for gas A as well piston A as well as for piston B and slowly the process of diffusion will start and slowly this piston would start moving back. It will take lot of time, but because the process is being reversible the process would start and therefore, meaning which the gas A will start getting combined with gas B that means the mixing process will start a reversible process of mixing will get start. First we have unmixed or separated A and B, but now slowly gas A will start coming in gas B start coming in and process of mixing will get started. Hence the mixing of the gases would move the pistons away and produce work. So, here with time the gas A will start coming in the gap between piston A and B and it will start pushing this piston A back and piston B back that means the work will now get produced by the system. Earlier we had put in the work we had work done on the system W A plus W B. Now the same work will get created by on piston A and piston B by the gas because now the gas is going to work on the piston. So, here what you can see is piston A is moved back piston B is moved back and slowly a mixture of A and B getting formed in this space. So, this is basically the reversible process which is what you are seeing here occurring now. The work produced in this mixing process is same as the work done to separate at constant temperature T m. The final condition is system with a mixture of gas A and B at P m and T m. So, ultimately I get a mixture of gas A and B at P m and T m. So, what you see here now in which again now because the work is kind of expansion of the gases I am supplying Q R the temperature being kept constant the surrounding is now putting some heat inside. So, that the T m is kept constant temperature is maintained and W A and W B is the work done by the system work which is produced by the system during mixing. Also the partial pressure of gas A and gas B will be P 1 A to P 1 B. So, ultimately in this mixture it will have partial pressure of P 1 A and P 1 B as it was previously. So, ultimately I will get now a mixture of gases first I had unmixed the mixture and again mixture occurred automatically and if I put in the work W A and W B again the process would repeat. So, this is what we call as a reversible process. We are consuming a system and wherein we can call this particular process as a reversible process of gas separation. So, this is my initial condition when I got a mixture A and mixture B gas A and gas B in a mixture form. This is my final condition where I got mixture separated in gas A and gas B. The moment I say they are reversible process I should have the arrows like this also. So, I can go from initial to final and final to initial in both the cases temperature T m is maintained constant. So, basically the gas A is getting compressed from P 1 A to P m when I am pushing this piston and gas B is getting compressed from P 1 B or the partial pressure of B to P m. In other words thermodynamically each gas is compressed reversibly and isothermally from its partial pressure to the mixture pressure. So, what is what I am doing basically while separating this mixture I am compressing reversibly and isothermally gas A and gas B from its respective partial pressure to the mixture pressure. This is what basically I am doing in separating this gas mixture. In order to understand the processes of compression say gas A from P 1 A to P m the following analysis is done. Now, basically I would like to calculate the work of separation of gas A and gas B. When I say work of compression is nothing but work of separation and this will give me ideal work of separation. This is what I am going to calculate now. So, now let us see the ideal separation system. Let us do the calculations to understand what is the work of separation if one goes for a ideal separation system. Let the molecular weight of gas A and gas B be denoted as mole A and mole B respectively. The number of moles of gas A will be given by this is very simple mixture family which I am going to just kind of revise and go ahead. So, if I want to calculate number of moles it will depend on how much grams of gas A is there divided by its molecular weight that will give me moles. So, N A is a mole of moles of gas A in a mixture of A and B. So, N A will be given as M A upon molecular weight of A. Similarly, N B will give the moles of B that is M B upon molecular weight of B. The total number of moles in the mixture N M suppose the mixture number of moles in a mixture are N M you know N M is equal to N A plus N B. And the ratios now I call as Y A which is basically mole fraction of gas A Y B is a mole fraction of gas B. The mole fraction of gas A is called Y A which is N A upon N M and Y B is N B upon N M. So, Y B and Y A are basically the mole fractions of gas B and gas A respectively. Now, this is what our initial condition was of A and B when they were a mixture at T M and temperature of T M and P M pressure and the partial pressure were P 1 A and P 1 B. The corresponding volume occupied by the mixture was V total and I can apply now simple gas laws which is P 1 A into V total. This is for gas law ideal gas law for gas A and similarly I will have ideal gas law for gas B. So, what is the pressure of gas A P 1 A into V total is equal to N A R T where R in this format is what is we call as a universal gas constant similarly I will write the same equation for gas B also. If I combine them together P 1 A plus P 1 B into V total is equal to N A plus N M N B R T M. So, P V is equal to N R T is basically I am applying for gas A and gas B and also a mixture of gas A plus gas V. Note that the volumes are always the same because A and B together are occupying the same volume and individually also are occupying the same volume meaning which P 1 A plus P 1 B is equal to P M that is what we have been talking about. And now is a final when I got a separation has occurred and I got now particular volume of V A associated with gas A and particular volume of V B associated with gas B while the temperature is same as T M. And now the pressure of the gas A here is P M the pressure of the gas B here in V B volume is P M. So, again I apply the gas law the ideal gas law for gas A and gas B now the pressure of gas A in this volume of V A is equal to P M. So, P V is equal to N R T for gas A which is P M into V A is equal to N R T same thing I do for gas B. And if I now see that if I divide this by this what I get is V A upon V B. This will basically give the ratio of how much volume is occupied by V A and V B. It will basically give a function depending on what is the ratio of molecules of N A moles of N A divided by moles of N B. So, how many moles of N A are there how many moles of B will decide what is my V A and V B are. The volume occupied by each of the gas is directly proportional to the number of moles that is what we have just seen. From the earlier lectures the work requirement for unit mass of gas compressed isothermally is given by W I by M dot which is T M into S 1 minus S 2 minus H 1 minus S 2. This is what we had done we had derived this equation in the gas liquefaction cycle. So, basically W I is a ideal work done of compression basically because as we earlier said as earlier pointed out that separation is nothing but compressing the gas from partial pressure to the mixture pressure P 1 A to P M or P 1 B to P M. So, W I by M is equal to T M into S 1 minus S 2 which is entropy difference minus H 1 minus H 2 enthalpy difference. The net ideal work requirement of the separation process is the sum of the ideal work requirement for gas A and gas B. So, I can write mathematically that W I is equal to W I A plus W I B. The total work of separation is equal to work of separation of gas A plus work of separation of gas B. And dividing by M dot I get this W I by M M M divided by W I A M M plus W I B divided by M M. M M is nothing but the mass of the mixture. So, this is what we have just seen the total mass of the mixture M M and is the sum of gas A and gas B. We can write M M is equal to nothing but M A plus M B absolutely clear till now rearing in the terms. Now, I am just writing this terms in a different format as given here. So, W I by M M is equal to W I A upon M A this is nothing but work of separation for gas A multiplied by M A by M M plus work separation of gas B divided by M B W I B upon M B into M B by M M. So, I am just writing M M as W I M A M A by M M M A gets cancelled what I guess W I by M M which is what you see here. So, what I am basically you know writing this expression as W I by M M is equal to work of separation of M A into this component plus work of separation of B into this component. Here M A and M B are the masses of the gas A and gas B respectively. So, this is what we have just seen. Now, I can write an expression for W I by M A also. The work requirement for each of the individual gas is given by following equation. So, W I upon M A is equal to T M into S 1 A minus S 2 A minus H 1 A minus H 2 A. This is the expression we have just seen of work of separation when I compress the gas from 1 A to 2 A position for gas A. And similarly, I can write the same expression for gas B also. Substituting and rearing in the term if I put this value over in this equation I will get a very long equation. So, substituting this by this and substituting W I B upon M B by this what I get is this very big expression over here. T M is the temperature of mixture which is remaining constant. Now, let us find some expression for this entropy difference and enthalpy difference. So, this is my expression. What you can see from here that the T M which is a mixture temperature figuring out here and this is a work of separation of the gas. And as you can see that if the T M is lowered or the mixture temperature is lowered W I by M will be the work of separation will be very less. That shows directly that the temperature is very low or in the cryogenic temperature region the work of separation will be very very less. It is clear that work requirement decreases with the decreasing temperature. Hence, the separation of mixture at the cryogenic temperature is most economical and one point is proved mathematically from here that if T M is less if a mixture temperature is less the work of separation will be very very less. The subscripts 1 and 2 denotes the initial and final condition respectively. What does it mean? What is the condition 1? It means that for each gas S 1 and H 1 are at the partial pressure before separation because each gas is being compressed from its partial pressure to mixture pressure. So, one condition the S 1 and H 1 are basically conditions prevailing at partial pressure before separation while S 2 and H 2 are at the mixture pressure because the gases are being compressed from their partial pressures to the mixture pressure after the separation of the mixture. For the sake of understanding let us first evaluate the entropy and enthalpy terms for this. So, let us see if you could substitute by some simple expression for this enthalpy and entropy differences for gas A and gas B respectively. For an ideal gas the specific entropy S and specific enthalpy H are expressed as below. So, S is equal to C p log T minus R log p plus some reference entropy term. Similarly, enthalpy is equal to C p into T plus this reference enthalpy term basically is a any temperature any pressure one can select and what we mean to say that it should remain constant throughout calculations where S R and H R are some reference values they could be at 77 Kelvin or boiling point or 1 bar and 0 Kelvin or whatever because there are various temperature entropy diagram which can have different references values and what we mean to say that whenever we refer to such different diagrams we should keep the same reference values. Hence, S and H for each gas A are given as if I put now this general expression for gas A I will get this expression well similarly I will get for S 2 A that is after at a temperature T m and mixture P m I will get this expression alright. Similarly, enthalpy values at 1 A and 2 A conditions are given by this expression. The entropy and enthalpy terms are given what I want to basically find out is a difference in S 1 A and S 2 A minus H 1 A and S 2 A and if I put those values over here I get this expression this is for entropy difference and this is for enthalpy difference and if you see now because the temperatures remaining same from 1 A to 2 A this terms can get cancelled. So, this term will get cancelled similarly the reference values are same they will get cancelled over here and in this case again the temperatures remaining constant these terms will get cancelled. So, what remains here basically Ra log P 1 by P 1 A minus sign plus Ra log P m and it is basically nothing but logarithmic difference which is nothing but Ra into log P m upon P 1 A. So, entire S 1 A minus S 2 A minus enthalpy difference gets reduced to Ra log P m upon P 1 A. Similarly, I can do the same thing for gas B and I will get expression S 1 B minus S 2 B minus H 1 B minus H 2 B is equal to R B log P m upon P 1 B. So, I have done a major calculations and whatever expressions we have first got to calculate the work of separation wherein this entropy difference and enthalpy difference figure this complete term now will get replaced by this for gas A and gas B complete term will get replaced by this. Now, let us find what is this P m upon P 1 A mean or what is P m upon P 1 B mean. So, this is what we got an expression now this is what we got for gas B. So, I can put this expression back in my W i by M m is equal to this and my expression will be now replace this term by this in this expression and replace this enthalpy entropy difference by this and my final expression therefore, will be like this. So, I get W i upon M m is equal to T m into M a upon M m like this this entire term get replaced by Ra into log P m by A plus M b upon M m comes over here and this entire term of enthalpy entropy difference get replaced by R B log P m by P m. So, my expression becomes very simple and now I would like to understand what is my P m upon P 1 A is what is my M a R upon M m is I would like to understand those quantities. Since the process occurs at constant volume V m using ideal gas equation which is P v is equal to N R T I can write this expression which we have done earlier P m into V m is equal to N m R T m. So, I can write the same expression for gas A and gas B. So, this is a mixture equation for the mixture equation for gas A and equation for gas B dividing one over the other what we get is P m V m upon P 1 A V m is equal to N m R T m upon N a R T m R T m and R T m gets cancel V m and V a get cancel. So, P m upon P 1 A is equal to N m upon N a is equal to 1 by Y a we know that N a upon M M m basically moles of a divide by moles of mixture is equal to mole fractions of gas A. So, P m upon P 1 way ultimately gets reduced to 1 upon Y a. Similarly, I can do for ratio of mixture and gas B also and I get the same thing that P m upon P 1 B is nothing, but N m upon N B is equal to 1 upon Y b this is what we get from here now. So, I have got a ratio of P 1 by P 1 A figuring in the earlier expression I can replace that by 1 by Y a where Y a and Y b are the mole fractions of gas A and gas B respectively. The ideal gas equation can also be expressed in terms of mass of the gas as shown below. So, I can write this expression P v is equal to N R T for gas mixture gas A and gas B I can write this N m or the molar fraction as the mass of gas mass of mixture divided by molecular weight of mixture this is basically definition of N m I am just replacing this N m by mass of M m upon molecular weight of M. So, P m V m is equal to M m upon molecular weight of M R T M and same thing I am doing it now for gas A and same thing I am doing for gas B here M a upon molecular weight of A here M b upon molecular weight of B will figure. So, why I am doing all these things basically I know now that P m V m is equal to M m R T now here R by molecular weight of M is nothing but now specific gas constant. Here universal gas constant R divided by molecular weight of A is specific gas constant of A and here also similarly the universal gas constant divided by molecular weight of B become specific gas constant R B as shown over here. So, basically R upon molecular weight of M or universal gas constant R upon molecular weight of M will come as this R which is specific gas constant similarly I will do for gas A similarly I will do for gas B why in general R A is equal to universal gas constant R upon molecular weight of A where universal gas constant is 8.314 joules per mole Kelvin. So, basically now I would like to express the express have the earlier expression in terms of mass of gas A and mass of gas B also alright. So, universal gas constant R and specific gas constant R are the universal gas constant and specific gas constant respectively. From the earlier slide choosing the ideal gas equation in terms of gas mass what we get is this. So, I am now writing P by M V M is equal to M M R M T M this is what we just saw P V is equal to M R T P 1 B M is equal to M B R B into T M dividing one over the other I will get P M V M upon P 1 A V M now I cannot cancel both R and T because T M and T M will get cancelled over here V M and V M will get cancelled over here what I get is P M upon P 1 A is equal to M M R M upon M A R A which is nothing but 1 by Y A because P M upon P 1 A is 1 by Y A we have just now earlier seen that what does it mean same thing I am doing it for gas B P M upon P 1 B nothing but 1 upon Y B. So, what I get from here is equal to 1 upon Y B ultimately I will get M M R M upon M M M A R A upon M M is equal to R A into Y A if I just rearrange this I want an expression basically for this which is what is figuring in my expression for work similarly for gas B. So, putting all these things P M upon P 1 is equal to 1 upon Y A P M upon P 1 is equal to 1 upon Y B and M A R A upon M M is equal to R M Y A and M B R B M M is equal to R M Y B I will put this expression in my this expression which I had derived earlier substituting all these so I will this replace P M by P 1 A by 1 upon Y A and here I will replace this similarly I will replace this term by R M Y A similarly I will this replace this term by R M Y B. So, if I put those expressions I will get this equation. So, W I upon M M is equal to R M T M Y A log 1 by Y A plus Y B log 1 by Y B the expression looks very simple now I have got a RT mixture temperature and R M Y A upon 1 upon log 1 upon Y A plus Y B log plus 1 upon into 1 by Y B where P V is equal to if I just replace by R M T M and M M here and want to convert it to the molar form I can write work of separation per mole of gas mixture W I by N M and here immediately universal gas constant will figure please understand this please understand the relationship between the universal gas constant and specific gas constant because this is a work of separation per gram of mixture per mass basically and this is a molar expression. So, you got a mass expression and a molar expression also as soon as the molar expression comes here universal gas constant comes as soon as the mass comes over here specific gas constant comes over here. So, the ideal work of separation per mole of mixture gas A gas B will be given as the molar expression ideal work of separation per mole of mixture that is W I by N M is equal to RT M Y A log 1 by Y A plus Y B log 1 by Y B this is for a mixture of gas A plus gas B I can extend this to A B C or A B C D we can have different components in a mixture and similarly I can show an expression for A B and C mixtures on the similar line if the mixture is composed of three different gases gas A gas B and gas C the ideal work of separation per mole of mixture is given by W I by N M RT M Y A log 1 by Y A plus Y B log 1 by Y B plus Y C log 1 by C C if you have got a D we will get Y D log 1 by Y D what are Y A Y B and Y C some molar fraction of gas A gas B and gas C in the mixture what is T M temperature of mixture universal gas constant. So, this is basically work of separation of per mole of the mixture work of separation per mole of mixture what I am doing is basically find out work of separation is nothing but work done to compressed each of this component gas A and gas B and gas C from their respective partial pressure to the mixture pressure as we have seen in earlier reversible gas separation process and this is what ideal work of gas separation will be the actual work of separation. However, as you know will be different and therefore, we introduce a term in the next slide called figure of merit which we have already introduced in a gas liquefaction also. So, generalizing the above equation for a mixture of N constituents we just showed for A and B we also showed for A B and C, but if the mixture which has possibly has got N constituents what will the formula be now? The formula will be W i upon N M that is work of separation per mole of gas mixture is equal to R T M into sigma j is equal to 1 to N sigma Y j log 1 upon Y j. So, if I got j is equal to 1 to 2 N equal to 2 I get Y 1 log 1 by Y 1 plus Y 2 log 1 by Y 2 if 1 and 2 are respective gases I can have N gases and therefore, I can have log 1 upon Y 1 plus Y 2 plus Y 3 plus Y 3 same terms will repeat and the summation of all this thing will give the work of separation per mole of gas mixture where Y j is the mole fraction of the j th component. So, basically this sigma expression comes normally we will have 2 component or 3 component or maximum 4 components not more than that. So, from the same thing this is the ideal work of separation I define one more term as I said earlier similar to the liquefaction system the figure of merit or FOM is also defined as figure of merit is equal to ideal work divided by actual work. The ideal work could be in molar basis or on a mass basis also say W i by N M divided by W i by N M you know the negative sign is basically the work done on the system that is why. Similarly, W i by N M which is on a mass basis now divided by actual work of separation actual work of separation must be much higher than the ideal work of separation and therefore, the figure of merit will always be less than 1. So, this is what we have learnt we have basically got an expression for work of separation we understood what a figure of merit would be of a given system. So, just to summarize what we did in the first lecture of this topic of gas separation I just take you through a some summary points. We just discussed different techniques employed for gas separation which are synthetic membranes adsorption, absorption and distillation. The separation can be done at both room temperature and cryogenic temperature when I use room temperature technique I use normally the method of adsorption or I can use membranes or I can use absorption while the distillation normally done at cryogenic temperature. In an ideal system all the processes are reversible and the work requirement in an ideal gas separation is called ideal work. So, we just got an expression for ideal gas separation a work done in order to or a work requirement for ideal work requirement for gas separation and what is that expression the ideal work requirement per mole of gas mixture to separate a mixture with n constituents is given by this general formula. So, general formula tells you W i by N m is equal to R T m sigma y j into log of 1 by y j when j is given from 1 to n constituents n equal to 2 then and therefore, a and b are the two gases then y a log 1 upon y a plus y b into log 1 upon y b where y j is the mole fraction of the j th component. We got a small self assessment exercise to given please go through that thing and kindly assess yourself for this. Based on whatever we have learnt in this lecture I am devoting more time in the lecture 2 for solving tutorials. So, that you understand the relevance of the formula which we have just derived in order to calculate the ideal work of separation and this is a small self assessment test please go through it and assess yourself there are answers given at the end. Thank you very much.