 So, welcome to the 21st lecture on the cryogenic engineering under the NPTEL program. Just to give a brief of what we did in the last lecture, I will just summarize some points and then we will go to the topic of our gas separation. Last time we had studied Gibbs phase rule which is basically given as f is equal to c minus p plus 2, where f is degrees of freedom or the intrinsic properties requirement to be specified for any mixture, c is a component and p is number of phases. Also, we understood about the temperature composition diagram and it gives a variation of mole fraction that is y with temperature t at constant temperature. So, when we talk about any mixture, the number of phases are 2. If we talking about 2 phase mixture or mixture separation by lowering of temperature, we got 2 components and therefore, we require some degrees of freedom or the values of f in there. If you got a 2 components, 2 phases or a single phase, we require a particular number of properties to be specified. Whenever we talk about a mixture, what we require is a mole fraction y to be specified of a component a and component b and these are very important diagram. These are very important relationship which have to be known in order to specify a mixture. For example, 2 component, 2 phase, we need to have 2 values to be known to us. We need to have 2 degrees of freedom which could be pressure and mole fraction or temperature and mole fraction for a given pressure. When a mixture condenses or a boils that means when a change of phase happens, there is a change of temperature and this is a non isothermal process which we saw last time that during condensation of a single component, the temperature remain constant. During the phase, the temperature remain constant which is true with condensation as well as boiling. However, for a mixture we have seen during the last time when I had presented the temperature mole fraction diagram or T y diagram, we have found that during the change of phase, the temperature decreases which means that the process of condensation or boiling is not a isothermal process. It is a non-isothermal process which is a very important deviation from what have we been learning from normal one component system. We also found that the repeated rectification of a mixture enriches the liquid and vapor phases with high and low boiling components respectively. We had seen that when the mixture is being condensed or vaporized repetitively, what you get ultimately is a high boiling component liquid, liquid at high boiling component and slowly we get on the repeated rectification process, we get the vapor phase getting richer and richer in the low boiling component. For example, we had seen for oxygen-nitrogen mixture that when you go on rectifying in the repetitive manner, the liquid will get richer in oxygen or what you get is a liquid oxygen at one end which is 100 percent oxygen then and the vapor phase will get richer in nitrogen gas, nitrogen vapor. So, ultimately what we do is basically condense this vapor in order to get liquid nitrogen but what is important is to understand that the one phase, the liquid phase gets richer in a high boiling component which is oxygen for a nitrogen oxygen mixture and the vapor phase get richer in low boiling component which is nitrogen in this case. So, in the earlier lecture we have seen the temperature composition diagram and we also introduced those diagrams how they can be exploited to understand the rectification process which is happens in actually rectification column. These diagrams form the basis of rectification process. The molar concentration of vapor and liquid phases of a two component, two phase mixture change with temperature this is what we saw. At every point during condensation during the phase change, we saw that a vapor phase get richer in low boiling component while the liquid phase getting richer and richer in high boiling component. This is what we just studied in the last slide also and this is what exactly happens as the temperature gets reduced the vapor phase will still get richer in the low boiling component and the liquid phase will still get richer in a high boiling component. Hence, because these diagrams are also critical because these mixtures have to be ultimately separated from in the rectification column what is important is to understand the basic laws which govern this mixture because these are going to be used every time. So, hence what we understood is there is a need to study various laws governing the properties of the mixture and therefore, in this particular lecture what we are going to study is mostly the properties which govern this mixture outline of this lecture is we will have to study the Dalton's law partial pressure which possibly you must have studied that we will have to repeat this again for mixtures now. Then very important law is the Raoult's law and then we got a Dalton's law plus Raoult's law also then we will also got a Gibbs Dalton's law and then we come to understand what is the basic significance of a distribution coefficient concept. What is this? So, basically a very important role in a rectification column theoretical analysis and finally, all these things could be understood better when we got some tutorials. So, I got two tutorials where you can understand the application of these laws from gas separation point of view. So, let us come to Dalton's law the Dalton's law was formulated by an English chemist John Dalton in the year 1801 for gas mixture. So, it is a very old law you can see that is almost 200 year old law basically and it has definitely been applied over a period of time for gas mixtures. It relates the partial pressure of an individual component of the mixtures, mixture has got various components and various components will be in the mixture in different proportion the molar fraction will be different and the partial pressure therefore, will be different of each component in a given mixture Dalton's law relates basically this partial pressure to the total pressure of the mixture and we will see now how it does. So, this relates the partial pressure of an individual component of the mixture to the total pressure of the mixture and to its mole fraction is a very simple law I am sure you must have studied this law earlier, but mathematically I will relate this law to the total pressure of the system. It is applicable only to non-reacting ideal gas mixtures so, this is a very important thing that we will not come across non-reacting ideal our mixtures will be real mixtures. So, this will basically govern the ideal gas mixtures, but we can formulate it for real mixtures also. So, simple illustration is consider a mixture of two non-reacting ideal gases so, ideal gases being gas A and gas B at a temperature T. So, I have just shown a small cylinder where gas A has got some component gas B has some component and in this mixture as you can see in a given enclosure that we have got some component of gas A and some component of gas B. Let the total pressure of the mixture B total and the partial pressure of gas A and gas B B P A and P B because they are present in partiality the total pressure is P total and depending on the presence or amount of gas A and gas B it will have is partial pressure which could be P A and P B denoted respectively. Also the Y A and Y B are the mole fractions of gas A and gas B. So, how much mole fractions if they got a 0.2 mole fraction of Y A and 0.8 mole fraction of Y B or thing like that depending on how much quantity of gas A and gas B is there in a given enclosure which will decide the molar fraction of Y A and Y B. By Dalton's law the total pressure of the mixture is equal to the sum of the partial pressure of the individual components. So, it is a very clear and simple law that P total will basically depend on the sum of partial pressure of individual component that is P A and P B. So, mathematically I can write P total is equal to P A plus P B. This is the total pressure of the mixture is equal to the sum of the partial pressures of gas A and gas B as you can see from here. Also the partial pressure of each component is directly proportional to its mole fraction. So, if I want to compute the value of P A the P A will basically depend on what is the value of Y A, P B will depend on what is the value of Y B and mathematically P A is going to be equal to P total into Y A and P B is basically equal to P total into Y B. So, what we can say is P A is directly proportional to Y A, P E is directly proportional to Y B and the constant of proportionality could be P total in both the cases. So, once I know what is my mole fraction of Y A and Y B or a gas A and gas B and if I know the total pressure of the mixture I can compute P A and P B directly and that is what the Dalton's law of a gas mixture it will give us conversion of molar fraction to P A and P B for the partial pressures and then what you can compute is a P total. Extending it to general statement to the mixture of n components we will have following that the total pressure P total is equal to sigma P j when j is ranging from 1 to n for n component mixture I think this is very simple and clear. And the partial pressure of the j th component is given by P j is equal to P total into Y j. So, depending on the value of mole fraction of a j th component and depending on the value of P total or the pressure of the total pressure of the mixture you can compute the partial pressure of j th gas j th mixture j th component. So, where Y j is the mole fraction of the j th component. So, one can find out the partial pressure depending on the molar fraction and if we know the partial pressure we can find out the total pressure of the mixture. Now extending further from Dalton's law having understood the basic of Dalton's law now we will go to Raoult's law which is also a very important law to understand the properties of mixture. So, let us come to now Raoult's law. Now consider a simple one component two phase mixture in a thermodynamic equilibrium as shown here. So, again in an enclosure we got a two phase system that means we have got some liquid the component is A or a gas A and we got a some liquid which has only gas A as a component you will have some vapor pressure basically you will have some gas A component above this liquid in an enclosure. So, what you have is basically a two phase mixture in a thermodynamic equilibrium each other, but it is a one component mixture it is not a mixture actually it is a one component system which has got only gas A as a component. The vapor above the liquid fails exerts a pressure and this is called as vapor pressure as you know a simple vapor pressure basically is a function of the temperature. So, I find out the temperature I can compute the vapor pressure for a two phase system. So, as I just said it depends only on the temperature of the system if I know the temperature entropy diagram I can compute if I know the temperature I can find out on a dome what is my saturated vapor pressure at that point on a dome this will give me the vapor pressure of that component for a given temperature we will see this again in two tutorials later. So, similarly now we have just seen a system of one component let us now look at a system of two components and two phase mixture in thermal equilibrium as shown. So, here you can see the upper illustration is only for gas for component A, but now I have got two components here which are A and B and again it is a two phase system and therefore, this both the components are in liquid phase as well as on a vapor phase above the liquid and they exert a vapor pressure on this liquid. So, they are in thermodynamic equilibrium with each other, but it will have different vapor pressure depending on the temperature of the mixture. So, the components are assumed to be ideal that means there is no intermolecular forces and chemically non-reacting. So, non-reacting is a case must, but what is important in ideal consideration is there is no intermolecular forces. Such a mixture is called as an ideal mixture or a perfect mixture. The Raoult law basically holds good for perfect mixture or ideal mixture. The real mixture will deviate away from the Raoult's law and therefore, this deviation has to be understood because in the real mixture will have intermolecular forces and this have to be considered if you want to really compute to a very high degree of real analysis if you want to do in that case we will have to take into consideration all the intermolecular forces. It is clear that the vapor above the liquid has both the components. So, in liquid condition also we got A and B. In the vapor condition also we will have A and B. The Raoult's law was formulated by a French chemist Francois-Mary Raoult in the year 1882. This law also is a very old law almost again 130 years or something like that and this law also has been applied over a period of time and therefore, its importance has been well understood. So, consider a perfect mixture of gas A and gas me in thermodynamic equilibrium at temperature T and let gas A as following parameter. So, the parameter associated with gas A or P A which is this partial pressure x A is a mole fraction in liquid phase. Now, you have got a liquid phase coming into existence till now in the Dalton's law we talked only about the gas mixture itself. Now, we are talking about both liquid and vapor over here and therefore, the component what we identify in liquid is x A. So, x A is a mole fraction in liquid phase while y A is a mole fraction in the vapor phase. So, because we have got both the phases over here we got a concepts of x and x A and y A. Similarly, we will have x B and y B for gas B we will not talk about that right now we are talking about only gas A and pi A is a very important parameter we have just talked about this also earlier which is a vapor pressure at temperature T. Now, this is not for a mixture we are talking about this is independent or a characteristic of any component for a gas A if we know temperature we know the vapor pressure at that particular temperature. So, pi A is the one which gives me vapor pressure of gas A at temperature T alright. So, you got a four component which is a partial pressure mole fraction in liquid phase mole fraction in vapor phase and pi A is a vapor pressure at temperature T. Now, basically I would like to relate all these things not only for gas A, but for gas B also how do I relate that? Routh's law relate this parameter and therefore, let us see that in next slide. So, the statement of Routh's law is that the partial pressure of a component in the vapor phase is directly proportional to the mole fraction of that component in liquid phase. So, what is the partial pressure of a component in vapor phase? It is P A and this is directly proportional to the mole fraction of that component in liquid phase we just found that x A is the mole fraction of that component in liquid phase. So, basically the Routh's law relates the partial pressure in the vapor phase which is of this component is a gas A above the liquid. The vapor pressure will be for only for the vapor which is above the liquid and the vapor pressure exerted by gas A or gas B is related to the mole fraction of that component in the liquid phase. So, vapor pressure P A is going to be related to x A, x A is the mole fraction of that component in the liquid phase. So, my mathematical statement therefore, becomes P is a function of f x which is what we just stopped which is P is the partial pressure and function of x and what is x? x is the molar fraction of that component in the liquid phase. So, for gas A now, if I apply this law for gas A I will get P A is equal to pi A x A. It is clear here P A is related to x A and what is that constant of proportionality? It is the vapor pressure at that temperature at the mixture temperature. So, P A is equal to pi A x A is a very important result as to which is coming out from this Routh's law statement. I am relating the partial pressure of a vapor in a given mixture to the molar fraction of that component in liquid phase and it is basically getting related by the vapor pressure of gas A which is a characteristic pressure at a given temperature of mixture alright. So, if I know x A I can compute its partial pressure exerted by the vapor of that component in a given mixture or in a given enclosure at temperature T. Similarly, for gas B if I apply the same rule for gas B I will get partial pressure exerted by vapor component B on this mixture of A and B and it is related to the component x B and corresponding to gas the value of pi will be pi B which is the vapor pressure of gas B at the temperature T. This is a very important relationship between x A and P A and x B and P B extending it to the mixture with n components. We have just seen what is the value of P A and what is the value of P B. If I talk about n components I will have partial pressure of a j th component given as P j is equal to pi j x j same thing not A and B anymore, but I am applying it basically for any j th component if you have got n number of components in a given mixture where x j is the mole fraction of the j th component pi j is the vapor pressure of the j th component at temperature T it is clear I am just generalizing what we studied earlier. Now we have just studied Dalton's law we have now studied Raoult's law if we combine both the laws for a given mixture what we get is a very important relationship. Now if the vapor above the liquid phase is assumed to be ideal because you know Dalton's law is basically meant for gases vapor is not gas, but if I assume that vapor above the liquid phase is assumed to be ideal then combining the Dalton's law and the Raoult's law what we get is a two things I will just put in Dalton's law I will then combine with it with the Raoult's law. If I say Dalton's law what I know is a P total is equal to P A plus P B the total pressure is equal to the sum of the partial pressure which is a P A plus P B and what I get from Raoult's law is the value of P A and P B. So, P A is equal to pi A x A and P B is equal to pi B x B where x A and x B are the molar fraction in the liquid of component A and component B respectively while pi A and pi B are the vapor pressure of those gases at a given temperature. So, now I can basically relate the P total to this P A and P B taking the value from P A and P B. So, what my expression will be so P total I will just this P B P A by pi A x A plus pi B x B. So, I am now what I am doing is basically I am relating the total pressure to the x A and x B respectively to the molar fraction of component A and component B in a given mixture in the liquid x A and x B. So, extending further for the liquid phase the following statement now holds good which is we know that x A plus x B is equal to 1 all right the molar fraction sum in a given mixture is going to be equal to 1 and P total what we have just seen is equal to pi A x A plus pi B x B. If I just now replace this x B is equal to 1 minus x A and put it here in this expression what I get is P total is equal to pi A x A plus pi B into 1 minus x A obvious and therefore, I can do the otherwise also I can put x A is equal to 1 minus x B in the same statement. So, my P total is equal to pi A into 1 minus x B plus pi B x B I am just replacing the value of x B in this expression and the value of x A in this expression. And now I would like to put this graphically and this is what in short the root law will be. So, you can see that I have got 2 axis on which the P is mentioned and I got 2 end points of x axis on which I have given a molar fraction of x A. So, if I go from x A is equal to 0 to x A is equal to 1 which is 100 percent A and I got x B is equal to 0 and x B is equal to 1 here at 2 ends at in between values I got some value of x A for example, here I got x A is equal to 0.25 and x B is equal to 0.75 here I got x A is equal to 0.5 x B is equal to 0.5 here at the end I got x A is equal to 1 and x B equal to 0 that means 100 percent A exist over here 100 percent B exist over here. What I am plotting on this is basically P A and P B. So, P A is nothing but pi A x A which is given in a red color over here. So, the vapor pressure partial pressure of P A at A is equal to 0 is going to be 0 while A is equal to 1 is going to be maximum. So, P A ranges from 0 to P A max here similarly, the partial pressure of B will range from 0 to max here and the total pressure is at every point is P A plus P B. So, at the two end points the total pressure is equal to in fact, P A is equal to 0 and therefore, P total at this point will be equal to P B while the P total at this point will be equal to P A and at an every intermediate point we will get algebraic sum of this vertical heights which is given by P A plus P B and this is my P total. So, P total at every point is going to be some of the partial pressures of P A and P B which is what we have understood from Dalton's law and Raoult's law. So, this in actually this graphical representation of the Raoult's law and the Dalton's law alright. So, I think just to understand in more graphical format this will help and the values of P A and P B are related to x A and x B by the Raoult's law which is what we just saw. Now, going ahead from here I have got one more law actually just a deviation of the Dalton's law and it is called as Gibbs Dalton's law. Again assuming the vapor pressure above the liquid to be ideal, this is what we did earlier also Gibbs Dalton's law is application of Dalton's law to this vapor. What is this showing basically I am applying Dalton's law for this vapor assuming that the vapor above are behaving ideally like ideal gas basically. So, Gibbs Dalton's law is nothing but application of Dalton's law to this vapor. So, let P A and P total be the partial pressure of gas A and total pressure of the mixture respectively which is what we have seen earlier also. Let Y A is the mole fraction of gas A in a vapor phase which is what we have same nomenclature has been used as what we have done earlier. The application of Dalton's law to the vapor above the liquid we have got following now was what we are doing now basically relating P A to Y A as that what we have done earlier. For a gas A in a vapor phase what we know is P A is equal to P total into Y A this is what we have done. The partial pressure of gas A is related to a smaller fraction in the vapor phase and is the constant of proportionality is P total. So, P A is equal to P total into Y A similarly for gas B in the vapor phase we write P B is equal to P total into Y B this is what we have done earlier Gibbs Dalton's law is basically applying the same thing assuming that the vapor behaves as ideal gas extending this to the n component mixture and for the j th component what we know is a P j is equal to P total into Y j this is the same thing applied for any j th component in a n component mixture. Now if I combine Gibbs Dalton's law and Raoult's law together to the vapor above the liquid phase I get the following. So, if I go by Gibbs Dalton's law I got P A is equal to Y A into P total alright and the Raoult's law tell me that P A is equal to pi Y A x A. So, here I am applying the law for the vapor here I am applying the law for the liquid component. So, you understand that ultimately I am relating the vapor component Y A to x A which is a liquid component is a very important therefore, I will get the vapor composition I will get the liquid composition for gas component A. So, from here I get Y A is equal to pi A upon P total into x A or pi A x A upon P total it is a very important thing that I am relating Y A which is a vapor component vapor fraction of gas A and x A is a molar fraction in a liquid component in the component A in a liquid form. So, basically I am relating Y A and x A alright similarly for a B component we got a P B is equal to Y B into P total and correspondingly Raoult's law gives me that P B is equal to pi B x B exactly on the same line as what we did for component A. So, if I relate this thing to Y B now I will get Y B is equal to pi B x B upon P total. So, basically again I am relating Y B and x B in this case. So, if I just summarize whatever laws we have studied and then we can go ahead from there the Dalton's law relates partial pressure of non-reacting ideal gases. Raoult's law relate the vapor pressure with the liquid mole fraction of a component the Gibbs-Dalton law application of Dalton's law to the vapor above the liquid phase and we found that P A is equal to P total into Y A is basically applied to the vapor now assuming that the vapor is behave like a ideal gas. And lastly what we did was we combined Raoult's law and give the Dalton's law together and it establish a relationship between the vapor and the liquid fraction of a component. This is very important law because ultimately we wanted to come to this that Y A is getting related to X A. Now going ahead from here we define much more constant now which are very very important from mixture perspective. As derived earlier consider an equation with Y A and X A for gas A which is Y A is equal to pi A X A upon P total. Rearranging this what we get is Y A upon X A is equal to pi A upon P total alright. What is pi A is a vapor pressure of component A at a given temperature T of a mixture. What is P total is a total pressure of the mixture. So actually to say they are basically both constant for a given temperature and pressure of a mixture. If I know the temperature of a mixture if I know the total pressure of a mixture Y A and X A the ratio of Y A and X A essentially remains a constant alright as long as the temperature and the pressure of the mixtures are defined. So what we can understand from here is for a given temperature and pressure I have got Y A upon X A coming as constant and this is a very important finding alright. So the ratio of Y A X A Y A to X A is basically a constant and it is called as distribution coefficient. It distributes at a given temperature the value of A in vapor phase and the proportion or molar fraction of A in a liquid phase and this is called as distribution coefficient and is denoted by constant as K A. Similar value we have for Y B and X B also which will denoted as K B. But what is important to understand that Y A and X A have a ratio which is Y A upon X A called as distribution coefficient and given as K A. If I want to define that thing it is the ratio of the mole fraction of a component let us say gas A in vapor to liquid phases in a mixture at a given temperature. So if I know the temperature if I know the pressure of a mixture then Y A upon X A at a given temperature is called as distribution coefficient and it is constant alright. It is constant given by K A similarly I will have value of K B. So extending general to the j th component I have got a K A I have got a K B and if I go for a K J for example for a n component mixture I will have K J is equal to Y J upon X J and this will equal to pi J upon P total. So pi J is a vapor pressure of a j th component at a temperature T P total is a mixture pressure basically given as P total. So again this is going to be a distribution coefficient for the j th component which is going to be constant. So Y J is equal to K J X J which is what we know the distribution coefficient K for an ideal or a perfect mixture is determined using the boy equation. So if I know if I know the temperature if I know the pressure I can calculate the K J value and this is an ideal or a perfect mixture. As I say the Raoult's law basically holds good for ideal mixture or a perfect mixture or a perfect solution wherein there are no intermolecular forces acting alright but the non ideal or real mixture it is determined experimentally. So these are two ways of doing that thing one is to get experimental values which are also distribution coefficient obtained experimentally or by empirical correlations while one can calculate based on the ideal gas law etc ideal assuming that there are no intermolecular forces and therefore one can get the distribution coefficient theoretically also. We got one tutorial later to study if you could calculate the value of distribution coefficient theoretically and also experimentally and compare these two values and therefore you will be able to see what different does it make if I calculate it theoretically and if I get the value experimentally. So now for an ideal two phase mixture wherein we have got a gas A and gas B the relationship between K A and K B also has to be made clear alright. So K and K B also get related and the liquid mole fraction X A and X B. So ultimately I would like to relate in a given mixture where gas A and gas B are present how are K A and K B are related to X A and X B because they are having a mixture they will basically contribute to the calculation of value of X A which is a molar fraction of gas A in a liquid phase and X B is a liquid mole fraction of the component B. So how do I do that I know Y A is equal to K A X A I know Y B is equal to K B X B and I know that Y A plus Y B is equal to 1. So if I apply the same law for vapour we have already have done that for X A we know X A plus X B is equal to 1 similarly we know that Y A plus Y B is equal to 1 and Y A and Y B values are given over here. So if I write K A plus K A X A I just replace this K Y A by K A X A and Y B by K B X B I know K A X A plus K B X B is equal to 1 and this is what relates the value of K A and K B and X A and X B. So we also know that X A plus X B is equal to 1 and if I therefore replace the value X B here as 1 minus X A I get a K A X A plus K B into 1 minus X A is equal to 1 and therefore from here I can compute the value of X A is equal to 1 minus K B upon K A minus K B. So if I know the distribution coefficient of gas A and gas B that is K and K B for a given temperature and pressure of a mixture then I can compute theoretically what is the value of X A. Similarly, I can do the same thing for X B also here if I put X A is equal to 1 minus X B in the above expression I replace this value of X A as 1 minus X B and compute this I will get the expression for X B which is X B is equal to 1 minus K A upon K B minus K. If I know the distribution coefficient for K B and K A I can compute for a given temperature of pressure what are the values of X A and X B respectively by this formula. So very important thing theoretically I will know the value of K A and X B I can compute those value as I said earlier theoretically or I can get the value of K A and K B which are available the experimental data is available for a given mixture in that case I can theoretically compute what the value of X A and X B will be there for a given temperature in a mixture. So, the distribution coefficient K A is equal to Y A upon X A which is equal to pi A upon P total as mentioned earlier K is a ratio of the mole fraction of a component in vapor and liquid phases in thermodynamic equilibrium this is what we have defined K as it is meaningful and defined only to two phase region normally K A is given when component gas A and gas B will be together and then only there is a there is basically requirement to calculate the X A and X B Y A and Y B respectively for a mixture. So, normally the values are going to be defined from the mixture component gas A and gas B in a two phase region that means from boiling point of A up to the boiling point of B above which there is no component left as it is basically. So, we define it normally at a given pressure between the two boiling points and therefore, the two phases of the regions. For example, the mixture of nitrogen and oxygen at one atmosphere exists in two phases between 77 Kelvin at 90 Kelvin the boiling point of nitrogen and the boiling point of oxygen at one atmosphere. Hence, normally the K is defined in this interval only the same thing could be at two atmosphere or five atmosphere or whatever pressure we should talk about and corresponding to those pressures will have the boiling points of nitrogen or oxygen or component A and component B and K value will be defined during that temperature range. The values of K are determined experimentally at some reference pressure P O. Now, when I talk about experimental determination of these values of K it is normally given with a reference pressure P O which could be one atmosphere a two atmosphere or five atmosphere and wherever these values are given the value of P O will be given to you and therefore, your actual mixture pressure could be different which is just P or P mixture, but the values will be given for that pressure with some reference pressure which has to be taken into account to calculate the value of K. So, in the literature the values of K are normally given in logarithmic form and it is always given in this form log of K into P mixture upon P O which means that if my mixture pressure is one atmosphere the value one over here and if I having P O also is defined as a reference point is a one atmosphere then these two will get cancelled in that case the P mixture is going to be equal to P O and the log K value will be given in that table in that case. So, normally it is given in the logarithmic form from which and once we know this logarithmic value the value of K has to be recalculated from this expression alright. So, this is what experimental values will be given and the standard tables are available which I will show you in the next slides. The data for nitrogen, oxygen and argon are given in the next slide and a tutorial has been solved later to understand this conceptually. So, how does this data look like? The data is given in log K P mixture upon P O for oxygen nitrogen as given below. So, this is what you can see. So, I got a data for our nitrogen which is at one atmosphere P O is one atmosphere that means my reference pressure is one atmosphere. Similarly, I have got data for oxygen again the reference pressure is one atmosphere. However, my mixture can vary from one atmosphere to atmosphere, five atmosphere and whatever I have just shown 1, 2 and 5 and corresponding to this mixtures. Now, I got a temperature range as I said the temperature range for one atmosphere will be from 77.4 to let us say 90 because that happens to the boiling point of nitrogen and oxygen respectively. So, the data is given as ranging from around 77 to 90 and as you can see that the high boiling component when it goes on to low boiling component basically this is nothing but representative value of y by x. So, as you see that as you go from high boiling component to low boiling component the vapor component y by x go on less and less and therefore, the value of y by x or logarithmic value goes on getting reduced and normally what we have found is around the boiling point it comes nearly to somewhere around one which is what we can see later. Similarly, this value has been given for two atmosphere normally at two atmosphere the boiling point of the nitrogen will get increase and therefore, the table has been given from 84 onwards which may go up to 94 which may happen to the boiling point of oxygen at two atmosphere. The boiling point does not figure in this range for five atmosphere. So, it comes in the next page. Similarly, for oxygen what you can see is for one atmosphere in this case because it is below its boiling point the logarithmic value happens to be in a negative range that means the values are less than one in this case. The values of k are normally around one at the boiling point which is what we will see later the values of k happen to be around one. So, whenever there are negative points they happen to be less than one in that case. So, at two atmosphere again I got different table I got a five atmosphere because the boiling points get raised I got a different table. So, here you can see that the five atmosphere figures in a higher temperature range 94, 2108 for both oxygen and nitrogen respectively alright. So, this is what the values will be given as if I just want to against similar thing is shown for Aragon also and these values have to be taken from this table 1, 2, 5 atmosphere and I think this is for oxygen. So, if I want to just show it as a value of k what you have seen in a table was logarithmic value, but if I want to show this on a graph it is just the value of k with temperature at different pressure. So, if I say nitrogen at one atmosphere you can see this is a red line and this ranges from 90 Kelvin up to its boiling point which is 77 Kelvin and you can see that at one atmosphere at 77 Kelvin the value of nitrogen or a value of k at 77 is hitting around one while it has gone up to 4.2 at 90 Kelvin. Similar thing holds good for oxygen water atmosphere where you can see that oxygen value at around 90 Kelvin coming closer to one while below 90 Kelvin as I said earlier it is coming less than one and therefore, the logarithmic value of k will be shown as negative which is what we have seen earlier. If I go for nitrogen 2 atmosphere because at nitrogen 2 atmosphere the boiling point will be increased and therefore, it will possibly go from 84 Kelvin up to 96 Kelvin and in between around this point around 90 or 84 will come the value equal to 1 which happens to the boiling point of nitrogen at 2 atmosphere. If I apply the same thing for 2 atmosphere oxygen now it will range from 84 and go up to 96 and around this what we will have is a boiling point of oxygen at 2 atmosphere and below that the value of logarithmic value is going to be shown as negative or the value of k is going to be less than 1. So, this is just to give a understanding of what is the comparative value of k at different temperatures and pressure. The as I said the thumb rule is if you are near the boiling point of that corresponding to that pressure of nitrogen or oxygen it is going to be around 1 below that is going to be less than 1 above the boiling point is going to be more than 1 and the logarithmic value of this less than 1 is going to be shown as negative in this case. So, variation is shown here for nitrogen oxygen we found that the k decreases with the decrease in temperature for any given pressure. So, decrease the temperature decrease the value of k the component with lower boiling point has a higher k. So, this is clear that because of the boiling point being the minimum in this case the values of nitrogen the k values of k for nitrogen is going to be more as compared to other gases in this particular expression. For any component the value of k approaches to be 1 for a given boiling point for its boiling point also the value of k is less than 1 when the temperature is below the boiling point of 1 this is what I just talked that below 77 Kelvin the value of k is going to be less than 1 or it is going to be logarithmic value is going to be negative in this case. So, when k is less than 1 the logarithmic value of k is negative and that is what we saw in the table. So, with this background whatever we have understood till now about the concept of y a x a the distribution coefficient k a k b in a given mixture of gas a and gas b. In order to clear some concepts I would like to take two tutorials and once you solve this you will lot of concept will get clear. So, my tutorial one is consider a mixture of nitrogen and oxygen at 5 atmosphere and temperature is 100 Kelvin. The mixture temperature is 100 Kelvin the mixture pressure is 5 atmosphere. What I want to calculate? Calculate the distribution coefficient of nitrogen and oxygen also calculate the vapor and liquid composition using the obtained k values. So, basically I want to calculate the distribution coefficient and once you know the value of k calculate the vapor and liquid compositions using the values of k that is the problem. Use the data from the table given in the earlier slide. So, working pressure is 5 atmosphere temperature is 100 Kelvin mixture is nitrogen oxygen. This is what the problem definition this is what the problem is given as. So, what do I want to calculate? If I want to convert that to my understanding of the nomenclature which we have been using till now is I would like to calculate k n 2 which is the distribution coefficient of nitrogen. I would like to calculate k o 2 which is the distribution coefficient for oxygen which is what we termed as k a and k b earlier. I am now want to give a values to it I want to give the nomenclature to it which is k n 2 and k o 2. Also corresponding to that I have got a x n 2 x o 2 which are the mole fractions of nitrogen and oxygen in liquid phases which were earlier called as x a and x b. Similarly, I have got y a n 2 and y o 2 as the mole fractions of nitrogen and oxygen in vapor phases earlier they were called as y a and y b. So, basically I want to calculate k a k b x a x b y a y b as k n 2 k o 2 x n 2 x o 2 and y n 2 y o 2 in this problem and they have to be done at 5 atmosphere working pressure and temperature at 100 Kelvin. I first want to get the value of k n 2. So, I go through the table and get the value of k at 100 Kelvin. So, if I go to the table I get the value of 100 k value which is given in a logarithmic form log of k p max p mixture upon p o for nitrogen. I go to the table I get a temperature and for 5 atmosphere and locate the value of t is equal to 100 Kelvin and this logarithmic value happens to be 2.004. Having done this for nitrogen I would like to calculate what is the value of k from here. So, putting the value of k n 2 removing the logarithmic term what you get it p o upon p max p p mixture e to the power 2.004 alright and then what are the value of p o and p mixture p o is 1 atmosphere p mixture is 5 atmosphere. Putting the value of e is 1.5 e to the power 2.004 I get the value of k n 2 as 1.483 this is my distribution coefficient for nitrogen at 100 k and pressure of 5 atmosphere. Similar thing I will do for k o 2 also for the distribution coefficient for oxygen. So, for k o 2 I take a table of oxygen at 5 atmosphere and I locate the 100 Kelvin in that. So, logarithmic of log of k mixture upon p o for oxygen at 100 k and 5 atmosphere is 1.042 from where I will calculate the value of k o 2. k o 2 is given as p o upon p mixture e to the power 1.042 and putting the value of p o and p mixture in that I can again the value of k o 2 is equal to 0.567. So, I got now k n 2 I got k o 2 once I now k n 2 and k o 2 I will have to calculate the value of x n 2 and if I relate earlier expression I got x n 2 is equal to 1 minus k o 2 upon k n 2 minus k o 2 and similarly I have got the relation for x o 2 also which I will exploit in the next slide. So, once I know the two distribution coefficient for k n 2 and k o 2 I can calculate x n 2 put in the value of k n 2 and k o 2 as what we have just calculated I will get the value of x n 2 is equal to 0.472 clear I think it is very simple and straight forward having done this I can do the same thing for x o 2 where I know x n 2 plus x o 2 is equal to 1. So, 1 minus x n 2 will give a value of x o 2 just put those values and x n 2 is equal to 0.472 which we just calculate value of x o 2 therefore, is 0.528. So, I got now x n 2 and x o 2 and I will get value of y n 2 and y o 2 also that comes automatically when I know distribution coefficient. So, y n 2 related to x n 2 by y n 2 upon x n 2 is equal to k n 2. So, if I know the value of k n 2 and x n 2 I can straight away calculate the value of y n 2. So, just put 1.483 into 0.472 is equal to 0.66699 which is nothing but y n 2 and similarly, y o 2 is nothing but we can calculate in this way also we can compute by y o 2 is equal to k o 2 you know into x o 2 which is also known to us just do 1 minus y n 2 I get y n 2 is equal to 0.699 and for y o 2 is equal to 0.304 and therefore, I got all the values of y o 2 and y n 2 and x o 2 and x o 2 which is a vapour molar fraction in the vapour for oxygen and nitrogen and molar fraction in the liquid which is x n 2 and x o 2 which is what the tutorial asked for. Having done this now, let us go to the next tutorial wherein we talked about a different requirement now. So, what is the next tutorial now? This is the first tutorial just to calculation of k n 2 and k o 2 basically not calculation taking the value of k n 2 and k o 2 from the table and relating it to y n 2 y o 2 x n 2 and x o 2. The second tutorial is now consider a two phase mixture of nitrogen and oxygen again at a pressure of 2 atmosphere use the T s diagram for the vapour pressures of n 2 and o 2 at 86 Kelvin. So, what I want to do is get a vapour pressures at n of n 2 and o 2 at 86 Kelvin and determine the liquid and vapour composition of the mixture if the temperature of the mixture is 86 Kelvin. Now, I cannot use the table this is what is basically mentioned over here that I should use the vapour pressure of nitrogen oxygen at 86 Kelvin and this is basically in order to calculate the value of k n 2 and k o 2 theoretically. As you remember in the theoretically we do not consider real gas values operate vapour as real vapour whether no intermolecular forces are considered theoretically and that is where the this is coming from that I would like to take a vapour pressure value of nitrogen and oxygen at 86 Kelvin. Also calculate the value of k n 2 and k o 2 that means get the value of k n 2 k o 2 from here and compare that with the experimental data. The problem definition is compute k n 2 k o 2 theoretically and compare them with the experimental available value of k n 2 and k o 2 very interesting so that you can understand how to calculate it theoretically which we have already seen actually in principle how to get the k n 2 and k o 2 values and compare these values with the experimentally obtained k n 2 and k o 2 which are basically obtained from the given table earlier. So the given is working pressure is 2 atmosphere temperature is 86 Kelvin for a mixture of nitrogen and oxygen and what do I have to calculate? I want to calculate x n 2 x o 2 which is a mole fraction of nitrogen and oxygen in a liquid phase y n 2 and y n 2 which is a mole fraction of nitrogen and oxygen in the vapour phase and k n 2 and k o 2 which are the distribution coefficient of nitrogen and oxygen respectively for given temperature 86 and pressure 2 atmosphere. So how do I go? How do I start? As I said I am told in the problem or the tutorial that I have to take a vapour pressure of nitrogen let us say at 86 Kelvin. So corresponding to that I will get these values from the T s diagram and if I look at the temperature of 86 Kelvin in these corresponding to this value I can read the pressure. So this is basically saturated vapour curve. So this is a basically 2 phase equilibrium with the gas and the 2 phase the coming vapour and the liquid are coexistent there in thermal equilibrium on this saturated vapour curve and therefore wherever this horizontal line hits or intersect this saturated vapour curve corresponding to that whatever is the pressure is the vapour pressure at 86 Kelvin and therefore this happens to be at 2.517 atmosphere and this is the vapour pressure of nitrogen at 86 Kelvin and this is what I require which is my value of pi n 2. This is a function of only temperature and vapour will exist only up to the critical temperature of nitrogen. So therefore vapour pressure is given normally below the critical temperature of a given component or gas. Following the similar procedure for oxygen we have the vapour pressure for oxygen also which happens to be 0.64 atmosphere. So pi n 2 is 2.517 atmosphere and same thing is repeated for oxygen pi o 2 is 0.640 atmosphere. So having done this I can now calculate the value of k n 2 as p total pi n 2 x n 2 pi o 2 into 1 minus x n 2 these are the values which have to be theoretically applied x n 2 therefore is equal to algebraically I will get these values putting these values I will get x n 2 as 0.724 alright. So what I know is vapour pressure of nitrogen vapour pressure of oxygen at 86 Kelvin I know the value of x n 2 and I can calculate therefore the value of x n 2 depending on p total and the value of vapour pressures over here. So I get x n 2 is equal to 0.724 similarly x o 2 is 1 minus x n 2 I get x n 2 as 0.724 and x o 2 as 0.276. I will get the value of pi n 2 in the same way where I know the total pressure I know the pi n 2 I know the x n 2 having done this I will just put those values over here and y n 2 is calculated as 0.911. So once I know the pi n 2 upon p total which is what we have calculated I can relate that y n 2 with x n 2 similarly I can get y o 2 as y n 2 plus y o 2 is equal to 1 and 1 minus y n 2 will get y o 2 and therefore y o 2 is 0.089. I can get the value of k n 2 by just finding out the ratio of y n 2 upon x n 2 because the definition gives k n 2 putting those values over here I will get the value of k n 2 to be equal to 1.2583 understand that the value of k n 2 has been completed at 86 Kelvin which is avoid boiling point and therefore value of k is more than 1.25 here and similarly the value of k o 2 is going to be 0.089 upon 0.276 which is 0.3224 again understand that the value of k o 2 is completed at 86 Kelvin which is less than its boiling point and therefore it is happens to be less than 1 which is 0.3224 alright. So calculated if I want to see the value of k calculated which is y upon x and I get calculated value of k n 2 as 1.25 and k o 2 as 0.3224 and if I take the similar values from experimental now I can see the experimental values of k n 2 are 1.2335 and k o 2 is 0.36. So you can see that the difference between the two values theoretically calculated experimentally calculated difference is not too much but there is a difference and therefore sometimes this experimental data can be taken sometimes this calculated data can be taken but it is not possible to every time you know calculate the values because some values may not be available or it is not possible sometimes to have all the experimental data available in your table and therefore the compromise may have to be done for a given range of temperature. So if I compare this to you the difference is basically owing to the ideal values differed from the experimental value by small amount and this is because the effect of intermolecular forces is neglected in the ideal mixtures and this is what I have been talking about earlier. So summarizing my lecture today what we understand is Dalton's law relates partial pressure of non-reacting ideal gases. Raoult's law relates the vapor pressure with the liquid mole fraction of a component in a mixture. So y a is getting related to x a gives Dalton's law is an application of Dalton's law to the vapor about liquid phase which is what we understood and we found that distribution coefficient k is a ratio of mole fraction of component in vapor to liquid phases. So k is equal to y by x it is meaningful and defined only in a two phase region of a mixture alright. This is what we did I got a small assignment for you which is for a nitrogen oxygen mixture at one atmosphere and at a T Kelvin. Please find out the vapor pressure and compare the data with experimental value and this is the answer for that. The problem is given to you these are the answer please solve the problem for yourself. Thank you very much.