 So, today we will have 2 hours of lectures on combustion. So, before we start on combustion I mean this is some subject which is slightly peripheral to thermodynamics it will be used mostly when you are doing applied courses like power plant engineering or IC engines, but there are a few concepts that probably need to be reiterated. I mean many of the concepts are related to chemical reactions and there is of course the concept of adiabatic flame temperature which must be told and if this is known and reasonable knowledge is given I think it will be very useful in the future courses where power plant engineering or IC engines is taught. So before we begin combustion let us see what fuels we normally encounter and we will realize I mean I would not spend too much time on fuels. So, as engineers or as mechanical engineers we will encounter solid, liquid and gaseous fuels. So, the solid fuels are mostly used in external type of combustion engines. So, this would be you know most of the plants which are run on Rankine cycle they would use for example, coal as the main fuel of course there are lot of plants nowadays which use biomass and garbage, but coal is the most standard fuel that we use in our power plants and maybe you know some background students already have you know that coal is formed due to high pressure on plants and animals which have been buried for long and there are various forms of coal which are you know in the formation region like peat or lignite and lignite is a reasonably well formed coal that is used often and the most common coal that we use in Indian power plants is bituminous coal. One thing you can emphasize is that Indian coal is actually has a lot of big percentage of ash and the better coals are what are called as anthracites, but these are mostly used for metallurgical processes where pure coal is required otherwise in power plants we normally go for bituminous coal. So, if we come to liquids you all know that crude oil is obtained from various oil fields and this crude oil is later refined and undergoes a process called fractional distillation at different temperatures different fractions will liquefy out and finally, whatever remains in the gaseous form will probably be there at in gaseous form at room temperature and you know that most of the fuels that we encounter are going to hydrocarbon. So, coal is more or less purely carbon with some ash and other things, but most of the liquid and gaseous fuels are purely hydrocarbon and anything with more than C 5 is usually would be and above would be in liquid form. So, for example, you know what is normally called as gasoline or petrol the most common compound which is associated with gasoline is iso octane, but in general petrol consists of everything between around C 5 to C 9 a huge range and the iso octane is one of the biggest components that even thing goes in there. However, it just mixture of many compounds if you go slightly higher towards C 10 you get what is called as kerosene or sold as kerosene and if you go slightly higher towards C 12 you get what is called as diesel and if you go even higher you know the liquid starts becoming thicker and these are then no longer used in iso engines and used in furnaces and you get what is called as furnace oil you know close to C 20. And beyond this if the hydrocarbon as per molecule more than 20 carbon atoms the thing starts becoming stickier and thicker and you know at very high carbon content C 60 and all you will get what is called as star. So, these are the things that we obtain from crude oil and you know most of these are used in either iso engines or power plants. So, this probably needs to be told and finally, we have gases. So, you know anything below C 5 would be a gas. So, for example, you know butane and propane these are gases that remain after fractional distillation and butane and propane together are liquefied and sold as LPG that is liquefied petroleum gas. It is a combination of butane and propane and may be some ethane in there and when we come to methane it is mostly sold as natural gas that is because in most marshes etcetera the gas which comes out naturally after fermentation is methane and hence by default methane is called as natural gas. So, whatever is sold as natural gas is actually around 90 percent methane and the remaining could be ethane and butane and so on. So, and natural gas is used in CNG engines etcetera. So, this is roughly how the fuels would be and of course, you I think once this background is given regarding fuels we can come to the combustion of these fuels and I think here you must remind students that you know mostly it is finally a chemical reaction and they have learned chemical reactions in eleventh and twelfth standard and the first thing that we can probably tell is balancing of reactions. So, the simplest species to take normally and that is what we would start with is start with methane. Methane is C H 4 we require some amount of O 2 let me not put how much and I will get C O 2 plus H 2 O. So, this is when you can tell them to start balancing this reaction there is they know what balancing means and basically every element on the right hand side and left hand side has to be balanced and the reason is that we are not looking at nuclear reactions where atoms of one element will transmute into atoms of other elements. So, every element the atom should be balanced on either side. So, the carbon is one on the left side it is one on the right side. So, that is pretty good hydrogen you know there are four on the left side two on the right side. So, if I just add two here it will be four. Now, if I look at oxygen C O 2 consists of one O 2 and twice H 2 O will have another two O's. So, totally I require four O's on the left side. So, I will just put here too. So, this is the simplest balancing that I could think of, but here you must now tell that know whenever we burn fuel and we require some oxidant and we have just put oxygen here and of course, it must be emphasized that fuels need not be only hydrocarbon I mean rocket fuels do not consist of hydrocarbon and oxidants need not be purely oxygen I mean they can consist of other elements, but as far as we are concerned we look primarily at hydrocarbons and oxygen because this is what is used in all the heat engines that we operate and this is what is more important for us as mechanical engineers. So, oxygen is not just available directly, but it is available only in the form along with nitrogen in air and air is the most common oxygen that we will always use. So, at this point you must emphasize that all our calculations for IC engines power plant etcetera will involve the presence of nitrogen and it is very important to account for this nitrogen because it affects the net temperature of whatever we are using and at this point you can say see all these are on a molar basis here because I am saying 1 C H 4 means 1 mole of C H 4 twice O 2 would mean 2 moles of O 2, 1 C O 2 would mean 1 mole of C O 2, 2 H 2 O would mean 2 moles of H 2 O. So, in the atmosphere you know by volume there is around 21 percent oxygen and remaining 79 percent of nitrogen. So, moles and volume correspond to each other because equal number of moles occupy equal volume irrespective of the species. So, that means for 1 mole of oxygen there are 79 upon 21 moles of N 2 which roughly turns out to be 3.76 moles of nitrogen. So, hence what you must emphasize is that this reaction though is not was initially written only as a balanced chemical reaction as far as we are concerned it must be written as such C H 4 plus twice O 2 plus twice into 3.76 N 2 you see O 2 plus twice H 2 O the nitrogen does not take part in any reaction and I will just write it as twice into 3.76. So, on both sides I have added nitrogen and it is remaining unburned. Of course, you will realize that when we talk of very high temperatures we may have to account for dissociation of nitrogen or nitrogen reacting with oxygen, but we will keep it simple now and say that nitrogen does not react and we will add it on both sides and we will see that this is especially needed the nitrogen in calculating stoichiometric air fuel ratios and also the adiabatic flame temperature. So, this is what we write. So, balanced reaction we have balanced every atom of an element on either side. So, you also realize that if the same number of atoms of an element are on both sides the atoms are not going to lose any mass. So, the mass on the left and right side will also balance out. So, there is also a mass balance again this is not a nuclear reaction in a nuclear reaction you would have lost mass and you would have gained energy due to the loss of mass. Here you do not gain energy because of loss of mass, but purely chemical reaction. So, there is mass balance and mass on either side is the same. Now, you must emphasize that this does not mean that moles will balance. So, moles do not balance or need not balance as it was in some cases they will. So, for example, in the reaction that we have just written here on the left side there is 1 mole of C H 4 plus 2 moles of oxygen and plus 7.52 moles of nitrogen. So, this makes it 9.52 moles and this makes on the reactant side. If I look at the right side I see that there is 1 mole of C O 2 twice 2 moles of H 2 O and still 7.52 moles of nitrogen and I have 9.52 moles on the product side also. But if I choose a different species. So, for example, I can choose a Heptane in C 7 H 16. So, I will just write oxygen plus nitrogen giving. So, you will see that I will get 7 CO 2 plus 8 H 2 O. So, I will require 7 of O 2 here and 4 of O 2 here. So, I will have 11 O 2 here which means this is 11 times 3.76 which means I will have 11 point sorry 11 multiplied by 3.76 H 2. So, if you see the left side the reactants you will realize that there is 1 plus 11 plus 11 multiplied by 3.76. I will not do this. So, I will have to do this multiplication is just 12 plus 11 multiplied by 3.76 and products there is 7 plus 8 15 plus 11 multiplied by 3.76. So, you see that product side there is 3 moles extra because this is only 12 plus 11 multiplied by 3.76 where as it is 15. So, the number of moles will not balance that is because atoms of different elements combine and form bigger or smaller species. In this case Heptane was a very big species and that was just one mole of Heptane, but we found far more moles of carbon dioxide and water and hence the moles do not balance, but of course you will see that atoms of every element have to balance and the mass has to balance. So, as long as this is told I think we can proceed further. What we would do next is that you emphasize the fact that we will always be using 3.76 as a factor for nitrogen and this is something that they must remember now that 1 mole O 2 corresponds to 3.76 moles of N 2 in air. This is assuming air is being used and which is what we will always be using and which means that 1 mole of oxygen corresponds to 4.76 moles of air because air is composed of those 3.76 moles of nitrogen plus the 1 mole of oxygen. So, for every mole of oxygen there is 4.76 moles of air. So, this is something that probably needs to be emphasized this is a number that is you know people expect that you would often remember as far as you know mechanical engineers are concerned whenever we are going to use air the 3.76 is a number that we often remember. So, once this is done what we can next go to is again a concept which is known to students which is what is called as an endothermic and exothermic reaction. So, basically if there is a reaction A plus B giving C plus D and if C A plus B is that some standard state and again you can emphasize and we will keep on emphasizing that first we will define something called as a standard state and the standard state will always be 1 atmosphere 25 degree C or 298 Kelvin. So, let us say A and B are at standard state and you they react and form C and D and C and D are brought back to standard state you will realize that to carry out the reaction at least some energy was required or some energy is liberated. Now, if energy was required to get C plus D it means that you have to provide energy and that energy is absorbed during the reaction and it now is belongs to C and D. So, that means that the energy I will put the symbol U of C plus D minus U of A plus B which is energy of C plus D minus U of A plus of products minus energy of reactants this is greater than 0. If this is greater than 0 the products are at a higher energy than the reactants and this means that the reaction needed extra energy this is what is normally called as endothermic reaction and I mean it is something that everyone is aware of, but it is just that we are emphasizing it again and vice versa if you know energy of products is less than energy of reactants then this will be an exothermic process that is because now the energy will be released out the products are at a lower energy and normally people show it like this A plus B is here and either C plus D is here or C plus D is here. So, if you go up then you require energy and this is endothermic if you go down you are throwing out energy this is exothermic. Now you realize that all reactions take place at constant pressure in fact this is what we maintain either in a combustion chamber we maintain the pressure constant or we let things happen in the atmosphere. So, the reaction is at atmospheric pressure and hence the expansion work is inbuilt. So, things are happening at constant pressure and hence what happens is that rather than working with you everyone works with H that is because the constant pressure expansion work has to be accounted for and inbuilt into whatever we are using. So, instead of using you you will see that people will use H. So, H products minus H reactants is what is called as the delta H of the reaction or the heat of the reaction in olden days. So, we will call it as the enthalpy of the reaction. So, if delta H R is negative then you know that the products had a lower energy and this is an exothermic reaction delta H R greater than 0 is an endothermic. So, this is just the reaction. Ensuring that now everyone knows we are working in enthalpy and what the heat of reaction means in this case. So, I will give just an example we will write for example, carbon in some grapitic form plus oxygen in gaseous form gives CO 2 I am not including here any time now nitrogen or anything. So, the heat of reaction here delta H R is minus 393.5 kilo joule. So, this means that the products have a lower energy than the reactants and this is an exothermic reaction and 393.5 kilo joule have come out as a result of 1 mole of carbon burning up with 1 mole of oxygen giving 1 mole of CO 2. So, the thing that must now be told is that whenever we are using fuels fuel plus oxidant better be an exothermic reaction otherwise there is no point in using it as a fuel we are using something as a fuel only because it will burn and release energy. If it is going to absorb energy then it is not a fuel maybe we are using it for something else, but if something is a fuel then it better combust or react with the oxidant and release energy. So, carbon all hydrocarbons react with oxygen and they will release energy and hence this all these set of compounds are all classified as fuels. So, fuels are anything which will react with the necessary oxidant and release energy that means they will go undergo an exothermic reaction process. So, that is the key to being called a fuel otherwise you cannot call it a fuel. So, once this is done here again I must emphasize that this delta H R is obtained when both these are evaluated at the standard state. So, you have to get this back at 1 atmosphere and 25 degree C this is initial this is also at or you should get it back to the same temperature, but all tabulation is always done at the standard state and this delta H R will be given a symbol delta H R naught that is this is the enthalpy of reaction at the standard state. If we had carried out the reaction at another temperature that is both the reactants and products were at a different temperature then sorry I mean they are different than 25 degree C, but they are equal to each other. It would have been the enthalpy of reaction at that temperature not the standard temperature, but then we will normally never tabulate such values all tabulated values are always at the standard state and it is assumed that both the reactants and products are brought down to the standard state. So, whatever enthalpy of reaction is there this is once the enthalpy of the product is calculated at the standard state. So, once we get this thing straightened out the other thing that you can tell is that you know there was there is something called as the Hess law or the Hess's law and this had a reasonable and it was it made sense to teach this earlier it just said that if I needed the enthalpy of a particular reaction or earlier it used to be called the heat of a reaction. If I did not know or if I could not get that reaction straight in a straight forward manner I could use other reaction to get the heat of reaction of an unknown reaction for example, you know C plus O 2 gives CO 2. I can burn carbon with oxygen and get carbon dioxide and you know the heat of reaction for this is easy to obtain, but if I wanted the heat of reaction for C plus O 2 giving 2 CO. So, that means this should be 2 here the heat of reaction for this is very difficult to calculate that is because if I try to burn carbon with oxygen and try to obtain CO it is very difficult even if I try my best to ensure that complete combustion does not occur because some CO 2 forms by default. So, I cannot figure out whether the energy is entirely because of reaction which leads to CO whereas, I can really try to ensure complete combustion. So, CO 2 forming is far more easier. So, in this case we can use the fact we can use Hess's law as follows C plus O 2 this is bound to be an exothermic reaction I will go lower here and I will get CO 2 and the delta H R as I said earlier was 393.5 kilo joule. I need to know this that is C plus O 2 giving I will put it as C plus O 2 giving CO plus half O 2. So, I will balance it out like this because I want to start with the same set of reactants. So, CO is formed so half O 2 is remaining. So, you can think of it in terms of moles where you say one mole of C and one mole of oxygen were there. So, one mole of CO have formed you have utilized all the CO and half mole of oxygen is remaining, but as I said you know so this is how the reaction is. I say now I have gotten this I do not know what is the heat of reaction here, but if I ensure the reaction of CO plus half O 2. So, I get hold of CO and I burn it with oxygen I will get this CO 2. So, I just do this reaction CO plus half O 2 and I will carry out this reaction and here since this is now entirely the end product is guaranteed to be CO 2. You can easily measure the heat of reaction or the enthalpy of reaction and it will be 2 minus 283.0 kilo joule. So, now I had to start with one mole of CO C and one mole of O 2 I could get CO 2. Now if I wanted this I could not do it directly, but I realize this is enthalpy that is the state function I when I subtract the enthalpy of the reactants on the product I get my enthalpy of reaction. So, here all I had to I realize is that you know if I went in steps finally the net enthalpy change should have been the same. So, if the enthalpy change here is 283 and the enthalpy change here would only the two enthalpy changes here must add up and give this enthalpy change. So, here it should be delta H R is equal to minus 110.5. So, this is what how Hess's law was used they said that if you could get to a certain set of products in steps and if you did not know the heat of reaction as they called it then of a particular step you could use the heat of reactions of other steps plus the heat of reaction of the net reaction to get the heat of reaction of a particular step. So, this is what we have done C plus O 2 I can go via two steps or I can go via single step for the single step I know the enthalpy of reaction for the two step reaction I can get the enthalpy of reaction for one of the steps. So, I will be able to get the enthalpy of reaction of the other step. Now in all this one must ensure that one is balancing the reaction everywhere and ensure that you are starting with the same number of moles and ending up with the same number of moles of products irrespective of the path you take. So, that balancing must be done and then you can go ahead. So, but then this is you know earlier this was thought over this is a very nice method to get it, but I think now probably it should be obvious because once you know that the enthalpy is the state function is built it is a property of those that gas C O plus half O 2 at one atmosphere and 25 degree C you know what is the enthalpy here you know what the enthalpy here is I know that the difference will directly give me this I know the difference will directly give me this. So, this addition giving this should be pretty obvious now and this is and now no longer people use this people what they do is they just try to tabulate the enthalpy of formation for each of this species. So, that brings us to something called as the enthalpy of formation. So, what you do is I need the enthalpy of every species at the standard state and this is what is tabulated. So, I need to somehow get this and tabulate this and this is an in fact you would have we have uploaded one sheet on the website noting down the enthalpy of certain set of species. So, what here we have to follow some ground rules and things are standardized here. So, the standards that people employ are they say that let us start with giving reference enthalpy of zero elements occurring in their natural form at standard conditions. So, if we want to assign enthalpies to all elements or sorry all species we will have to come up with some reference or some standard where we decide that you know some guys or some species will have particular enthalpy at a particular temperature we need to do this assignment and then we will find out the enthalpy of other species which are formed from those species. So, the standard or the most reference standard reference that people use is that they assign zero enthalpy to elements occurring in their natural form at the standard conditions. So, what does natural form mean? It means that for example, if I am considering a species oxygen and oxygen occurs as O 2 and not as O. So, O 2 at one atmosphere and 25 degree C is assigned zero enthalpy. So, you notice that the standard or references used in combustion are much different from let us say the references we used in moisture calculation where we use zero degree centigrade as the place where we wanted to set zero enthalpy, but in combustion calculation all the zeroes for naturally occurring elements are set at 25 degree C which is considered as the standard state. So, nitrogen occurring again as N 2 at these standards is assigned zero. Carbon coming in graphitic form graphitic form is considered the standard state. So, this is solid at one atmosphere and 25 degree C this is assigned zero and so on with other species that we will encounter. So, we are talking only of elements here. So, now if the elements occur in non-standard form for example, O or N then we come up with an enthalpy for these species, but we use a regular reaction to come up with an enthalpy for this. So, for example, I will just now write down I do not have the value here, but let me just tell you that I have oxygen giving 2 O. So, let us say that I have obtained the species using the standard reaction and let me see where I have the value for this. I have the value for let us say H 2. So, H 2 again here as one atmosphere and 25 degree C will be assigned zero. So, let us say that I need to form from H 2 I have to go to twice H and both of them. So, I now will find out the delta H R for this reaction and it will turn out that I will need to provide energy to actually get this reaction going and that is because I need to break the H 2 bond and form 2 hydrogen atoms. So, if I have to provide energy it is an endothermic reaction and the product will have higher energy than the reactants and the delta H R for this reaction turns out to be 436 kilo joules. So, this is a positive number and what we do now is we say that you provided 436 kilo joules for 1 mole of hydrogen, but what you actually ended up is getting 2 moles of H. So, we say we assign half the energy to each of these moles of H and we define something called as the enthalpy of formation for H or the standard enthalpy for H as 436 by 2 or 218 kilo joule per mole. So, the enthalpy of formation for H is 218 kilo joule per mole. Similarly, you will realize that for oxygen I will need to provide energy I will get 2 O. Let us say I will need to provide x amount of energy I do not have the value here right now, but if I provide x amount of energy then again I get 2 moles of O and I will assign x by 2 as the enthalpy of formation for oxygen and this is all at standard state because the heat of reaction or the enthalpy of reaction is calculated such that all species are brought back to the standard state which is 1 atmosphere and 25 degree Celsius. So, what and then let me see for example, if I now go to a compound let us say I go to H 2 O then the enthalpy of reaction then the enthalpy of formation of the species is defined on the basis of the heat of reaction for forming that species if from the elements which form the species and which should occur in that standard state. So, for example, H 2 O is formed from H 2 and O 2. So, I should consider H 2 and O 2 in their naturally occurring form at standard state I should get the reaction which forms H 2 O and then I should find out the heat of reaction necessary and using that I will get the enthalpy of formation for H 2 O. So, I will write this reaction as such H 2 gas. So, this means 1 mole of H 2 plus half mole of oxygen. So, remember I am putting half mole here we could have always written it as a balanced reaction by saying 2 H 2 and 1 O 2, but if I write this reaction like I get 1 mole of H 2 O. So, this is all 1 mole and these are at 1 atmosphere 25 degree C. So, the elements should occur in the naturally occurring form at the standard state and if I form the species that is of interest and bring it back to the standard state and I find delta H R here the delta H R will turn out to be minus 241.8 kilo joules that means there is a release of energy. So, now I see that H 2 O 1 mole of H 2 O when it is formed you know has to the heat of reaction as minus 241.8 kilo joules and since I have now assigned 0 enthalpy at standard state to both H 2 and O 2 it means that the H of products minus the H of reactants should have given me delta H R. So, now since reactants is 0 H of products is nothing but delta H R and is equal to minus 241.8 kilo joules and I say that H F for H 2 O and I put this dot here to denote standard condition is minus 241.8 kilo joule per mole. Remember that if I had written the reaction as twice H 2 plus O 2 giving me twice H 2 O in a regular balanced form that delta H R for this reaction would have been twice of this number and it would have been 483.6 minus. So, then I would have said 2 moles of H 2 O will require 483 moles or you know this many 483.6 kilo joules are liberated and you will get the same for 1 mole of H 2 O you will have minus 241. So, in this way what we do is for every species that is of interest to us we note down H F. So, for example O 2 N 2 H 2 C in graphite form these are all assigned 0. Then I will find out what is the enthalpy of formation for H 2 O C O 2 C O and any other species of interest. So, for example if you are doing high temperature calculations you will realize that nitrogen will react with oxygen to give N you know nitrogen plus oxygen let me just write here. I can get all sorts of products N O 2 N O O N 2 O. So, when reactions involving these are necessary you realize all of these are some kind of compounds they will not have a 0 enthalpy of reaction at standard state, but they will be formed from nitrogen and oxygen and we will list H F 0 for these species too. So, here we are talking only of relevant species to us you will realize that relevant species to us will all be formed from the compounds H C N and O because N and O come from air and almost all our fuels are hydrocarbons. If you are using some other chemistry for some other kind of fuels then you will require the relevant species there and you should have a tabulated form for all those species, but this is what is done standardly. Then you realize that you will require H F 0 for all the fuels. So, there will be H F 0 for C H 4. So, C H 4 is formed from carbon in graphitic form plus twice H 2 will give C H 4 and you will realize that there is a delta H R for this and that delta H R now you assign here to C H 4. Similarly, C 2 H 6 and all other species you can think of using C N and H M it could be alkene, alkene or it could be any of those alcohols or ketones etcetera, which would involve oxygen, but they will be formed from all of them will be formed from the standard sorry the elements in the naturally occurring form at the standard state. You find out the heat of reaction for that and accordingly assign the heat of formation for all these species. So, once this tabulation is done then at standard state you have H for every species that you are every species of interest not every general species because we are we will tabulate only for whatever is of interest every species of interest. So, once this is done then delta H R for all reactions that we know of can be easily found out. So, if we have a reaction C 2 H 6 plus O 2 giving twice C O 2 plus 3 H 2 O and I will just balance it twice O 2 and 1 and half. So, 3.5 O 2 I will at standard state I will get the H of this H of this is 0 H of this is tabulated H of this is tabulated. So, I get the H of the products I will get the H of the reactants I will just subtract this and get the H of the reaction and hence this is how I will proceed to get delta H R for reactions of interest. You will realize that this is delta H R 0 which is delta H R at standard state. Now, the one thing that we will realize is this is at standard state we need to find out the enthalpy of species at other temperatures and this is especially necessary when we need to do calculations regarding adiabatic flame temperatures etcetera. Because at that time we need to know what kind of temperatures we reach. So, though the heat of reaction is all defined at the standard state we require H for all species at higher temperatures and we realize that for every species as you start providing the energy its enthalpy will start increasing and for all species for all elements to though we started gave 0 at 25 degree C. So, let me just draw it here let us say this is temperature this is 25 degrees this is H I have given 0 here. You will realize that as temperature increases H will increase and this is what is bound to happen for every species. I have CO 2 CO 2 has a negative enthalpy of formation. So, that means at 25 degree C I start somewhere here and as I start heating it its enthalpy will go on increasing and at some point it will cross 0. It does not matter at all where it crosses 0 and this is because we have arbitrarily set the enthalpy of formation to be 0 for certain species at 25 degrees. So, at 25 degree certain species of positive enthalpy is some certain species of negative enthalpy certain species which have negative enthalpy at some higher temperature they will cross 0. So, this is really of no concern to us because finally, we will be just interested in the differences in enthalpy to figure out whether we have an exothermic or endothermic reaction and what kind of energy is there. But if we want to find out how this enthalpy increases you will realize that if I just write d H is equal to C p d t I will need to integrate I will need if I know the C p for every species then I can get the H for every species as a function of temperature. Now, it turns out that C p is a function of temperature. So, it is not a constant in fact it is going to be a constant only for noble gases. So, all monatomic gases will have C p constant, but any other species oxygen. So, let me not write oxygen, nitrogen, CO 2, H 2 O for all of these the C p itself if I plot C p it will have some shape like this it will go on increasing with temperature. So, it will not be a constant and we can more or less find this out. So, if we do statistical mechanics roughly it will turn out that diatomic gases. So, for example, monatomic gases the C v would be 3 by 2 and C p would be pi by 2, but if I go to diatomic gas I will have 6 degrees of freedom and C v will be 3 R t sorry 3 R and C p will be just 4 R it will always be C v plus R. So, this is if you if the entire diatomic gas is at very high temperature, but normally what happens is at room temperature all degrees of freedom have not yet kicked in. So, you will get a certain C p at lower temperature and as you go higher in temperature you will attain a certain C p. For example, for a diatomic gas I will start somewhere here and then I will plateau out like this. So, it will just between two values it will vary smoothly, but if I have a triatomic gas I will have far more degrees of freedom and every time an extra degree of freedom kicks in till all the degrees of freedom have you know I mean populated properly. I will keep on varying the C p till I reach a maximum here. So, this is what happens with all C p and they keep on varying as a function of temperature and what was done was that all these C p values were tabulated as a function of temperature or there were polynomial fits which were given to these. And we will use something which is called as a NASA format. So, a lot of this work was done initially for reactions and not only that for reactions of air to form N 2 O 2 etcetera you know whenever not only for burning of fuel, but whenever you know space vehicles re enter the space all kinds of reactions regarding nitrogen and oxygen will occur in the atmosphere and then hence a big database was created for the C p values of many hydrocarbons as well as many species involving oxygen and nitrogen. And if you go to the NIST website which is US government website it is called the National Institute of Standards and Technology you will get a whole lot of C p fits. So, we will use what is called as the NASA format. So, in the NASA format C p is a polynomial with 5 coefficients. So, I will write it as follows C p is equal to a 0 plus a 1 t plus a 2 t square plus a 3 t cube plus a 4 t 4 and so a 0 a 1 a 2 a 3 a 4 these are 5 coefficients. So, it is you see it is the fourth order polynomial with one constant and usually the temperature has to be input in Kelvin. So, when I put this T is T as 300 I will get the C p at 300 if I put T as 1000 I will get the C p at 1000 and so on. Now, using only a fourth order polynomial it is not so easy to get the fit in a huge range and what the range of interest which was of interest sorry I mean the temperature range that was of interest was mostly from somewhere between 0 sorry 25 degree Celsius and going up to some 3000 Kelvin or even up to 5000 Kelvin because such temperatures could be anticipated. And you will see that most of the fits are given between 300 Kelvin and around 3000 or 4000 Kelvin and it is not so easy to fit one single fourth order polynomial in this huge temperature range which is nearly of the order of 4000 Kelvin. So, what they have done is they have fitted two different ranges and usually they will tell that the ranges are. So, 298 Kelvin which corresponds to 25 degree C up to 4000 Kelvin is fit, but this is in split in two ranges. So, 298 Kelvin up to let us say 1500 Kelvin and then 1500 Kelvin up to 4000 Kelvin or some such split. So, it is not always going to be 1500 Kelvin the split temperature may be 2000 Kelvin 3000 Kelvin. In fact, I have given you the sheets and you must download them because we will use them in the afternoon and you will see in those sheets. Now, I am not going to project those sheets, but you will see that I have given sheets for the CP coefficients and let us say the coefficient for CO is given as CP upon R U this is universal gas constant. So, these are all molar specific heat and it is given as A 1. So, I have written A 0 plus A 1 plus A 2. So, they instead write it as A 1 plus A 2 t plus A 3 t squared plus A 4 t cube plus A pi t 4 and you will see that they are split between 300 and 1000 is one range and 1000 Kelvin to 5000 Kelvin is second range 2. So, in fact, for all the species that are given to you in this table CO CO 2 H 2 H 2 O N 2 the range is 300 to 1000 and 1000 to 5000, but it need not always be like this if you go to the website you will find that all kinds of different fits are given sometimes it is from 300 to 1500 and then 1500 to 3000. So, whatever is available is given in a certain format and you will realize that CP has been proper values or well established values of polynomial fits have been provided and it is a very good source for all our calculations. Now, this is CP now if I want H you require you realize that we could have just written d H is CP dt I can integrate d H which means I can integrate CP dt. Now, since CP is a function of temperature I cannot take it outside the integral and I will get in their format it will be written as a 1 t squared by 2 plus a 2 sorry I think they gave it as a 1 plus a 2 sorry I will just rub this here. So, I will just write it properly. So, if I want this integral of d H is integral of a 1 plus a 2 t plus a 3 t squared plus a 4 t cube plus a 5 t 4. So, I will get this as a 1 t plus a 2 t squared by 2 plus a 3 t cube by 3 plus a 4 t 4 by 4 plus a 5 t 5 by 5 and if I just write this as H then you will realize I will have to put a constant of integration. So, let and they call it as a 6 and you will realize that this a 6 will be will have to be some function of H f that is because if I now put t is equal to 298 Kelvin in this if I put t is 298 Kelvin. So, I will all the values have been put here. So, a 6 should be such that once I add up all these values it should give me H at 298 Kelvin which is H f 0. So, a 6 is in a way it is put in the constant of integration is put in such a way that once you put 298 Kelvin you will get the correct H at 298 Kelvin and it will depend on what the enthalpy of formation of that species is like. So, this is a 6. So, you will realize that in the table they have written it in this form H 0 by R t is equal to a 1 plus a 2 t plus a 3 t squared by 3 plus a 4 by 4 t square t cube plus a 5 by 5 t raise to 4 plus a 6 by t. So, they are just taken R and t on this side you just multiply by t and you will see that it is the same form. So, if you want to get H you first put the t there calculate the expression and multiply this R u which comes here and you will get your H at that particular temperature. Now, assuming this is an ideal gas H will be only a function of temperature. So, in their formula you will realize that they have put H 0 and again it means at standard pressure really and this is assuming one atmosphere pressure, but if the gas behaves as an ideal gas it is irrespective of pressure you should still get the same enthalpy. However, a s is not so straight forward. So, I will write d s or rather I will write t d s is d u plus p d v. So, for any simple compressible substance I could have written this t d s is d u plus p d v and of course, this will now be applicable to an ideal gas and if p is constant. So, if p is constant and equal to one atmosphere I can write this as d H that is because d H should have been d u plus p d v plus v d p, but v d p we are putting as 0 because p is constant. So, t d s is d H this is assuming this and d s would be just d H by t which means it is just c p d t by t which means it is just a 1 plus a 2 p plus a 3 p square plus a 4 p cube plus a 5 that means just upon t d t is there. So, if I integrate this s would be a 1 by t if I integrate it will be a 1 log t plus a 2 the t would go when I integrate again I will get back it a 3 is t square by 2 plus a 4 is t cube by 3 plus a 5 is t 5 by sorry t 4 by 4 and I will need to add a constant of integration which they call as a 7 and you will realize that they have written it in forms of s 0 let me just take a 2 minute break here.