 In the previous lecture, we showed that a single component gas for instance something like this or a mixture of gases and a two phase mixture or all pure substances. So, what we will do in this lecture is to see how to calculate properties of ideal gases and ideal gas mixtures. We will take up the calculation of properties of two phase mixtures in the next lecture. So, let us look at ideal gases first, single component ideal gas then we will look at ideal gas mixtures. As many of you know from your physics classes, the equation of state for an ideal gas may be written as Pv equal to RT. So, the V here is the specific volume and T is the temperature in Kelvin. So, you have to keep that in mind, T is the temperature in Kelvin or here is actually called the particular gas constant, it is not the universal gas constant, it is called the particular gas constant and it is equal to the universal gas constant divided by the molecular weight of the gas. So, in engineering thermodynamics, it is quite common to use this form of the ideal gas law, where R is the specific or particular gas constant. Since the specific volume is the reciprocal of the mass density, you may also write this equation as P equal to rho RT and since rho the mass density is nothing but the mass divided by the volume, I may also rewrite this as Pv equal to M RT where M is the mass of the gas and V is the volume occupied by the gas. So, if you have the gas in a certain vessel, then V is the volume of the vessel itself. So, this is ideal gas equation of state. If you recall, we said that when we wanted to calculate the change in total energy of a system, the only thing we needed to figure out was how to calculate change in internal energy. So, that is what we are going to look at next, internal energy of ideal gases. And once again, if you recall the two property rule where we said that the state of a thermodynamic system which consists of a simple compressible substance is determined by two properties. So, internal energy is a property and internal energy we can say is a function of two thermodynamic properties namely temperature and specific volume. So, we say that specific internal energy is a function of temperature and the specific volume. In reality though, the dependence of the internal energy on the specific volume is very weak for gases and so, we can generally neglect this. When we move on to two phase mixture next, we will see that this is not the case for two phase mixtures. The dependence of specific internal energy on the specific volume is quite significant there and it will be accounted for. But for gases, the dependence is very weak. So, we can neglect it and essentially say that the specific internal energy is just a function of temperature alone and such gases are called thermally perfect gases. So, for gases for which the specific internal energy is a function of temperature alone are called thermally perfect gases. Now, the specific enthalpy as you know is nothing but U plus PV. And if I substitute for PV from the ideal gas equation of state, I get this to be U plus RT. So, what this is telling me is that you know for a thermally perfect gas, the enthalpy is also a function of temperature T alone, both the specific internal energy and the specific enthalpy or functions of temperature alone. Now, we now get into the details of the internal energy of a gas. You may recall that we said that the internal energy of a gas is associated with the molecular motion and hence it is equal to the kinetic energy of the molecules. So, when we have a container, the molecules are moving in a random manner with all possible velocities in the three coordinate directions. So, the kinetic energy associated with the molecules is what is being accounted for as internal energy of the system. Now, depending on the nature of the gas, this accounting may be slightly different. Let us take a closer look at this. So, if we have a monatomic gas for instance, then let us say we have a container like this. So, we have the container filled with monatomic gas, then the molecules inside the gas are free to move in any of the three coordinate directions. And it turns out that because they are monatomic, they actually do not have any other degrees of freedom or any other modes in which energy can be stored. So, for monatomic gases, the three directions in which they can move x, y and z directions or all the degrees of freedom they have. So, for monatomic gases, we have only translational degrees of freedom and the translational degree of freedom is 3 because we have three coordinate directions. It can move in each one of the three coordinate directions. So, for monatomic gases, as we keep increasing the temperature, the molecules keep moving with higher and higher velocity. There are no other energy storage modes possible for monatomic gases. On the other hand, if we have diatomic gases, then some other modes are possible. For instance, for if we model a diatomic molecule as a dumbbell sort of structure. So, this is a dumbbell model. So, we have one atom here, another atom here and they are connected by a bond. So, for a molecule like this, additional modes of energy storage are possible. First of all, the molecule itself the with the two atoms on the bond can itself move in translate in the three coordinate directions. So, the molecules can each molecule can move like this or it can move like this or it can move like this. So, it has three translational degrees of freedom. So, that is applicable to the diatomic case. Now, in addition to translation, these molecules may also rotate actually. So, there is an additional mode of energy storage here. So, they may rotate like this about this axis. So, this is the axis. So, they may rotate like this or they may also rotate like this, which is this axis and they may also rotate like this. So, there are three axis of rotation possible and the molecules may actually rotate in all these three axis, which would then suggest that, you know, we have to add three more degrees of freedom to rotation. But it turns out that the moment of inertia of the molecule about this axis of rotation or this mode is very small. So, the amount of energy that can be stored in rotational modes like this is very very negligible. So, essentially the rotation in this in this fashion or rotation in this fashion contributes to additional modes in which significant amount of energy can be stored. So, rotation adds two and not three degrees of freedom. So, this degree of freedom actually is not very helpful because we can store very little energy in this because of the low moment of inertia about this axis. Now, as we keep increasing the temperature, it is also possible that at higher temperatures, the molecules which are connected through a bond can also store energy in the form of vibration of the bond. So, if we model this bond as a spring, then this spring begins to vibrate at higher temperatures. So, energy can be stored in the form of vibrational energy in the bonds. So, that is that is an additional degree of freedom. So, depending on the number of vibrational levels that are excited, you could keep storing energy in each one of these modes. Although I have indicated this as one, what I mean here is that this is one mode in which energy can be stored. So, the degrees of freedom will actually depend on how many vibrational levels are excited. That is a little bit more complicated. So, for our purposes, for this course, we will say that vibration adds one more degree of freedom in which energy can be stored. This is for the sake of simplicity. Now, let us see how much energy we can store in each one of this mode. Equipartition theory states that for each degree of freedom. So, this comes from equipartition theory in classical mechanics. So, for each degree of freedom, the specific internal energy is one half RT. So, translation brings in three degrees of freedom. So, we get this to be three halves RT. Rotation brings in two degrees of freedom. So, this comes out to be two times one half RT equal to RT itself. Now, vibration, energy that can be stored in vibration cannot be calculated using classical theory because vibration is quantized. Strictly speaking, translation and rotation are also quantized. But the spacing between the levels is so small that classical theory itself gives very good expressions for the energy that can be stored. However, for vibration, because the levels are not so closely spaced, we need to use quantum theory to calculate the amount of energy that can be stored and it comes out to be something like this. So, this is the amount of energy that can be stored in the vibrational mode. Now, we also have a concept called excitation temperature associated with each of these modes. What do we mean by excitation temperature? Now, anytime gas is at a temperature above 0 Kelvin, the molecules actually start moving that is they start translating in any one of the three coordinate directions. So, for as we said for monatomic gases that is the only freedom that is available for diatomic gases, they do this. So, they are moving in one of the three coordinate directions until the temperature exceeds 5 Kelvin. 5 Kelvin is called the excitation temperature for rotation. So, what this means is at temperatures above 5 Kelvin, translation is fully excited and if I increase the temperature beyond this, the molecules do move with higher velocity in the coordinate directions, but now they can also actually start rotating until we attain 5 Kelvin temperature, the molecules only move, they do not rotate. So, once we reach 5 Kelvin, translation is fully excited which means that now the molecules can actually start storing energy in the rotational mode. So, now they start rotating and as we keep increasing the temperature, molecules keep moving at higher velocity and they also rotate with higher velocity. So, more and more energy is stored until we reach a temperature of about 600 Kelvin or so, which is called the which is the temperature at which rotation is fully excited and vibration energy storage in the vibrational mode starts. So, once we reach a temperature of 600 Kelvin, the molecules are translating, they are also rotating like this, like this and like this. In addition, the bonds once we once I go above a temperature of 600 Kelvin, the bonds also begin to vibrate slowly at first and as I reach a temperature of about 2000 Kelvin, vibration is fully excited. That means the maximum amount of energy that can be stored in vibration can be stored. If I increase the temperature beyond this, dissociation will start taking place, the bond will be broken and other things start taking place. These are not degrees of freedom or my energy storage mode. So, we are interested only up to these sort of temperatures and we can consider them to be degrees of freedom. Actually, the theory is a little bit more complicated for than this, but for a first level course, this should be sufficient. So, basically what we mean by this excitation temperature is that is a temperature at which the gases can switch from storing energy in one mode into storing energy in an additional mode. So, up to 5 Kelvin, the molecules are for a diatomic gas, the molecules are only translating. Beyond 5 Kelvin, they are translating and also rotating like this and like this. Now beyond 600 Kelvin, in addition to these two, the molecules also begin to vibrate. So, this is called the excitation temperature. In fact, 2000 Kelvin is called the characteristic vibration temperature. 2000 is typical for most diatomic gases. So, we have said that for monatomic gases, translation is the only degree of freedom that is applicable. Whereas it can store energy only in translation. Whereas for diatomic, energy can be stored in translation, rotation and vibration modes. But for the purpose of this course, we will assume that the temperature remains below 600 Kelvin. Even it goes above 600, we will assume that vibration is not excited. We will assume that only translation and rotation are active. And based on that, we can easily see that the dependence of U on temperature is linear. So, this we can see clearly here. Specific energy, internal energy is 3 halves RT plus RT 5 halves RT for diatomic gas and just 3 halves RT for a monatomic gas. So, the dependence of specific internal energy on temperature is linear in this case. And such gases are said to be calorically perfect. You must remember that we define thermally perfect to be those gases for which specific internal energy depends only on temperature. But the dependence on temperature can be complicated. It could be a polynomial in temperature quadratic or even polynomials of higher degree. But if it depends only on temperature, such gases are called thermally perfect. In addition, if the dependence is linear, such gases are called calorically perfect gases. We assume gases to be calorically perfect in this course throughout. So, the specific internal energy for a diatomic gas as we just saw is 5 halves RT and for a monatomic gas, it is a 3 halves RT. CV, which is partial U partial T is nothing but 5 halves R and 3 halves R for diatomic and monatomic respectively and so on. Gamma, the ratio of specific heats comes out to be 7 fifths for diatomic gas and 5 thirds for a monatomic gas. What is that? We can also, it is very easy to see from here that CP minus CV is equal to R. And in fact, we may write CV as R over gamma minus 1 and CP as gamma R over gamma minus 1. So, once I know the molecular weight of the gas and the nature of the gas, monatomic or diatomic, I can calculate CP and CV directly. Remember, this R is nothing but the universal gas constant divided by the molecular weight of the gas. So, once I know the molecular weight, I can calculate the particular gas constant. And once I know the nature of the gas, I know gamma, I can calculate CP and CV, which is what we will do in all the examples that follow. So, you may recall that our objective in looking at pure substances and the now ideal gases is to develop a method for calculating changes in internal energy. We have that method now. So, U is nothing but 5 halves R T for diatomic gas and 3 halves R T for a monatomic gas. So, delta U, when we apply first law, delta U can be calculated using this relationship. Let us now move on to a mixture of ideal gases. For all intents and purposes, mixture of ideal gases may essentially be treated as single component ideal gas with an effective molecular weight, effective CP, effective CV and so on. So, we develop the theory now that effective, quote unquote effective means how do we need to know the details. In principle, it is effectively treated as an ideal gas, but we need to know or quantify what we mean by effective. So, let us say that we have a gas mixture, which is composed of a certain number of gases. So, it could be oxygen, nitrogen, carbon monoxide and so on and so forth. And so, we have Mi kilogram of each component and we know the molecular weight of each component. Now, the number of moles of each component is very easy to determine. We simply take the mass of each component divided by its molecular weight, we get the number of moles of each component. So, the total number of moles in the mixture is simply a sum of the number of moles of the individual species. So, this is the number of moles of the individual species, this is the total number of moles in the mixture. We define a term called mole fraction of a species as Yi as the number of moles of that particular species divided by the total number of moles. The concept of mass fraction is also useful, this is defined as the mass of each species divided by the total mass of the mixture. So, notice that just like total number of moles, the mass of mixture is the sum of the mass of the individual species. So, we can now define a mixture molecular weight, we know the number of moles in the mixture, we know the mass of the mixture. So, the molecular weight of the mixture itself is mass of the mixture divided by the number of moles and we can easily write using these expressions that this is equal to sigma Yi Mi or 1 over sigma Xi over Mi. So, this is quite easy to derive, I leave this as an exercise to you to derive. So, when we said effectively, so what this means now is we are making things clear now. Effectively, a mixture of gases will have a molecular weight equal to this. So, this is the mixture molecular weight, which means that the mixture behaves like an ideal gas with an equation of state equal to, with an equation of state given by this, which is what we wrote down for a single component gas also. We simply replace the R here with the effective molecular weight, which is based on the molecular weight of the mixture. So, the equation of state for the mixture still remains the same, it is as though it is a single component gas with a molecular weight equal to the mixture molecular weight M. So, for the entire mixture, we may write PV equal to M R T or PV equal to N R T, where here N is the number of moles of the mixture R is the universal gas constant. We need to also look at the components, because each component is an individual, I mean each component in the mixture is itself an ideal gas. We need to have a way by which we can go from components to the mixture and vice versa, we should be able to go back and forth. So, for this purpose, two models are very widely used, there are other models, but two models are very widely used, they are the Dalton's model and the Amagat's model. Now, in the Dalton, so let us consider mixture of gases in a container like this at a pressure P, volume B and temperature T. So, we look at a mixture of gases in a container like this. Now, the Dalton's model assumes the molecules of each component of the gas are free to actually move about in the entire vessel. So, occupy the entire volume and it is at the same temperature as the mixture, but it exerts a lesser value of pressure or a different value of pressure, lesser value of pressure than P. If there are two components here, so component one, it is as if component one is at a pressure P 1, volume V, which is equal to the volume of the mixture and temperature T, which is equal to temperature T of the mixture and similarly for component two also. So, this pressure of the individual components is called the partial pressure of that particular component. So, partial pressure, so P 1 refers to partial pressure of component one, P 2 refers to partial pressure of component two. So, that is the Dalton's model. So, obviously, if I add up all the partial pressures, it should be equal to the pressure of the mixture. So, that is quite obvious from here. So, sigma P i should be equal to P. So, it is as if, so let me repeat, each component as the freedom to occupy the entire volume of the vessel and be at the same temperature as the mixture, but exerts a lesser pressure than the mixture pressure. That is how Dalton's model handles mixture of gases. In Amagat's model, as you can see from this visualization, the Amagat's model says that each component is at the same pressure as the mixture, same temperature as the mixture, same pressure as the mixture, same temperature as the mixture, but it is as if it occupies a volume V 1, which is less than the volume of the vessel. So, it occupies a smaller volume, but it exists at the same pressure and temperature as the mixture itself. So, obviously, when I add up these two volumes or volume of these two boxes, it should be equal to the volume of this box. So, that means sigma V i equal to V and these V is, V 1 and V 2, they are called obviously partial volumes. So, V 1 and V 2 are called the partial volumes. So, V 1 is the partial volume of mixture 1, V 2 is the partial volume of mixture 2. So, that is the difference between Dalton's model and Amagat's model. One assumes that the pressure of each component is different, pressure alone is different from that of the mixture and if I add up all the partial pressures, I get the total pressure. The other one assumes that the volume of each component, volume alone is different from the volume of the mixture and if I add up all these partial volumes, I get the total volume of the mixture. So, this then allows me to write for each component, I may write the ideal gas equation of state like this. So, for component i, it exerts pressure P i, but occupies the entire volume of the vessel and exists at the same temperature as the mixture temperature. So, I can write this as P i V equal to M i R i T i. On the other hand, for Amagat's model, I write it like this. So, we use either Dalton's model or Amagat's model. So, sum of partial pressures is equal to the mixture pressure in the case of Dalton's model and sum of partial volume is equal to mixture volume in the case of Amagat's model. And we may write actually, if I take this equation P i V equal to M i R i T and divide that by the equation for the equation of state for the mixture, then I get the following P i over P equal to M i over M times R i over R. So, I may write this as X i times M i M over M i, where M is the mixture molecular weight. Similarly, V i over V may be written as being equal to Y i or X i times M over M i. In fact, this also may be written as equal to Y i. Now, we may write for instance P i times V for each component as N i times universal gas constant from which it follows that P i over P is equal to Y i. So, if mole fraction information is given, then we can use this expression. If mass fraction information is given, then we can use this expression. So, I can write a similar thing for the, for Amagat's model also. Partial volume divided by total volume is equal to the mole fraction. So, this allows us to handle the individual components as ideal gases and also the mixture as an ideal gas and we know how to go back and forth between the two. Now, how do we calculate the specific internal energy or C p or C v or gamma of the mixture, which is what we are going to look at next. So, any quantity phi, any property phi or any quantity phi may be written as on a mass basis like this. So, basically, let us say that this quantity is a specific quantity, let us say specific internal energy. So, the mixture internal energy is nothing but M times u. So, that we can write as the contribution from individual components. So, this is the specific internal energy of component i, this is the mass of component i. So, if I take this M to the denominator here, I may write the specific internal energy of the mixture as sigma M i over M times ui, which is then equal to sigma xi times ui. So, any specific property on a mass basis may be written like this. So, we can replace the phi with u, h and so on. C p may be written like this, C v may be written like this. Notice that for the mixture also C p minus C v equal to r and gamma equal to C p over C v. Notice that gamma must still be calculated as a ratio of the specific heat. So, gamma cannot be calculated as xi gamma i, sigma xi gamma i. Gamma still has to be calculated as specific C p for the mixture divided by C v for the mixture. Many students make the mistake of writing gamma just like all these expressions as sigma xi gamma i that is not correct. If we want to write things on a molar basis, there are applications you know for instance in combustion and so on where we would like to write things on a molar basis. So, we replace this on a molar basis. So, let us say that the total internal energy of the mixture is n times u bar where u bar notice that this has units of joule per kg. Whereas, this u bar will have units of joule per K mole. So, m has units of kg so that the internal energy mixture internal energy total energy comes out to be joules and similarly here n is kilo mole that is the number of moles in the mixture. So, n times u bar which is the mixture internal energy may be written as the sum of n i times u i bar. This quantity here is a specific internal energy on a molar basis of component i. So, once you do this we can write each one of these quantities on a molar basis also u bar then becomes equal to sigma y i times u i bar and so on h bar equal to sigma y i h i bar and so on. Other quantities may also be written in the same way and we can actually write gamma also as C p bar over C v bar which will be the same as it will be the same as C p over C v. So, in this module we actually saw how to calculate specific internal energy for calorically perfect gas and also ideal gas equation of state. So, the calorically perfect gas along the ideal gas equation constitutes the so-called perfect gas model. We also saw how to calculate specific internal energy and other quantities like partial pressure or partial volume of a mixture of ideal gases. We also saw how to calculate the effective mixture molecular weight so that a mixture of gases may be quote unquote effectively treated as a single component ideal gas. So, this shows us how to calculate delta u for ideal gases and ideal gas mixtures. What we will do in the next lecture is develop methodology for calculating properties of two-phase mixtures. For two-phase mixtures no closed form equations like what we have had so far is possible. We need to resort to tabulated data for properties and we will show how to calculate properties of interest for two-phase mixtures. The two-phase mixture could be water and water vapor or as I said before refrigerant liquid and refrigerant vapor.