 Hello and welcome to lecture 6 of this lecture series on Introduction to Aerospace Propulsion. In the previous lectures, we have looked at different aspects of very fundamental thermodynamics like defining very basic terms which are to be used throughout this course of thermodynamics. So, in this particular lecture, we are going to take it forward and look at some more aspects of basic thermodynamic interactions. So, let us take a look at what we are going to get introduced to in this particular lecture. Now, in this lecture, what we are going to look at are the following. We shall at the beginning understand what is meant by specific heat. We shall define specific heat and also define specific heat for two different cases or specific heat in two different forms known as specific heat at constant pressure and specific heat at constant volume. Subsequently, we shall look at different forms or modes of energy interactions. One of them is heat transfer. We shall understand the thermodynamic meaning of heat transfer and also look at the types of heat transfer. And then, we shall look at the other type of energy interaction that is known as work. We shall understand the basic meaning or thermodynamic meaning of work and also look at different types of work interactions. Now, as we are already aware that it takes different amounts of heat or energy to raise the temperature of a given substance or identical masses of different substances. For example, if you would like to raise the heat of water by a particular temperature and if you would like to the same amount or same mass of let us say iron to be heated to the same temperature, then we know that it takes different amounts of heat. So, based on this property, we shall define what is known as specific heat. And so, specific heat is a property that helps us in understanding how much is the amount of temperature or energy required to raise the temperature of identical masses. So, therefore, it is desirable that we have a property that will enable us to compare the energy storage capabilities of different substances. And this property is what we shall define as specific heat. So, specific heat by definition is the energy required to raise the temperature of a unit mass of a substance by one degree. And in general, this energy depends upon how this process is executed. And based on this, we can define two kinds of specific heat. One is known as specific heat at constant volume and the other is known as specific heat at constant pressure. Now, if you were to look at specific heat as a property which we shall see very shortly, that specific heat is basically a property of a system. As we have already seen, properties are certain characteristics that define the state of a particular system. Now, let us look at an example of how we can define specific heat for a particular substance for a particular process. Now, what is shown here is a particular mass which is let us say 1 kg's and we have to increase its temperature by delta t which is equal to 1 degree Celsius. And if this is the case, then we will need 5 kilo joules of energy for example, to accomplish this that is to raise the mass of 1 kg by 1 degree Celsius. And therefore, specific heat here would be the amount of energy here required to raise the temperature of a unit mass which is 1 kg here by 1 degree Celsius in a specific way. And therefore, the specific heat for this particular process will be 5 kilo joules per kilogram degree Celsius. And so, this is how one would define specific heat for a particular process. And now, let us now consider a fixed stationary mass of a closed system undergoing let us say a constant volume process. As I had already mentioned that specific heat can be defined in two different ways depending upon how the process is carried out. One of them is known as specific heat at constant volume and the other is known as specific heat at constant pressure. So, let us now define specific heat at constant volume and also find specific heat at constant volume in terms of known properties. Now, for this purpose let us now consider a stationary mass of a particular system of a closed system which is at constant volume. Now, we already know that you perhaps are aware of the conservation of energy principle. Now, if you were to apply conservation of energy principle which we shall see in much detail little later as well. Then, conservation of energy principle for this process is expressed in a differential form as shown here. Now, this differential form of expressing the conservation of energy principle leads us to this equation which is delta E in minus delta E out is equal to d u. As we have seen earlier that for a stationary system the net change in energy for the system because it is stationary will be equal to the change in its internal energy because kinetic energy and potential energy changes are 0 and therefore, as we had seen in the previous lecture delta E will be equal to delta u. So, this is the same equation that we have expressed in a slightly different form. Now, let us look at this equation once again d E in minus d E out is equal to d u. Now, this left hand side of this equation as I was just mentioning represents the net amount of energy transferred to the system because it is delta E in minus delta E in out which is essentially the net amount of energy which is transferred to the system. And therefore, the left hand side is actually equal to C v times d t because C v is specific heat at constant volume and as per the definition of specific heat it is the amount of energy required to raise the temperature of a unit mass of a substance by unit degree rise. And therefore, left hand side of the equation is equal to C v times d t which is equal to d u. And therefore, specific heat at constant volume which is denoted by C subscript v is equal to del u by del t at constant v. So, specific heat at constant volume is defined in terms of two properties of a system which is internal energy and temperature and del u by del t at constant volume gives a specific heat at constant volume that is C v. Now, the other property which I had mentioned as specific heat at constant pressure can also be derived in a very similar fashion. So, for deriving C p we would be considering a constant pressure process it could be either an expansion process or it could be a compression process. So, if you were to do a similar exercise as we did for C v then we get C p d t is equal to d h or C p is equal to del h by del t at constant p. Again as you can see here specific heat at constant pressure is again a function of two properties of a system that is enthalpy and temperature defined at constant pressure. Specific heat at constant volume is a function of two other properties of a system the internal energy u and the temperature t and since both of these depend on properties of system it is obvious that C p and C v have to be properties of a system. They are properties of a system because they depend on different properties of a particular system for example, C p depends upon the enthalpy and temperature and C v depends upon the internal energy and temperature which means that both C p and C v are properties of a system. And both C p and C v definitions though we had used or we had defined these properties for constant pressure and constant volume these are valid for any other processes it can be shown that C p and C v would be valid for any other process. And it is also a fact that specific heat at constant pressure C p is always greater in terms of its magnitude than specific heat at constant volume C v. Basically the reason for this is that if you were to carry out a process at constant pressure we have to allow the system to expand which means that you have to spend additional energy for this particular expansion of a system which is not there in the case of a constant volume process. So, if the volume were to be constant then there is no more expansion or compression of a system. So, the system does not require additional energy for expanding against the atmosphere. And therefore, because in a C p process that is constant pressure process you will need additional energy to keep the pressure a constant to allow for the system to expand. And therefore, C p is always greater than C v and specific heat of a substance will change with temperature which also will be clear in one of the examples which I shall show little later. Now, as I was mentioning C p is always greater than C v let us look at why is that so. Now, on the left hand side you can see a system which is enclosed in a chamber and therefore, its volume is constant. And let us say its mass is 1 kg and we have to raise the temperature by 1 degree Celsius. And for that to happen we have to supply this is just an example we have to supply for example, an energy equivalent to 3.12 kilo joules. And therefore, the C v would be 3.12 kilo joules per kilogram degree Celsius. Now, if the same system one of the boundaries let us say the upper boundary were to be movable that can convert the system into a moving boundary closed system. And if this were to be the case as you transfer energy to the system the system will expand which means that its volume will keep increasing, but the pressure would remain the same we can keep the pressure a constant. Now, in this case in addition to transferring energy to raise the temperature by 1 degree Celsius you need additional energy to account for this expansion process. And therefore, that is reflected in the additional energy which you can see here that in this example we have an energy of 5.19 kilo joules. Therefore, C p would be 5.19 kilo joules per kilogram degree Celsius. And this is why specific heat at constant pressure will always be greater than specific heat at constant volume. Now, the other point I was mentioning was that specific heat of a substance will change with temperature. To illustrate this let us look at another example let us consider air which is enclosed in a closed volume which has a mass of 1 kg. Let us say we have to raise the temperature of air from 300 Kelvin to 301 Kelvin. So, for this process we would probably require an energy of about 0.718 kilo joules. Now, if we were to consider the same system which is initially at a higher temperature of let us say 1000 Kelvin and we have to raise it to 1001 Kelvin we will need additional amount of energy that is amount of energy required to raise this unit mass of 1 k g by 1 Kelvin which was initially at a higher temperature than the previous case will require more energy that means the specific heat in the second case will be greater than the specific heat in the previous case. Therefore, specific heat of a substance will always change with temperature specific heat is a function a strong function of temperature. And in many of the practical engineering calculations this has to be taken into account because for simplistic calculations we normally assume that specific heat does not change with temperature but that is restricted to very simple calculations as we go towards more and more complex calculations which are actually required in engineering applications. We have to take this variation of specific heat with temperature into account and perhaps in later lectures we would take a look at some examples where we shall see how much difference we get by using specific by assuming specific heat to remain the same for different temperatures. And also looking at how specific heat changes with temperature for different cases and we shall compare the difference between the results we get from these two calculations and we shall see that for practical applications it does make a lot of difference. So, what we have defined now are two different properties of a system specific heat at constant pressure and specific heat at constant volume and we shall see little later that these are very important properties of a system which have lot of practical significance and we shall be using both these properties very frequently when we do engineering analysis of different systems. So, what we shall see next is two different energy transfer mechanisms one of them is known as heat and the other is known as work. So, energy can cross the system boundaries of a closed system in two modes one mode is known as heat and the other mode is known as work. So, what is shown here as an illustration is a system a closed system which has a mass equal to constant there can be two modes of energy transfer or energy interaction across these system boundaries. And both of them are as we shall see little later known as path functions we will define path functions little later. So, there are two different energy interactions possible one is known as heat and the other is known as work. And we shall see in the next lecture that for an open system there is one more form of energy interaction that is through mass flow rate. So, for a closed system there are two forms of energy interactions now let us look at the one of the forms of energy interaction known as heat. Now, as we have or as you are probably aware or we refer to heat on a day to day basis for example, we say some on a particular day that if it is a hot humid day we say the heat is very high and so on. So, heat is a term which is very commonly used in our day to day lives, but thermodynamically it might mean slightly different it might be used in slightly different context. So, heat is if you were to define thermodynamically is a form of energy that is transferred between two systems or a system and its surroundings by virtue of a temperature difference. So, the energy interaction that takes place between either two systems or between a system and its surroundings by virtue of its by virtue of a temperature difference is known as heat. And therefore, energy interaction is classified as heat or is qualified as heat only if it takes place by virtue of its temperature difference. And therefore, it is obvious to state that heat is energy in transition and therefore, heat is recognized only as it crosses a system boundary. So, thermodynamically heat is a form of energy which gets transferred between two systems or between a system and its surroundings by virtue of a temperature difference. And therefore, that is the thermodynamic definition of heat. Now, let us look at an example here. Now, we have three different systems well it is the same system at three different temperatures. In the first case we have the system at 25 degree Celsius we may consider the system as let us say a cup of cold coffee or let us say a can of cold drink or something like that which is in the first case let us say the system has the same temperature as the ambient. Let us say the ambient air or the room air is also at 25 degree Celsius. If the system is also at the same temperature as that of the room then there is no heat transfer which takes place between the system and the surroundings. Now, if the system were at a lower temperature of let us say 15 degree Celsius then there is definitely a heat transfer from the surroundings to the system because the surrounding air temperature is higher than the system temperature and therefore, there is a heat transfer from the surroundings to the system. And if the temperature of the system were even lower let us say 5 degree Celsius there will still be a temperature well heat transfer from the surroundings to the system. You might notice here that I have written 8 joules per second in the first case and 18 joules per second in the second case. Well these are just examples, but the point I wanted to state here is that because heat is recognized as an energy interaction by virtue of temperature difference temperature difference is the driving force for heat transfer which means that larger the temperature difference the more is the rate of heat transfer. So, larger the temperature difference between a system and its surroundings higher is the temperature well higher is the heat transfer. And which is why in the first case where the system was at 15 degree Celsius and the surroundings was at 25 degree Celsius we had heat transfer of let us say 8 joules per second whereas, in the second case because of the higher temperature differential between the system and the surroundings the rate of heat transfer is higher in the second case as compared to the first. And therefore, the point here is that rate of heat transfer is a very strong function of the temperature difference. The primary reason being heat transfer is basically by virtue of a temperature differential between system and its surroundings or between two different systems. And therefore, the higher the temperature difference or higher the temperature gradient between a system and its surroundings the greater or higher would be the rate of heat transfer. And so, in thermodynamics we would always refer to heat basically meaning heat transfer. So, whenever we refer to the term heat it actually means heat transfer. And as we have seen earlier heat transfer is energy in transit by virtue of a temperature difference. And if in a process there is no heat transfer across the system boundaries then that particular process is known as an adiabatic process that is a process during which there is no heat transfer possible is known as an adiabatic process. And there are different modes of heat transfer or different mechanisms of heat transfer. We shall not go into the details of these modes of heat transfer in this particular course as it is out of the scope of the syllabus. But I guess you would have already heard about these terms that heat transfer mechanisms there are basically three mechanisms of heat transfer. One is known as conduction the other is convection and the third is radiation. So, heat transfer can take place through three different mechanisms conduction convection and radiation. We shall not discuss details of these mechanisms of heat transfer you may refer to any text book on heat transfer for more information on these mechanisms of heat transfer. Now, heat transfer as I mentioned earlier is recognized as an energy interaction between a system and its surroundings only as it crosses the system boundaries that is a system may possess as we commonly refer to a system may have particular amount of heat. But thermodynamically a system is supposed to have had an energy interaction with the surroundings only as the heat which the so called heat which the system has crosses the system boundaries. So, let us look at an example where I shall illustrate this point in little more detail. Now, the system that we are referring to here in this example let us say has a thermal energy of about 5 kilo joules just for an example. So, this is a system boundary as you can see and it is surrounded by a surrounding air. Now, if the energy that is contained within the system crosses the system boundary then that particular energy interaction which the system has had with the surroundings is referred to as heat. So, as long as the system does not interact well as long as the system does not transfer energy with the surroundings that is not thermodynamically referred to as heat. So, heat is considered to have or heat transfer is recognized as an energy interaction only as it crosses the system boundary. Now, if you were to look at this one small example which I would like to illustrate that we have a system here which is bounded by an insulation. So, there is an insulation which is bounding a system and what is contained inside the system it is let us say a candle which is burning. So, a candle is contained within the system boundaries and the system is insulated which means that there cannot be any energy interaction between the system and the surroundings. So, for this particular system in question is there any heat transfer which takes place during this burning process. So, that is one of the questions that you may have and the other question is that is there any change in the internal energy of the system. Now, let us try to analyze these questions a little deeper. So, we have here a candle which is burning within a system volume and the system is insulated. So, because the system is insulated there is no energy interaction which is possible between the system and the surroundings and therefore, obviously there cannot be any heat transfer which can take place between the system and surroundings. So, thermodynamically there is no heat transfer taking place in this process and the second question was is there any change in internal energy. Well, you may wonder that because the candle is burning there should be a change in internal energy of the system. Well, the answer to that question is that there is no change in internal energy as well. The reason is that as we have seen earlier that for a system and the surroundings the net change in energy of that particular process if it is stationary will be equal to the change in internal energy because kinetic energy and potential energy are 0. So, in this case we have a system wherein delta E is equal to delta U. Now, the question was is there any change in delta U? Well, there could be a change in delta U only if there is a change in delta E and the first question which we answered was there is no heat transfer taking place between the system and the surroundings. So, there is no change or net change in energy of the system and the surroundings which means that delta E is equal to 0. And therefore, delta U that is internal energy change will also be equal to 0 which means that for an adiabatic process there is no change in the internal energy of the system as well as there is no heat transfer taking place between the system and the surroundings. Now let us now define the other mode of energy interaction which we had classified or qualified as work. Now, work is any energy interaction which takes place between the system and surroundings other than temperature difference. So, any energy interaction which takes place between this system and the surroundings by virtue of something which is other than temperature difference is classified as work which means that any energy interaction which takes place between a system and its surroundings by virtue of any other property which is not by associated with temperature difference is known as work. And thermodynamically we shall define work in the next slide work is basically the energy transfer associated with the force acting acting through a certain distance. So, if we were to look at work and heat transfer there is a certain sign convention that we follow for heat transfer processes as well as for work. And in thermodynamics heat transfer to a system and work done by a system are usually taken as positive that is we attach a sign to heat transfer and work done heat transfer to a system and work done by a system are taken as positive. And therefore, heat transfer from a system and work done on a system are negative. So, the interaction between the system and the surrounding is illustrated here the two modes for a closed system are heat and work. So, heat transfer to a system let us say Q in and work done by a system that is W out these two are usually taken as positive. And heat transfer from a system and work done on a system that is Q out and W in are usually taken as negative. So, this is a sign convention which we normally following thermodynamics. Now, let us look at both heat and work interactions in little more detail both heat and work as we know now are both energy interactions which are by virtue of energy interaction between the system and the surroundings. So, both are usually referred to as boundary phenomena because these are both energy interaction which take place across the system boundary. And therefore, you also referred to as boundary phenomena and a system can possess energy, but it cannot possess heat or work as we have seen heat or work are energy interactions between system and surroundings. And therefore, they are also referred to as boundary phenomena. Therefore, a system cannot essentially possess or contain heat or work well they can contain energy, but not heat or work. And both system both heat and work are associated with a process and it is not associated with a state and unlike properties heat and work do not have any meaning at a particular state. And therefore, both of these that is heat and work are also known as path functions that is their magnitudes depend upon the path followed during a process. And as it also depends upon the end states, but unlike properties which are point functions both heat and work depend on the path taken during the process as well as the end states. So, you might wonder what are path and point functions. Let us look at what are path and point functions. Well path functions are those functions which depend upon the path of the system as the name suggests. And both all path functions have in exact differentials and these are sometimes designated in different textbooks it is designated in different ways. It is sometimes designated by symbols delta Greek symbol delta or D with a cross. Examples are delta Q or D cross Q or in the case of work it could be delta W or D cross W instead of D Q and D W. But you might also come across some textbooks where it is also referred to as just D Q and D W, but I think it is already implicit that these are path functions. And point functions on the other hand like properties of a system these are exact differentials and they are designated by symbol D. And therefore, change in pressure or differential change in pressure is denoted by D P differential change in volume by D V and D T and so on. So, these are properties of a system and therefore, also referred to as point functions. Now, let us look at an example to explain the properties which are point functions and path functions. Now, in this pressure and volume diagram P V diagram we can see two processes process A and process B taking place between states 1 and 2. So, in the first process that is process A we have change in volume delta V because the states end states are the same in both the processes. The change in volume delta V is equal to 4 meter cube as the initial volume was 1 meter cube and the final volume was 5 meter cube. The delta V change in volume is 4 meter cube for both the processes. But if you look at work which has occurred during process A and process B it will be different basically because from the P V diagram work done during this any process will be the area under the curve in a P V diagram. So, for the first process that is process A, W A will be area under this process A and for the second process W B will be area under process B. So, as you can see W B should be greater than W A and therefore, work is a path function that is it depends upon how the process is carried out and or the path taken by the process unlike properties like volume here. It does not depend does not matter how the process was carried out it depends only on the end states and which means that if you want to integrate volume between states 1 and 2 integral 1 to 2 d V is equal to V 2 minus V 1 because it is an exact differential. Whereas, you cannot integrate 1 to 2 d W because it is not equal to W 2 minus W 1 and the other aspect of property is that if you have a cyclic integral of a property it obviously will be equal to 0. Therefore, properties all properties of a system like volume, temperature, pressure etcetera are point functions whereas, heat and work are path functions. Now, let us define what is meant by work. So, thermodynamically as it is defined work done by a system and its surroundings is basically referred to as a work interaction if there is the sole effect of this interaction can be viewed as raising of a mass through a distance against gravity. That is work done work is said to be done by a system on its surroundings or vice versa if the net effect can be considered to be raising of a particular mass through a certain distance against gravity. So, this is the thermodynamic definition of work. Now, we look at an example to understand what this definition basically means. Now, in this example that I am going to show you we shall consider a system which is having certain interaction with its surroundings and we shall see how we can understand work interaction for this particular system with its surroundings. Now, we have here a system which is bounded within system boundaries. We have an electric motor which is driving a fan and does the system do any work on the surroundings? Well, the motor is driven by a battery which supplies the required power for the motor and it is driving a fan. So, if you were to consider that this system is doing work on the surroundings as per the thermodynamic definition this should reduce to this. That is if we can replace the fan with a pulley it means that we can attach a pulley to instead of a fan to the motor and that can and as the motor rotates this would raise a weight against gravity. And therefore, it satisfies the condition thermodynamic definition of work that the net effect can be considered as raising of a weight against gravity. Therefore, this example which has a system driving a fan definitely does work on the surroundings because we can always replace this fan with a pulley and therefore raise any weight against gravity. And therefore, the system does work on the surroundings as per the thermodynamic definition of work. Now, we shall look at some examples of different work interactions what are the different types of work interactions that are possible between the system and its surroundings. So, there are different ways in which work interactions can be carried out between a system and its surroundings. And we shall look at one of these particular work interaction which is of interest to us as engineers aerospace engineers and that is known as the displacement work. So, we will look at displacement work in lot more detail than the other types of work interactions. So, displacement work is one of the forms or modes of work interaction between the system and the surroundings. The other types of work interactions are for example, electrical work that is if you were to heat a resistor then that is known as electrical work because you can always replace the resistor with a motor and therefore, drive a pulley and raise a weight as we have seen in the previous example. Besides this the other type of work interactions are shaft work that is rotation of a shaft to generate work output or it could be paddle wheel work or spring work or stretching of a liquid film etcetera. So, there are different modes of work interaction which can take place between a system and the surroundings. And one of them as I have mentioned earlier that is displacement work is what we shall be interested in analysis little more. Now, let us illustrate these different modes of work interaction through these examples. So, shaft work that is rotation of a shaft to generate work output as is shown in this example it is a motor car on an automobile which has an engine which generates a shaft work output and that it is that shaft which drives the car the automobile forward and therefore, it generates a work output. The other possible mode of energy interaction or work interaction is spring work example shown here we have a spring which is at rest and in a particular position and if you exert a force on the spring then it expands and gets displaced by a certain distance. So, the work done during this process is referred to as spring work then stretching of a liquid film that is if you were to have a liquid film which is held between the let us say this rigid wire frame and a movable wire. So, there is a surface of film which is formed in between the rigid wire frame and the movable wire. So, as you move the wire it causes stretching of the liquid film here. So, the work done during this process is also one another mode of work interaction electrical work would involve heating of a resistor that is if you have a system which houses a resistor you supply a certain electrical power to the resistor it causes heating of the resistor and. So, there is a work interaction between the system and its surroundings because of heating of this particular resistor. So, these are some examples of different modes of work interaction other than the main form or main mode of work interaction which would we would be interested in that is known as the displacement work. So, displacement work or moving boundary as it is also referred to is of a significant interest to engineers because there are several engineering systems which can be approximated to be operating in the displacement work mode that is they generate work output through the displacement mode. And for example, if you were to look at the engine of an automobile it generates work output because of the movement of the piston during the power stroke of the engine. And therefore, that is displacement work or if you look at steam engines or different pumps many of these pumps and steam engines all operate in the displacement mode that is P times d v mode. So, displacement work as I mentioned there are so many applications engineering applications for displacement work we shall analyze displacement work in little more detail for different types of processes. Now, let us look at what we actually mean by displacement work. Now, in this example I am showing here we have the piston and cylinder arrangement and there is a certain amount of gas which is which is housed within this cylinder volume. And if we were to supply heat to this gas it as we know it would expand and during expansion the piston moves against the atmospheric pressure. And therefore, it exerts a certain force and therefore, this gas does a differential amount of work which is delta w b where b subscript b denotes boundary work. As it does a work it does work equal to delta w b because it forces the piston to move by a differential amount d s. Therefore, delta w b which is the differential work done during this process will be equal to force times the distance. And as we know force is pressure times area because pressure is by definition force per unit area. Therefore, force can be expressed as pressure into area. And as we are also aware that area times this distance will be the volume that the piston has moved or displaced during this process. Therefore, we can replace a times d s by d v. Therefore, the delta w b that is delta work done during this process is equal to p times d v or if you are looking at work done from state 1 to state 2 w b will be integral of p times d v. So, this is why it is known the displacement work is also referred to as p d v work that is pressure times differential volume work p d v work. Now, the same process we can explain in terms of the p v diagram as well. Now, in this p v diagram we can we have the same piston cylinder arrangement here. And as we supply energy to the cylinder the piston moves generating work output that is the process would would take place between states 1 and 2. And this is an expansion process and so the process path is indicated here. This is the process path for this particular process. And if you were to look at one small differential strip in on this process path then the area of that differential strip will be equal to pressure average pressure during this process times d v. And therefore, area is equal to integral d a which is integral p d v. And therefore, area under the curve on a p v diagram will be equal in magnitude to the work done during this quasi static or quasi equilibrium process which could be either a compression process or it could be an expansion process. And it is basically for a closed system. Therefore, area under the curve in a p v diagram should be equal to the work done during this process which obviously is quasi equilibrium or quasi static. And the process may be either compression process or it could be an expansion process. Therefore, if we if we can plot the pressure versus volume diagram for a particular process we can actually calculate or determine the work done during the process simply by measuring the area under the p v diagram. And therefore, for any work interaction process a p v diagram plays a very significant role in analysis of work interaction during a particular process. So, we shall be using p v diagram very often in thermodynamic analysis. And similar to p v diagram we also have what is known as temperature entropy diagram which we shall see in detail later on that is T s diagram. T s diagram also plays a significant role in thermodynamic analysis of engineering systems. Now displacement work as I mentioned because work is a path function it depends upon the path taken by the process and not just the end state. It depends on the end states, but also it depends upon the path taken during the process. Therefore, if you were to look at states 1 and 2 the infinite number of possibilities of carrying out a process between states 1 and 2. The properties at states 1 and 2 remain unchanged, but you could do the process in different ways. And depending upon how you do the process the work done during the process will change. That is if you can do this process by a b and c the work done during each of these process will be different. And as we now know the work done is the area under the curve. And therefore, process a work done during process c will be the least because the area under this curve is the lowest followed by b. And the highest work done would be done during process a because area under this particular process is the highest. Therefore, just an example shown here W a could be let us say 12 kilo joules W b 10 W c is 7 kilo joules and so on. Therefore, the work done during a process depends upon the path as well as the end states. And the net work during a cycle is basically the difference between the work done by the system and work done on the system. That is if you were to consider this as a cycle that is the process starts at let us say state 1 and takes path a and reaches state 2 and then takes path b to come back to state 1 again. The net work done during this process will be area between the curve that is W net will be equal to area under a which is work done by the system. And area minus area under b that is work done on the system which is equal to W net. So, W net will be equal to W a minus W b that is the net work done during this particular cycle. So, net work done during a cycle will be the difference between work done on the by the system and work done on the system. We shall now analyze displacement work done during different processes. So, you can carry out the process in different ways. It could be either a constant pressure process, it could be a constant volume process or it could be a process wherein P v product P v is a constant which happens to be an isothermal process because as per state equation P v is equal to m r t. And therefore, if P v is constant it means that temperature is a constant. So, an isothermal process wherein P v is constant or it could be what is known as a polytropic process wherein we have product P times v raise to n is a constant where n is a number which can depend upon the type of process and that is known as a polytropic process. So, we shall now calculate the work done during these different processes. A constant pressure process, a constant volume process and then we shall consider a polytropic process. P v is equal to constant is a subset of the polytropic process. Now, let us first look at a constant pressure process wherein the pressure remains constant the volume keeps changing during the process. So, if you were to look at the P v diagram for such a process wherein P v is well pressure is constant this process takes place between states 1 and 2. Since pressure is a constant you can see that P 1 is equal to P 2. So, what is the net work done during this process it is basically the area under this curve and how and it is very easy to find the area under this curve because it is just the pressure as it is constant P times v 2 minus v 1 which also comes from integral v 1 to v 2 P d v. So, if you were to integrate P d v between states v 1 and v 2 you get P times v 2 minus v 1. So, work done during a constant pressure process is the product of the pressure times the difference in the volume during the process. So, P times v 2 minus v 1 will be the net work done during this particular process. Now, what happens if the volume was constant? So, if you have a process where the volume is constant that is it is a closed system and the volume is also constant, but the pressure can change is there any work done during this process. Now, since v is equal to 0 if you were to integrate P d v between states v 1 and v 2 since v 1 and v 2 are the same in this particular process the work done will be 0 because there is no change in volume of the system and therefore, for a constant volume process the net work done will be equal to 0. So, there is no work done during a constant volume process. Let us look at a polytropic process that is a process in which P times v raise to n is a constant where n can take different values 0 1 2 3 anything. So, we have here on a P v diagram different process are shown here n equal to 1 n equal to 2 n equal to 3 and the limits of this particular process are n equal to 0 when n equal to 0 what happens is that you will have a constant pressure process. We have already seen what is the work done for a constant pressure process which is equal to pressure times the difference in the volumes and if n was equal to infinity then you get volume is equal to constant where in the work done will be equal to 0 because area under the curve is 0. Now for a P v raise to n process work done is integral 1 to 2 P dv which is equal to C times v raise to minus n times dv because P v raise to n is constant let us say the constant is C. So, P v raise to n is equal to C. So, we can replace P by C times v raise to minus n. Now, we also know that since P v raise to n is constant P 1 v 1 raise to n is equal to P 2 v 2 raise to n is equal to C and so if we were to substitute these values into the work equation here that is for if you were to replace v in terms of the pressures what we would get is work done from state 1 to 2 will be C times v 2 raise to minus n plus 1 minus v 1 raise to minus n plus 1 divided by minus n plus 1. So, since we know that P 1 v 1 raise to n is equal to P 2 v 2 raise to n is a constant this expression reduces to P 2 v 2 minus P 1 v 1 by 1 minus n. Therefore, work done during polytropic process is basically equal to the product of P 2 v 2 minus P 1 v 1 divided by 1 minus n. So, suppose n was equal to 2 then the work done during this particular polytropic process will be equal to basically depends upon the properties at the n states P 2 v 2 minus P 1 v 1 by 1 minus 2 and so on. So, you can actually calculate the work done during a polytropic process if you know the n states as well as the exponent of the polytropic process. So, we shall see later on that when n is equal to 1.4 where 1.4 raise to n is equal to 4 for normal temperatures is also referred is also equal to the ratio of the specific heats. In such a process we usually referred to this process as P v raise to gamma is equal to a constant where gamma is the ratio of the specific heats C p divided by C v. So, this process we shall see in detail in later lectures. Now, from this particular polytropic process if you were to assign different values of n you can calculate the work done during these different processes and as a subset of this process would be n equal to 1 where in P v is a constant. So, when for P v is equal to constant you can actually integrate we obviously need to integrate P d v for between states 1 and 2. So, if you look at this particular equation if you substitute n equal to 1 then that is not the right way to do it you will still have to go back and integrate for the work equation substituting for in the same way as we did for this case where P v is equal to C and so on and the work equation will come out to be P 1 v 1 log v 2 by v 1. So, I leave that as an exercise for you to calculate the work done during a P v is equal to constant process. So, let me now recap what we have discussed during this particular lecture. We started the lecture by defining what is known as specific heat for a process and we saw that the specific heat is basically the energy required to raise the temperature of a unit mass of a substance by unit degree rise in temperature. And there are two different types of specific heat specific heat at constant pressure and specific heat at constant volume. Subsequently, we discussed about energy interaction modes between a system and its surroundings for a closed system. And for a closed system there are two modes of energy interaction heat transfer and work. And we discussed about heat transfer the meaning of heat transfer basically the thermodynamic meaning of heat transfer and also mentioned about different modes of heat transfer. We discussed about work in lot more detail. We defined thermodynamic definition of work and also the different types of work interactions possible between a system and surroundings. And then we also discussed the displacement work that is PDV work in lot more detail. We also calculated work required for different displacement processes, a constant pressure process, constant volume process and a polytropic process. And somewhere in between the lecture we also discussed about path and point functions. Basically path functions are those which depend upon the path of a particular process as well as the end states. And point functions depend only on the end states. All properties of a system are point functions, heat and work are path functions. So, this is what we discussed in today's lecture. And in the next lecture what we shall do is to solve some problems based on what we have discussed during lectures 4, 5 and 6. Primarily we shall solve some problems related to calculation of work done during different types of processes. So, we shall solve these problems during the next lecture.