 So, let us start with the first example. So, here we have a fixed vessel like this. It contains a saturated mixture of R134A initially at 100 kilopascal. The volume of the vessel is also given. We now supply heat to the vessel until the contents reach the critical state. We are asked to determine the mass of the mixture and the heat transferred to the contents of the vessel. So, as we can see this vessel rigid, so this undergoes a constant volume process. The final state is the critical state that means the property values are actually known. Let us see how this comes about. So, here we have properties of saturated R134A pressure table. So, here we can see that this is the critical state. So, the specific volume of the liquid, specific volume of the vapor are both the same at the critical state and that is the information that we will use. So, you can also see that the specific internal energy of the liquid and the vapor are the same. So, specific volume at the final state is known. We can make use of this information to carry out the analysis. So, the specific volume at the final state is 0.00197 meter cube per kilogram and the specific internal energy may also be retrieved from the pressure table. Since it undergoes a constant volume process, V1 is equal to V2 which is already known. So, the mass of the mixture may be evaluated from the given volume of the vessel. So, M is equal to V over specific volume which gives us the mass to be 10.15 kilograms. So, if we take our system to be the contents of the vessel, the system looks like this. So, delta E equal to delta U equal to Q minus W. No displacement work or any other forms of work. So, W is equal to 0. So, we can now evaluate Q from this, Q is equal to delta U equal to M times U2 minus U1. U2 is already known to us, U1 has to be evaluated. So, at the initial state the specific volume is known and it is at a pressure of 100 kilopascal. So, we go to the pressure table corresponding to 100 kilopascal. We retrieve the specific volume of the liquid, saturated liquid and saturated vapor. And using this we can calculate the dryness fraction, let us see how this is done. So, you can see here that using the specific volume of the saturated liquid and saturated vapor, we can evaluate since a specific volume at the initial state is known, we can evaluate the dryness fraction to be equal to 0.0065. So, the mixture specific volume is known at the initial state, mixture specific volume is known and we know the pressure, so we know Vf and Vg. So, we can calculate the dryness fraction of the mixture using this expression. And once X1 is known, the specific internal energy at the initial state may be evaluated as Uf plus X1 times Ug minus Uf. Now, Ug and Uf may be retrieved corresponding to 100 kilopascal from the table. So, notice that this is Uf at 100 kilopascal, this is Ug at 100 kilopascal. So, we simply retrieve these values and calculate U1 because X1 is also known, X1 is also known, so we get U1 to be equal to this, so Q is equal to M times U2 minus U1. So, the heat supplied comes out to be 2186.54 kilo joules. So, the important concept here is that once we know the specific volume at the initial state, we can calculate the dryness fraction and the property values of the mixture. So, basically we located the state, the final state was given to be the critical state. So, we were able to locate that state initially, so if I draw a PV diagram, so the final state was given to be the critical state. So, this is the final state and because the volume of the vessel remains constant, it is a rigid vessel, the working substance has undergone a constant volume process, so this is state 1 and the initial pressure is 100 kilopascal. But the dryness fraction and the initial state is not known. So, specific volume V1 is equal to V2, so once I know the specific volume here, I can evaluate dryness fraction from this expression and I can also evaluate the specific internal energy from this expression once the dryness fraction is known. So, we have discussed this earlier how to calculate properties in the two phase mixture region. If you have any questions, I suggest that you review that material and then go through the problem one more time. That is why in the initial part, in the previous module, we talked about locating the state on a PV diagram like this for a two phase mixture and once you locate the state, we went through the procedure of how to calculate property values like specific volume or specific internal energy and so on. Let us look at the next example, frictionless piston cylinder device contains 5 kg of saturated R134A vapor initially at 25 degree Celsius. The vapor is compressed slowly until the volume is reduced by two thirds. So, please pay close attention to this volume is reduced by two thirds of the initial volume. Heat transfer during the process with the ambient at 25 degree Celsius keeps the temperature of the contents of the cylinder constant, determine final pressure, work done on the heat transfer. So, basically we have a situation like this. So, this contains R134A. So, this is initially at initially I am sorry it initially contains 5 kg at 25 degree Celsius and it is compressed slowly until the volume is reduced by two thirds. Notice that it initially contains saturated R134A vapor. So, the initial state is saturated vapor and it is compressed until the volume is reduced by two thirds of the initial volume temperature also remains constant. So, we can take this to be our system obviously. Now, the initial pressure is not given, the initial pressure is not need not be explicitly given because it is stated that initially it contains 5 kg of saturated vapor. That means the contents will be at the saturation pressure. So, we go to the table. So, corresponding to 25 degree Celsius. So, here we go to the temperature table for saturated R134A. So, corresponding to 25 degree Celsius, their pressure is 665.80 kilo Pascal. So, the initial pressure is this known. So, initial pressure P1 is equal to P sat of 25 degree Celsius because it is initially a saturated vapor. So, initial state is saturated vapor at 25 degree Celsius mass is also given. So, from the temperature table we can retrieve the following values pressure is also known and the pressure is equal to 665.8 kilo Pascal. Now, if you actually look at the process that the R134A undergoes, we may draw it like this. So, so this is the initial state and the initial pressure is known. So, this is if I draw the isotherm 25 degree Celsius and the initial pressure we retrieve to be 665.8 665.8 kPa. Now, the saturated vapor is compressed while keeping its temperature constant. So, heat transfer takes place. So, its temperature remains constant. What is that for its temperature to remain constant? It has to actually follow the isotherm corresponding to 25 degree Celsius and because it is compressed it has to go in this direction since we are trying to increase its pressure. So, basically what happens is as we compress it because it is a saturated vapor it condenses and when it condenses there is a decrease in the volume occupied by the mixture because the specific volume decreases as you can see. So, as we compress it there is condensation and reduction in volume. So, the pressure remains constant because of the condensation. You see it is undergoing a phase change just like what we saw before when we moved the piston up some of the water evaporated and occupied the additional space that was available. In this case the opposite is happening when we compress it there is a reduction in the volume and some of the R134A condenses and causing the which is able to accommodate the reduction in volume. So, consequently the pressure also remains constant. That is the important part. So, as the vapor is compressed isothermic it condenses. So, the reduction in volume due to the compression is accommodated by the reduction in volume due to the condensation and conversion to liquid. So, this continues until all the vapor has condensed. We are given that the volume reduces by two-thirds which means the final volume is one-third of the initial volume. So, V2 is equal to V1 over 3 which is equal to this. Now, this is greater than Vf corresponding to 25 degree Celsius which means that the final state is going to be somewhere over here otherwise as we compress it all the vapor could condense entirely and as we continue to compress the final state could have ended up somewhere over here. In this case because V2 is greater than Vf the final state is still a two-phase mixture inside this otherwise the process curve or process line would have looked like this. So, it comes like this then it would have gone up this isotherm because we are maintaining the temperature constant and we would have finally reached a compressed or sub cold liquid state. Since V2 is greater than Vf we do not reach that state we still remain in the saturated mixtures region. So, X2 may be evaluated as V2 minus Vf over this. So, X2 comes out to be 0.3148. So, here we have retrieved the values corresponding to T equal to 25 degree Celsius. Let us just take a quick look. So, corresponding to T equal to 25 degree Celsius we can get Vf, Vg, Uf and Ug from the tables. So, the final specific internal energy may also be evaluated. Notice that the specific volume at the initial state is simply equal to Vg and the specific internal energy at the initial state is simply equal to Ug. We can calculate the displacement work there is no other form of work. So, displacement work is P delta V and the pressure remains constant. So, we may evaluate the work interaction as minus 68.644 kilojoules. Clearly work is being done on the system and if you apply first law to the system we have delta E equal to delta U no change in Ke or Pe equal to Q minus W. We are asked to calculate Q. So, we rearrange this expression and get Q to be minus 609.084 kilojoules. So, as we do work on the system it is clear that. So, as we do work on the system and compress it it is clear that heat is being lost to the surroundings. So, the temperature of the R134A remains constant because it is undergoing a phase change because we started with saturated vapor because it is undergoing a phase change the pressure also remains a constant during this phase. The next example that we are going to look at involves frictionless piston cylinder device like this. So, this contains R134A set amount which is actually 2.6 kg of saturated R134A mixture at an initial pressure of 500 kilopascal and occupying a volume of 0.1 meter cube. Actually we know the specific volume we know the pressure. So, with this just with these two property values we should be able to establish that it is a saturated mixture of R134A. Anyway the information is given in the problem statement. So, we can make use of that. So, the mixture expands according to the relation PVE equal to constant until it becomes a saturated vapor and we are asked to determine the final pressure work done and heat transferred. So, if I actually try to show this on a PV diagram. So, initially we have a saturated mixture at a pressure of 500 kilopascal. So, this is 500 kilopascal this is state 1 and the mixture expands according to PVE equal to constant which means the pressure is going to decrease and finally it becomes a saturated vapor. So, the final state is going to be on the saturated vapor line the final pressure has to be determined we do not know what the final pressure is that needs to be determined. So, the process undergone by the by the two phase substance looks like this. So, this is given to be PV equal to constant. So, the initial specific volume may be evaluated as 0.0385 and the pressure is also given we can actually retrieve the required property values from the table Vf and Vg may be retrieved from the pressure table and the specific internal energy may be calculated by using the value of x1 which is calculated by using the fact that the specific volume at the initial state is known. So, x1 comes out to be 0.934. So, we may evaluate the specific internal energy at the initial state as 227.824. Now, the mixture expands according to PVE equal to constant since mass of the mixture remains a constant. So, it is given that PVE equal to constant. So, if I divide both sides by the mass then I may write this as P times V over m equal to again another constant since the mass remains a constant. So, we may alternatively write this as PV equal to constant where V is the specific volume which means P2 V2 equal to P1 V1. Now, remember we need to have or know two independent properties to locate a given state. So, if I look at the final state I know that it is a saturated vapor. So, I have one piece of information. The second piece of information comes from this equation P2 V2 equal to P1 V1. Now, it may appear that there are actually two unknown quantities here P2 is not known V2 is also not known. However, you should remember that it is a saturated vapor. So, V2 is equal to Vg of P2. So, that means there is really only one unknown in this, but that needs to be determined iteratively. So, if you do trial and error you should be able to get the final value of P2 as 100 kilopascal. So, basically the trial and error procedure would go like this. We know that the pressure decreases. So, we guess a certain value for pressure and we get the corresponding Vg. We evaluate the product and see how close it is to P1 V1. So, basically this is what we would do. So, this is the pressure table. So, initial pressure is given to be 500. This is the initial pressure. Now, we know that the pressure decreases as the vapor expands. So, we may take our initial guess for the final pressure to be 200. So, if we take the final pressure to be 200, then the specific volume of the saturated vapor is known. So, we take P2 multiplied by V2 and see how close that is to P1 times V1. If it is not close, then we reduce, we adjust our guess. So, we may actually decrease our pressure to some other value. Again check the product, see how well it matches P1 V1. So, we do this by trial and error which is until we get a final value which is actually acceptable. In this case, we can actually get the final value to be 100 kilopascal. So, the specific volume and the final state is known and the specific internal energy, remember U2 is equal to Ug of P2. So, that is also known. Now, the work interaction for the process may simply be obtained from integral PDV. Since the relationship between P and V is known, we can evaluate this in a close manner and get this to be equal to P1 V1 natural log of 5 which gives us the work interaction to be plus 80.472 kilojoules. As the working substance expands, it is doing work on the surroundings which is why we have a positive sign. Now, if we apply first law to the system consisting of the R134A. So, basically our system is this. So, if we apply first law, we have delta E equal to delta U because there is no change in KE or PE. So, delta E is equal to delta U equal to Q minus W and delta U itself may be written as M times U2 minus U1. So, if you substitute the numerical values, we get Q to be 47.676 kilojoules. So, we are supplying heat to the working substance, it then expands until it reaches a saturated vapor state. One important thing here, again one common mistake that students tend to make in this particular problem is that the moment you are given PV equal to constant, students assume the temperature to be constant. But you have to keep in mind that in this case, the temperature does not remain constant, but that is actually quite clear from this PV diagram. Notice that we are actually moving from one isotherm to another. So, for instance, we go from this isotherm to this isotherm. So, the temperature definitely does change during the process, but that is because it is not an ideal gas, it is a two phase mixture. So, that is something that you should keep in mind. So, it is not an ideal gas.