 We're now going to solve an example problem involving the first law applied to closed systems. So what we will do is we will begin by writing out what is known within the problem statement We know that we have a mass of 0.2 kilograms and it's a piston cylinder device with R134a inside of it. We know that at state 1 we have a pressure of 200 kilopascals and 75% is in the liquid phase. So from that we can immediately determine the quality and recalling the definition of quality we said that it was the mass of the vapor to the mass total. So that would mean that if 75% is in the liquid phase 25% is in the vapor phase. So the initial quality for this problem is 0.25. Now the process that is underway is one of heat addition and we were also told that it was at constant pressure. And state 2 what we were told is that given its constant pressure the pressure will remain at 200 kPa. And we were also told that it was when all of the liquid had turned to vapor which would imply that we've moved to the saturated vapor line where the quality at state 2 would then be 1.0. So x2 equals 1.0. What are we asked to find? So taking a look at this on a process diagram. So we have isotherms given that we're looking at a PV diagram. We're starting at state 1. We are going through heat addition with a constant pressure process and we end at the saturated vapor line. So we end here at state 2. So that's what our process looks like on a PV diagram. What we'll do now is we'll go through analysis determining the initial volume, the work done and the total heat transfer. So we know state 1. We can then go in and determine property data using tables in the back of your textbook. And we're in the two phase region and so it turns out that the temperature at 200 kPa for R134A in the two phase region has a saturation temperature equal to minus 10.09 degrees Celsius. So that is the temperature of our system at state 1. And while we're in the table we might as well extract out information in terms of the properties of both the saturated liquid and saturated vapor lines. So let's go about and do that now. Now I'm not sure if we are going to need all of these but given that we're in the table I like to write these down. And that way we can refer to them later on in the analysis in the event that we need them and we don't have to go back into the table again. And the units for the specific volume, internal energy and enthalpy at the saturated vapor will be the same as they were for the saturated liquid so I won't write them out. Okay so that's some of the property data. Now what we're going to do, let's begin by determining, we want to find the initial volume v1 and we know the mass. And so if we can find the specific volume at state 1 we can then directly calculate the volume at state 1. So that's the next thing that we will do. So the specific volume at state 1 we're in the two phase region so we have a mixture. We use our equation that enables us to determine that. Then in order to determine the volume, we know the volume in kilograms we multiply it by the specific volume at state 1. And that enables us to determine the answer to the first part of the problem. The second thing that we need to determine is the work done in the process. Recall it's a piston cylinder device at constant pressure and the volume is changing and consequently work will be done due to boundary work. And so what we need to do now is determine the form of the first law that we want to use as well as the properties. The properties that will be important for us will be the internal energy at state 1 as well as the enthalpy at state 1. So let's go ahead and calculate those now. Okay, so that is the internal energy as well as the enthalpy at state 1. Now in order for us to calculate the amount of work being done what we'll need to do we need to take a look at the first law. So the first law, the form that we want to deal with, okay so looking at this form we know that our piston cylinder device is not moving, moving macroscopically of the entire unit and consequently what we can do is we can neglect kinetic energy nor is it moving in some sort of gravitational reference frame and consequently we can remove the potential energy as well. So those two terms disappear. The next thing that I'm going to do is I'm going to rewrite the work in terms of remember we referred to work other plus work boundary. Now what we can also say is that there is no other form of work, the only form of work taking place is boundary work and consequently with that what we're left with is heat is equal to the change in internal energy plus the boundary work WB. Well that is also equal to the change in enthalpy. So what we're going to do is we're going to call that equation number one or we can also write and we'll call that equation number two. So we can see we can use two forms of the first law, one using enthalpy and the other one using internal energy. So what we'll do is we will try solving the problem using both of those forms of the first law. Now in this equation I should say let's write out what the boundary work is. So we have boundary work. Now it's a constant pressure process and consequently what that means is that we can pull the pressure outside of the integral and we're going from state one to state two times the change in volume of our piston cylinder device. And I will call that equation number three. So that is an expression for the boundary work using the change in volume. So let's go and see what's going on in state two now and get some of the property data for that because you can see in the forms of the equations we're looking at we need to know state information of both state one which we already have as well as state two. So taking a look at state two it's the same pressure as at state one and we're told that we're saturated vapor. Consequently the quality there is 1.0. So going back to the data that we collected for our tables that we pulled out earlier when we were looking at the two-phase region we can write out the specific volume, the internal energy and the enthalpy. And those are the property values for state two. First thing we can do is calculate the volume of two just like we did for the volume of one. We know the mass and we know the specific volume at two. With that we can then calculate the boundary work using equation three. Now you have to be careful here because when you do this calculation you will get a value in joules but our equation is expressed in kilojoules with all of our properties. So be careful when you calculate the boundary work. Your number will come out in joules but you need to convert it into kilojoules. You divide by a thousand. Now what we're going to do is try to solve for the heat flow. So actually that there is the answer to the second part of the problem. It told us to find the work done and so the work done is the boundary work. Now we can move into the third part which is calculating the total heat transfer. And what we're going to do we're going to use both approaches. The one with enthalpy as well as the one is using internal energy. And if all goes well we should get the same answer for both of the techniques. So what we're going to do we're going to use equation two to begin with. So the first approach we get thirty point six seven. Another thing I should comment on the use here are large use. It's not per unit mass and that's why I brought in the mass into this term here. Because the little two or the little use are the specific internal energy. That's kilojoules per kilogram. So you need to multiply it by the mass and that's why I made that change. So that is the answer using the boundary work and the internal energy. Let's try using equation one. So h2 minus h1 is the difference in enthalpy between state two and state one. And when we do the calculation here again we get thirty point six seven kilojoules. So the fact that this and this are in agreement is good. Because that's what we would have expected both of the techniques will work. One you use internal energy and the other you use enthalpy to solve the problem. So that concludes that example problem.