 Hello and welcome to another video of understanding thermodynamics. My name is Adrian and today we will be looking at boundary work. Now what is boundary work? This is for example work that is done in an internal combustion engine and we said that in thermodynamics we study the conversion of energy and an internal combustion engine the heat is released upon combustion of the fuel is used to power the wheels of the car. You see this happening visually on the slide in the picture on the left you can see ignition as just occurred of the fuel where the spark ignites the air fuel mixture and this creates a high pressure inside the combustion chamber. The high pressure forces the piston down and transfers its energy to the rotating crankshaft below. The gas is still contained within the combustion chamber and this is called boundary work and you can see from the first picture up onto the middle one to the last one how the volume inside the boundary expands. Therefore the work done in a closed system is called boundary work. Now we will consider the work done in three special processes. The first process is called a constant pressure process. The second process we will look at is a constant temperature process and then the last process we will look at is an adiabatic process. Now adiabatic process means that there is no heat transfer across the system boundary. Now let's see what is the definition of work first before we look at these three processes. In physics work is defined as the product of force and displacement. Now let's consider a piston cylinder setup. The gas inside the cylinder exerts a force on the piston head. When we add heat to the gas it will expand and the piston head will move upwards. As the weight of the piston does not change and the ambient pressure remains constant outside the cylinder the pressure inside will remain the same. When the heating stops the piston moved upwards through a distance s meters. Now we know that the pressure inside is equal to the force exerted by the gas on the piston divided by the cross-sectional area of the piston as well and the displacement s multiplied by the cross-sectional area is equal to the change in volume. Therefore we can say that in a constant pressure process the work done is equal to the pressure multiplied by the change in volume. Now let's look at an example problem. One kilogram of water at 25 degrees Celsius and 100 kilopascals is heated at constant pressure to a temperature of 200 degrees Celsius. Calculate the work required. Now you can pause this video and have a go at this problem on your own. So the equation that we need to solve is the work done equals the pressure times the change in volume. We know that the pressure is equal to 100 kilopascals but we need to determine the values of V1 and V2. Now the phase of water at 25 degrees Celsius and 100 kilopascals is compressed liquid. We assume that liquid water is incompressible and we can therefore use the specific volume of saturated liquid at 25 degrees Celsius and we get an answer of 0.001003 cubic meters per kilogram. Now the phase of water at 200 degrees Celsius and 100 kilopascals is superheated vapor and its specific volume is 2.4062 cubic meters per kilogram. You can find these values in the steam tables. Now knowing all these values we can go and calculate the work done and we get an answer of 240.52 kilojoules. Next let's look at a constant temperature process. Consider the piston cylinder set up with a number of weights on the piston. When one of the weights is removed the gas will expand and the temperature and pressure will drop. Now heat can be added to keep the temperature constant and we can perform this expansion process gradually and slowly add heat at a sufficient rate to keep the temperature constant. Now for an ideal gas the relation between pressure and temperature is given by this equation. M, R and T are all constant in this scenario and in order to calculate work for this process we integrate the differential form of the work equation. We can use the ideal gas law to substitute the variable P here and we can get a much more workable equation and we end up with an equation for work at a constant temperature process as shown here. All right let's do an example problem for a constant pressure process. The problem states a two kilogram of air at 25 degrees and 87 kilo Pascal is compressed slowly to 200 kilo Pascal. The air releases heat at a sufficient rate that the temperature remains constant. Calculate the work. You can pause this video and have a go at this problem yourself. So the equation that we need to solve is the one that we've just shown in the previous slide and looking at it we need the values of v1 and v2. We can calculate the values of v1 and v2 using the ideal gas law because we know mass, r and temperature are constant and we've been given the two pressure values that we can use to calculate volume 1 and volume 2. We then end up getting an answer of minus 142.5 kilo Joules. This just indicates to us that it is work required to get that air from 87 kilo Pascal compressed to 200 kilo Pascal. Lastly we can have a look at the adiabatic process. Now for monoatomic gases such as helium, argon, neon and krypton it is safe to assume that the ratio of the constant pressure specific heat to the constant volume specific heat does not vary with temperature. Now such gases are often called perfect gases and the ratio between cp and cv is called k or even gamma. For a specific adiabatic process in other words a process that does not exchange heat with the environment pv to the power of k remains constant and we can denote that constant value as c. We can then use this equation to get rid of the pressure variable in the work equation and the result is this equation at the bottom. Now let's do a problem using the equation for an adiabatic process. The problem states one cubic meter of helium at 25 degrees Celsius and 200 kilo Pascal in a piston cylinder arrangement is expanded adiabatically to twice its volume so two cubic meters. Calculate the work. Pause this video and have a go. Now if we look at the equation that we derived we only need the value c for this problem as we have been given volume two and volume one. We also know the pressure at point one so we can get the value of c. Now this is done by just multiplying the pressure at state one by the volume at state one to the power of k and this value of k you can get from your tables in your handbook that's 1.667 and we get a value of c of 200. Now we can substitute that value in the equation we derived and we get a value for the work done is 111 kilo Joules. Now just for interest sake because we've got the value of c we can actually go and determine the pressure at state two with saying that the relationship of state one between pressure and volume is the same at state two and then the only unknown in this equation is p2 so we can get an answer for the pressure at state two is 62.98 kilo Pascal. So in summary we have seen that the definition of work is the integral pdv for an expansion process work is done and your value is greater than zero for a compression process we require work and thus the value of work done is smaller than zero. For an isothermal process and an ideal gas we can calculate the work analytically and we can also calculate the work analytically for an adiabatic process and a perfect gap and that's it hope you've enjoyed the video the course notes which these videos are based on is available on my blog audienceblog.com i'm also on twitter my twitter handle is at asvn90 if you want to connect with me or ask any questions i am more than happy to answer them thank you very much for watching and i will see you in the next video bye