 So, in this and the next lecture, we will be looking at examples, which require us to carry out first law analysis of systems. We will do examples involving both ideal gases and two phase mixtures of water as well as R134A. So, let us start with a few examples involving ideal gases. The first example actually involves liquid and the solid. So, we have 2 kg of ice in a container. So, the container initially, so let us say we have a container like this. So, this initially contains liquid water at 20 degree Celsius and we have certain amount of ice which is 2 kilogram. So, 2 kilogram of ice at 0 degree Celsius is dropped into this container of liquid water and then we also add 3500 kilo joules of heat to the container. So, assuming that there is constant pressure and there is no loss of heat to the surroundings, we are asked to determine the final temperature and mass of water vapor formed, if any. We do not know whether any of the water is converted to vapor or not, but if it is then we are asked to calculate the amount of water vapor formed. So, for us the system in this case is relatively easy to identify. We take the ice and the liquid water in the container to be the system. So, when the ice is dropped into the water, it will melt and perhaps all of the both of them would evaporate together, but this system will contain the same amount of mass. So, we take that to be our system and we apply first law, but what we would do is since we do not know how much of the water if any is going to evaporate, let us calculate the amount of heat that is required to take this system to a final temperature of 100 degree Celsius. We are assuming that in everything is at atmospheric pressure. So, we require the I mean we calculate the amount of heat required to take the system to 100 degree Celsius and if the supply heat exceeds this, then we know that some of the water is going to evaporate. We can calculate the amount of water that evaporates after that. It is a very logical way of approaching the problem and there are other shorter ways perhaps, but this is a very logical way of approaching this problem. So, let us do it that way. So, we apply first law delta E equal to Q minus W. In this case for the system under consideration, there are no kinetic energy changes. There is no, there are no potential energy changes. Only the internal energy change is present. So, we write delta E equal to delta U equal to Q minus W. Now, none of the system boundaries deform during this process, which means displacement work is 0 and no other form of work interaction is also there. We do not have electrical work or steering work or any spring work anything like that. So, we take W to be equal to 0. So, which means that the internal energy change in internal energy of the system is equal to the heat that is supplied and the system itself is comprised of ice and liquid water. So, delta U ice plus delta U liquid is equal to Q. So, if you focus on the ice, then the ice at 0 degree Celsius melts to form water at 0 degree Celsius and it is then heated to 100 degree Celsius. So, the change in internal energy for this process comes out to be the latent heat that is required to take the ice from ice at 0 degree Celsius to water at 0 degree Celsius and then the amount of heat that has to be supplied to take it to a final temperature of 100 degree Celsius. Now, for the liquid water it is initially at 20 degree Celsius. So, the change in internal energy for the liquid water would be m liquid times a specific heat capacity of the liquid times a change in temperature which is 100 minus 20. So, the sum of these two would be the amount of heat that is required to take the system to a final temperature of 100 degree Celsius. So, if you substitute the numbers we get this to be 3188 kilojoules. So, we need to supply this much heat to take the ice at 0 degree Celsius and the water at 20 degree Celsius combined to 100 degree Celsius. The heat actually added is 3500 kilojoules so that is more than this. So, we know that some of the water will definitely evaporate. What we do next is as I said there are shorter methods for doing this but let us just look at the logic behind doing you know logic behind solving this problem. So, if all of the water evaporates then the amount of heat required for that is given by m ice plus m liquid times L prime where L prime is the latent heat of evaporation and if you substitute the number that comes out to be 15820. So, which means that this is not sufficient to evaporate all of the water. So, the amount of water that will that will evaporate can be calculated by taking 312 which is the excess heat divided by the latent heat. So, that comes out to be 0.1381 kilogram. So, the analysis proceeds in almost all examples along the same lines you identify the system and then you go through the analysis by applying first law in a consistent manner and then you go through the analysis that is most important identification of the system applying first law to the system and then proceeding. But there will be increasing degree of complexity as we go through the examples this is a very very simple example. Let us go to the next example. Here we have a frictionless piston cylinder mechanism. So, this initially contains air and the air in the cylinder is compressed from 100 kilopascal to a final pressure of 800 kilopascal and the process is given as PV raise to n equal to constant where n is equal to 1.27. Such a process is usually called a polytropic process with n being called the polytropic or index exponent. We are asked to evaluate the work done and the heat interaction. So, the assembly is not insulated it is only frictionless which means there is going to be some heat interaction between the cylinder and the surroundings. So, it is quite clear that we take the air in the cylinder as our system. So, here we are looking at polytropic processes for different indices on a PV diagram this is for ideal gas. So, it is very important to keep that in mind. Notice that n equal to 0 represents an isobaric process where P is constant and n tending to infinity represents an isochoric process for which the volume remains constant. All other processes fall in between n equal to 1 is an isothermal process. So, this is isobaric and this is isochoric and n equal to gamma for an ideal gas as we will see later is an isentropic process where entropy remains constant. n equal to 1 is isothermal where temperature remains constant all the other processes lie in between. So, if your initial state is 1 then states on the second quadrant are obtained by compression processes and states on the fourth quadrant are obtained as a result of or at the end of expansion processes. So, we take the air as a system and the work interaction remember we have only displacement work in this case that is integral of PDV. So, integral 1 to 2 PDV we substitute the process equation we know how P is related to V that is explicitly given in this example. So, we can get this to be equal to this and since we assume air to be an ideal gas we can take PV equal to MRT. So, we finally end up with an expression that looks like this. For air we take the molecular weight to be 28.8 and gamma to be 7 5th. So, we can calculate the specific gas constant or particular gas constant as the universal gas constant divided by the molecular weight and CV is nothing but r over gamma minus 1 from Meijer's relationship we derived this earlier. So, we will be making use of these two. Now, PV raise to n equal to constant may also be rewritten as P raise to 1 minus n over n times T equal to constant once we use the equation of state. So, we can apply this to the initial and final state and get the final temperature to be 467 Kelvin. So, once I know the final temperature I can actually substitute these values into the expression for displacement work and get it to be equal to minus 357.11 which means that since air is compressed the negative sign is consistent with the fact that work is being done on the system to increase the pressure. Now, we apply first law to this system. So, delta E equal to Q minus W. In this case also there is no change in potential energy of the system and there is no change in the kinetic energy of the system. There is no work except displacement work. So, we can write delta E equal to delta U equal to Q minus W which may then be rewritten like this. So, delta U is as we said before MCV times T2 minus T1 and if you substitute the values Q comes out to be minus 116.06. The negative sign indicates that heat is lost by the system to the surroundings as the air is compressed. Now, at this point students at least quite a few students have a question which is that since CV is specific heat at constant volume should it not be applied only to constant volume processes how are we justified in using CV for this process. One must keep in mind that when we wrote the expression for total energy you may recall that we derived the first law for a cyclic process and then we identified E as a property and you may recall that we wrote E is equal to U plus PE plus KE. We know that KE is a property, P is a property, E is a property. So, U is also a property of the system which means that U depends only on the initial and the final state and not on the process by which we arrive at that state. And you may also recall that we further wrote the specific internal energy U to be equal to 5 halves or T depending based on the degrees of freedom that a diatomic gas had correct. So, we wrote it to be 5 halves or T notice that T is a property. So, which means that U is also a property and it depends only on the initial and final state from this we identified. So, we wrote this as CV times T and we equated CV to be 5 halves or for this case per unit mass. So, this is independent of the process. So, the fact that delta U equal to M times CV times delta T is independent of the process and CV is equal to 5 halves or again it does not depend on the process. It is called specific heat at constant volume because as we wrote earlier if we go back a little bit as we wrote earlier as we wrote earlier CV is equal to partial U partial T and remember we said U is a function of T comma specific volume it has to be a function of two properties because we are dealing with the simple compressible substance. So, CV when we take the partial derivative it is understood that V is a constant here because we are taking partial derivative with respect to T V is a constant which is why it is called specific heat at constant volume. But it does not mean that it is dependent on the process the constant at constant volume refers to the fact that we have to take a partial derivative because this is a function of two variables. So, that completes this example let us move on to the next example. So, here we have an insulated vessel like this notice that the insulation is indicated here through this hatching. We have an insulated vessel and side A is initially evacuated we have a stirrer in compartment B and there is also membrane which will rupture if the pressure exceeds 1 megapascal in compartment B. So, we have a certain amount of air in compartment B. So, work is transferred externally through the shaft to the stirrer and the stirrer is turned until the membrane ruptures. So, once the membrane ruptures the air fills the entire vessel we are asked to calculate the final temperature or temperature in B when the membrane ruptures work done by the stirrer and the final temperature and pressure once equilibrium is attained remember this is evacuated initially which means that this this will be an there will be an unresisted expansion when the membrane ruptures. So, that is a non-equilibrium process what we are asked to calculate is a final equilibrium temperature and pressure. So, we take the entire vessel to be our system as shown here and as we discussed earlier. So, let us proceed with the example. So, the initial temperature may be evaluated in B may be evaluated by using equation of state in this manner to get the initial temperature to be 303 Kelvin. Now, the membrane ruptures when the pressure in compartment B is 1 MPa. So, the temperature of the air in compartment B when the membrane ruptures may also be calculated quite easily using the equation of state. So, once it reaches 433 Kelvin the membrane ruptures. So, this process is labeled 1 2. So, starting with the initial state until the time when the membrane incident when the membrane ruptures is we identify that as process 1 2. So, we can apply first law. So, we write delta E equal to delta U because there is no change in PE or there is no change in KE. There is no heat interaction because the vessel is insulated and W 1 2 consists only of sterile work. There is no displacement work, displacement work is 0 because there is no deformation of the system boundary. We have identified the system to be to contain the entire vessel. So, there is no deformation of system boundary. So, delta U is for the system is simply M times Cv times T2 minus T1 for the air in compartment B. So, we may evaluate this as minus 187.64 which means work done by the sterile is 187.64 kilojoules. So, let us say that the final equilibrium state is numbered 3 and we apply first law for process 2 3. Again delta E is equal to delta U is equal to Q minus W. Q is 0 because vessel is insulated. In this case for 2 3 W displacement is already 0 because there is no deformation of the system boundary. The sterile is also stopped once the membrane ruptures that is given in the problem statement. So, there is no work interaction for process 2 3 which means that the temperature remains constant. So, the final temperature may be obtained as 433 Kelvin. And knowing the final temperature and the final volume remember the air now occupies the entire vessel we can calculate the final pressure to be 333.33 kilo Pascal. The next example is somewhat involved straight forward, but quite involved in terms of work that we do. So, we have 2 cylinders here of different cross sectional areas. The bottom cylinder as indicated here is completely insulated both the piston and the cylinder are insulated. Now, heat is supplied to the air. So, it this contains air this also contains air. So, heat is supplied to the air in compartment A or cylinder A and the piston moves slowly down by distance of 2 meters. We are asked to calculate a bunch of quantities here as you can see areas are given total distance is given mass of the piston piston assembly is also given. So, let us proceed with the analysis. So, first quantity which is asked is mass of air in cylinder A. So, we can calculate it using ideal gas equation of state all the values are known. So, that comes out to be 0.2 kilograms. Now, at any instant during the process of force balance on the piston looks like this. So, this is the differential pressure. So, if you look at this. So, the weight of the piston acts in the downward direction. The pressure at A acts in the downward direction, but the atmospheric pressure on the lower face of this piston acts in the upward direction. So, the net force in the downward direction is P here minus P atmosphere times the cross sectional area. Now, the pressure of the air in cylinder B acts in the upward direction atmospheric pressure acts in the downward direction on this piston. So, the net force is P B minus P atmosphere times area in the upward direction. So, the force balance may be written like this. And if you rearrange this, then and use the known values take all the known quantities to the right hand side, we can get the pressure in compartment B to be initial pressure in compartment B to be 325.45 kilo Pascal. So, once I know the pressure, I can I am given the initial temperature also. So, I can use the equation of state to evaluate the mass of air in the in cylinder B as 0.15 kilogram. Now, work interaction for atmosphere basically is P atmosphere times delta V for the atmosphere, because atmospheric pressure remains constant. Now, delta V for the atmosphere has to be calculated carefully because this piston assembly moves down by 2 meters. That means, the volume occupied by atmosphere decreases on this side and it increases on this side. So, delta V on the A side actually is negative delta V on the B side is positive for the atmosphere. So, we calculate. So, delta V on the A side is minus 2 times A cross-sectional area on that side, because the piston moves down by 2 meters and delta V on the B side same height times the cross-sectional area with the positive sign. So, the work interaction for the atmosphere comes out to be minus 786 joules, which means work is done on the atmosphere. Now, the piston work comes out to be positive based on the elevation change because the piston moves down. So, which means that there is a reduction in potential energy of the piston, which means the piston is doing work on these aerotics or whatever it is interacting with. So, that is plus 200 joules. Now, if I consider the air in B as my system at any instant, I can apply first law to the air here and write like this for a differential change. Here we are using the differential form of the first law because we are applying it at an instant. De equal to du no change in ke or pe equal to delta q minus delta w, delta q is 0 because B is insulated as we already mentioned and the only form of work is displacement work. So, minus P dv. Now, if I substitute for du, I can write it as mcv dt and this may be written as in terms of specific volume as minus m times P times dv. Rearrange and integrate. We finally, get this equation which describes the process undergone by the air in cylinder B, which is simply nothing but T times specific volume V raised to the power gamma minus 1 is equal to a constant. So, this is the equation that describes the adiabatic fully resisted process. Remember the air, the cylinder and the piston are insulated on the B side. So, and the process takes place slowly. So, the process undergone by air in the B side is an adiabatic fully resisted process, which is described by this. So, I can calculate the final pressure in B using this equation. The final volume is V initial volume minus 2 times the cross sectional area. So, we get the final pressure in B to be 442.84 kilo Pascal. Final temperature in B may also be evaluated using equation of state as 327 Kelvin. We know the final volume. We know the final pressure. We can easily calculate the temperature. Now, if I apply first law to the air in the cylinder B, remember previously we applied first law to the air in cylinder B at an instant. So, we use the differential form of the law. Now, we are applying first law to the air in cylinder B for the entire process. So, we write delta E equal to delta U equal to Q minus W. And so, the displacement work for the air in B comes out to be minus 2923 joules. Clearly work is being done on the air in cylinder B and the magnitude is 2923 joules. Now, I know the final pressure. We have already calculated the final pressure in cylinder B. So, using our force balance equation, I can now evaluate the final pressure in cylinder A. In the beginning, we use the force balance equation to calculate the initial pressure in B. Now, we are using the final pressure in B to calculate the final pressure in cylinder A and that comes out to be 258.7 kilopascal less than the pressure in cylinder B. The final volume of air in cylinder A is higher by 2 times the cross sectional area because the piston assembly moves down by 2 meters. So, that may be evaluated quite easily. And using the equation of state, we may evaluate the final temperature of air in cylinder A to be 458 Kelvin. Now, if we apply first law to the air in cylinder A, we get delta E equal to delta U equal to Q minus W. Q is not 0 here because we are supplying heat to the air in cylinder A. So, we can rewrite this expression as delta U A plus W A. Notice that the work interaction for the air in cylinder A plus work interaction for air in cylinder B plus W piston plus W atmosphere is equal to 0. So, if I take the air, then the work done for the system plus work done for surroundings is 0. For the air in cylinder A, surroundings it is interacting with consists of air in cylinder B, piston plus atmosphere. So, the sum of all this is equal to 0. So, I can replace W A with a negative of W B plus W piston plus W atmosphere. Substitute the values, we get heat supplied to be at 26.315 kilojoules. The problem is involved, but conceptually not very difficult to follow. You need to do this systematically, identify the system, apply the first law, use the equation of state and then proceed based on the information that is given in the problem.