 So, problem 3, 2 kg of water, so m equal to 2 kgs at 25 degrees, T1 equal to 25 degrees in degree in a frictionless piston change arrangement as in the figure. The mass of the piston is such that it exerts a pressure of 100, so that is directly the piston pressure is given, adhesive pressure also is same. So, that means, when the piston has to float, the pressure should be at most the pressure plus piston pressure that is 200, at least it should be this, when the piston has to float. So, at state 1, the pressure should be 100 plus 100 is equal to 200 kilo Pascal's, so that is the thing. Heat is slowly added, when the piston just touches the stops, so let us say that is state 2. State 2 is when piston just touches the stops, so that means I will say V2, sorry P2 will be the same 200 kilo Pascal's and V2 we can calculate, so that we can ensure this, because at till that point it will be a constant pressure process. Then after that, state 3 here temperature finally reaches, heated until the temperature finally reaches 1000 degrees centigrade, so T3 will be 1000 degrees centigrade, so these are the states which are given to us. So, here when the piston just touches the stops, T2 is also given, 240 degrees centigrade temperature. Pressure should be constant at this point, then pressure should increase, because piston cannot move upwards, the pressure will increase, constant volume will occur then, so this is the problem. So, let us see state 1 properties, state 1, state 1 is 200 kilo Pascal's, 25 degrees centigrade. So, we have to go to the tables, so here if you go to the tables, 200 kilo Pascal's is 2 bar, so we are in 2 bar now, 2 bar and we will see that the temperature is for 2 bar saturation temperature is 120.2 ok, at 2 bar T sat saturation temperature is 120.2 degrees centigrade, but T1 is 25 degrees centigrade, so the state is sub cooled liquid, so sub cooled liquid state is the state 1. So, this means properties of saturated liquid at the given temperature can be taken. So, what I am saying is V1 will be equal to Pf at 25 degrees centigrade, U1 equal to Uf at 25 degrees centigrade, so that is the main, because it is sub cooled liquid ok. So, what we are saying here is, here the saturation lines are like this in a TV diagram, so the pressure will be like this, but for the given pressure of say 2 bar, you are here somewhere, this is 25 degrees centigrade and this is 120.2 degrees centigrade. So, you should not take the value corresponding to the pressure, but take the value corresponding to temperature, this is 25 degrees centigrade. So, for sub cooled liquid, the properties of saturated liquid at the given temperature should be taken, so that is what we are trying to get here. So, V1 will be equal to, so we will see the properties again. So, go to the, now we should not take the entries from the saturated pressure tables, but go to the entries in the saturation temperature based tables. Here you can see that for 25 degrees, the value of Vf, Vg are given, but we have to take value of Vf, which is this and the Uf ok. So, now, V1 equal to 0.001, 0,032 meter cube per kg, very small value volume. Similarly, U1 will be equal to 104.9 kilojoule per kg. So, these are the initial state properties. Process 1 to 2 occurs at constant pressure, because the distance is free to move. So, P2 will be also equal to the same as P1, so 200 kilopascals. T2 is given as 240 degrees centigrade. So, T sat at 200 kilopascals equal to 120.2 degrees centigrade. Now, we see that T2 is greater than T sat at P2 implies the state 2 is superheated state. Do you understand? So, like this, in the previous case this is 120, but given for state 2 it is 240. So, you are in the superheated state. So, state 2 is in superheated state from superheated tables for 2 bar. We will see the entries now. So, here this is the table. We are talking about 2 bar table, temperature is 240. So, here you can see that we have to interpolate between these two entries, 200 and 250 are given, but given temperature is 240. So, we have to interpolate. So, we know that at 200 the specific volume is 1.08, at 250 it is 1.199. So, for 240 what will be? So, that we can take here. So, interpolating between 200 degrees centigrade and 250 degrees centigrade entries, we can get V and U for 240 degrees centigrade at 2 bar. So, by doing that we get V2 equal to 1.175, 2 meter cube per kg, U2 equal to 215.6 kilo kilo joule per kg. So, these are the values. So, state 2 is fixed. Now, state 3 process 2 to 3 occurs constant volume. Why? Because the piston is now hitting, it stops. So, it cannot move upwards. So, constant volume. So, here we can say V3 will be equal to V2 equal to 1.175, 2 meter cube per kg. So, one value is fixed and what is given is temperature is given, final temperature is given as 1000 degrees centigrade. So, these are the two state, two state properties available. That is at 1000 degrees centigrade, the volume is 1.152. So, what is this state? So, 1000 degrees centigrade. So, the state should be superheated. Already, it is superheated in the previous state. So, further heating has done, has been carried out. So, it should be superheated state. State is superheated vapor. But now the problem is, I do not know the pressure. We know that all the superheated tables are given in terms of pressures. So, once the pressure is known, we can go to the particular table and do. But now we have to go to a particular table where the temperature is 1000 degrees. And the specific volume is 1.1752 meter cube per kg. So, we have to go to the table and get the value of pressure and u from that table. So, we go here 1000 degrees. So, if we say, for example, 2 bar, it is actually it should be more than 2 bar because already we are in 2 bar in the state 2. So, now 3 bar if we take, 3 bar I find the volume is still high. 4 bar it is 1.469. What is the volume required is 1.175. So, now we can see these two are higher. So, go to higher pressure where the volume reduces. 5 bar it is for 1000, it is 1.175. So, it is very close to our value. So, this is the state for us. The pressure has reached 5 bar and the temperature has reached 1000 degrees where the specific volume is 1.175 which we have calculated. So, we can take the u value as 4052. So, here 1000 degrees 1.175 pressure is 5 bar and u3 is 4052 kilojoule per kg. So, this is the value. So, now all the states are fixed. Ok, now what you should do is you can apply the first law q122 equal to w122 plus u2 minus u1 delta ke equal to delta pe equal to 0. So, from this I can say, first we will calculate work. Work is when there is a change in volume which has occurred in the process 1 to 2. So, I can say work here is p into 2 into 200 this mass into p into v2 minus v1. So, I will write the value such as 1.175 2 minus 0.0010032. This is v2 minus v1. So, that will be equal to 469.6 kilojoules. So, delta u122 will be equal to m into u2 minus u1 which is 5221.4 kilojoules. So, q122 can be calculated as 5691 kilojoules. So, now for process 2 to 3 delta v equal to 0, so w equal to 0. So, q223 equal to u3 minus u1 or u2 u3 minus u2. So, that means you will see this is m into u3 small u3 minus small u2 which is equal to 8363.8 kilojoules. So, now let us try to draw the diagram, tv diagram. Saturation lines we have to draw for all the things. First is 200 kilo pascals or 2 bar 200 kilo pascals. So, here initially 25 degrees centigrade is the temperature. So, here 25 degrees centigrade. So, state was somewhere here sub cooled then that is state 1. Now, state 2 was 240 degrees. So, maybe this is this is 120.2 degrees centigrade saturation then say 240 degrees somewhere here this is 240. So, this is the state 2. So, the state 1 first the liquid sub cooled liquid goes to saturation saturated liquid then plate in heat is added then it becomes saturated vapor then it becomes super saturated vapor. So, that is the state 2. Then what happens? Pressure increases volume remains constant. So, volume remains constant from now. So, this is the volume V2. V2 can say this is V1. So, now what is the final pressure? 2 to 3 final pressure is 5 bar. So, something here only qualitatively I am showing here something like this. So, this is actually 1000 degrees centigrade may not be the scale please understand 1000 degrees centigrade actually should be much higher than the critical point. The critical point temperature is 370. So, I will draw it properly. So, this is 1000. Okay, now constant volume. So, 2 to 3 here is this process superheated to superheated. So, this is 3 states. So, this is the way first constant pressure process sub cooled liquid at 200 kilo Pascal which because temperature is 25 degrees less than the saturation temperature. Now, it is heated becomes saturated liquid at 200 kilo Pascal then later it is added becomes saturated vapor then it goes to superheated vapor in final state constant pressure process occurs till that. Then constant volume process occurs where the volume remains constant and pressure increases. So, this is 5 bar. So, 2 bar to 5 bar final pressure is 5 bar we have estimated correct. The temperature is given was given. So, corresponding to this temperature and this V2 was 1.175. So, based upon this we have calculated this pressure as 5 bar. So, this is the state diagram. Always in the draw state diagram you have to show the saturated lines. Saturated this is called saturated liquid line and this is the saturated vapor line. Saturation line should be shown. So, this completes this problem. Now, this fourth one a cylinder in which a piston is retained by a spring. So, as a spring is having this is retained by this always the spring is connected to the top push of the piston. This contains water. Water means it can be liquid in vapor initially contains 0.01 meter cube. So, V1 is 0.01 meter cube. The mass of the water is also given 0.5 and the cross sectional area of piston is given AP is 0.05 meter cube. Mass is 0.5 kg. Okay. Now, mercury manometer is connected to the cylinder initially reads height difference between its limbs this and this. It has 37.5 centimeters as shown in the figure. Water is now slowly heated ok heated until the piston rises through a distance of same 37.5 centimeters. So, from this it goes to some height. So, this also is 37.5 37.5 centimeters. So, now at any instant during this process bottom surface of the piston this bottom surface of the piston and the mercury meniscus in the openly more at the same level. So, when the piston goes here mercury meniscus also will be in the same level. So, this will be the meniscus level that is the from this point the meniscus also would rise to 37.5 centimeter. Similarly, this will drop to 37.5. So, the final state we can say somewhat here and here could be meniscus levels. Okay. Assume that the pressure excited in the left limb. So, here the left limb pressure excited in the left limb of the manameter due to the pressure presence of water above the mercury column. So, this is the darker region is mercury and this is water. So, this water has some height and it has density of 1000. So, it can also put some pressure on this point correct, but it is told that assume that the pressure excited in the left limb of the manameter due to the presence of water above the mercury column is negligible. So, you need not take into account of that because the mercury density is 13600 kilogram per meter cube water density is 1000. So, we can neglect it. So, 13 times more density. So, we can neglect it. We can also include it if you want, but this clarity is shown. So, given that no, don't need to take into account of the water hydrostatic pressure in the left limb. Now, G is 9.81 and the atmospheric pressure is 1 bar. So, this is the problem. So, the pressure at every point can be calculated as rho G H plus P atmosphere here of this opening here the atmospheric pressure acts that is 1 bar. Okay. Now, what find the initial and final pressure of water relationship between pressure and volume the heat and work interactions. Okay. So, this is the problem. So, let us do this solution initial pressure equal to atmospheric pressure plus density of mercury into G into H1 which is equal to 1 into 10 power 5 because 1 bar plus 13600 into 9.81 into 37.5 into 10 power minus 2. So, that will be equal to 150,000 Pascals or 1.5 bar. Okay. So, that is the initial final pressure P2 will be equal to P atmosphere plus rho HG G into H2. Now, what is H2? H2 will be when the when the piston moves to 37.5 the miniscule in the right limb will also go up to the height of 37.5. But the left limb will also descend from the given position to 37.5. So, the total H will now be 2 times the 2 times of 37.5. Okay. So, that we have to take into account. So, when the piston goes up by a height of H the difference between the miniscule levels in the right and left limb will go by 2 H. So, that we will take into account and calculate this as 1 into 10 power 5 plus 13600 into 9.81 into initially the difference was 37.5 plus due to the descent of mercury by in the right limb by 37.5, left also will go down by 37.5. So, there is increase of 2 H. So, that into 10 power minus 2. So, this is the H2. Now, that is our delta H you can say. So, now, that will be equal to 250,000 Pascals or 2.5 bar. So, the initial and final pressures are got from pressure based saturation tables for P equal to 1.5 bar. We can take Vf equal to 0.001053 meter cube per kg Vg equal to 1.159 meter cube per kg as we have done in the last time. Similarly, Uf in Ug, Uf equal to 466.9 kilojoule per kg and the Ug equal to 2520 kilojoule per kg. So, these are the values. Now, how will we calculate the value? V1, first we have to say V1, we have to calculate as what? 0.01, this initial volume V1 meter cube divided by M. M is 0.5 kgs in kg. So, you will get this as that is initial specific volume as 0.02 meter cube per kg ok. Now, you see that the Vf is less than V1 is less than Vg correct. Because you can see V1 is more than Vf less than Vg. So, that means, state 1 is saturated mixture of liquid and vapor that is it ok. So, now, I can find the quality. What is quality at state 1? It will be equal to V1 minus Vf divided by Vg minus Vf which is equal to 0.0163625. So, that is the quality. So, from then I can find U, U1 equal to Uf plus x1 into Ug minus Uf. So, that will be equal to 500.5 kilojoule per kg ok. So, state 1 and the properties are fixed now. State 2 P equal to 2.5 bar. Now, again for this what is V2? We can calculate V2 as V2 that is total volume is V1 plus A into delta H which is equal to 0.01 plus 0.05 meter square area into 37.5 into 10 power minus 2 because piston goes up by 37.5 centimeters. So, that is the piston that in centimeters 37.5. So, convergent meters I put this. So, V2 will be equal to 0.02875 meter cube. So, I can find the specific volume as V2 by m which is equal to 0.0575 meter cube. Now, from saturation table for 2.5 bar I can find Vf equal to 0.001067 meter cube per kg and Vg equal to 0.746 meter cube per kg. So, now, we can again find that V2 Vf is less than V2 is less than Vg. So, the state 2 is also a saturation mixture of liquid and vapor ok. So, I can again find the quality. X2 will be equal to V2 minus Vf divided by Vg minus Vf. Vg Vf etcetera are from 2.5 bar which we have written earlier. So, from this I can find the X2 equal to 0.07575 and now Uf equal to 532.8 at 2.5 bar. Similarly, Ug equal to 2537 kilojoule per kg at 2.5 bar. So, that means, U2 can be calculated as Uf plus X2 into Ug minus Uf which is equal to 684 for 0.71 kilojoule per kg. So, now, we can apply the first law and define the values that is for work and heat interaction. So, what is W equal to integral p dv ok. For that I need for doing this I need p versus v relationship ok. So, now, for doing that let us write pressure at any point is equal to p atmosphere plus 13600 into 9.81 that is rho hg into g into delta h which is equal to 37.5 plus 2h into 10 power minus 2 ok. Because from the initial point when piston goes up by a height of h then the difference between the meniscus in the right and left limbs will become 2h. So, the total height difference delta h will be 37.5 by 2h centimeters or 37.5 plus 2h into 10 power minus 2 meters. So, that will be the pressure at every point. Now, I will connect volume, volume at any instant after the initial state v minus 0.01 this is v1. So, that will get written as 0.05 this area of the piston into delta h into 10 power minus 2. So, that means, I can say h equal to or delta h equal to so, I will say h itself h is better. So, h equal to what 2000 into v minus 0.01 ok. So, now, we can see that I am writing h in terms of v. So, then I can write pressure in terms of v directly. So, pressure is written as substituting the value 100 kilo Bascals. So, 10 power 5 minus 3335.4 into sorry plus 5336640 into v substituting this I get this value. So, that is the relationship between pressure or I can say which is equal to 96664.6 plus 5336446640 into v. So, that is the relationship between the pressure and this. So, now, I can integrate this because pressure and volume relationship is going. So, I can calculate w1 to 2 as integrating this. Instead of pressure you put this volume based expression here and get w1 to 2 as 3751 joules or 3.751 kilo joules. Similarly, q1 to 2 can be calculated as w1 to 2 plus m into u2 minus u1 delta kepe equal to delta ke equal to 0 ok. So, now, this will be equal to 3.751 into mass is 0.5 into u2 is 684.71 minus u1 is 500 500.5. So, that will be equal to 95.856 kilo joules ok. Here you can see the pressure increases linearly with volume and state 1 say to both are saturated states. So, this is the problem number 4.