 Then, the fourth problem, 5 kg of nitrogen is kept in a piston cylinder at 800 kilopascals and 127 degrees centigrade. We will draw this also, so there is a piston. Now, at this state, the piston rests on the stop. So, there is a stop here. The piston is resting on the stop. Now, heat is transferred slowly to the nitrogen until its temperature reaches 527 degrees centigrade. A pressure of 1500 kilopascals is required to lift the piston. So, the mass of the piston is such that a pressure of 1500 kilopascals is required to lift the piston. So, atmospheric pressure is 100 kilopascals they can they can assume it may not be required. So, determine the work and heat interactions considering that nitrogen is an ideal gas obeying P v equal to 297 T and C v for nitrogen is 743 joule per kg Kelvin. So, solution P 1 equal to 800 kilopascals. Okay, now T 1 equal to 127 degrees centigrade, T 2 equal to 527 degrees centigrade, mass of nitrogen is 5 kg. So, these are all the data given. Now, we can find V 1 the initial volume that is what using this. So, this will 5 into 297 into 127 plus 273 divided by 800 kilopascals. So, 800 into 10 to the power minus 3 say 10 to the power plus 3. So, that will be equal to 0.7425 meter cube that is V 1. So, now until pressure increases from 800 kilopascals to 1500 kilopascals the piston will not lift. Okay. So, during this volume remains a constant. Do you understand? So, let us say that it be that this let this state where P just reaches 1500 kilopascals as state 2. Okay. So, the constant volume. So, V 2 will be equal to V 1 equal to 0.7425 and P. So, this is actually final pressure I will say this is final temperature T f. So, this P 2 will be equal to 1500 kilopascals. So, that is the state at which the piston will just lift after. So, till now what is the heat added we will see. So, this is the condition constant volume process. So, that means work 1 to 2 equal to 0. Okay. So, what is the temperature now? T 2 we can find now that using the state of equation of state that will be 1500 into 10 power 3 into 0.7425 meter cube divided by 5 into 297 that will be equal to 750 Kelvin T 2. Now, I can find delta U 1 to 2 equal to M C V delta T which is equal to 5 into 743 into 750 minus 127 plus 273. So, 750 in Kelvin. So, we have to keep the initial temperature also in Kelvin. So, this will be equal to delta 2 delta U will be got. So, now I want what is the Q for this Q 1 to 2 can be found as W 1 to 2 which is 0 plus delta U 1 to 2. So, delta U 1 to 2 will be equal to 5 into 743 into 750 minus. Okay. So, that is the value. Now, final temperature is is 527 degree centigrade. So, this means further heating is happening because now the temperature is only 750 Kelvin. So, this is what this is will be equal to 800 Kelvin. Okay. So, some 750 to 800 still heating is happening. So, but this time. So, we can say further heating has continued. So, now P remains constant at 1500 kilo Pascal because now it can be lifted. So, now it just get lifted and goes out. So, pressure will not change now volume will increase final temperature is known. So, we can say P 3 or the final pressure will be equal to P 2 equal to 1500 kilo Pascal that is known. Okay. So, now volume you have to find final volume. So, state 3 P 3 equal to 1500 kilo Pascal T 3 equal to 800 Kelvin. So, we can find V 3 V 3 equal to what 5 into 297 into 800 final temperature divided by 1500 into 10 power 3 which will be equal to 0.792 meter cube. Now, we will see that work done 2 to 3 will be equal to 1500 into the final volume minus V 2. V 2 is 0.7, 0.7425. So, that will be equal to 74.25 kilo joules. So, similarly delta U 2 to 3 can be found as 5 into 743 into 800 minus 750. Okay. Now, what is Q 2 to 3 will be equal to W 2 to 3 plus delta U 2 to 3 equal to 74.25 plus 5 into 743 into 800 minus 750. Okay. Now, this is Q 2. So, what is total Q equal to Q 1 to 2 plus Q 2 to 3 which is equal to 74.25 plus 5 into 743 into 800 minus 400. So, which is equal to 1560.25 kilo joules. So, you can understand that in this problem the initially pressure increases in order to float the piston. Once the pressure reaches that value of 1500 kilo Pascal where the piston gets floated then a constant pressure process occurs. Okay. Now, fifth problem here a piston cylinder device has 0.4 kg of air initially at 1500 kilo Pascal under temperature of 800 Kelvin. The cylinder has stopped such that its volume cannot increase below 0.04 meter cube. Now, air is gradually cooled to 300 Kelvin. To determine the final volume and pressure, work and heat interactions. Okay. Now, P 1 equal to 1500 kilo Pascal T 1 equal to 800 Kelvin. These are the two values. Now, what is M 0.4 kg? And again the equation of state P v equal to 287 T and a C v value is given. Okay. Now, what is the initial volume? Initial volume we can find now as we are using this. So, mass is 0.4 into 287 into T 1 is 800 Kelvin divided by pressure is 1500 into 10 power 3. So, that gives the initial volume as 0.061226 meter cube. That is the initial volume. Now, it says so we can draw the figure now. There is a stop here. The piston is initially here. So, the volume initially is 0.061226. Now, if you cool this, what happens? The piston descends down. The piston cannot descend down beyond the stop. So, finally, I can say the position of the piston will be as shown by the dashed line. It should be on the stop. In this situation, in this state, the volume will be 0.04. After that, the piston cannot descend down. Okay. So, initially when it descends down by cooling, it is a constant pressure process. Pressure remains constant as the volume changes from 0.061226 meter cube to 0.04. So, this in this state. So, this is state 2 and this is state 1. Okay. Now, state 1 volume is known. Volume of state 2 is 0.04 meter cube. That is also given because below this, the piston cannot descend down. So, when the volume decreases from 0.0612262.04, pressure remains constant. So, that means that I can say it is free to move a piston. Now, after that, what happens? The pressure will decrease, but volume will remain constant after that. So, let us see this first. So, as the air is cooled, its temperature decreases. Piston descends at constant pressure. Okay. Now, state 2 is fixed now. When the temperature state 2, when the piston is just sitting on its stops, then what is T2? T2 will be equal to 1500 into 10 power 3 into volume is 0.04 divided by 0.4 into 287. So, that will be equal to 522.65 Kelvin. So, you can see that the temperature has decreased from 800 to 522.65. Finally, the temperature has reached 300 Kelvin. So, further cooling happens. Further cooling happens under T final equal to T3 is given as 300 Kelvin. So, now, during this piston cannot descend down. So, V equal to constant. So, we can say V3 equal to V2 equal to 0.04 meter cube. From this, I can find the pressure. What is this? P3 equal to 0.4 into 287 into final temperature is 300 here divided by the volume 0.04. That will be 861,000 Pascals or 861 kilo Pascals. Now, you can see that the states are fixed. Now, as per the work interaction, it is there only when the volume is changing. That is in the first line. So, we can say W122 is P into delta V. Obviously, it is negative because volume is decreasing. So, that we can write as what? 1500 into 0.061226 minus 0.04. So, that value is the work. That will be minus 31.84 kilo Joules. In the second process, there is no work. So, W223 equal to 0. Similarly, I can directly calculate delta U from 3 to 1. Finally, that is the total change in this delta U. That is 0.4 into 717 into final temperature is 300. Initial temperature was 800. So, that is the total delta U. That will be equal to minus 143.4 kilo Joules. Now, what is Q? Q will be equal to delta U plus W. So, this is total Q. I can say total Q. What is this? Delta U is the total delta U. That is minus 143.4 plus the total work is minus 31.84 because for 2 to 3 it is 0. So, the total Q is minus 175.24 kilo Joules because heat is removed it should be negative. So, it is clear that in this problem first a constant volume process occurs where the volume decreases. Sorry, pressure process occurs where the volume decreases. Then volume remains constant and pressure further decreases. Temperature continuously decreases. Now, this sixth problem the rigid vessel of volume point, this is rigid vessel of volume 0.25 meter cube contains air initially at 100 kilopascals and 300 Kelvin. P1 equal to 100 kilopascals T1 equal to 300 Kelvin. A stirrer, a stirrer is driven by descending a mass. There is a mass connected to a pulley to a shaft and there is a stirrer. So, this mass is descended down. Slowly it is descended down. The mass is 20 kgs by descending a mass. I shown in the figure. So, when you pull it down slowly then this will turn the pulley and the shaft so that the stirrer also rotates. So, that is the work which is done from the surrounding or by the mass on the system taken as the air in the system. So, air is present in the system. Now, starting from the rest the mass descends through a distance of 15 meters in a time interval of 5 seconds. That means, what that is a it is not a constant speed descent. It is a so it is rest it is slowly traveling like the there is an acceleration here. There is an acceleration. How can how can it be calculated because it is given that 15 meters is covered in 5 seconds. So, initial velocity is 0. There is a final velocity so that there is an acceleration. And at this work of doing this descending of the mass, somebody has to there is a agent has to do this. You will make some work to be transferred to the air to the stirrer. So, that is the work transferred to the system. Now, during this time interval there is a heat loss occurring which is having a magnitude of 100 joules. So, heat loss occurs 100 joules. Determine the final temperature and pressure of the air in the vessel. Initial temperature and pressure is given. Take the value of G as 9.81 meter per second square and the air obeys P v equal to 2870 equation of state and the C v is also given. This is the problem. Now, so pressure volume and temperature are given. So, you can find the first you can find the for example, you can first find the velocity. So, given that initial velocity equal to 0 meter per second and the distance traveled by the mass s we can say is 15 meters. Then the time it takes to descent to the descent is 5 seconds. Okay. Now, from this information I can find the mass of the air. Mass of the air in the vessel will be equal to 100 kPa into the volume divided by 287 into initial temperature 300 which is equal to 0.29036 kg mass is formed. So, from this information I can say the final velocity v2 equal to v1 plus a into t. This is one equation I have. Okay. Similarly, another equation is s equal to initial velocity into time plus half acceleration into t square. Okay. Now, I can say this 15 equal to initial velocity is 0. So, 0 plus half a into phi square which implies the acceleration will be equal to 1.2 meter per second square. So, that means see I can pull the mass down at constant velocity also, but in this case it is not. So, it is having initial velocity 0, final velocity can calculate final velocity is what from the acceleration 0 plus 1.2 into 5 that is basically 6 meter per second. So, that is a kinetic energy change to understand for the mass. This is this is all for the mass mass. Correct. Here the mass of the block is equal to 20 kgs. So, we can see that there is a kinetic energy change, potential energy change, etcetera for the mass. Okay. Now, we have to understand this. What is the energy transferred to the we have we want to calculate what is the energy transferred to the stirrer. The energy transferred to the stirrer is the net energy change for this mass. Correct. So, what we have to do is for the block for the block of mass we can write delta E equal to delta U plus delta kinetic energy plus delta potential energy equal to Q minus W. Okay. Now, temperature of the block is constant. So, delta U will be equal to 0. Similarly, Q is equal to 0 for the block. There is no heat addition or heat rejection from this, because the block does not have any heat interaction with the surroundings. So, I can say Q is also 0. So, this goes this also goes. Okay. Now, from this I can find the W W. So, I can say W is a B output because this is W for the block. Okay. Now, what is W for the block will be equal to minus delta kinetic energy minus delta potential energy. Okay. So, that will be equal to minus mass of the block plus a point of M B V 2 square by 2 minus V 1 square by 2. This is the delta K E minus delta P E will be equal to M B G into H. So, we can write this as 0.5 into 20 into 0.5. So, 0.5 I put. No. So, this is not this 2 will not come. So, this is just 0.5 M B V 2 square minus V 1 square. Okay. So, now, this 6 into 6 minus 0 minus minus this M B here this will be 20. Why I am putting minus here? Minus or minus I have to put. Okay. So, this is sorry this minus 20 into 9.81 into minus 15. Why this minus comes? This is because the potential energy decreases. So, this will be equal to 2583 joules for the work. So, this is the work the block is transferring. Okay. Now, W R. So, R plus W block should be equal to 0 because the immediate surrounding okay see for the shaft which is connected to the stirrer the immediate surrounding is a block. So, block is actually doing the work. So, we can say this which implies W for the air will be equal to minus 2583 joules. Do you understand? So, that is it. So, please understand that in this problem the kinetic energy is what is bringing in the energy basically and the potential energy decreases. Kinetic energy increases you can see that the final velocity is higher. Okay. So, now with this I estimate that total energy change for this and negative of that will be given to the block. That is the work involved in the block. This work is actually imported to the air. So, air receives a work of or energy in the form of work transfer equal to 2583 joules. Okay. Now, for the air we can say first law Q air minus W air equal to delta U air which is equal to M air C V into delta T, T2 minus T1. Okay. Now, what is Q air? Q air is given 100 kilo 100 joules of in the problem if you go back you can see that here 100 joules of heat is lost to the surrounding. So, Q is minus 100. Okay. Then we have to W air we know that value here. So, we can substitute this. So, Q air is minus 100 minus of minus 2583 you substitute for W equal to M air is 0.29036 which we have calculated into C V is given as 717 into, but I do not know T2. So, T1 is known. So, from this I can find the final temperature of the air as 311.9 Kelvin. So, that is the final temperature. So, we can see that the first law is used to find the state final state of the air because you know work transfer heat transfer etc from then I find the final state. Since volume is constant we can say P2 by P1 equal to T2 by T1 which implies P2 equal to 103.966 kilo Pascals. So, initial pressure was 100 kilo Pascals. So, you can see that the slight rise in the pressure slight rise in the temperature due to the small work which is transferred. So, this is the final pressure. Okay. This is the final pressure.