 Good morning. Welcome to tutorial 3 which is on the first law of thermodynamics. So, first law of thermodynamics for systems. The first problem, a system undergoes a process from state 1 to state 2, where its pressure linearly changes from 1000 kilobascals to 200 kilobascals and volume increases from 0.05 meter cube to 0.2 meter cube. So, during this process, the heat input of 65 kilojoules is given to the system. Then the system returns adiabatically from state 2 to state 1. Determine the total work transfer. So, this is the problem solution. P1 is given as 1000 kilobascals, P2 is given as 200 kilobascals, V1 is given as 0.05 meter cube and V2 is 0.2 meter cube. And it is given that pressure linearly increases. So, we can write pressure as a plus b into volume and substituting for these two pressures 1000 equal to a plus b into 0.05. This is one equation. Similarly, 200 equal to a plus b into 0.2. So, from this I can get value of a as 1266.67 kilobascals and b equal to minus 5333.33 kilobascals per meter cube. So, this is the value. So, now we can integrate for the process 1 to 2, where the pressure increases linearly. The corresponding work input work output can be written as integral 1 to 2 p dv which is equal to integral 1 to 2 sorry 1266.67 plus sorry minus 5333.33 into v dv. So, when you integrate this we can get the work as 90 kilojoules for the process 1 to 2. Now, first law will apply q 1 to 2 minus w 1 to 2 equal to delta u equal to u 2 minus u 1. So, here we can find delta u 1 to 2 equal to q 1 to 2 is given as 65 kilojoules. So, 65 minus this work is 90. So, equal to minus 25 kilojoules delta 1 u to 2. Then for the process 2 to 1 q 2 to 1 equal to 0 since adiabatic. So, process adiabatic. So, we can say minus w 2 to 1 equal to delta u 2 to 1 equal to u 2 minus u 1 ok. But we know that delta u 2 to 1 equal to minus delta u 1 to 2 which is equal to minus of minus 25 kilojoules. So, delta u 2 to 1 equal to 25 kilojoules that means w 2 to 1 will be equal to minus delta u 2 to 1 equal to minus 25 kilojoules. So, total work total work we can find total work easily by analyzing the heat input. See for the cycle for the cycle you know that cyclic integral of del q will be equal to cyclic integral of del w. So, here the first process the heat is 65 kilojoules which is positive heat input to the system. Second it is 0. So, the net q will be equal to 65 kilojoules. So, we know that net w will also be equal to 65 kilojoules ok. Now, we can calculate by what we have already calculated 90 minus 25 which is equal to 65 kilojoules. So, it is verifying that the first law of the thermodynamics for a cycle for a system undergoing a cyclic process is cyclic integral of del q equal to second law del w. So, that straight away gives the answer, but anyway the illustrative procedure gives the same answer. It is a verification for the first law for the cycle. The second problem a certain quantity of a pure substance undergoes a cyclic process A B B C C D and D A as shown in the figure here in the PV diagram. Now, A B and C D this A B and C D these are isothermal processes at this is isothermal process T equal to 600 Kelvin. This is an isothermal process where T equal to 300 Kelvin ok. Now, these are the two processes then B C is a constant volume process as it indicates B C is the constant volume process and D A is an adiabatic process. So, D A is an adiabatic process ok. So, q equal to 0 here. And ethos also told that for the process D A it obeys PV power 1.4 equal to constant. So, that is also given. The net work from the cycle is 50 kilo joules. Now, we need to calculate the work and heat interaction for each leg say for example, A B what is the work and heat interaction etcetera for each leg of the cycle. The pure substance obeys the equation of state given by PV equal to 289 T where P is in Pascal's B is in meter cube and T is in Kelvin. And the internal energy depends only on the temperature like an ideal gas. So, obviously the pure substance is an ideal gas. So, say this is say T A equal to T B because at this when this is constant temperature process temperature that A and B will be same. So, T A and T B will be 600. Similarly, T D and T C will be 300. Now, we also have for the process D A D to A we can write P D V D power 1.4 equal to P A V A power 1.4 that is the process which is given. And equation of state is also given. Okay now let us do the calculations. First is we have to find the fixed states. So, for that we can say V D by V A will be equal to T A by T D power 1 by 0.4. Now, T A and T D values are known to me which is equal to 600 divided by 300 power 1 by 0.4 which implies we know V D by V A equal to 5.66. Next is we will try to find the work interaction we need to do for all the process. So, for A A to B, A to B is an isothermal process. So, we can say that the work interaction for A to B will be equal to P A V A natural logarithm of V B by V A. Why? Because P V equal to constant for an isothermal process. Since this is an ideal gas, so you can see that P V equal to 289 T. If T is constant, then P V also will be constant. So, that is the reason we are writing P V equal to constant when we integrate this. So, we can say that P equal to constant by V. So, P A V A equal to P B V B. So, using that I can write the constant itself as P A V A into natural logarithm of V B by V A. So, that is the expression I am getting here. So, we know we do not know the values here. So, let us keep this equation. Now, similarly for B to C delta V equal to 0. So, W B to C will be equal to 0 because the constant volume, so there is no change in the volume. So, integral P D V will be equal to 0. Now, W C to D. C to D again is an isothermal process. So, you can say same thing you can write P C V C natural logarithm of V D by V C. So, that is the expression. Now, here I can further write this in terms of temperature as 289 289 T A because P A V A equal to 289 T A. So, that you can write into natural logarithm of V B by V A. Similarly, here also I can write 289 T C P V P V in the state C is equal to 289 T C natural logarithm of V D by V C. That is it. Now, finally, we have W D A. W D A will be equal to adiabatic process. So, I can say P D V D minus P A V A divided by 1.4 minus 1 because it obeys P V power 1.4 equal to constant. So, that when you integrate for that work, you will get this expression. So, in terms of temperature, you can write this as 289 T D minus T A divided by 0.4. So, now, you know the answer for this because it is equal to 216.75 kilo joules. This is because T D value you know T D and C are 300 kilo Newton and T A is 600. So, we know this answer. This is negative adiabatic compression. So, the work involved in the adiabatic compression is negative. So, these are the values you get, but you have to fix the systems. So, let us do that now. So, what we will do next is it is given in the problem that the work network from the cycle, network from the cycle equal to 50 kilo joules. There is a positive work is delivered by the cycle. So, that we can say as W AB plus W BC plus W CD plus W DA equal to 50. Now, we know that for the BC, the constant volume process W BC is 0. So, this is 0. Now, we know the value of W DA. W DA is minus 216.75. So, this is minus 216.75. So, we can say that we can say this is basically in kilo joules in kilo joules. So, everything will write in kilo joules. Now, this we can write as 0.289. I am just dividing by 1000 into 600 into natural logarithm of V B divided by VA plus C. W CD is again 0.289 into 300 into natural logarithm of V D by VA, V C. We are written now V D by V C here, V D by V C. Now, we have to. So, this will be equal to this you take it as a side. So, 50 plus 216.75 this is in kilo joules. So, now we will see that what is V C? What is V C? V C is equal to V B that you can substitute. So, this V D by V C equal to V D by V B because V C equal to V B. Now, this can further be written as V D by VA into VA by V B. Do you understand? So, this is what we can write. So, now I can write this totally now WDA is known. So, here this expression now becomes 0.289 into 300 into logarithm of V B by VA square into V B by VA into VA by V B. This is equal to 266 it is adding these 2.75. So, now because you can see that in this case 300 is taken here into 600 I have to put 300. So, the square has come come to the V B by VA. So, the V B by VA square appears because I have now taken only 300 outside this bracket. So, this square term appears and this is same whatever I have done this is actually V D that this is V D by VA into VA by V B. So, that is the expression I have here. Do you understand? So, this square term appears because out of 600 I have just taken 300. So, I can write 300 into 2 long the 2 long either this 2 are brought up to the square of this. So, V B by VA square into here V D by VC is written as V D by V B that is again written as V D by VA into VA by V B. So, that term I have written like this ok. So, now this term alone can be now evaluated because these are all known values. So, I can write natural logarithm of V D by VA into now one of this gets cancelled. So, that into this I will have V V by VA that is it correct. So, that will be equal to 266.75 divided by 0.289 into this I am taking out other side 300. So, which is equal to 3.077 that is it do you understand? So, now I can find V B by VA. So, now what is this V D by VA V D by VA V B by VA both we can easily understand. So, we get V B by VA equal to 3.83. So, from that I can find W AB equal to 232.92 kilojoules. Similarly, W C D equal to 33.88 kilojoules. So, all the work has been done. Now, heat transfer Q AB, Q AB equal to W AB plus delta U AB, but this is 0 because delta T AB equal to 0 isothermal process. So, this is known Q equal to 232.92 kilojoules. Now, Q C D equal to W C D because again this is an isothermal process. So, this is equal to sorry here this will be equal to W C D is 33.88 kilojoules. Now, Q DA equal to 0 because adiabatic process. Now finally, Q BC can be calculated as minus of Q AB plus Q C D plus Q DA plus W net because sigma Q will be equal to sigma W the first law for the cycle supplied here. So, we will get this equal to 200 minus 216.75 kilojoules. So, this is the problem. So, you can see that it is a cycle constituted by four processes, two of them are isothermal AB and C D are isothermal. One is a constant volume process, one is an adiabatic process obeying PV per 1.4 equal to constant. So, you have to fix the state, understand that the gas obeys PV equal to 2893 T and the network is given as 50 kilojoules in the cycle. So, network is worked from each process added together. Similarly, finally, we apply the first law for the cycle. Sigma Q equal to sigma W and we get the answer, ok. So, this is the problem number 2. Now problem number 3, you have a piston cylinder device which is restrained by a linear spring. It contains 2 kg of air at initially at a pressure of 150 kilo Pascal and temperature of 27 degrees centigrade. It is now heated until its volume doubles and temperature increases to 527 degrees centigrade. Determine the work and heat interactions during the process. Air obeys PV equal to 287 T where P is in Pascal, T is in Kelvin, V is in meter cube per kg and the CV for the air is given as 717 joule per kg Kelvin. So, we will see the solution for this. Mass equal to 2 kgs. Similarly, P1 equal to 150 kilo Pascal, T1 equal to 27 degrees centigrade and T2 equal to 527 degrees centigrade and also given V2 equal to 2 times V1. So, since PV equal to 287 T, I can write this as P capital V equal to M into 287 into T. So, that means I can find the volume 1. So, V1 will be equal to 2 into 287 into 273 plus 27. Please understand that the temperature is in Kelvin in this equation. So, you have to add to the degree centigrade 273 divided by the pressure. Pressure is 150. So, 150 into 10 power 3. So, that will be the volume which is equal to 1.148 meter cube. So, this implies V2 will be equal to 2.296 meter cube because V2 is 2 times V1. So, it is a linear spring. It is compressing this. So, that means pressure linear spring causes pressure to vary linearly as it is compressed. So, that means I can say P equal to a plus Bv. Now, before then I can find P2 because state 2 is known. So, P2 will be equal to what? P2 will be equal to 2 into 287 into T2. T2 will be 527 plus 273 divided by volume. Volume is 2.296. That will be equal to 200,000 pascals or 200 kilopascals. So, that will be the pressure. So, now, this pressure from 150 to 200 linearly varies. So, I can assume P as a function of volume given as P equal to a plus Bv. Now, I know V1, V2, P1, P2, I can find a and b, ok. Using P1, V1 and P2, V2, we can find a comma B as a equal to 100 kilopascals and a B equal to 43.554 kilopascals per meter cube. So, this is one way to do this. So, now, when I do this, then work interaction is integral P dv which is equal to 100 into 2.296 minus 1.148 plus 43.554 into 2.296 square by 2 minus 1.148 square by 2. So, that will be equal to 200.9 kilojoules. Now, delta, say T1, T2 is known. So, delta U will be equal to m Cv delta T which is equal to 2 into 717 into 527 minus 27 which is equal to 717 joules or 717 kilojoules. Now, applying first law Q equal to delta U plus W which is equal to 717 plus 200.9 which is equal to 917.9 kilojoules. So, this is the solution for this. So, we can also do because we can see that now atmospheric pressure is 100 P atm equal to 100 kilopascals. So, those infusions if they are given mass of the piston if it is m P, if the information is given then we can find these spring constant and do this problem also ok. So, in this case since pressure the states are fixed by temperature itself straight away and pressure causes basically the volume to double, the pressure increase causes volume to double in this case. So, we can see that this is a simple approach.