 7. Air is contained in a piston cylinder assembly initially at a pressure of 5 bar and 300 Kelvin. A paddle wheel like in the previous problem a stirrer or a fan within the cylinder is rotated using an electrical motor till a work of 75 kilo Newton per kg specific work is done on the air walls of the piston cylinder assembly are cooled using cold water such that the temperature of the air does not change. However, the piston is allowed to expand so that the volume occupied by the air initially is doubled ok to determine the amount of heat rejected to the cold water. So, you can see this is the situation I can say there is a stirrer like in the previous problem. So, some work is given to the stirrer and here you can see there is a piston which is free to move. So, that it moves to another location the volume is doubled basically due to this stirrer this work is transferred due to stirrer and there is a heat rejection to the cold water Q. So, that T is constant in the for the air ok. So, this is the so R. So, we can say that P v equal to m R T where R of the air is 287 kilo Newton given that. So, this is the problem ok. So, to solve this given data P 1 equal to 5 R T 1 equal to 300 Kelvin which is given and isothermal process. So, T 1 equal to T 2 and specific work of the paddle paddle or fan is given as minus 75 kilo joules per kg why minus because work is done from the surroundings to the system. So, air is considered as a system for the air as a system work coming is negative. Similarly, it is known that v 2 equal to 2 V 1. So, these are the given data and we know that P v equal to 287 T ok. So, I will find a first specific volume at state 1 V 1 equal to 287 into 300 divided by P 1 5 into 10 power ok. So, that will be equal to 0.1722 meter cube per kg that is the specific volume 1. So, that means, v 2 will be equal to 2 times V 1 because it is a mass is constant because of the system mass is constant. So, we can say this is 3 3 4 meter cube per kg that is it. So, these are the two things. Now, delta u equal to C v T 2 minus T 1 equal to 0 since T 2 equal to T 1 isothermal process. So, we can say q 1 to 2 equal to equal to w 1 to 2 plus delta u 1 to 2. So, this is 0. So, we can say q 1 to 2 equal to w 1 to 2 where w 1 to 2 is written as w paddle plus there is a expansion here correct. This expansion occurs at constant temperature. So, that means, the expansion should obey P v equal to constant. So, when you write this it will be P 1 v 1 into natural logarithm of v 2 by v 1. So, this will be the first law. Do you understand? So, P 1 v 1 is known, v 2 v v 1 is also known. So, from this I can find this q implies q 1 to 2 should be equal to minus 1 5 3 2 0 joules. Heat is rejected. So, you can see negative sign here. Work is also coming in from surrounding. So, work is also negative. This also is negative ok. But please understand the paddle work is negative, but this expansion work is positive because volume doubles. So, there is a positive delta v that causes. So, the total work basically is the combination of the work which is coming in due to the rotation of the paddle and the one more is the expansion work which is done by the air. One is done on the air, one is done by the air. So, this is the problem number 7. Now, go to problem number 8. Here the cylinder is divided into two compartments. A and B are shown in the figure and there is a frictionless thermally insulated piston. Please understand this is frictionless thermally insulated. So, heat cannot cross these two A to B or B to A and there will be no heat transfer. But this is free to move because it is frictionless and it is free to move. Now, the cylinder is well insulated except at the right end where there can be a heat transfer. So, compartment A if you see it is totally insulated because the sides, the walls surrounding the compartment A is insulated further the piston is also insulated. So, compartment A will be fully insulated. Compartment B will have piston, the three sides insulated one side there can be heat transfer compartment B. So, initially compartment A contains 0.1 kg of air at 100 kilopascals and 40 degrees. Compartment B contains 1 kg of air at the same temperature and pressure initially. So, I will say P A 1, pressure in the compartment 1 is a state initial state will be equal to 100 kilopascals which is also equal to P B 1. Similarly, I can say T A 1 equal to 40 degrees centigrade that is 313 Kelvin equal to T B 1 this is also known to me. Then mass of air in the compartment A is 0.1 kg, mass of the compartment B is 1 kg, ok. That is it. So, these are the given data for proceeding further we should understand what is happening in the problem. Heat is now slowly added after the initial condition is achieved, heat is slowly added until the. So, we can see obviously, it is added to the B from the phase where it is not insulated and the final pressure is 300 kilopascals. So, P final see please understand that if pressure of B reaches 300 automatically the pressure of A also should reach 300 because for the piston to stay at a particular location if there is a pressure difference between A and B the piston should have moved to should move to left or right correct based upon the pressure difference. So, the final pressure is common for both chambers that is 300 kilopascals. So, we can say P A 2 and P B 2 are same and the pressure is equal to 300 kilopascals. Now, P V equal to 288.68 T and C V equal to 133 these are the given information about the equation of state and the C V value. So, now how to solve this problem? First we will find the volumes V A 1, V A 1 equal to P A 1 is known. So, mass in A is 0.1 into 288.68 the R value into temperature is 313 correct. So, that divided by 100 kilopascals. So, 100 into 10 power 3. So, that will be the volume which is 0.935 meter cube. Similarly, V B 1 can be found V B 1 equal to 1 kg into 288.68 into same temperature divided by the pressure also is same 100 into 10 power 3. So, which will equal to 0.9035 meter cube. So, now one thing we should understand that the piston now moves to the left because of the heat addition the piston moves to the left and the pressure temperature increases and B pressure also increases. So, the air in A undergoes a compression which is adiabatic, adiabatic compression. So, we will try to do the first law for A. For A we can write del Q equal to 0 because all the sides are adiabatic here piston also is adiabatic 3 walls are adiabatic. And we can say minus del W equal to du which is equal to m Cv dt. Similarly, del W can be written as p dv. So, I can say minus p dv equal to m Cv dt. So, now pv equal to 288.68 times t this is given. So, that means I can say p equal to mass into 288.68 into t divided by the volume v because I want to use this for the integration ok. So, I can say minus m into 288.68 into t by v. So, now dv by v I will write dv by v this will be equal to m into Cv into dt ok. From this I have written this. So, now I can further write this as dv by v I keep dv by v here and take all other terms there mass cancels ok. Now, I know the value of this Cv also. So, I can write this as 7 what is Cv 7 1 7 ok. So, go back and see this Cv 7 33 ok. So, 7 33 the right hand side divided by this I will bring 288.68. So, I will put a minus n also here minus because this minus n goes there into I will say dt by t ok. So, separating the variables I separate the variables. Now, I integrate integrating what I will get is natural logarithm of v 2 by v 1 will be equal to minus 2.5391 into natural logarithm of t 2 by t 1 or I can say v 2 by v 1 will be equal to t 2 by t 1 whole power minus 2.5391 or t 1 by t 2 power positive 2.5391. So, we know that using the equation of state v 2 by v 1 can be written in terms of p 2 by p 1 ok. So, that means, v 2 by v 1 equal to p 1 t 2 divided by p 2 t 1 which implies I can say p 1 by p 2 will be equal to t 1 by t 2 power 3.5391. So, this is the condition for the for a. So, now from this I can find see what is given is final pressure is given initial pressure is known. So, p 2 for a p a 2 by p a 1 is 300 by 100 that is known initial temperature is known for a and b. So, a is 313 from then I can find the final temperature in a correct. So, which implies t 2 will be equal to t a 2 which is equal to 427 Kelvin. So, by applying this first law and writing this for a adiabatic process I can say that the final temperature of the a is this 427 Kelvin. So, that is done. So, from this I can find the final volume in a what is final volume in a we know the final pressure. So, this is again mass of the vessel a 0.1 this 288.68 r value into final temperature is now known that is 427 which we have calculated divided by the final pressure that is 300 kilo Pascals. The substitute does I get 0.041088 meter cube. So, you can see that initial volume of a is 0.09 it has reduced to 0.04 because of the compression. Now, what is the volume? So, this this is about the a vessel a I have found the states that is whatever be the initial final condition I have found. Now, to proceed I need to find the state of the a also. So, we know that go back to the figure piston is known at a particular position. The volume occupied by a are in a volume occupied by the air in b if you add them it should be same wherever the piston moves correct. So, initially there is a volume occupied by a which is calculated here 0.09035 volume of b is 0.9035. Now, it is compressed to 0.04 something. So, this would increase, but the total should remain the same. So, we note that initial volume in a plus initial volume in b should be equal to the final volume in a plus final volume in b. This you understand because of the where are the piston moves ok one volume should increase one volume should decrease. So, using that I can find now I know v a 1 v a 2 v b 1. So, I can find v b 2 equal to v a 1 plus v b 1 minus v a 2. So, that will be the final volume that is 0. You can see it is increasing 95262 meter cube. So, once I know the final volume I will find get the final temperature in b. The final pressure is known 300 kilopascals. So, that will be equal to 300 into 10 power 3 pressure into the final volume 952762 divided by 1 kg in the b that is 1 kg of mass into 288.68. So, that will give you the value of 990.1 Kelvin. So, this is the temperature. So, finding the state you have to understand that there are two important things here adiabatic compression occurs in a adiabatic compression occurs. So, in order to calculate that I have to do the first law and integrate or I have to find relationship between pressure ratio and the temperature ratio when there is a adiabatic compression ok. Now, then I find the temperature final temperature of the A using that volume is found. Now, you use the balance here that is the initial volume and final volume of two chambers should be same ok wherever be the piston. So, using that concept we have found the final state for the b also pressure only was known. So, one more property should be known that is vb after that equation of state is used to find this. So, now I can find I can now take a and b as system ok. So, a plus b as system because now I know what is the exact q w will be equal to 0 because for a and b the total volume should be the same. So, the what is no volume change ok. For a and b separately there may be volume change, but together if you take the volume change will not be there. So, w will be 0. So, I can say q equal to delta u. So, now what is delta u delta u for a plus delta u for b? Now, we know the initial and final temperature in A initial temperature is 313 in both final temperature in A is 427 final temperature in B is 990. So, I know do this. So, this is m 0.1 into 733 that is the cv value given correct cv into a for a it is 427 minus 313 plus 1 kg in the B same cv because same material into here 990 0.1 minus 313. So, this will give q value as 504670 joules this is the answer. So, fixing the state and applying the first law to find this. So, that is the 8th problem. So, 9th problem now solution for this first the problem is again there is a two compartment here one compartment as 1 kg of air at 5 bar and 350 and second compartment as carbon dioxide 3 kg of it initially at 2 bar and 450 confined at opposite sides of a rigid well insulated container the it contains well insulated. There is a partition this is the partition which is initially held by a pin this is a pin which is a holding the partition here this is thermally conducting in the previous case was insulated no this is thermally conducting. Now, the pin is removed the one because it is held by the pin it is static here, but when the pin is removed it is free to move thermally conducting as well as it is free to move. So, when when the pin is removed then the piston will move so that you see that the gases will be attaining an equilibrium state final state will be equilibrium state. Please understand that since there is no heat transfer from the air to the surroundings or CO to the surroundings only the heat transfer will occur between air and CO2 that means, you will see that that initial temperatures can be different by the final temperature should be the same. So, equilibrium condition mechanical equilibrium the piston should go to such a location where PF is common for air and CO2. Similarly, since it is thermally conducting thermal equilibrium also will be attained so that there is a final temperature which is common between air and CO2. So, that is the thing. So, how to solve this problem? So, here air and CO2 are assumed to be pure substances air obeys PV equal to 288.68 T CV is 73 for CO2 PV will be equal to 189 T and CV is 750 that is it. So, when you handle so there is this I call A this I call B. So, when I handle air I have to use corresponding equation of state this PV equal to 288.68 and the CV is 733 for CO2 I have to use the equation of state as PV equal to 189 and CV equal to 750 that is it. So, the initial pressure I will write in the same rotations P A1 equal to 5 bar T A1 equal to 350 Kelvin and mass of A equal to 1 kg. Similarly, P B1 equal to 2 bar that is 2 into 10 to the power 5 pascals and T B1 equal to 450 Kelvin mass of B will be equal to sorry 3 kgs it is given. Now, same thing will obey here. So, you can find volumes correct V A1 from the equation of state for the air that is you can find this as point volume 1 1 into 1 kg into 288.68 into 350 divided by 5 bar 5 into 10 to the power 5 equal to 0.20207 meter cube. Similarly, V B1 can be found as 3 kg into 189 into 450 divided by 2 into 10 to the power 5 that will be equal to 1.27575 meter cube. So, we also like in the previous problem I can say V A1 plus V B1 will be equal to V A2 plus V B2. So, this equation is important for us. So, now taking A plus B as the system okay now thoroughly insulated correct. So, if you go back you can see that it is totally insulated. So, totally insulated then volume change for this rigid vessel correct rigid vessel also. So, no change in volume for the combined system of A plus B plus no heat transfer also. So, it is simple for me Q minus W equal to delta U. So, this is also 0 work also is 0. So, delta U equal to 0 that is what we are getting. So, what is delta U? Delta U will be equal to delta U equal to delta U for A plus delta U for B which is equal to for A mass in A is 1 kg its CV is 733 okay. So, I can say 1 into 733 into that is the final common temperature we told you it will be equilibrium temperature that minus initial temperature of A is 350 substitute that 350 plus for B it is 3 into 750 CV is 750 750 into Tf minus 450. This will be equal to 0 from which I can find Tf as 425.4 Kelvin. I can find this okay this is first one. Second one is see what is asked here is to find the final temperature and final pressure determine the final temperature and final pressure. Final temperature is got by energy balance applying first law how to find final pressure we have to apply this equation. So, we know that Va1 plus Vb1 will be equal to 1.47826 meter cube that is we have calculated these two in the initial condition that should be equal to Va2 plus Vb2 okay. Now, how to calculate Va2? I know now the final temperature, but pressure is not known okay. So, we have say pV equal to mRT. So, I can say V equal to mRT by p that I will apply here. So, what I will apply here for A2 is A2 as a mass of 1 into R is 288.68 into T is T final okay. I can also substitute this as 425.4 divided by I do not know the pressure. So, I just say pF similarly for Vb2 it is what is the mass 3 kg and R is 189 correct. So, the substitute is 3 into 189 into same temperature same pressure. So, I know the left hand side this should be equal to this only unknown is pF. So, this implies pF will be equal to 246312 pascals or 2446.3 kilo pascals. So, this is the way. So, there are two things here one is the energy conservation taking the total rigid vessel as the system. There is no so rigid vessel so no volume change well insulated that means no Q. So, Q and W becoming 0 delta U equal to 0 that energy balance will give you the final temperature and that this conservation of volume initial volume total initial volume should be equal to total final volume in the both chambers that will give you the final pressure okay.