 In our textbook cyclic process is mentioned as a unique type of process but in reality cyclic process is a mathematical outcome of or you can say cyclic process is defined mathematically rather than you know physically what is going on okay so basically in a cyclic process please write down cyclic process so in a cyclic process state of this system repeats after every few intervals of time okay so basically whatever was a state earlier the same state is achieved again and again okay so what happens is a cyclic process in PV diagram looks like a closed loop okay so please write down cyclic process can be represented by a loop in a graph okay so a loop can be like this this is a loop it goes like this like that and like this now it doesn't happen only once this thing keeps on happening again and again and again same thing repeats alright so if you have to study the cyclic process you don't need to study all the cycles it keeps on happening till infinity so whatever happens to one cycle of repetition same thing will happen for all the cycle of repetitions okay so I just need to analyze what is happening in one of the cycles and that is true for complete this thing okay so basically machines follow cyclic processes so cyclic process please note down another point is nothing but is a collection of multiple processes to generate a closed loop okay simple now can you tell me in a cyclic process if you start from here and end also there what would be the change in internal energy for entire cyclic process zero zero because delta T is zero you started from same temperature and ended at the same temperature delta U for entire cyclic process is zero getting it now I'm going to ask you a simple question tell me that first law of thermodynamics which states delta Q is equal to delta U plus W that first law of thermodynamics is it valid for one type of process at a time or can I apply first law of thermodynamics for multiple processes together okay what I mean to say is this for example this is the plot let us say this is P and V and suppose the gas undergoes isochoric process and then isobaric process like this okay so point one then two then three okay now tell me the first law of thermodynamics delta Q is equal to delta U plus W by the way first law of thermodynamics you apply between the two points apply between the two points okay whereas PV is equal to NRT is applied at a single point single point all the variables here you are applying between the two points delta Q is equal to delta U plus W now tell me can I apply delta Q is equal to delta U plus W between one and two and then two and three or can I apply for one to three directly just tell me can I apply first law of thermodynamics between one and three directly or not just take the poll don't say anything can I apply first law of thermodynamics between one and three directly between one and three directly or do I need to break it like one to two then two to three you don't need to be careful in taking the vote whatever you may feel take it it is an anonymous poll I will not get to know who has answered what okay so this is what you can see okay the answer is that first law of thermodynamics is one of the fundamental theorem in the universe okay so it does not matter between which two point you are applying it is independent of that so when we have studied the first law of thermodynamics I never told you that you can apply only between this and that this should be happening there is no constraint as such that you can only apply when this happens okay so you can apply first law of thermodynamics between any two points okay please write down delta Q is equal to delta U plus W can be applied between any two points okay but you have to take delta Q between those two point delta U between those two point and work done will be between those two points only whatever you are taking but what happens is when you apply delta Q is equal to delta U plus W for different different kinds of processes separately then not only you can apply this equation but also you can use the process equation for example if you apply this equation between one and two which is isochoric apart from first law of thermodynamics you can use P by T is constant between any two points okay similarly when you apply first law of thermodynamics between two and three you can use volume by temperature is constant okay by the way temperature should be Kelvin just letting you know not sure whether you should be careful okay so you can use between any two point based on what you are dealing and what are you trying to solve do not do not feel any constraint in utilizing the first law of thermodynamics okay and suppose I am using first law of thermodynamics for entire cycle for full cycle suppose the cyclic process I am using delta Q is equal to delta U plus W for the complete cycle or cyclic process delta U is zero so basically delta Q will be equal to W only okay now delta Q is what anyone what is delta Q in a cyclic process quick skanda what is delta Q in a cyclic process doesn't matter does it matter in a cyclic or not what is the area inside the loop in a cyclic process I don't use things which I haven't yet discussed whatever I have discussed based on that tell me what is delta Q delta Q is heat absorbed yes or no now in a cyclic process you have to be careful delta Q is net heat absorbed it is not just heat absorbed so it is heat absorbed minus heat released write it down at times we only consider heat that is absorbed we ignore heat that has rejected okay so make sure this and in cyclic process there will always be a portion in the cycle which absorbs the heat and always always there will be a portion in the cyclic process which releases the heat heat release can never be zero if there is work done in a cycle that comes from the second law of thermodynamics that we are going to study later on not now any doubts till now anything that you want to discuss or ask no doubts okay so I will take numericals solve this numerical anybody got the answer you can type in in the chat box don't speak you get the answer by the way the value of gamma is 5r by 2 sorry not 5r by 2 gamma is 7 by 5 okay the value of gamma is given to you directly so gamma is 1.4 gamma is 4 only for diatomic gas what are you saying sir gamma has a different value for monoatomic and diatomic gas yes yes that we are going to learn in kinetic theory of gas right now I am directly giving you the value of gamma now work done will be positive or negative first tell me that work done will be positive positive positive okay which process it is adiabatic isothermal isochoric what it is adiabatic it is adiabatic because the boundary of the system do not conduct any heat with the surrounding to the surrounding so that is the reason why it is the gas inside this undergoes the adiabatic process and I know that work done is equal to nr delta t divided by 1 minus gamma we have derived it right so I am directly writing it down so number of moles is 2 2 into 8.314 delta t is 400 minus 300 which is 100 divided by 1 minus 7 by 5 so when you substitute everything you will get negative of 4157 joules anybody got this answer yeah 500r correct 500r that is correct negative of the 500r see gamma is always greater than 1 so denominator will be negative okay so what happens in the adiabatic process adiabatic process internal energy goes down to do the work because from outside there is no heat supplied so when internal energy goes down temperature also goes down because internal energy is directly proportional to the temperature so that is the reason why when adiabatic process there delta t is negative work done is positive okay you can think like that or if you want to mathematically see it delta q is equal to delta u plus w see whenever you see a numerical on the thermodynamics you must write the first law of thermodynamics expression it will drastically reduce the silliness delta q is 0 so w should be equal to negative of delta u so delta u proportional to delta t so w is negative of delta t