 please write down a little bit about reversible or irreversible processes see basically the first statement of the second law of thermodynamics told us that we cannot have 100% efficiency so now i need to find out or we need to find out if 100% efficiency is not possible then what is the maximum possible efficiency of the heat engine okay in order to understand what is the maximum possible heat efficiency we need to understand what is reversible and irreversible process okay so we have already learned about the quasi-static process quasi-static process is the process in which system is in equilibrium with the surrounding okay at every moment so if surrounding changes the system also accommodates with the surrounding slowly now the reversible and irreversible processes okay they are quasi-static process only first of all reversible processes are quasi-static process in which please write down reversible processes are the quasi-static processes in which the frictional losses are absent i'm telling you a very crude definition no frictional loss in a reversible process there is no frictional loss in a reversible process okay so definitely when we are talking about the highest possible efficiency we are going to take only those processes in which frictional losses are absent otherwise every time a process happens some part of energy will be lost due to the friction and that cannot be reversed if i trace back the frictional losses will not be able to come back it will be gone now fine so that is the reason why we like to have reversible processes fine so that is what is about the reversible processes and then we are now going to discuss about the Carnot cycle okay so Carnot cycle basically is the heat engine having the maximum possible efficiency so Carnot was a person he's a he's a french engineer who who was given a task by Napoleon to find out what is the maximum possible efficiency and he he came back with a quantification of the heat engine formula all right and that actually helps us to understand you know that how much maximum efficiency is possible otherwise you know if you don't know what is the maximum possible efficiency possible then unnecessarily we'll be comparing it with 100 percent efficiency suppose Carnot tells Carnot efficiencies let's say 50 percent okay then even if i build a engine which has 45 percent efficiency it's a very good heat engine but if i don't know that maximum possible efficiency is 50 percent i'll think that 45 percent is a very bad efficiency because 100 percent should be maximum right so that is why the discovery of Carnot engine is extremely crucial so that it gives us a benchmark against which we can compare the efficiency of every other engine write down Carnot cycle so please write down Carnot cycle is the heat engine Carnot cycle is the heat engine operating between two temperatures and has highest possible efficiency okay now we know that the heat engine should absorb heat at constant temperature and release the heat at constant temperature so the absorption of heat and the release of the heat should happen at the constant temperature all of you agree this let's say that the higher temperature line is this and the lower temperature line is that so the cycle should operate between the two lines these two lines these are the two isothermal lines okay now i need to complete the cycle also when it is absorbing the heat it should expand and when it is releasing heat it should contract but cycle should be completed there should be a line from here which joins the line there and from here it should join there what these lines should be anyone adiabatic because the heat exchange is not allowed when you're going from here to there or when you're going from here to there okay heat exchange is not allowed heat should be only exchanged at the reservoir temperature please draw this this is the Carnot cycle and now we are trying to derive the efficiency of Carnot cycle okay sorry one small error we have made is that adiabatic will be below and isothermal will be above that we have discussed already so this is what the process will look like which is adiabatic the lower one is adiabatic okay the up this one is isothermal so let's say over here pressure is p1 volume is v1 and temperature is t1 over here pressure is p2 volume is v2 and temperature will be what will be the temperature p1 only it is an isothermal so it will go like this this is the cycle Carnot cycle this will be let's say p3 v3 and t2 let us say this point will be p4 v4 and t2 because this line is isothermal this one and this is also isothermal the upper one and lower one okay now i know that heat is exchanged at a constant temperature so at a higher temperature by the way which one is higher temperature t1 or t2 look at the diagram t1 t1 is higher right because p into v should be nrt so multiplication of pressure into volume will be highest at the upper portion lower in the lower portion so heat is absorbed there so you can draw in the diagram like this so this is q1 and heat will be released at this temperature t2 t2 is a reservoir temperature so you can say this is q2 okay and we have already arrived at this expression that efficiency will be equal to 1 minus q2 by q1 okay right now i what i'm trying to find out here is that what is the efficiency in terms of t1 and t2 okay so all of you please tell me what is the value of q1 equals to and q2 equal to think of it and let me know it is an isothermal process okay 1 2 2 and 3 2 4 isothermal process q1 and q2 find out see it is an isothermal process so if you apply first law of thermodynamics between 1 and 2 delta u will be 0 change in temperature is 0 delta q will be equal to work done of an isothermal process work done in the isothermal process will be nrt1 log of v2 by v1 isn't it this is delta q which is my q1 only right so q1 is nrt1 log of v2 by v1 now tell me what is q2 q2 is what minus nrt ln v4 by v2 see actually it should be final volume do it by initial volume only q2 if you take if you find out it will be nrt2 log of v4 by v1 v4 by v3 sorry but i know that this quantity is negative getting it but in the expression i already know that q2 is released okay and when i substitute in this relation q2 should be made positive that is where there is a minus over here there should be nrt2 log of v3 by v4 if you take v4 by v3 that is not wrong because the heat released should be negative but when i substitute here i already know q2 is heat released i want to know how much heat is released so you should put a positive number over there i hope it is clear to all of you nrt2 log of v3 by v4 so efficiency can be written as 1 minus nrt1 divided by nrt2 log of v2 by v1 divided by same natural log of v3 by v4 okay nr and nr gone so i can say it is this 1 minus of that okay now tell me have we used till now the fact that 2 to 3 is adiabatic and 4 to 1 is adiabatic have we used that till now no sir no sir no so let's try to use that how can we use that uh where should i write okay this small space all right so for adiabatic p into v gamma is constant i want a relation between volume and temperature because temperature and volume are correlated here and the expression also temperature and volume is there so that's why i want between temperature and volume i know that p into v is equal to nrt okay so p is equal to nrt by v so that into v to the power gamma is constant and that's the reason why t into v to the power gamma minus 1 is constant n into r is anyway a constant so that into constant is another constant so for adiabatic process t into v raised to power gamma minus 1 is a constant all right now i'm going to apply this equation t v raised to power gamma minus 1 between 2 and 3 and 4 and 1 so what will be the equation if i apply between 2 and 3 i'll get what temperature at point 2 is what t 1 so what will i be writing there t 1 into v 2 gamma minus 1 v 2 gamma minus 1 is equal to what t 2 into v 3 gamma minus gamma minus 1 this is between 2 and 3 between 4 and 1 what can i write v 2 v 4 gamma minus 1 is equal to t 1 v 1 gamma minus 1 correct t 1 v 1 gamma minus 1 is equal to is equal to t 2 v 4 gamma minus 1 okay now if you divide this first equation i'm intensely writing here only so that everything is visible otherwise it will not make much sense so there are two equations if you multiply left hand side of this equation with the left hand side of that equation t 1 is gone so you're going to get v 2 by v 1 raised to power gamma minus 1 is equal to v 3 by v 4 raised to power gamma minus 1 okay so this is what you're going to get if you divide it v 2 by v 1 gamma minus 1 is equal to v 3 by v 4 power gamma minus 1 gamma minus 1 power if you remove v 2 by v 1 is equal to v 3 by v 4 right so that's the reason why this can be cancelled away from that have you understood all of you yes okay so efficiency of a Carnot cycle can be simply written as 1 minus t 1 by t 2 sorry t 2 by t 1 where t 1 is the hot reservoir oh i have actually numerator should be this t 2 with the numerator and denominator got exchanged please correct that cylinder so this is v 2 by v 1 still it will cancel out numerator and denominator is at 1 minus t 2 by t 1 t 2 has to be colder reservoir otherwise t 2 by t 1 will become greater than 1 and efficiency will come out to be negative so that is i mean that's another way to remember the numerator should be lower than the denominator when you write the temperatures okay so towards the end a small proof is there just five minutes i will take to prove that Carnot cycle has the maximum possible efficiency okay the proof of Carnot cycle being the maximum possible efficiency is done by contradiction they assume that suppose such cycle exists which gives more efficiency between the two temperatures than Carnot cycle so let's say t 1 you absorb the same amount of heat okay and you are able to extract more work okay but then you're releasing the same amount of heat t 2 please draw this two cycles q 1 dash w 1 dash and q 2 this is original and this is Carnot and this claims to have more efficiency this cycle let's say claims to have more efficiency than Carnot if okay so basically you are able to convert actually it should be w 1 dash and q 2 dash you're absorbing same amount of heat but you're extracting more work all right so basically w 1 dash should be greater than w 1 this is what it means for the same q 1 all right now if this is true then both should be reversible 1 and 2 reversible so i can reverse the cycle 1 i can reverse the original Carnot cycle and this is how it would become it'll become a refrigerator now t 1 and now it will give the heat back but it will need what to be done now it'll take heat from the colder reservoir t 2 and this is q 2 see we already know Carnot cycle is reversible so i reversed it okay so Carnot engine becomes a refrigerator right and then i'm coupling i'm coupling 2 and 3 when you couple 2 and 3 then basically whatever is a heat this q 1 and q 1 you can couple and this w 1 which is work done on this cycle can be taken from this modified cycle all right so entire cycle will become like this t 1 this is little tricky to understand so it will take time to sink in but initially someone need to tell you an entire thing first so this is t 2 this will be q 2 minus q 2 dash okay and this will be equal to w 1 dash minus w okay so clearly w 1 dash we already assumed is more than w 1 so w 1 dash minus w 1 w 1 dash minus w 1 should be greater than 0 okay and since more amount of heat in the modified cycle is converting into work q 2 dash should be less than q 2 so that is the reason why even q 2 minus q 2 dash is greater than 0 so now if you look this look at this what is happening here what is happening here is that you're taking heat from the colder reservoir okay you're taking heat from the colder reservoir and converting entire heat into the work and you're not rejecting any heat is that possible no sir can you just repeat that last thing what I'm saying is that ultimately when you couple the modified cycle with the reverse of the Carnot cycle what comes out is that you're extracting heat from a colder reservoir okay and converting entire heat into work that is against the second law of thermodynamics so that is the reason why there cannot be heat engine which has more efficiency than the Carnot cycle okay so this is this proof is done by contradiction so that's why it'll be little tricky to understand you just need to spend a little bit more time to visualize what has happened all right so sir coupling means like we're adding the engines and adding the heat and actually taken and taken out the work and also coupling means whatever work is required by the reverse Carnot cycle is taken by taken from the modified Carnot's engine which has more efficiency and the whatever heat that is supplied to the modified Carnot engine is taken by the reverse Carnot it is is given by the reverse Carnot engine so what this earlier what was happening is that reverse Carnot engine was giving heat to T1 rather than giving to T1 the same heat can be taken directly by the modified Carnot engine so that is what coupling means okay so sir the one in the blue box is the net engine of both of them sir the modified engine is doing work on the reverse Carnot modified engine is doing work on the modified engine you can't say it is Carnot or not it is combined there cannot be any process it doesn't talk about engine there cannot be any process whose sole result is to convert heat into work completely does not matter one engine two engine five engine does not matter all right