 okay welcome everyone waiting for a few more to join in what's going on in your school everyone what are they teaching currently heat transfer in kormangala and in hsr what is happening hsr in the other places rishabh what is happening in nps hsr shm oscillations so you guys are doing oscillations and you you guys are done with thermodynamics or that is something is you oh that is done kandik theory is also done in nps hsr kandik theory is not done fine still lesser number of people have joined oh that is also done nps hsr is very fast in terms of completing the syllabus so you have anything going on in your school currently or any major exam coming up i've seen some not some messages we're circulating in the group some midterm right it's a board exam grade 11 you have a board exam i i am sure as a for pu it is for pu it is a state board it is a board exam okay next week and in nps when is the midterm going to start in nps when is the midterm midterm over did i ask your marks have i asked okay i already asked anyways so i think in the previous session last class it was very theoretical okay we have covered many things and i'm sure you guys are not as comfortable with those concepts yet as you want it to be right so in today's session what i plan to do is a series of problem practice and these questions i will take it school level ct level or edmax neat or mainstay there will not be any advanced level questions okay so i plan to finish them off and how many of you have already done the assignment and submitted anyone has completed the assignment which was given because of the exams you did okay there was no submitting option in the classical is it true everyone let me just check i'll extend the date if that is so what we can do is that i will anyway do a lot of questions with you today so maybe you can use that in knowledge and complete the assignments oh there is no submitting of sorry about it i'll immediately create it i'll keep a due date of sunday fine sunday is a due date the assignment is already given to you fine now it will show on your log in fine so you can submit it before sunday ends now you'll be able to see it fine all right so today we'll be doing questions but before doing the directly jumping on the question i'll just quickly you know summarize what all the small small things that we should i mean i'm just trying to revise it for you whatever we have done in the previous class before we get into the questions right so only theory related to the problems this is what first law of thermodynamics delta u plus w okay this is the first law of thermodynamics it is used for a system okay you define a system then you have to follow some sort of sign convention if heat is supplied it is positive or negative delta q will be positive negative if everyone should participate okay delta q will be even though it is a trivial thing if you participate in the trivial things then the difficult thing will also come out naturally okay fine now if work is done by the system then w is positive or negative good it is positive negative it is in chemistry understood in chemistry the formula itself is different in chemistry delta u is equal to q plus w okay the formula is reversed because the sign convention is different otherwise first law of thermodynamics is same for chemistry as well as for the physics anyways but this is what we follow in physics what is the reason we are following work done by the system as positive because our focus is work done right that is why we are studying this chapter we are studying this chapter so that we know how to convert heat energy into work so what do you want we keep it positive work done we find like this integral of pdv for the flutes okay and then if you know cp and cv are basically molar is specific heat capacities of the gas at constant pressure process or constant volume process okay so let's say you give heat in a constant pressure scenario then you can write the heat given in the constant pressure process as n cp delta 3 okay and if you see that during the process temperature is changing and volume is constant then it is a constant volume process so then you have to write heat supplied at constant volume will be n cv delta t fine so there can be there can be infinitely many molar specific heats for the gas okay as it depends upon the process okay so remember this then what we have learned is that for isochoric process we had learned the work done is zero for isobaric process the work done is p delta v this can also be written as what in terms of temperature what you can write work done look at your notes good n r delta t fine then you have the adiabatic for adiabatic process the work done is negative of change in the internal energy what is the formula for the internal energy change you remember that the formula for change in internal energy is what change in kind energy but what is the formula for that change in internal energy what it is you have done it in chemistry also and cv delta t okay even though you are using a specific heated constant volume but this formula is same for all the processes if you are finding the change in the internal energy because the change in internal energy is the state function okay and for adiabatic even the work done comes out to be a state function because work done is nothing but negative of the change in internal energy fine so in adiabatic of course heat exchange from the surrounding is zero now these are the work done's apart from that there are few process equations that connects one point of a process with the other point of the process fine so for example isochoric process p1 by t1 is equal to p2 by t2 this is i guess charles law right then isobaric process you have v1 by t1 is equal to v2 by t2 okay i think i have missed one more process here which is isothermal isothermal what is a work done you remember that what it is isothermal work done nrt natural log of v2 by v1 this is also equal to nrt natural log p1 by p2 all right so this is galuset law this is the charles law and for the isothermal we have boyle's law p1 v1 is equal to p2 v2 then for the adiabatic what is the process equation everyone you remember for adiabatic what is the process equation all of you t1 v1 gamma is equal to p2 v2 gamma all right then t1 v1 you remember this or not type in what is gamma what is gamma gamma is what cp by cv this is greater than one because because cp minus cv is equal to r mayor's relation okay so you know you just need to remember these few things and anyway when you solve numericals you don't even have to make any efforts to remember all of this you automatically tend to remember just like i am remembering it now few things related to the graph also you should keep in your mind while solving numericals okay let's say you have this pressure volume curve you go like this from this point number one to point number two the work done is the area under the curve now looking at the process can you find out whether the work done is positive or negative whether the gas is doing work or work is done on the gas how will you find out in this process work done is positive or negative when you go from one to two correct work done is positive if the volume increases it is very simple the formula for work done is integral of pdv dv is what change in the volume so if change in the volume is negative final is less than the initial change is negative so work done will be negative that way okay then one last thing about the cyclic process because machines runs on the cyclic process so when you learn the heat engines today you'll see cycles is heat engine removed in your school in hsr have the covered heat engines did i already ask you that question one you did not know setu setu menacha car not car not engine is done in your school or not no but you are in kormangala right in kormangala who is from hsr okay it is deleted car not engine is deleted but heat engine is done have you done refrigerators heat engines refrigerators okay no actually it is officially not deleted okay so heat engines and refrigerators if you go to the official site cbsc.nic.in it is there and that is the reason why it is there in all the complete exams you're going to write so i'm not going to delete anything we are going to learn everything okay and you may not do it for your uts but then use it like a practice kind of thing okay because engine is no new concept is there in engine whatever you have learned you're using it to understand engines okay for example this this could be car not engine cycle whatever i have drawn a cyclic process one of the cyclic process can be car not engines okay okay so you do not know let me just show you cbsc syllabus for grade 11 where is the official site this is the official site okay dot ni c.in they are the official sites for physics this is physics look at this heat engine refrigerator is there or not so i'm not showing you so that i mean i'm my intention of showing is that you should understand that if you leave it it will come and haunt you in the company exams it is not deleted officially okay anyways if it is cyclic process what is work done work done is the area of what area of the loop okay and you can also find the work done like this work done in 1 to 2 plus in 2 to 3 then 3 to 4 then 4 to 1 like this okay and what else how will you find out work done is positive or negative everyone total work done total work done in the cycle is it positive or negative how will you find out correct if loop is in quadrant one all pv graphs are in quadrant one only pressure cannot be negative volume cannot be negative so the clockwise is positive the clockwise cycle this is like a thumb rule if you remember anti clockwise negative okay the logic is that the process which is happening at the top of the cycle it will have the largest area under the process so the top process will dominate overall if top process work done is positive net net will be positive only so it comes out to be that if it is clockwise volume will increase on the top side so you can't go like this if work done is positive but anyways you can remember these thumb rules and for the cyclic process change in terminology is what delta u is what for one full cycle 0 you came back to the same point all right so if you use first law of thermodynamics for entire cycle then delta q which is net heat that is gone in net heat that has come out which will be equal to w because delta u is 0 all right so this is the summary of the last session we had let us see some of the questions we'll take the basic ones only because I know we are just learning something completely new and you have not done the assignment also do this yeah so some of you know good I don't want to announce it Siddharth is saying something so work done in A to B is what everyone what kind of process A to B is isochoric so isochoric the volume is constant so work done will be zero it is simple right so simple work done in A to B is zero because the volume is constant and what is the process B to C what kind of process it is isothermal constant temperature what is the formula for the constant temperature process work done n r temperature of that process t2 log of final volume by initial volume is what it is okay now can you guess what is this c2a process assuming actually it is not very clear assuming that this line goes through the origin like this okay imagine the line get this is a straight line only what process it is c2a correct shadowy got it others isobaric what is the reason slope of line is what volume divided by temperature it's a constant and when that happens what is the isobaric process okay so work done in c2a is p times v1 minus v2 right the going from c why did you say going through the origin it is going through the origin this not given in the question but i'm telling you it is they should have mentioned it so work done is p times final volume which is volume at a minus initial volume that is the volume at c okay if it doesn't go through the origin then it can't be isobaric okay but then that's a that's a very good question by the way okay can there be a straight line in this volume temperature graph which will not pass through the origin is it possible to have some temperature at zero volume no why as in c it is not mentioned in the question okay i'm telling you assume it it is an incomplete one okay so pressure into final volume minus initial volume all right now the volume is given but pressure is not given so i'm going to use nr t1 minus t2 over here n is equal to 1 so r times t1 minus t2 okay so this is actually the one of the questions which i had given in the assignments and in the assignment this is the answer what they have given assuming it is isobaric process all right so that is why i know what is the answer and it can't be that if it is not isobaric and Rishu Shadli do this one so that knows the answer others okay great so total work done is equal to what work done in a to b plus work done in b to c okay now which part the work done is negative a to b it is negative or b to c it is negative or both are positive only a to b negative right so this is area under a to b it's a trapezium so negative of the area under a to b so that is half of some of the parallel side to p0 plus p0 into v0 then b to c work done is positive so area under that you have to take half of this distance between them to v0 okay so half p0 v0 minus 3 plus 10 this will be 7 by 2 p0 v0 so option c all right so this is yes i'm sharing my screen all of you can see my screen yeah Siddharth what happened Siddharth is looking somewhere else don't open the browser okay okay no problem over here this one for this we'll have poll don't answer after two minutes we'll have a poll okay just type in that you're done nobody's typing done done no hint nothing hint will be given after the poll is done okay so the poll is coming Shaduli wants to see the poll take the poll Siddharth are you able to see the poll currently okay once the poll is gone probably then the screen will come all right so i'm ending the poll oh you don't know the option ending it this is what you guys have selected fine majority there is no majority actually full of confusion but between a and d is what the correct answer is d for deli okay let us see how to solve some issue with your laptop maybe Siddharth not sure which laptop using macbook or some windows laptop tab oh tab you have to again and again zoom out the tabs then select the tab which you are this thing which tab the ipad samsung i don't know how to okay so between initial and final i and f what is that one thing which is same between the initial and final points pressure is same okay other than pressure see it appears that pressure is same need not be getting it it is not written it's not written that exactly it is same it might happen that from initial the pressure goes something like this a slightly different the fs so there is no guarantee pressure is same forget about that but what is the guaranteed thing which is same between the two points same correct change in the internal energy delta u do you all agree for both the cases delta u will be n cv temperature at final minus temperature at initial it does not matter what process it is do you all understand this type n okay delta u is same work done is more where work done is more in a or b whose work done is more both are positive only work done in a is more because area under a is more all right so delta q of a is delta u plus work done in a so w a is more so that makes q a more right delta q b is delta same delta u but lesser work done so naturally delta q a is more than delta q b is it clear to everyone why this d option is correct fine let us move forward okay some theoretical questions school level stuff tell me a b c d reservation of work what is that no no you already got negative marking once you pick an option okay fine no problem it is conservation of energy i think i have told it's so many times that i don't need to even discuss it everyone should answer okay so heat is supplied so delta q is positive and as per the first law of thermodynamics this is delta u plus w what is delta u for isothermal delta u is what if temperature doesn't change delta u is what d zero okay temperature is constant n c v delta t is delta u right delta t is zero so delta u is also zero delta q is equal to w and both are greater than zero so work is done by the system the gas will do the positive work clear to all of you fine so such kind of theoretical questions are asked in lot of theoretical questions are asked in the NEET exam and in j-main also at least 20 to 30 percent questions are theoretical in nature this is the kind of question they ask okay this one i have to draw the graph for you so it's the volume this is the temperature this temperature is t1 this is t2 this volume is v1 and v2 okay do it this line if you extend will pass through the origin it'll go through the origin arrow you want arrow is actually not required but then let us assume this is the arrow the part you can name the process this is a this is b this is c done all of you everyone tried type n have you tried all right let us do it now we'll take pressure on the y axis and the volume on the horizontal axis okay so a to b is what process what is a to b tell me what is the process a to b everyone correct isobaric right pressure is same all right how can I say pressure is same because the slope of this line is v by t and that is constant this happens for the isobaric process all right so it is an isobaric process a to b and the volume of b is more than volume of a so i'll just draw a horizontal line over there this is a horizontal line i have drawn in the middle of this so i'm saying this is point a that is point b so i have to go from a to b right the arrow is from a to b now a to b b to c is what b to c is constant b to c is isobaric right everyone at least b to c is isobaric type n you can see that it is a horizontal line so volume is constant okay so b to c is an isobaric process this b to c now c to a is what what kind of process c to a is all of you c to a is what kind of process isothermal isothermal you have seen it is a curve so between a and c you have to connect a curve like this okay this is how you do it and look at this in volume temperature plot it is an anti-clockwise cycle in pressure volume it becomes clockwise cycle okay so there is no rule as such that if it is anti-clockwise will remain anti-clockwise in volume temperature graph also and so on okay nothing like that why the curve have to be bulging out it is like this only isothermal curves are like that only the this is the kind of isothermal curves will be it is a rectangular parabola right pressure into volume is constant for isothermal this will do at the end leave it he come back cp and cv values are given part one shaguli got something others answer for the part one yeah shaguli get in terms of p naught v naught right how are you getting a number over there is p naught v naught given oh they're not given right siddharth got something shalini got something walk them by the gas will be positive or negative in this case clockwise cycle positive will be area of this loop loop happens to be a right angle triangle there is a best part half base into height half of base it base is what this length which is v naught two v naught minus v naught this is the altitude three p naught minus p naught two p naught the answer is p naught v naught this is for the a right do it for the b heat rejected in c to a there is a b c this is a and this is c in c to a it is an isobaric process so and cp delta t cp is given five r by two use that all right so let's do it delta q is equal to and cp delta t temperature a minus temperature at c okay since it is heat rejected it should come out to be negative and it will because t a is less than t c okay then n is equal to one cp is what five r by two into t a minus t c fine then at point c p naught into two v naught is equal to n r t c n is equal to one so i'll keep it that way so t c is equal to two p naught v naught by r at a p naught v naught is equal to r ta so temperature at a is equal to p naught v naught by r so t c and ta are this so when you substitute it over there you are going to get delta q as minus of five by two p naught v naught so you can ignore minus because they're anyway saying it is rejected so the answer is five by two p naught v naught okay now get the answer for the c very similar tell me what is the answer adiabatic no it is not adiabatic who said a to b is not adiabatic isobaric a to b is isobaric a to c yeah it is isobaric correct a to b is isocoric here so delta q would be equal to n cv temperature at b minus temperature at a okay so n is equal to one this is three r by two temperature at b is three p naught v naught by r temperature at a is p naught v naught by r can use ideal gas equation at b and a fine those who are using shortcuts will get it completely wrong so the answer is three p naught v naught follow all these steps okay this is what you will get till now everything is clear this question has multiple parts but each part is if you look at just one part you can see it is a school level question fine so there are like four questions in a single question now find out for the d d for deli last one which process is b to c can you tell me what is the process b to c temperature is constant why why you say temperature is constant what is the process b to c everybody it is not any standard process okay it's not any of those standard processes which you have learned it is some process which you have not named so now do the d part you can still find out kind of how you get it how you get that yeah both added b and c then is that it no no you created some rule in your head doesn't make sense tell work done then do what do what total work then is pv okay then what okay so all right setu will speak he wants to speak how setu how to do it sir can you hear me yes everybody can hear me so so total heat is equal to delta q total is equal to delta q total which is zero plus total work done which is p0 v0 and delta q total will be qca plus qab plus qbc correct that is how you did good and one of them will be negative if heat is given to the system it is positive it is released it is negative so what setu said absolutely correct for entire cycle delta q when you write delta u plus w for full cycle delta u is zero so delta q a to b delta q b to c plus delta q c to a is equal to total work done which is p0 v0 which you have already found out okay so you can't use any formula for b to c because this is not a standard process so that is how you are doing it but c to a and a to b are standard processes so this is minus 5 by 2 that is 3 so minus 5 by 2 p0 v0 plus 3 p0 v0 plus delta q of which process b to c b to c is equal to p0 v0 this is how you get the answer 2 p0 v0 minus 2 p0 v0 plus 5 by 2 p0 v0 so that is half off okay okay anyone has any doubts the last part of the question was you can say it is good the other parts were just direct formula substitution type in is this clear to everyone not it okay so shall we do slightly difficult one one difficult question the all these type of questions you have done till now will take care of your school take care of your ct neat j main in by some extent but let us take slightly tricky one do you all agree shall we take or should we keep on solving like this only we'll come back to these kind of questions after this one do you all agree no one is answering only one person is should we take a difficult one slightly difficult not very difficult not very difficult I mean I was anyway going to take that but then don't consider it to be advanced level this is not advanced level okay this is a derivation from your textbook which in your school they have deleted can you guess what it is this is a Carnot cycle which we are studying very important I don't want to skip it pressure and volume we are treating it like a numerical okay so there are two isotherms isothermal lots let me first completely draw it okay then you can draw now you can draw this is a b b to c c to d and d to d okay okay over here you can say p1 v1 p2 v2 p3 v3 p4 v4 okay so a to b is given isothermal process at a temperature of t1 b to c is adiabatic process c to d is again isothermal at temperature t2 and d to a is adiabatic okay now let's say a p1 v1 p2 v2 p3 v3 everything is given to you let's say everything okay I want to find out the okay by the way in b to c process how much heat will be absorbed by the gas in b to c zero d to a also zero okay so a to b heat is absorbed or released can you tell me it's an isothermal process in a to b heat is absorbed or released a to b circle is saying zero by area under a to b is the work done in a to b this is positive work done or not work done is positive delta u is zero so delta q is also positive delta q is equal to work done for isothermal okay so heat is absorbed heat comes in from here this is the cycle that runs inside your car or bike broadly you can say this is the cycle that runs in which a to b what happens in a to b process the fuel is burning so the gas is absorbing heat from the fuel all right now in c to d heat is released q out okay there is no heat absorbed or released from any other point or any other process so I want you to simply find out q in value q out value w value in of the cycle these three things find out you can use t1 t2 v1 v2 anything done in direct formula to use here q in for a to b q in is equal to delta u plus w in a to b right delta u is zero temperature is constant and it is an adiabatic process so n r t1 log of v2 by v1 there is a heat that is going in okay then c to d heat that goes out this is equal to what with the same logic it is equal to the work done in c to d fine that is equal to n r t2 natural log of v4 minus v3 remember this is a negative quantity okay this is negative v4 is less than v3 okay now work done is what heat absorbed if you use the cyclic process and first law of thermodynamics into it delta q is equal to delta u plus w for the entire cycle delta q is heat absorbed in a to b plus b to c plus c to d d to a delta u is zero it's a cyclic process that is equal to total work done okay so i know that b to c adiabatic d to a adiabatic right so the total work done is simply some of these two which is n r t1 log of v2 by v1 plus n r t2 log of v4 by v3 till now it is clear to all of you we have got q1 q in q out and w okay will there be a negative sign it is already a negative quantity this is a negative quantity if you want to put a minus there first take a mod of there when you take a mod you become v3 by v4 are you getting it there's the work done this is q in now if you have to find the efficiency of this cycle efficiency we have seen in the previous class that it is equal to the total work done divided by heat that has gone in right total work done is n r t1 log of log of v2 by v1 plus n r t2 log of v4 by v3 divided by heat that is given heat given is this only okay n r t1 log of v2 by v1 all right so the efficiency can be equal to one plus t2 by t1 log of v4 by v3 divided by log of v2 by v1 i'll put a minus here t2 by t1 log of v3 by v4 i've reversed whatever is inside the log in the numerator this is it understood till now ask if you have any doubts ask if you have any doubts last step last step see v4 by v3 i have reversed log of x is negative of log 1 by x you know this or not is it yes or no all right now what you can do is look at this graph here so b2c is what b2c is adiabatic process okay temperature of this point is t1 and that point is t2 right so since b2c is adiabatic process i can say that t1 v1 gamma minus 1 sorry t1 v2 at point b volume is v2 this is equal to t2 v3 to the power gamma minus 1 b2c is adiabatic process so i can use a process equation all right similarly i can say that d2a is an adiabatic process so i can say that at point a temperature is t1 v1 gamma minus 1 this is equal to t2 v4 gamma minus 1 all right and if you divide these two equations divide these two equations you're going to get v2 by v1 is equal to v3 by v4 is this clear this one is clear what your last two step which i have written everyone type in it's it was a very important derivation in your textbook it got deleted all right deleted in your school not from your syllabus actually and it would have been better if would have been there then you know what will happen if there is an exam from this chapter only this will come nothing else so the earlier times it used to be like this only so if there is a school exam and thermodynamics chapter comes they ask only car not cycle efficiency derivation nothing else anyways so v2 by v1 is equal to v3 by v4 so that is why i can cancel out this and that and say that efficiency is 1 minus t2 by t1 let's see we have done so much circus at the end we are getting such a simple expression okay this one this is the efficiency 1 minus t2 by t1 okay and this is the car not cycles efficiency derivation this is the maximum possible efficiency a heat engine can have between the two temperatures maximum possible efficiency all right the higher temperature is t1 that is the flame temperature when the fuel is burning it is that temperature t2 is you can say is the exhaust temperature roughly it is the temperature of the atmosphere only okay so if you need to have an engine between the atmospheric temperature and the temperature of the fuel when it is burning the maximum possible efficiency will be 1 minus t2 by t1 you cannot devise any heat engine any engine it can never have efficiency more than this between the two temperatures okay so that is the importance of the car not cycle fine all right so since talking about the those who whose focus is only schools we are done with the school curriculum okay everything we are done now we are going to learn about heat engines and refrigerators properly okay because it is very much part of all the exam that you're going to write later on so let us talk about the heat engines I have a lot of other numericals that I will do once the theory gets over fine so let us cover up the theory of heat engines and refrigerators it will not take much of time probably before the break we are done with entire theory and then again we'll take up the numericals we'll be taking the simple numericals now after the theory write down heat engines the sole intention of learning the chapter other than getting the marks in the school exam the sole intention is to learn how to create work from the heat energy okay so if you do not learn how engines work which is actually converting heat energy into work then this chapter is useless fine so we must learn basics of you know how the machines work when it comes to converting heat energy into the work fine so all the heat engines they follow cyclic process that is like given thing it will be a cyclic process fine then all the heat engines will consist of a working substance working substance is nothing but for example in in our car engines the working substance is air plus fuel mixture let's say petrol engine is there so petrol is a fuel and with fuel you need to inject air also you need to inject oxygen otherwise it will not burn so the mixture they're calling it as working substance fine write down the next point the next point is the working substance absorbs it absorbs heat at some higher temperature high temperature is nothing but the flame temperature when this air and fuel burn chemical reaction happens and temperature goes up so it absorbs the heat due to that chemical reaction and it releases heat at some lower temperature lower temperature is the atmospheric temperature so every heat engine will have an exhaust to reject the this thing heat alright third it will have a hot reservoir reservoir just a name we are saying reservoir because we are saying that it is a reservoir because we are assuming that you can absorb as much heat as possible without changing the temperature of the reservoir that is why without changing the temperature of the source that is why source is called reservoir you can it is like huge you know even keep on absorbing heat temperature will remain constant so to keep the temperature constant you need to continuously burn the fuel right that is how the temperature remains constant a hot reservoir T1 from which heat is absorbed okay it needs to have a cold reservoir T2 from which heat is released fine okay fifth point fifth point is working fluid does the work and it get transfers transferred to other components get transferred to the other components as in wheels will get rotated and so on and so forth right let me just show you something this is a diesel engine I showed a petrol engine my job is to make college easier because students have a lot of fun all of you can you hear the sound also okay so I'll keep on pausing and explaining to you curious actually it's a syringe a car is made up of wheels seats as it adds a body and an engine combustion engine works in four stages first a valve opens and the piston lowers letting a mixture of fuel and so I don't know why it is so animated this valve is open so the air and fuel comes in from here all right this is what it is saying it's like a combustion chamber second as it comes up the piston is closed sorry not piston the valve is closed and this piston is moving up as this piston is moving up the pressure over here is increasing this valve is closed so the air cannot escape now fine and now a blast will happen as in some chemical reaction will happen so now the chemical reaction happens pressure suddenly increases because of the combustion and this piston is pushed down and after the burning has happened the fuel is gone because it has burned so that has to go out so this is where the exhaust go out okay let me complete it now piston down again let us hear it from the wheels seats as it adds a body and an engine internal combustion engine the engine works in four stages first a valve opens and the piston lowers letting a mixture of fuel and air into the combustion chamber second as it comes up the piston compresses the fuel air mixture third a spark from the spark plug ignites the mixture which pushes the piston down again fourth the exhaust valve opens and as it comes up the piston pushes the spent gasses out and the cycle starts over again intake compression combustion exhaust intake compression the movement of the engine is transmitted to the wheels for the driver's delight so I hope you have understood how everything works all right so this is how all the cars right now are running all right not just cars all the automobiles everything so for any cycle if the cyclic process I can say that delta q is equal to this delta u is zero is the work done so q in minus q out this is equal to the work done all right and for heat engines what is what is there in the heat engine it absorbs heat at the higher temperature as it is an isothermal process it absorbs heat okay from the higher temperature and releases heat at the lower temperature so both absorption of heat and release of heat they are isothermal processes okay and you are not allowed to absorb or release heat from any other process okay so in order to complete the cycle let me show you you need to absorb at a higher temperature release at a lower temperature you cannot absorb or release from anywhere else so in order to complete the cycle you need to connect to adaptive processes that is what Carnot cycle does okay the clear now fine so there is a block diagram you can show it by using some sort of block this is a popular way of representing that at a higher temperature t1 you absorb heat you're absorbing heat q1 and you are releasing heat q2 at a temperature t2 this is a working fluid the fluid does the work this is how you represent it a better representation actually okay we have already found out that efficiency is equal to work done divided by heat that is supplied work done is q in minus q out divided by this is the efficiency so efficiency is 1 minus q2 by q1 which we have already derived is equal to 1 minus t2 by t1 okay this is like the max possible efficiency Carnot cycle efficiency is this t1 and t2 in kelvin okay fine so this is the heat engine okay should not have been deleted but then anyways just delete it so delete it next we are going to learn how refrigerator works right so if you are interested in learning in detail like you want to construct the heat engine all right that you can do if you take automobile engineering or mechanical engineering in your as your major later on right now it is just theoretical stuff next is refrigeration cycle how refrigerators work when I say refrigeration cycle same thing is happening in the air conditioner also so in a refrigeration cycle what happens is what is the basic function of the refrigerator what do you want to do with the refrigerator you want to you want to keep you want to maintain the colder temperature okay how can you maintain a cold temperature of a system tell me inside the refrigerator how you maintain the lower temperature broadly what should be the process if you do not maintain the temperature inside then if you do not maintain the temperature somehow inside it will go up why the temperature inside the refrigerator goes up when you switch it off tell me why temperature goes up suppose you already cool let's say refrigerator is already cold okay you have maintained the coldness now you switch it off okay it is perfectly isolated from outside then why temperature still inside the refrigerator will go up because when you put something some let's say vegetable or anything inside inside the system what will happen is that there might be some staining process happens that releases heat okay that is one very obvious thing other one could be that let's say the refrigerator is already cold now you are putting something hotter inside it that will also release heat okay so you need to constantly take heat out from the enclosure of the refrigerator and also nothing is perfectly insulated there you might have maintained a very good insulation but still heat from the surrounding may go inside the refrigerator all right so what happens in the refrigerator is that it takes heat away it takes heat away at a lower temperature or at a higher temperature there are two temperatures in which refrigeration cycle works so you take heat out at a lower temperature or at a higher temperature at a lower temperature good so refrigerator takes heat from lower temperature and releases heat at a higher temperature okay here's what it does all right and the work is done on the system system doesn't do the work so the gas doesn't do work work is done on the gas have you seen compressor in a refrigerator behind the refrigerator the older refrigerators had a black box sort of thing have you seen it behind the refrigerator when you were kid i'll show you have you seen something like this in your refrigerator this nowadays they cover everything up so you have entire covered refrigerator is it you don't see anything behind so in case your refrigerator stops working for some reason just try to break it open you will see this black object over here you know what it is any guesses what it is it is written here it's a compressor it is a compressor so work has to be done on the system right when work is done by the system it expands and when you are doing work on the system the system should get compressed the volume should decrease in order to decrease the volume you need a compressor to compress it and that is why you need electricity electricity 99% of the electricity will be consumed by the compressor other than that there'll be some lights you fit inside the refrigerator and other maybe pump and other things but the majority of the electricity is consumed by the compressor to do the work on the system all right so the cycle is like this it is exactly reverse of the heat engine what heat engine does in a heat engine that this heat is absorbed from a hot reservoir and is released to the cold reservoir that is a heat engine in refrigerator heat is absorbed from the lower temperature and release at the higher temperature okay so if you draw the refrigeration cycle this is t1 that is t2 it is reverse exactly and the work is done on the system the blue q1 q2 got it and here also we can use first law of thermodynamics it's a cyclic process all right so q in minus q out should be equal to minus w this w is going into the system system is not doing the work so minus of it q in minus q out so q1 minus q2 let me name this as q2 only because in Carnot cycle also we have name for t2 q2 for t1 we'll have q1 so q2 minus q1 is equal to minus w or q2 plus w is equal to q now tell me how you will define efficiency for the refrigerator any guesses for engine it was worked and divided by heat that is given that's what it is work done i mean what do you want from the refrigerator you want work to be done what you want divided by what you give is the efficiency right what you want divided by what you give so what do you want out of refrigerator cold out of these three things q1 q2 and w what do you want refrigerator what is the main thing w now don't tell me in words tell me which one of these three q1 q2 or w what do you want from the refrigerator from the heat engine it was w for the refrigerator what is it the output is what it should be q2 you want heat to be absorbed from the cold place right so q2 and what do you give what is given what is given q2 is taken out from this temperature what you give you give w okay so we call it coefficient of performance we don't call it efficiency for the refrigerator the reason is it is more than 100 percent efficiency cannot be more than 100 percent that is q2 divided by w is what q1 minus q2 this can be written as uh one divided by okay is it clear to all of you coefficient of performance this is okay so little bit of theory is still there so right now we will take a break after the break we will continue little bit of theory is left and then the remaining numericals all right all right fine so let us continue from where we left off this is a refrigeration cycle or heat pump that we have discussed just now now did you learn about the second law of thermodynamics in school that is also deleted go on and bits it just have second law of thermodynamics have you done it reshame second law of thermodynamics was done in your school or not it was done fine so let us talk about the second law of thermodynamics so if there is one law that you have to say that is the most sacrosanct laws of entire maths physics chemistry bio put together that would be second law of thermodynamics okay so it is it is a law which is so fundamental that uh you know if you find let's you can find exceptions to almost every law but if you find exception to second law of thermodynamics you are wrong fine so that is just an exaggeration of the same point that it's uh let's say I can say that this is a law that comes directly from the natures or god's book fine so I'll just talk about one example and make you understand what is this law about so for example let's say there is this block okay which is lying on the floor fine so why doesn't this block jump on its own why it does not jump on its own and move up you'll say okay energy will be increased so energy is required so I'm saying why don't this block absorbs energy from the surrounding let's say if I supply heat from the down let's say I'm burning something so it is getting energy so that energy can it automatically convert into work and jump off the answer is no it can never happen okay but energy conservation is correct I mean energy conservation is holding good here you can supply energy which is equal to the gain in the potential energy then also it will not convert and jump on its own second example when you have a hot cup of coffee coffee or tea or a milk which whatever you prefer the spontaneous process is that it will cool off on its own okay but when it cools off it cannot become hot again by absorbing heat from the surrounding from by itself it cannot become hot again all right so basically second law of thermodynamics tells you in which direction any process will happen okay conservation of energy or the first law of thermodynamics will tell you that if process happens then delta q will be equal to w plus delta u but second law of thermodynamics will tell you whether it will happen from point number one to point number two or two to one okay so basically it gives us direction in which things happen actually second law of thermodynamics is so fundamental that it is difficult to put it in words and explain okay so there are various examples and explanation to it but there is no statement as such so this is basically what that god or nature wants things in disorder okay so disorderness is what is a natural thing to happen fine for example if you just leave a room for many years and come back the room will be in disoriented conditions fine it will never happen that you can't expect that a room which is disoriented will become ordered by on its own fine so randomness is preferred in fact it is so fundamental that it explains why you know you age and why we take birth all right basically the randomness which is preferred by the nature or the universe the physical quantity of randomness is entropy basically okay entropy is nothing but degree of randomness just like temperature is degree of hotness entropy is degree of randomness have you heard of name entropy anyone right so how do you measure randomness you can't measure it you can device some scale for example it's like you're asking me how to measure a temperature okay so you can device a scale just like temperature scale is measured okay so here degree of randomness has to increase in every this degree of randomness or the disorderness has to increase in every isolated system if from outside let's say there is no influence let's say that there's a room you leave it for very long when you come back it'll be disordered but if you ask someone to take care of the room so you'll make make it in order after every few days so it'll be in ordered manner so I'm not talking about that I'm talking about spontaneous process that happens on its own in an isolated system okay so if you consider let's say if you consider yourself as an isolated system okay so you as a I mean we as a human body we have atoms molecule tissues and all those things okay so initially everything was very organized as in we went when it was like a cell kind of thing okay then multiple cells created so disorderness increases all right then organs grew so more and more as you age disorderness increases wrinkles will come up okay and you'll grow older and that is how the death happens okay so even you can explain I mean you will not see people aging backwards right you'll not grow younger you'll always grow older because that is as per the second law of thermodynamics fine so I mean there's a saying that as per the second law of thermodynamics the only purpose the human has is to come increase the entropy of the universe and die okay if if let's say that because of our existence if universes entropy cannot increase then we would not have existed our existence is to increase the entropy of the universe to create pollution to create the randomness in the overall scenario I mean that may sound funny but then this is how it is okay so it is a natural process for every natural or spontaneous process for an isolated system should be in such a manner that randomness entropy increases in the system so you can say that there need not be isolated system okay there will be someone who can come and arrange things but then if you include entire universe as a system then nothing is outside the universe so entire universe is definitely an isolated system so whatever happens whatever happens inside universe the entropy of the universe must increase if you are doing a process okay you may say that it is not a spontaneous process you are doing it all right but if you consider entire universe whether you do it someone else do it you're inside the system whatever you do for the universe it is a spontaneous process okay let's say this mobile phone one atom is doing something inside this mobile phone on its own that atom will say no no it is not isolated I am doing it but what I will say it is inside this mobile phone it is happening on its own if one atom is reacting from the other atom I'll say it is a spontaneous process atom can't say no no I am doing it is not spontaneous okay similarly what do you do you may be exercising in the gym you might be burning some fuel you might be running across whatever you do it is a spontaneous process for the universe and it will happen only when entropy of the universe increases fine so this is actually I have not given any statement as such because it is very difficult to bound the second law of thermodynamics with words all right so what various scientists they did they have tried creating their own statements as per their understanding fine so there are second law of thermodynamics statement from the kelvin plank second law of thermodynamics statement from the clausius okay there are many other statements are there but these two statements are mentioned in your ncrt so let us quickly talk about it and then the theory of the chapter gets over I think in chemistry there's a formula for the entropy also in physics there is nothing like that write down kelvin plank statement no process is possible whose sole result is to absorb heat from the reservoir convert it entirely into work it will never happen that you can convert the entire energy into work heat energy I'm talking about some energy will be released to the universe as a tax to increase the entropy okay so this is how it is you can never convert 10 joules of heat energy into 10 joules of work this is what this is as per the second law of thermodynamics okay second statement is clausius write down no process is possible whose sole result is to absorb heat from the cold reservoir transfer it to a hot reservoir it can't happen spontaneously that you take heat from the cold and transfer it to the hot reservoir fine so I mean if you find out some of the there are a lot of documentaries on second law of thermodynamics and many interesting things are there so I will send one link also in case you have some free time do watch those documentaries there are many interesting stories behind second law of thermodynamics fine so this is what this entire chapter is all about now we can solve the remaining numericals and by the way do you have any doubts anyone can you tell some documentaries documentaries I'll forward on the group itself no doubts let us solve the remaining questions do this in heat engine for piston to move up to increase pressure how is the energy supplied inertia it is due to inertia okay once the wheels start rotating it will rotate quarter circle then remaining part it would rotate on its own okay done anyone sharduli got it siddharth got it work done by the gas in b2c it is a positive work done or a negative work done everyone positive or negative it's negative right sharduli look at this volume is decreasing or not right so it cannot be positive all right so let us do this question we have a cyclic process abca two moles of idle gas a total of this much julie is withdrawn from a sample a to b is what process which process a to b is all of you agree it is isobaric right okay done by the gas during the process b to c and a to c is isobaric so what is given is this much heat is withdrawn total heat withdrawn is that much so q out minus q in well let us do it like this total heat withdrawn in a cycle what it is plus 1200 should I write or minus 1200 here for a cycle it is given as minus 1200 okay and this is equal to w or not for entire cycle right that is w so w is work done in a to b plus work done in b to c fine plus work done in c to a this is equal to delta q which is minus 1200 fine so c to a the work done is zero all right a to b is isobaric process for isobaric process the work done is p delta v properly do it in one or two step we'll get the answer there is nothing in this chapter other than first law of thermodynamics to use every time you have to use first law of thermodynamics only now here delta the volume information is not given temperature is given so instead of p delta v what should I write here because temperature I know what I can write correct n r delta t now you know that a to b volume is increasing so work done has to be positive so it will come out to be positive only but if you know beforehand that it has to be a positive quantity it will feel more confident about it so 1200 minus n is 2 r is 8.3 I hope you know that the value of r in si unit is 8.3 final temperature 500 minus 300 how much you're getting this anybody done like this anybody did like this no one three into two 200 minus 4520 joules but I was thrown out of the meeting because of the internet issue my voice is clear I'm using mobile phone network now the wifi has some issue now okay I'll keep my video off because it is the mobile phone network may see some lag right so heat is withdrawn withdrawn from the gas all right so it means gas is releasing heat is it clear sardar is it clear now let's move forward these are not difficult questions fine you must be comfortable with this you might be not very comfortable right now because you haven't done practice right so a little bit of practice will be comfortable do this something others cc is cubic centimeter cubic centimeter cube si unit is meter cube i'll do it now k pascal k is 10 is for 3 this is 50 into 10 is for 3 pascal oh sorry i have to switch off my video now my voice is not breaking right setu setu doesn't want to see setu are you able to hear me still it is breaking for everyone my voice is breaking all right i'm back with a better internet connection if somebody now says that my voice is breaking look at the speed okay so my voice should not break all right so let us solve this we have you're going from a to b through acb then through adb all right whether you go from acb or adb what is that something which will remain same delta u will be same delta u is same okay so for acb process all right for acb process delta q will be equal to delta u plus w w 1 let's say this is delta q 1 50 calorie of heat is supplied all right so 50 calories how many joules 50 into 4.2 this is equal to delta u plus w 1 this is your first equation okay then you need to find delta q 2 this is equal to delta u which is same over here also plus this w 1 is what w 1 is area a1 which is this area okay the other area is this blue one okay let us name these two areas a1 and a2 so when you go from adb the work done is a1 plus a2 all of you agree the sum of two areas how do i get 4.2 or if one calorie is how many joules say two one calorie is how many joules 4.2 you have to convert this in si units right you're using si units so i am interested in finding delta q 2 is this clear to everyone the way i have written here fine now subtract so 50 into 4.2 minus delta q 2 why i am subtracting because wait this is a2 a2 right lower area for acb this is for acb lower area why i am subtracting because delta u is not given to us when you subtract it will be gone this is equal to minus of a1 now delta q 2 would be equal to 50 into 4.2 plus a1 a1 is the area of that rectangle which will be equal to 155 minus 50 into 10 raised to power 3 this is that length multiplied by this which is 400 minus 200 200 cubic centimeter 10 is from minus 6 all right this is a1 which you can substitute here and you get the value of delta q 2 is this clear how you get the final answer should i tell you the final answer the final answer is 231 joules are you finding these questions difficult let me ask you that let me take a poll are these questions forget about that carnot engine derivation this question let's say this question is this question difficult yes or no this question only is it difficult all right ending the poll now so this is what i have some of you are still finding this kind of question difficult so what do you do you know i can give you very simple questions that will make you very happy and i know how to make you happy right i'll just give you so simple question you'll you'll be like oh i am able to solve everything you'll feel good about yourself but then when the exam comes whether you write ct level or j man or neat whatever level you write these kind of questions will come you'll not be able to solve them so it is better to do the these kind of questions okay so that you understand what you can expect later on no point doing the simple ones okay what is 1 plus 2 what is 3 minus 4 okay that that kind of stuff those days are gone all right so just follow the schedule do your assignments okay and pay attention in the class it's very easy like i can give you so simple question everybody can get the answer but that will not help anyone isn't it temperature of the gas at point b is what tell me the value tell me the value of temperature at point b answer you tell me that is wrong tell me the value i sadly got it i know what kind of cellular you guys will be making okay that is why i asked the values all right so let us see and you'll get minus one because you have made a basic mistake here of not converting Celsius into Kelvin and i know that this is a trap okay so the people who are setting the question paper they know you better than you know yourself just like what i know that kind of cellular you will that is why i was asking tell me the value you're telling two times of ta right this is equal to vb by tb so tb is equal to vb divided by va by ta into ta okay so vb by va is this two times ta now there is a trap here convert temperature into Kelvin it is 300 Kelvin so this is how much 600 Kelvin in degree Celsius what is the temperature 600 minus 273 okay so this is 327 degree Celsius few of you were telling 56 degree Celsius because you multiplied 27 by 2 anyways do the b part what we'll do is that b part one by one we'll find out first tell me heat absorbing a to b what it is assume cv to be 3r by 2 and cp 5r by 2 say 2 got it others 30 second the unit should be joules right time more as i you want that the second should i like not give you the assignment and tell you to complete the previous assignment only and i'll ship the due date will you all do that assignment do the previous assignment then completely all right let us try that those i'll do that the previous assignment since not many of you have done it do it i'll not send the solutions of those do that only let me change the due date of that assignment that is your next week's assignment also there it is so i'll put it as 10th but don't wait till 10th okay completed quickly you have anyway some exams coming up right all right so q ab it is what isobaric process all of you agree v by t is a constant it's an isobaric process so it'll be n cp delta t this is 2 cp is 5r by 2 delta t delta t is tb minus ta which is 600 minus ta 300 this is the answer 1500 times r anybody got this answer say 2 got it that good then b to c what is the heat absorbed in b to c chargely again and again isothermal process i think you're telling third time that heat absorbed with zero in adamanty heat absorbed with zero why are you telling isothermal heat absorbed with zero every time so that got something heat absorbed in b to c it's an isothermal process delta u is zero so whatever heat is absorbed is equal to the work done all right so this is equal to our temperature at b log of vt by vb now vt by vb is not known or is it known yes you can divide this equation from that one all right delta u is zero because delta t is zero it's an isothermal process 2 r tb is 600 log of 2 because if you divide this with that you'll get vd by vb as 2 that is 1200 r natural log 2 anybody got this so this third part is the homework try doing it yourself let me see if i have something new just do the first part that is enough for today cv is 3 r by 2 cp is 5 r by 2 that's given actually if you learn kinetic theory of gas just because it is given it is monoatomic so it is understood cp and cv values are this do you know the formula for the work done adiabatic process correct no no no minus and cv delta t minus so what what is not given to this temperature at b everything else is given so the question boils down to finding the temperature at b just tell me the value of temperature at b what it is the problem is solve them say to got some temperature and and that is not correct others say to check for the silly errors little time one minute given okay now let us do it quickly so what all things are given to us pressure at a is given pressure at b is given temperature at a is given temperature at b you have to find out so what we have is pv raised to power gamma is a constant okay even v1 gamma is equal to p2v2 gamma i want the relation between pressure and temperature okay then i can easily get it so i know that pv is equal to nrt so v is v is equal to nrt by v this is a constant so v is equal to nrt by p so t to the power gamma divided by t to the power gamma minus one so this is equal to t to the power gamma divided by t to the power gamma minus one all right so t2 which is tb is equal to p1 by p2 gamma minus one divided by gamma times t1 okay so this is how you get the value of t2 so i will tell you the final answer for the t2 t2 is where it is 850 kelvin is what you'll get t2 okay not a problem even though you got a wrong answer way better than not doing anything so do the assignment trust me if you do the assignment immediately like today tomorrow and day after in these three days if you complete the assignment you will have to never study this chapter till grade 12 gets over okay so these simple chapters you have to be a little aggressive like this all right so that's it from my side we will meet next week and see whether we have to do SHM or kinetic theory by that okay bye for now