 those who are online please type in your name on the chat box please type in your name on the chat box those who are online am I audible welcome all of you will wait couple of minutes for others to join in meanwhile if you have any doubts from whatever you have done please WhatsApp me your doubts I will be taking it up first okay in case you have any doubts please WhatsApp me and I'll be taking it fast okay meanwhile I'm waiting for others to join actually I had received one of the questions as doubt Snigda I think you have sent me the doubt can you send me another pick off your doubts because okay the audio is not very good all right now it will be okay who is then then this must be Lalitha and don't mesh you guys know right today we are going to revise thermodynamics and thermal properties of matter okay all right so nobody's asking any doubt okay all right fine so here is one of the question student has asked I'll just you know tell you what this question is and then we'll take it up okay so all of you please write down there is a hundred kg block okay hundred kg block and it started with a speed of the speed of this block was 2 meter per second okay on a long rough belt which is kept fixed in a horizontal position the coefficient of kinetic friction between the block and the belt is given as point to those who have etsy Verma this is question number three okay from etsy Verma so hundred kg block it is it has started with a speed of two meter per second on a long rough belt which is kept horizontally coefficient of friction is point two okay you need to find out consider the situation from a frame of reference of frame of reference moving at two meter per second so there is a frame of reference that is moving two meter per second along the initial velocity of the block okay so you have a reference frame that moves with two meter per second along the direction of the velocity of the block as seen from the block the block is gently put on the moving belt and in due time the block starts I have received only half of the question it starts with the belt at two meter per second so you have placed this block on a moving belt this is the moving belt you have placed a block on the moving belt and moving belt was also moving with two meter per second okay you need to calculate the increase in kinetic energy of the block as it stops slipping past the belt okay now since your frame of reference is moving with two meter per second you have to take the relative velocity okay with what velocity you are moving fine so with respect to that whatever is the initial velocity and whatever is the final velocity you can identify so it has to be a relative velocity when you are observing an object and you are also moving okay you can do like that okay welcome tapas se ho ja lau etia okay atmesh is asking fine good atmesh so this particular situation all of you please listen to this you have a bimetallic strip okay so this is a situation this is metal one okay and let's say this is metal 2 okay the coefficient of linear expansion for this metal is alpha 1 and for that metal is alpha 2 okay and if alpha 1 is greater than alpha 2 and you heat this bimetallic strip and you increase its temperature by delta t what will happen if you increase the temperature by delta t what will happen to this bimetallic strip let's say the strip length is l and this width is d and this is also d what will happen all of you is answer if I heat it if I heat since alpha 1 is more than alpha 2 this metal the upper metal will try to expand more and the lower one will try to expand lesser than the upper one okay so ultimately the situation will be like that okay so it will become an arc sort of scenario like this all right so we need to find we need to find what is the radius of curvature so this length is what this is what we need to find this is the radius okay how much is this all of you please try this then I will tell you how to go about this spend some time on it we need to find the radius of curvature of the bend of this bimetallic strip okay the hint is I mean you assume this to be an angle alpha okay then alpha times the radius of curvature will be the length now see here you can take the midpoint of the first rod okay so this length this length will be equal to r minus d by 2 okay so I can say that the length of metal 2 has gone from l to r minus d by 2 times alpha okay and for the second metal this is for metal 2 okay the lower one and for the second metal for metal 1 it will be r plus d by 2 alpha getting it so you have change in length for both of the metal rods okay and the force between these two the force between these two is the internal force okay so the stress the amount of stress they both will have same okay so sigma for metal 1 is equal to sigma for metal 2 because that is an internal force alright the whatever force the metal 1 applies on metal 2 that's the same amount of force metal 2 will apply on metal 1 okay so the force penalty area for both the rods will be same okay so now I guess you can do it from the definition of Young's modulus which is what stress divided by strain okay strain is delta l divided by l okay now there will be a small issue here because delta l is not the change in length from its original length okay delta l is a change in length from what will what should be its length at that temperature okay so it's time it's length at its length at the temperature when you have rose the temperature by delta t will be equal to l is equal to l not 1 plus let's say x delta t where x is coefficient of linear expansion fine so you measure delta l with respect to this length okay so from this length it has gone to that length delta l will be difference between these two lengths fine so you can use these equations to solve it okay I hope at least you have got some hint about how you should go about this particular question okay this is not a very straightforward one but one of the question which is you know which is like routinely you know discussed everywhere okay so now should I start with the previous here j questions or you have any doubt you can type in yes or no you have any doubts type in yes if you don't have type in no okay so let's take no previous year j question these are the actual questions which came in in j exams and these are like very very old also starting from 1970s you know from then on so let's take up this question from 1983 it refers to a diagram all of you please draw this this is p this is v this point is p comma v that is to p comma v this is to p comma to v that is p comma to v all right so this is a PV diagram an ideal monotomic gas is taken around a cycle a b c d says a b c and d okay the work done during the cycle is what work done in this process is how much all of you they start out work done by the gas in this process is what all of you getting PV great so the work done by by a gas okay work done by a gas in a cyclic process okay and if it is a PV graph will be the area of that cycle fine so you just find out the area of this loop that is the magnitude of the work done okay now whether that work done is positive or negative it depends on at higher pressure what is happening fine you can see that at higher pressure volume is increasing and lower pressure volume is decreasing okay so work then will be positive and its value will be equal to the area of this loop so it will be plus p into v okay so this came in 1983 so those kind of questions used to come set point in time okay next one from 1988 all of you one mole of monotomic gas okay one mole of mono atomic whose gamma is 5 by 3 which is given alright it mixed with one mole of diatomic gas you are mixing it with one mole of diatomic gas gamma for the diatomic gas is 7 by 5 alright you need to find a mixture when the mixture is getting formed what is the value of gamma for the mixture fine what is the value of gamma for the mixture how much it is you have four options 0.4 0.5 1.53 and 3.07 which one okay loyta is telling 1.53 others loyta that's not correct others okay so these kind of question first you need to find the CP for the mixture CP for the mixture is n1 CP 1 plus n2 divided by n1 plus n2 okay then you get the value of CV n1 CV 1 plus n2 CV 2 divided by n1 plus n2 so this is CP for the mixture and this is CV for the mixture so CP by CV for the mixture becomes equal to what it is n1 CP 1 plus n2 CP 2 divided by n1 CV 1 plus n2 CV 2 okay so n1 is equal to n2 is equal to 1 so you get what CP 1 plus CP 2 divided by CV 1 plus CV okay so this is how you get this and since gamma is given alright so for monotomic you can get the value of CP CP 1 is gamma r divided by gamma minus 1 and CV 1 is r divided by gamma minus 1 similarly here also I mean this is gamma 1 okay so CV 2 is r divided by gamma 2 minus 1 CP 2 is gamma 2 r divided by gamma 2 minus 1 I am sure many of you might have just taken the average of gamma okay that's not how you should do it you have to take weighted average of CV weighted average of CP and then take the ratio alright so now when you substitute CV and CP values you'll get the value 1.5 for this any doubts on this particular question kindly message while I am moving to the next question not that you're clear now let us move to next question all of you please draw this what is the value of r in CP and CV shares r is universal gas constant that is approximately equal to 8.31 okay if you want to write r in terms of SI units so r is 8.31 but when you substitute one good thing is that r will get cancelled out we're taking ratio so r will come in numerator as well in denominator so r will anyway get cancelled out now here is a situation three rods made of same material okay three rods made of same material and have the same cross-section fine so let's say conductivity is k so that same conductivity is there for all three rods okay so they are joined as shown in the figure fine each rod is of same length the left and right ends are kept at zero degree Celsius and 90 degree Celsius respectively the temperature of the junction is what what is a temperature of the junction anyone got the answer this is from heat transfer how the heat transfer will happen heat will be conducted from 90 degrees Celsius and we'll go like this okay let's say this is dq1 by dt this is dq2 by dt okay any other answer by anyone and let's say this is dq3 by dt so at the junction amount of heat that is coming in should be equal to amount of heat that is going away alright so dq1 by dt is a rate at which heat is coming in towards the junction plus dq2 by dt so this is total heat coming to the junction this should be equal to total heat going out from the junction fine so dq1 by dt will be equal to kA now of course T will be less than 90 degree because heat is coming from 90 and going towards 0 so in between temperature will be between 0 and 90 degrees fine so this is kA 90 minus T divided by L this dq1 by dt fine this plus dq2 by dt the expression for that remains same yes Rohan I guess that's correct 90 minus T this is equal to dq3 by dt now dq3 by dt is kA by L T minus 0 heat is flowing from this point to that point so this point temperature is T and that point is 0 okay so few things to notice here first of all we haven't converted temperature in Kelvin that is not required because we're dealing with delta T difference in temperature that is same in Celsius as well as in Kelvin okay and second thing is we always subtract lower temperature from higher temperature okay because we already know the direction fine so we don't care about positive or negative thing so like that if you solve it you get 3 T is equal to 180 degree and T will come out to be 60 degrees okay so this was asked in 2001 mains you can say okay alright so this was the third question right next question guys this is again from 2001 only for an ideal gas if T w is 0 for an ideal gas if the work done by the gas is 0 in that okay sir wouldn't we take the two arms to be in parallel okay lawyer see okay if you are using thermal resistance then it will be like this it will be like that this point is 90 degrees Celsius and this point is 0 degrees Celsius fine lot yeah so again the heat current that is coming from this side will be equal to the heat current going that side so probably you might have made some silly error okay the same situation yeah they are you can consider them to be parallel fine so coming back to this particular question so in a process dw is equal to 0 and T q is less than 0 fine then for the gas which of these four statements are correct delta T will be negative as in the temperature will decrease second delta v is positive as in the volume will increase pressure constant or I can say that temperature will increase which one can I say conclusively like that I hope your doubt is clear else we can message again also a others at me is saying peace constant fine so see if temperature changes one thing you can conclusively say that the internal energy will change because internal energy is a function of only temperature fine energy is a function of only temperature now if dw is equal to 0 and if I use first law of thermodynamics dq is equal to du plus dw since dw is 0 I can say that du is equal to dq and since dq is less than 0 du is also less than 0 fine and since du is less than 0 right and du is also equal to n CD date dt are you getting it this is a formula for idle gas it change in energy this remains unchanged doesn't matter what is the process fine so if du is negative dt has to be negative fine so that is the reason why temperature will decrease okay fine any doubt on this particular question then I'll take up what loyta is asking can you first message any doubt on this particular question question number four okay fine so what loyta you are saying that this is thermal resistance this is thermal resistance and this is also thermal resistance so these two thermal resistance are in parallel right the thermal resistance equivalent for these two becomes r thermal by two okay this thermal resistance is connected in series with that thermal resistance so total thermal resistance will be this so I know this temperature and I know that temperature so dq by dt if I use thermal resistance will be equal to change in temperature that is 90 minus 0 divided by 3 thermal resistance by 2 3r by 2 these two some okay this should also be equal to the heat that is flowing from here to there okay so this will be equal to t divided by r thermal also so again you are getting 60 degrees only from these equations fine loyta don't skip the steps all of the steps you are doing it correctly you will get the answer okay don't skip you can skip in the exam when you have practice told you loyta before itself guys it is very easy to score minus one okay even if you know how to solve you can easily go wrong and score minus one you may be feeling very happy while solving it but as soon as you come back home you find out the silly errors you'll you'll feel very very bad about it okay so be careful okay next question guys so after couple of questions I will start taking up slightly difficult type of questions also because the learning happens from the difficult questions only alright guys so please draw this diagram first this is volume and this is pressure okay this is 1 is 2 this is 10 Newton per meter square this line represents that fine this is how the process is going on is a that is B and this is C now listen to the question carefully the volume is in meter cube by the way so both are SI units only x and y axis this came in 2002 when I wrote J fine so this is a cyclic process a b c a if the net heat supplied to the gas in the cycle in the total cycle net heat supplied to the gas is five joules fine the work done by the gas is what this is supplied you need to find out work done by the gas how much it is work done by the gas in C2 a process guys sorry I missed it yeah I'm not finding total work done I'm finding work done by the gas only in the process C2 a okay anyone got the answer anyone should I do it you have four options minus 5 minus 10 minus 15 and minus 20 you look at C2 a process the volume is decreasing okay since volume is decreasing work done has to be negative alright usually we have pressure on the y axis but this time it is reversed no Rohan that's not correct see if you apply first law of thermodynamics to the entire cycle delta q is equal to delta u plus w okay if you apply this for an entire cycle delta u is 0 for the cyclic process right so delta q will be equal to total work done fine so this is work done in process a b plus work done in process b c plus work done in process c a okay now heat supplied to the gas is five so this is five okay and work done in b to c process this is 0 right because it's a constant volume process Sayuja this is like this cycle is made up of multiple processes a to b is a constant pressure b to c constant volume okay c to a is a different process okay so work done b to c is 0 and work done in a to b is easy to find how it just pressure into delta v it's a constant pressure process so work done in constant pressure process is the delta v this plus work done c to a okay and p delta v for a to b is what pressure into 1 it goes from 1 meter cube to 2 meter cube so delta v is 1 2 minus 1 so this is 10 joules the work done in a to b is 10 joules so work done in c a plus 10 is equal to 5 so work done in c to a process is equal to minus 5 joules okay now one more question this I guess is question number 6 alright so water there is water of volume 2 liters okay in a container that is heated with a coil of it is heated with a coil of 1 kilowatt at 27 degree Celsius okay the lid of the container is open and energy dissipates at the rate of 160 joules per second this much energy is going out because a lid of the container is open reading it in how much time temperature will rise from 27 degree Celsius to 77 degree Celsius okay you can take specific heat of water to be equal to 4.2 kilo joules per kg okay any doubts with respect to understanding the question please type in if you have any doubts if you don't have any doubts please start solving it every time you do a physics question make sure you draw some diagram heat is going out at a rate of 160 joule per second okay and from the heating coil the heat that is coming in is what heating coil is 1 kilowatt so that is 1000 joule per second heat is coming in okay now try solving it this is 2 liter 2 liter as in 2 kg of the water loyatia are getting 500 seconds others loyatia can you convert it into minutes because the options are in terms of minutes these are the options I'll write here 7 minutes 6 minutes 2 seconds 8 minutes 20 seconds and 14 minutes which one 500 seconds is like is it 8 minute others what are you doing others yes loyatia that's correct okay and others fine so let me now try solving it so water is absorbing heat at what rate the water is absorbing heat at a rate of what it receives minus what it rejects this is 840 joule per second water absorbs the heat fine and if water has to change its temperature from 27 to 77 then how much heat water has to absorb then it that will be equal to MS delta T mass of the water is 2 kg so 2 into specific heat which is 4.2 kilo joule so 10 is for 3 this is MS into delta T delta T is 27 to 77 that is 50 this should be in joules fine so if this happens in T seconds so this will be equal to 840 into T because this module per second is the energy absorbed fine so if you multiply that with time you will get total energy absorbed so these two should be equal because whatever energy is getting absorbed that will lead to change in temperature getting it all of you clear okay guys so let me take up next question I suppose no doubt from the previous one you might be able to see a question on your screen try solving this if you have any doubts on the previous question please type in straight away try solving this question should I do it all of you tried right like that don't don't watch on phone okay it will strain your eyes use laptop or desktop three dots of material x and three are of material y are integral length and cross section area but the conductivities are different let's say kx and ky and a is maintained at a temperature of 60 degree Celsius let us say okay and E is maintained at 10 degree Celsius calculate the temperature of junction B C and D thermo conductivities are given so you can you know convert this entire circuit into this you can draw parallel between heat transfer and the current electricity chapter fine so this temperature is 10 degree Celsius this is 60 degree Celsius this is let's say thermal resistance for y is R1 so this is R1 this is R1 and that is R1 and thermal resistance for x is R2 so this will be R2 R2 and R2 okay do you see a Wheatstone bridge over here yes or no do you see a Wheatstone bridge so you can remove this R2 fine so heat will go like this and it will transfer like that fine so correct so this will be equivalent to R1 and then two times R2 and this is two times R1 like this okay so you can just analyze you know you can say that dq by dt is like current and R is thermal resistance which is equal to L by KA all right so I hope you can do it now temperature temperature is like potential so temperature difference or potential difference you can analyze it like current electricity chapter and you'll get the answer now this temperature will be temperature of B all right okay lawyer got the answer 18.66 at B not exactly lawyer that's not correct so I'll just tell you the answer so you guys try your own and let me know if you do not get the temperature at B once you analyze it should come out to be 30 degrees Celsius temperature at C should be equal to temperature at D that is 20 degrees Celsius okay because these two temperatures are same that is the reason why there is no heat flow on this and that's why it is useless fine so let us move to next question guys all right so you can see there is another question on your screen attempt this this came in 1980 during those times J used to be a subjective paper and why we are doing subjective questions because one subjective question can test multiple concepts so solving one subjective is equal to solving let's say four or five objective ones so that's the reason why we are solving the subjective ones yes they represent change in state the first part of the question AB is a change state from solid to liquid and CD is from liquid to vapor okay you have to specify because from solid liquid state will come and then only the gaseous state will come like that impressive today others guys you guys are not answering only loyalties answering everything it doesn't matter where right or wrong participate all of you clear right that the hundred zone AB and CD since temperature is remaining constant since temperature is constant so AB and CD must be change in state okay A to B could be change in state from solid to liquid okay and C to D you can say from liquid to vapor fine now if CD is equal to two times AB if CD is equal to two times AB what does it mean same amount of substance absorbs more heat to completely convert from liquid to vapor this is more heat than this one right the width of this is the heat that is supplied so you need to supply more heat for this substance to go from liquid to vapor compared to if it has to go from solid to liquid that's what it means okay so since you need more heat to convert from liquid to vapor than from solid to liquid that is why latent heat of vaporization is more than latent heat of you know fusion okay so 2 is done 3 slope of DE what is DE this one is DE slope of DE now slope is what slope is this divided by that okay this is Delta T and this is the heat that is supplied Delta Q the slope is what Delta T divided by Delta Q yes loyalties you are in a hurry so you may go wrong so be careful so we know that Delta Q is heat capacity C into Delta T I am not talking about specific heat capacity specific heat capacity is per kg but if I just talk about heat capacity then I just have to multiply by Delta T to get the value of heat supplied okay so Delta Q by Delta T is heat capacity so Delta T by Delta Q which is slope represents inverse of heat capacity okay now the slope OA is more than slope of BC what does this indicate so it means that one by heat capacity of solid okay that is slope of OA right this is more than one by heat capacity of BC is liquid so I can say that heat capacity of solid is less than heat capacity of liquid so like this you have to solve this particular question let's take up the next one all of you please try the question number sixth this question I hope you are clear with the previous one if you have any doubts feel free to message me you can message on the group you can message me personally WhatsApp whatever don't just if you have any doubt don't sit just like that we are doing question number six please try this and message as soon as you get the answer you have to draw the diagram so you may not be able to message that diagram but just say done once you're done a to b process passes through the origin okay that's the hint okay people are messaging me over WhatsApp their answers okay are you guys done type in are you guys done trying at your end okay let's see now this is a VT diagram on an idle gas show the same process on a PV diagram so on a PV diagram we need to draw this these are the two accesses so let's take P on the Y access and V on the X access right good let's see so a to b is what let's start from a a to b is what process what kind of process a to b is this line it's a straight line right whose slope is constant a straight line slope is constant so and also this straight line passes to the origin and the slope will be then V by T so V by T is a constant it means that pressure is constant fine so a to b is a constant pressure process okay and you can see that the volume is increasing from V1 it goes to V2 all right hence I can draw this is a constant pressure process so let's say this one this line represents a to b this is a okay and that is B now B to C is also very easy to draw B to C is a constant B to C is a constant volume process okay so even B to C is easy to draw now one thing is little tricky over here tell me from when you go from B to C the pressure is increasing or decreasing so let's say this is a and this is B should I draw B to C this way upward or B to C downward that's what I'm asking you B to C which is constant volume process should be drawn upwards or downwards B to C the pressure is increasing or decreasing guys type in your response B to C pressure is increasing or decreasing okay so definitely temperature is decreasing right temperature is something which is decreasing and this is constant volume process okay is a constant volume process so I can say that pressure divided by temperature is a constant okay so P1 by T1 will be equal to P2 by T2 from B to C are you getting it now see what happens here the temperature goes down so T2 is higher temperature and T1 is lower temperature okay so let's say pressure at B is PB so this is PB by TB this is TA and PA sorry this is C right so I'll write it as C and this is TC so PC is equal to TC by TB into PB getting it okay and TC is less than TB so PC pressure here PC is less than PB I hope it is clear now see it's like this you take a gas and you decrease its temperature keeping the volume fixed its pressure will decrease isn't it pressure increases when you heat it in a constant volume so it will not happen that in a pressure cooker you're cooling the pressure cooker and the pressure inside the pressure cooker increases that'll never happen right so pressure has to go down I hope all of you know so the pressure goes down so from B I move down like this constant volume process I go down and I reach C okay I hope all of you have understood this now from C to A what it is a constant temperature process so constant temperature process I hope you guys know is a curve like this in a PV graph okay so it will be like this okay so a this is B and that is C all of you clear about this type in yes or no all right let's take up the next one so now let's go to 1990s enough of 1980s all of you please attempt this particular question it's there in front of your screen all of you tried what is the answer for C first tell me C what is the answer yes that should be 0 it's a cyclic process right for every cyclic process the net change in internal energy has to be 0 yes so since change in internal energy is 0 you know from first law of thermodynamics I can get delta Q which is equal to delta U plus W I will get delta Q is equal to W getting it so the answer for A and B will be same fine so you can get get the answer for B first and then say that is what the A is also from first law thermodynamics we know work done in various processes right constant here we have constant pressure process and constant temperature process for both the processes we know how you write the work done okay and mind you this is not PV graph you cannot find the area of this graph and say that is the work done area of PV graph is the work done okay this is P and T graph are you stuck should I do it sure I'll just write this and then I'll pause total work done is some of all the work done find them one by one individually add up A to B is what process it's a constant pressure process fine so work done in A to B should be P into delta V okay and since it is constant pressure process P delta V is also equal to Nr delta T from ideal gas equation fine so why I'm writing it as Nr delta T because delta T is given okay so it goes from 300 to 400 we have two moles are delta T this time you have to maintain final temperature minus initial temperature you cannot take the you know you cannot just take the positive value of the work done because work then can be negative so make sure you use a formula as it is don't take a mod of it so 2r into what 400 minus 300 right so this will be equal to 200 R work done in A to B process fine work done in B to C how will you get this okay Shreyas got the answer let me check no Shreyas that's not the exact correct answer anyways at least you guys are trying so work done B to C this how will you find Nrt ln Vc by Vb right final volume divided by initial volume and since it is constant pressure process no it is not 0 since it is constant pressure process we know that Pb into Vb is equal to PC into Vc why I am doing all this because I don't have a knowledge about volume but I have knowledge about the pressure fine so Vc by Vb is equal to Pb by PC fine so this will be equal to Nrt let's say this is T2 Nrt2 log of Pb by PC alright this will come out to be N is 2R T2 is 400 log of Pb is 2 PC is 1 so this is log of 2 this will come out to be 800 or ln 2 getting it this work done in B to C now work done in C to D it is again a constant pressure process this will be again P into delta V only alright and P delta V will again be equal to NR delta T fine and in this case delta T is negative you're going from 400 to 300 so 300 minus 400 is minus 100 all right so this will come out to be minus of 200 R fine so you can see that work done in AB is equal in magnitude of work done in CD because the temperature changes same okay now let's talk about work done in D2A there's a last one this will be equal to again I'll be using this formula and since it is D2A it will be equal to NR T1 log of Pd divided by Pa okay now Pd is 1 and Pa is 2 so it will come out to be N is 2 this is R and T1 is 300 log of 1 by 2 this will come out to be minus of 600 R ln 2 all right so when you add up everything so when you add up WAB WBC WCD and WDA this will get cancelled out and when you add these two you'll get the answer as total work done is equal to 200 R ln 2 the value of ln 2 is very popular everybody knows because it comes in the radioactivity chapter which we have done right this is 200 into R which is 8.31 into log of 2 is 0.693 okay so this is the answer for the work done and this is the amount of heat also you have to supply in this process okay anyone have any doubt if you please type in yes or no if you have any doubts or just type in no doubts or is it okay okay fine let us take a similar question now try this out all of you we are doing this question on the screen for adiabatic process I'm sure you know that V1 V1 gamma is equal to P2 V2 gamma and T1 V1 gamma minus 1 is equal to T2 V2 gamma minus 1 so these are the process equation for adiabatic process so I just thought I could write it down you can answer the first part initially okay if you get the answer to the first part please message I was hoping you guys will come with better preparation than this you guys are stuck should I do it okay so I guess I will do it fine so a to b is adiabatic expansion we need to find work done by the gas in process a to b so if I just take process a to b for process a to b okay if I apply first law of thermodynamics delta q is equal to delta u plus w if I apply this delta q is 0 because it is adiabatic process fine so work done will come out to be negative of change in internal energy fine and negative change in internal energy will be equal to N this is okay fine and CV for a gas is r divided by gamma minus 1 this into delta T right now you can further simplify it and write down it as N r delta T is what TB minus TA divided by gamma minus 1 okay now number of moles is 1 N is equal to 1 r is known gamma is is it known it's a monatomic gas so you should be knowing that gamma for a monatomic gas is 5 by 3 so gamma is known r is anyway known do we know TA TA is also known okay but TB is not known TB is something which we don't know right so if I get to know TB then I'll get the answer right but one thing I know that PB is equal to two-third of TA all right this is something which I have to use now you have two equations for adiabatic process all right now this is relation between pressure and volume this is a relation between temperature and volume okay so if I get a relation between temperature and pressure okay then I can relate temperature at a with temperature at B then I'll be able to get TB okay so how we can do we can say that PV raise to power gamma is constant let us say C1 so I can write down P raise to power 1 by gamma into V is a constant let us say C2 and TV raise to power gamma minus 1 is a constant let us say C3 so T to the power 1 by 1 minus gamma my gamma minus 1 into V is a constant let's say C4 okay now if I take equation 1 and equation 2 and divided what will happen volume will disappear and I'll get a process equation P raise to power 1 by gamma divided by T raise to power 1 by gamma minus 1 is a constant okay so PA TA this is equal to PB 1 by gamma divided by TB 1 divided by gamma minus 1 okay so from here I'll get the value of I'll get the value of TB all right and then I will use it over here to get the answer for the work done all right there's a negative sign also here make sure you take care of that fine any doubt on this anything you'll get the value of TB get TB from here and substitute it there okay we know that PA by PB is 3 by 2 so you can simplify this further and gamma is 5 by 3 any doubts anything please message any doubts any doubt guys no lord yeah any doubts have you understood please type in okay now part B the heat lost by the gas in the process B to C now B to C is what it's a constant volume process fine B to C is a constant volume process and I need to find how much heat is lost in constant volume process so in a constant volume process delta Q is equal to delta U because work done is zero right and delta U is equal to N C V delta T okay so again here it will be N R divided by gamma minus 1 times delta T so it will be TC minus TB okay so from here you'll get TB at around 850 okay this is 850 Kelvin you'll get all right now again here TC is not known you need to know what is TC in order to know how much heat is lost from B to C okay so this since this is constant volume process I can say that PB divided by TB is equal to PC divided by TC okay so I know that see from here we have PB in terms of PA and PC in terms of PA so if I divide PB and PC what will I get PB divided by PC this will come out to be two-third divided by one-third this will be equal to two okay so now I know the the ratio of the pressures fine so using that I can get the value of TC okay so you get TC from here and then substitute there to get how much heat will be supplied all right fine now the last part of this question to find the temperature TD fine so all of you after the class get the value of TD I'll send you this this sheet which I'm scribbling on so you you might have also noted down the question so get the value of TD and message me in case you have any doubts okay all right so we'll move to next question time just runs away we'll take this one 32 this came in J 2002 again when I wrote the J my time start doing this question all of you okay guys this is from kinetic theory of gases sorry for that we will take up a different question okay today we are doing only thermodynamics and thermal properties sorry for that we'll take next question give me some time okay this one all of you try out this one please message if in case you get stuck okay don't just sit party anybody got the answer for party lohita as usual again again again again okay three yes lohita you're correct this time congratulations others please answer all of you getting 600 TA is given we need to find TB okay so it's a straight line that passes through the origin so VA by TA is equal to VB by TB okay the way these guys test you I mean it's amazing that tests you at each and everything they will intensely write the temperature in degree Celsius so we'll just double it and you'll say that 54 is the answer okay which will be there in the options and that is wrong so TB is equal to VB by VA into TA now VB by VA is 2 so it should be two times of temperature at A but two times of temperature at A in Kelvin not in degree Celsius so 2 into 273 plus 27 this is 600 Kelvin okay okay now can you find out the heat release and absorbed in each processes this is A part we'll do the B part and C part will be the homework all of you WAB where it is asked WAB but I read the question it asks for heat absorbed and released not work done anybody got any of the I mean even if you got the process from A to B what is the heat absorbed or released you can type in your answer A to B or B to C whichever process you got the heat absorbed or released QC to D no lohitya QAB no okay guys listen here in A to B WAB is what WAB is a constant pressure process right so it is what W in A to B process should be equal to P into delta V and since it is constant pressure I can write it down as NR delta T remember it is two moles it's two moles C D no atmash all right so work done in A to B process is NR delta T I hope you have understood this so N is what 2 into R into delta T delta T is what TB minus TA so 300 to 600 so this will be this so 600 R is WAB plus delta U of A to B delta U is what N C V delta T okay which is equal to what this will come out to be 600 R plus N is 2 no they don't retract their answers that's fine divided by gamma for monotomic monotomic gamma is 5 by 3 so this is 5 by 3 so what do we do we'll get plus 900 R all right so this will be 300 R we just give what happened here back again at 15 8.13 this we have heat absorbed and as goes any doubts on this just look at it and tell me if you have on this question any doubts please message in case you have any doubts on this question guys I'll meet you in another five minutes okay I hope all of you are there am I audible please message am I audible kindly message okay thank you so no doubts on this particular question right no doubts so we will take up another question so give me moment I'll download the question read this question all of you please read this question then I'll write down what is asked okay guys so these are the two parts of the question that is asked see hot oil is circulated like this okay since hot oil is circulated like that there might be heat transfer from this side hot oil will transfer some heat to this and once it is getting transferred after that the heat this is the conduction that is going on okay and after that radiation goes on did you get the A part part A I think part A is something which is doable easy anybody got the answer for part A what is asked due to radiation what is the heat loss from the lid okay so the radiation is a surface phenomena the temperature of the top of the lid that is only the surface from where radiation can happen and the surface temperature is given fine so dq by dt because of the radiation should be equal to sigma ea temperature of the lid ratio power four minus temperature of the atmosphere ratio power four getting it so rate of heat transfer per unit area will be just equal to sigma e temperature the lid which is what 127 plus 273 that is 400 ratio power four minus atmospheric temperature 327 degrees Celsius right 300 ratio power four so just substitute the values you'll get the answer okay so when you substitute the values sigma 17 by 3 10 is power minus 8 e is 0.6 and this bracket term can be written as 400 square plus 300 square multiplied by 400 minus 300 into 400 plus 300 okay so you can simplify this and get the answer for heat transfer per unit area okay so this is the part a now can you try part b any doubts on part a you can message or start solving part b of the question you need to find temperature of the oil how much it is assume that temperature of the oil is unchanged okay because it we're talking about a steady state so the same temperature of the oil is circulated throughout assume that temperature of the oil is to okay so here the temperature is to okay and outside the temperature is 127 okay so although the heat loss is due to the radiation from this surface okay but before the radiation there's a conduction that is going on from the bottom surface to the top surface fine so dq by dt dq by dt due to radiation should be equal to dq by dt due to conduction right in order for it to be having a steady state it will be like this dq by dt due to radiation we have already got okay you just multiply this everything with a you'll get dq by dt okay so this is something which is known this will be equal to dq by dt due to conduction that is what k a temperature of oil minus temperature of the lid that divided by the thickness let's say thickness is t okay so a will come here also fine so this a will get cancelled out and you'll be able to find out the value of temperature of the oil okay any doubts on this question please type in yes or no or great so you got it we'll now move to next question see there's so many other question that we may not be able to take it so you have to solve many questions yourself i'll take another question this one start solving this question all of you lauette is getting the answer very fast let's see it is too fast smart lauette retract very fast yes it is 70 good that you got it very nice see here you have a cylinder of mass 1 kg that has been given a heat of 20 000 joules okay this much is the heat that is given at atmospheric pressure and initially the temperature is 20 degrees Celsius we need to find the final temperature okay now coefficient of volume expansion is given okay so coefficient of volume expansion will actually help you to find out how much change in volume has happened so delta v is equal to v0 coefficient of volumetric expansion into delta t fine and v0 is what mass which is 1 kg divided by density this into gamma into delta t this is delta v so work done by the cylinder in expansion against the atmosphere pressure is p into delta v fine so this will be equal to m0 p gamma delta t divided by rho so i can write down here delta q is equal to delta u plus work done right now you might have supplied 20 000 joules but some part of it is utilized to do the work against the atmosphere and only this much is used up to increase the temperature of the object fine so delta u which comes out to be delta q minus w okay this will be equal to m s into delta t okay so a delta t expression will come for w also so that you have to take right hand side and then you know you can get the value of delta t and once you get delta t the final temperature will be equal to the initial temperature plus delta t okay so i hope you have done like this the answer will come out to be 70 degrees Celsius now work done by the cylinder work done by the cylinder is this only which we have found out initially okay and change in internal energy of the cylinder which is heat supplied minus work done only work done you are getting 5000 atmesh let me check see this work done will come out to be so small it will come out to be so small that even if you equate delta q is equal to m s delta t then also you'll get 70 degrees Celsius only okay but then you know that's not correct so work done when you substitute everything you will get this to be equal to 0.0499 joules atmesh you might have missed a factor of maybe 10000 fine so this is how you do this particular question right so let us move to the next one you guys have any doubts type in yes or no this one all of you please try this okay how we get the work done in the previous question this one see work done is what work done is basically pressure into change in volume because it is a constant pressure process outside pressure is atmospheric and it's a very slow process so you can say that you know as you can assume it to be constant pressure process so work done is pressure into delta v now delta v you can get from coefficient of volumetric expansion when you change the temperature the object will expand okay so v is delta v is equal to v naught gamma delta t so just substitute here and p is atmospheric pressure okay so you'll get this expression for work done by the cylinder in the expansion okay so I hope I am able to clear the doubt in case the doubt persists please message me take up this question now all of you anybody got the answer anyone is able to get the answer see we don't know exactly what will happen so we will evaluate okay we'll evaluate how much heat can be lost by the steam if it has to go to zero degrees Celsius and how much you know heat is required by the ice which is at 250 degrees Celsius to go to water at zero degrees Celsius fine so ultimately what should happen that temperature of the mixture should become same fine so we'll evaluate how much heat is required and how much heat we have and depending on the evaluation we will arrive at what will be done as in what would have been the situation tell me what is the answer for the first one how much heat is lost by the steam to condense to water at 100 degrees Celsius all of you quickly how much is this okay the heat heat lost by the steam to come to condense at 100 degrees Celsius is mass of steam into latent heat of vaporization okay this will come out to be you have to take mass in kgs 1000 sorry 113 400 joules fine this is what you'll get and heat lost by 100 degrees Celsius water to go to zero degrees Celsius this will be what ms delta t so m specific heat 4200 into delta t that is 100 fine so this will come out to be equal to 21000 joules see the amount of energy that will be released if the steam at 100 degrees Celsius becomes water at zero degrees Celsius okay let's see how much energy is required for this particular process to happen you know right that the heat is getting transferred as in steam is losing heat and this heat is getting transferred to ice ice is absorbing heat okay stress is getting on 113.4 yeah that is kilo joules stress okay third third one is mass of the ice which is 0.45 into specific heat of ice which is 2100 into from 253 Kelvin which is minus 20 to 273 which is 0 degrees Celsius this 20 degrees Celsius is the change in temperature so this will come out to be 18 900 joules all right and heat required by this ice to go to water at zero degrees Celsius this will be equal to what 0.45 into 336 that is latent heat of fusion into this okay that comes out to be 151200 joules okay now look at these numbers and tell me what can you conclude from these numbers any observation that you can make out of these numbers you can type in what is observation guys condensation won't happen completely why why condensation won't happen completely see maximum how much heat the steam can provide if it has to go to zero degrees Celsius water it will be 113 400 plus 2100 zero okay so this much energy will be released okay so this much energy is sufficient to convert ice from 250 degrees Celsius to 273 Kelvin from 272 53 to 273 the ice will go okay it can absorb 18 900 joule out of this available energy okay then in order for ice to completely melt it will further require 151200 joules are you getting it so this energy is less than of course like you know ice will absorb 18 900 joule and will go to zero degrees Celsius and part of the ice will melt but ice will not completely melt guys because to melt the ice you require this much energy which is which steam is not able to provide by the time it provides this much energy the temperature of the steam which you have taken initially goes to the temperature of the ice and once that happens then they will not be able to exchange heat getting it so the final temperature will be what all of you message final temperature of the mixture will be what ice has partially melted it has not completely melted what should be the final temperature exactly zero degree Celsius okay it will be zero degree Celsius because in order to melt the ice itself completely you require one one like 51000 joules approximately okay and that the the steam is not able to provide so it will just partially melt the ice and it will remain at equilibrium at zero degree Celsius okay all right so that's it for today and for next class I'll just write down the topics name next class revision topics next class guys we are revising how to tell me one thing I are we done with these chapters atoms nuclei dual nature which one of these chapters we have not done a b or c okay okay we haven't done atoms as well as nuclei all right so next class probably we will you know we'll start the atoms chapter because we cannot ignore these topics so next class we'll start atoms fine so revision will put on hold right now okay so we'll start with atoms but meanwhile make sure you are done with kinematics fine kinematics I have already done a revision with another batch the video of the revision of kinematics is there on the youtube on our channel okay so meanwhile you please revise kinematics your own and watch that video which is there on our youtube channel okay and next class we'll start the atoms chapter fine yes we'll do all that okay so thank you very much thanks for joining in online and hope you have learned many things today and make sure you practice a lot of questions from thermodynamics and revise kinematics also and it's just a matter of two months guys two or three months that's it okay i'm not asking one year two year commitment this is just two months are there make sure you put everything whatever you got just do this only for two months three months and you'll see that your hard work will not go waste okay thank you very much we'll see you next week now