 we come back today and we start where we left off, we are slightly behind the schedule. So we will start with problem solving particularly related to work interaction but we will spend some time because there is a section here, section 4 problem solving in thermodynamics. So that is what we are going to look at. In the middle of a problem we will really talk of it as an exercise, exercise in understanding, exercise in assimilation, exercise in writing something down step by step so that first the teacher who will evaluate it and later on anybody else will understand what actually is happening. So we begin perhaps by looking at our exercises and my favorite happens to be 1.3, let us begin with 1.3 and later on we can extend that principle to others, 1.3 we would solve and discuss in detail. So there are a few steps which we would like to follow and those steps are essentially written down in the method of solution problem solving in thermodynamics item 4 and remember that when you teach someone not just the answer but the method of solution or how the answer has been obtained and how it is presented is more important. My TAs are sitting behind you, they will tell you how meticulous we have to be particularly with undergraduate students and the impression that answers by themselves, the number and all that do not really count should be emphasized by providing a low weightage to it. So basically 70-30, 70 for the procedure, 30 for the actual numerical, in fact I would prefer that a solution be presented in the procedure followed by computation part as far as possible do not mix up the procedure and computation but in the very first illustration it may be difficult to do because the school and the first year of engineering college it is usually formula, computation answer, formula, computation answer that type of methodology whereas what we will finally like to have is the whole formulation first followed by the actual numericals. Sometimes this is not possible to follow because in the formulation you will have 2 or 3 branches which branch you take will depend on which condition is satisfied or not satisfied. So you do up to that do the calculation then you decide I have to follow this branch and continue. We will follow a mix of this when we tackle exercise 1.3. If you look at 1.3 first a cool headed reading of any exercise is a must. So let us see what we have a system containing 5 kg of a substance it is stirred with a torque of some torque is given at some speed for some time. So first thing we should realize that a tau d theta or stirrer work type of interaction is involved then it says the system meanwhile expands from some volume to some other volume against a given pressure that means there is a expansion work also immediately one should realize that there are more than one mode of work. Once an expansion is involved something like a gaseous fluid, expansible fluid is supposed to be the substance which it talks about then it says determine the net work done in kilojoules. Now although it we ask the student to determine the net work done in kilojoules we should emphasize on the student that since the method is important he should he or she should calculate all intermediate answers and provide them. It is possible to finally write a big formula and nowadays calculators are very powerful you can write a 20 term formula and just one key at the end and you get the answer. The disadvantage in that is if there is any mistake anywhere you will just not notice that because the intermediate answers are not there you do not develop any feel for any answer. For example, you will get the final net work done but you would not get the feel for how much of that is the stirrer work, how much of that is the expansion work are both of them positive both of them negative or is one positive one negative and they are sort of cancelling out each other. Such questions will not be clear unless you do step by step. So we start off with the following for example we expect the student to provide a system diagram at this stage it is important to say that it is similar to the free body diagrams they have come across. We call it system diagrams here but in mechanics they have drawn by this stage time applied mechanics or solid mechanics a reasonable number of free body diagrams. In fact even in our fluid mechanics or heat transfer when we take that small element of dx by dy it is a small part of the system taken out and we have drawn a free body diagram showing stresses or flows or pressures around it. Then depending on the complexity this could be diagram or diagram similarly the process can be shown or must be shown on the requisite process diagrams that means this will be depiction of what happens in state space and usually a 2D projection it is sufficient at this time it is necessary to emphasize on the students that since thermodynamics has a number of variables we will usually be talking about pressure, volume, internal energy or enthalpy, entropy, drainage fraction what have you. Consequently when you talk of a 2D projection any pair out of this 7 or 8 properties is fair enough but one should tell them that if there is a expansion work involved it is usually PV diagram which is suitable and you should tell them that as we proceed you know sometime PV, sometime PT, sometime TS, sometimes HS, TV diagram is appropriate but one cannot restrict oneself to one particular diagram all the time it is not wrong to use just any XY but one should select the most suitable, most illustrative of those diagrams. Then we expect the student to do the analysis of the situation this would require assimilation of the information and mind you although I am listing these out one below the other things need not proceed in this fashion because unless we assimilate and understand what it is it may not be possible to sketch a part of the system diagram. Process diagram can be completed only perhaps as the solution progresses so at the end one should have all these things in some order it is possible that you expect some process diagram to begin with then as this situation progresses you realize that look it is significantly different my initial assumption was not right the real process diagram now should look like this assimilation then split into sub problems or steps and then proceed with solution this involves use appropriate laws governing equations of state it is important to make necessary assumptions which if possible one could justify but later on as we progress one should emphasize on them that there are certain good default assumptions in term which can be later on illustrated as we progress. Finally treat units and digits with respect actually this does not necessarily pertain to exercise 1.3 this will pertain to all exercises now let us come to exercise 1.3 proper coming back to our problem we have realize that there is stirring there is expansion so an imagination of the system could be a fluid in a cylinder piston type of arrangement and with a stirrer the moment you have show the interactions there is going to be W expansion there is going to be W stirrer since work done by the system is the positive direction we always show the work arrow going out of the system although we know that this could be positive or negative this will always be negative it is also a good idea because there are so many lines to show the emphasize the system boundaries by means of a dotted line out here it is obvious that this is our system but later on when we will have more complicated problems with more than one subsystems involved then it is better to show what exactly it is system A and system B by a set of appropriate dashed lines. So you have some substance though it says 5 kg let us write m equals 5 kg then what we have is there is a stirrer torque of whatever so when it comes to stirrer write down tau equal to 0.3 kg force meter omega is 1000 rpm and time is 24 hours for expansion we are given that there is a constant pressure of 4 kg force per centimeter square. So p is 4 kg force per centimeter square given constant initial volume 1 meter cube final volume 2 meter cube v1 is 1 meter cube v2 is 2 meter cube the process diagram or process diagrams if you want to sketch one would be for the expansion process one would be for the stirring process. Now in the expansion process this is pertaining to work of expansion this pertains to work of stirring remember work of expansion is of the type pdv both p and v are properties so work of expansion or the treatment can be exposed on a pv diagram we are given that the pressure is constant that is 4 kgf per centimeter square and initial volume 1 meter cube final volume 2 meter 1 to 2 meter cube since it is given to be constant let us show it as constant and then we say that since p equals constant this we can say implies that the process of expansion is quasi static similarly when it comes to the stirrer process it is minus tau d theta neither tau nor theta are properties of the system. So this will not be a state space this will be just a trace like a polygraphic trace something against time so when this thing happens let us show against time so the process takes from 0 hours to 24 hours and the tau is constant at 0.3 kilogram force per meter this is the tau which is constant and this is the speed which is also constant so on this diagram we are showing just tau and omega just as a function of time so notice that this is a process diagram this is just a time variation of the involved variables so it is a process diagram but it is not on a state space now we come to the next step and that is analyze the situation we have to determine total work but we know that the work is made up of two component there will be an expansion work component plus there will be a stirrer work component is the first step in analysis. The second step in analysis is the expansion work is written down as integral p dv from the initial state to the final state which becomes p into v2 minus v1 and here since p is constant then comes w stirrer will be integral initial state to final state minus tau d theta and this will be written down as minus tau omega and the total time over which it takes place again since tau is constant omega is also constant the formulation of the problem ends here after this it is only calculation taking care of units and as I said treat number of digits with respect now here one has to spend a reasonable amount of time to emphasize two things on to the students one is why is it that the calculators typically display between 8 and 12 digits but the teacher gets upset if you show all of those in your answer that is one thing and second one is there are units here which are kilogram force meter rpm hour centimeter finally we want the answer in kilo joules so they have an idea that look some conversion factors will have to be used for these dimensions and units will have to be converted but they do not really have a clear idea of how that is to be done so we have to emphasize on them something about conversion factor what is the conversion factor what is the numerical value of a conversion factor what does a conversion factor do convert some quantity from one unit to another unit when we use a conversion factor what is the what are the two things which are required when you use a conversion factor so to say convert say pressure from kilogram force per centimeter square to say bar are the dimensions changing dimensions of pressure will have to remain the dimensions of pressure is the value of pressure changing no but by using a conversion factor we are converting it from one set of units to another set of units since the dimension is not changing since the value is not changing the conversion factor should be a dimensionless number and the value of that number should be 1 now this is not very clear because usually we write conversion factors as 1 kilogram force is 9.81 Newton or 1 hour is 3600 seconds okay it is not clear to us that the it is dimensionless nor is it clear to us the value is 1 so as we proceed I will show to it how we should treat conversion factors and once you do that the students you will find that many many students lose a large number of marks because the conversion in between is not done properly okay and they are losing those marks unnecessarily their understanding may be perfect but their understanding of the nitty-gritty of using conversion factors is not right. So let us proceed now the next step is computation and let us do the computation of W expansion and I am going to use a long winded method just to emphasize the utility of conversion factors once they get used to it they can start using short forms W expansion we have decided is P into V2 minus V1 and now make it a habit that you write P in its full form P is given to be 4 kilogram force per centimeter square so write it as 4 and treat kilogram force per centimeter square at some as if it is 4a by b let this be say that assume this to be an algebraic symbol assume this to be another algebraic symbol square so let it be just 4a by b square let it be like that multiplied by V2 minus V1 that is 2 minus 1 it is meter cube meter cube so this is like something like C cube okay at this stage they can even proceed and say that this is going to be 4 into 2 minus 1 that is 1 so this is 4 what will be the units now the units will be do algebra kgf meter cube by centimeter square what we want is the final answer in kilojoules so this intermediate answer should also be in kilojoules you have to add two components of work so unless this is in kilojoule this is in kilojoule you cannot get this in kilojoule so it is a good idea to have both of these in kilojoules to begin with but this is not kilojoules what is kilojoule a kilojoule is kilo Newton per meter sorry kilo Newton into meter we have a meter here but we have something called meter square by centimeter square and we have a kilogram force and not Newton so at this stage we say we use conversion factor and we use two conversion factors here one to convert kilogram force into Newton and another is to convert meter square by centimeter square into nothing because it is the same dimension it is the same unit okay so they know that 1 kilogram force is 9.81 Newton but this is not in the standard conversion factor form I said conversion factor has to have value 1 dimension less like Reynolds number okay but at that time they are not used to Reynolds number but we are used to Reynolds number so convert this into conversion factor either as 1 by 9.81 kilogram force per Newton or 9.81 Newton per kilogram force this is a conversion factor this also is an equivalent conversion factor written the in the reciprocal both have the same value 1 both are dimension less since the value is 1 and the dimension is nil nothing happens if I multiply any quantity by any conversion factor nothing will change dimension will not change value will not change but of course there will be conversion factor of say meter per second divided by say light here per day but that is of no use here I want to convert kilogram force into Newton so maybe if I multiplied by this remember kgf and kgf are now considered to be algebraic symbols they can cancel out so this will be first used I will write this as 9.81 kilogram force sorry Newton per kilogram force so that would take care of kilogram force I should now multiply it by something which converts meter square by centimeter square into nothing so what is the conversion factor we know that 1 meter square is 10 raise to 4 centimeter square this is how they know the conversion factor but we should now insist that they should write this as 10 raise to 4 centimeter square by meter square or its reciprocal this is equivalent so multiply this by 10 raise to 4 centimeter square by meter square now are we able to cancel stuff out yes we are able to cancel out kgf and kgf we are able to cancel out centimeter square centimeter square and there is a meter cube here and a meter square here so this can be cancelled out this remains only meter so what remains is now 4 into 9.81 into 10 raise to 4 what remains here is Newton meter so if you are very fast ideas you should write this as Newton meter but what we want is kilojoule what is the link between kilojoule and Newton meter 1 kilojoule divided by 1000 Newton meter that is the conversion factor value 1 dimension nil so multiply this some students will laugh at you but that is good because they have understood what we are doing write this as 1 Newton meter is 10 raise to 3 sorry we want joule so 1 joule is 10 raise to 3 Newton meter 1 kilojoule so Newton meter cancels out and the answer without using the calculator as yet is 4 into 9.81 into 10 kilojoule take out your calculator 4 we have such a calculator even show it to the students 4 into 9.81 into 10 392.4 may be a good idea to write it as plus but that is a minor issue once they understand students will do it properly in no time but it is worth spending 5 to 10 minutes in the first few problems to emphasize what a conversion factor is and in fact we will find hardly any problem after this we will have such funny units like kilogram force per centimeter square kilojoule this has been included just to bring in the matter of conversion factor. Now our job is not over we have just evaluated the expansion part we have to evaluate the stirrer part so let us do that W stirrer we said is equal to minus tau omega t and what are the values given to us this minus size continues tau is 0.3 kilogram 4 meter into omega which is 1000 revolution per minute multiplied by t which is 24 hour. So if you expand it you are going to get minus 0.3 you can start converting from here but it is a good idea minus 0.3 10 raise to 24 into 10 raise to 3 let us see the units are going to be something funny kilogram force meter revolution hour per minute you need not combine them you can start conversion factor but nothing wrong if you take this route again you have to go to kilojoules kilojoules means this kilogram force will have to be converted into Newton once you go to Newton meter kilojoule is only a conversion factor of 10 raise to 3 revolution will have to be converted into radians and hour and minute will have to be taken care of. So this becomes equal to minus I can simply start multiplying it convert kilogram force into Newton again 9.81 Newton per kilogram force the one we have seen on the previous slide meter no objection to it revolution to radians it is 2 pi radians which will simply be dropped out per revolution we do not have to worry about radians it is just a name to a dimensionless unit of value 1 and hour and minute we know there are 60 minutes in an hour so 60 minutes per hour is the conversion factor now you will notice we do not have to worry about radians we have kilogram force kilogram force revolution canceling revolution hour canceling hour minute canceling minute. So we end up with Newton meter which is what we want radians we have to and of course multiplied by 10 raise to minus 3 kilojoule per Newton meter that is our final conversion. So what you will get now is minus 0.3 this 10 raise to minus 3 and 10 raise to 3 will cancel out multiplied by 24 multiplied by 9.81 multiplied by 2 pi multiplied by 60 kilojoule 0.3 multiplied by 24 multiplied by 9.81 I think I might have missed out that multiplied by 2 multiplied by pi multiplied by 60. Now at this stage students will write all this minus 26627.637 kilojoule what do you do if you write all this we should put a big red circle surrounding it and say all thus it tells me is that your calculator has an 8 digit display or if you write 12 digits all that it tells you is your calculator has a 12 digit display this is not proper why is it not proper because we are giving importance to unnecessary significant figures. Notice here 0.3 correct to 1 significant figure 24 may be 2 significant figures 9.81 3 significant figures 61 or 2 significant figures 2 and pi some ratio we do not have to worry about it significant. So no information here is better than 2 or 3 significant figure. So our answer cannot be better than 2 or 3 significant figures. So actually although it has so many digits we should write it down properly instead of this should write it may be only as minus 26630 kilojoule or better still because even this 3 has no significance we should write it as minus 26.63 4 significant figures multiplied by 10 raise to 3 and when you combine this with this answer this answer I will overlay it here. Now this is something into 10 raise to 3 this is nothing. So this should be converted for combining with this as equal to plus 0.3924 kilojoules into 10 raise to 3 right. So that 10 raise to 3 kilojoule or megajoule is common but here you only have 0.63 after the decimal point. So even these two are going to be of no consequence to us. See to it that even if it does this this 0.24 or this 2 4 are not added emphasize this this is not thermodynamics but this is good problem solving practice and good engineering practice because anyway we are teaching them engineering we are teaching them mechanical engineering. So they should respect that certain things are measurable and significant only to certain digits beyond that it is meaningless they will calculate the temperature rise of something as 3.6943594 degree Celsius but will be ever be able to measure beyond may be 0.01 Celsius and it is not worth talking about all that you are wasting time you are wasting paper you are wasting ink tell them that and I have found a neat weapon provide a quiz after this and significantly deduce marks if they do such mistake you have to do it once. Moment they realize that you know misbehavior like this gets heavily penalized they will automatically fall in line if they do not write units straight away give 0 marks without units what is the value you have to do it in one quiz and the student will not mind if there are a number of quizzes so that that small mistake does not really penalize them but the fact that the small penalty has been imposed is good enough. Now that brings us of course that last step I have not done I will leave it to you to combine these two you are mature enough to do it properly but any discussion on this yes sir page number 2 system diagram generally we are saying that the work input to the system considered to be a negative work done by the system is positive work given to the system is also surrounding to the system is also is negative here in the case of stirrer work work is coming out from the system right. So that is the when we show an arrow we always show an arrow in which the positive direction of work exist so we will show an arrow for the expansion work to be outside we will show an arrow for the stirrer work also going out of the system knowing fully well that as I said at that time that this could be positive or negative this will definitely be negative but knowing it is a negative work I will not put it inside because if I put it inside then I will have to change the sign here as negative and it is not good to here we know because it is a stirrer work but there will be some problems in which maybe some stirring some heating some interaction takes place and we are asked to determine what is the expansion work and what is the change in volume I have been issue I do not know whether the system is going to expand or contract. So I cannot put an arrow this way or that way it is always proper to put arrows in which the positive direction of work exist. So even if there are four components each one of them could be positive or negative show the four components are going out some of them will turn out to be algebraically positive that means positive work is being done if it is expansion the piston is moving out it is expanding if it is electrical and if it turns out to be negative we say that the cell or the electrolyte is getting charged rather than discharged but we know the stirrer work is a negative number but we will not use if we calculate the stirrer work and if it turns out to be positive then we know something is wrong somewhere either in the specification or in our computation. So and this is the symbolism we will generally be using till we come to cycles because in cycles we have a special treatment of heat absorbed and heat rejected heat rejected is a mandatory so that we will be showing in the wrong direction but otherwise till we come to cycles everything will be in the proper direction in which default arrow in the positive direction. So heat transfer to a system will always be shown inside even though it is negative work done by a system will be shown outside even though it is negative and it is a good question because in this case we know it is negative but in many cases we will not know whether it is positive or negative it is better to do the standard nomenclature because then this is plus, plus, plus, plus otherwise if it is stirrer then minus h t but then expansion we do not know positive or negative so what will you write plus or minus write the arrow both ways that is confusing. Any other question? Yes. Page number 3 sir. Page number 3 that 4 kg per centimeter square into meter cube like that so simply we will be introducing 9.81 Newton in place of kg it will be little bit easy to introduce. If your students understand that as 1 kg f is 9.81 that is okay but I find that this is a much better method for any complicated stuff this can be used. Okay sir. What you say is algebraically equivalent but that way you are not emphasize the basic fact that the value of any conversion factor is 1 and its dimensionless is nil. If I write it I wrote it first now we know that 1 kg f is 9.81 Newton but this is it is like saying 3a equals 6b does not they no special meaning comes out of it and I would like to emphasize the characteristic that the value of a conversion factor is 1 its dimension is nil and that is why this is the answer 4 kg f meter cube per centimeter square Yes sir. But the unit is not convenient otherwise mathematically it is a perfectly okay answer and what I am doing is I am just multiplying it by 1 multiplying it by 1 I am not changing the value because the numerical value is 1 I am not changing the dimension because the dimension is nil. A little bit it was lengthy process like Let the process be emphasized like this after that see once they understand it so long as they do correct conversion even if they do it in their mind it is okay what we want them is to do correct conversion because you know the correct answer has been obtained here while lose mark in writing it in the proper units and in thermodynamics we have not very complicated units when you go to heat transfer when you have to determine something like Grashof number or Rayleigh number there are so many units naturally nowadays most of the tables are in SI units so the convergence are only some powers of 10 but earlier we had the CP in some centigrade heat units or kilo calorie per kilogram Celsius something else in some other units viscosity in kilogram force units all sorts of conversion factors there. So do this as a habit if they continue with this nothing wrong but if somebody looks through this and start using shortcuts it is okay sir there in second slide it may be funny as far as the stirrer is concerned stirrer work is positive as far as the if we if we consider the sister stirrer as the system okay stirrer work is the positive agree it is doing work on something else yeah suppose if we tell the stirrer work on the system then it will be negative work done by this system of fluid on the the surrounding system which happens to be the stirrer is negative our system is what is shown by the dotted line so this is work done by our system this is also work done by our system this is also work done by our system this is work done by our system by the expansion mode on whatever is holding the piston on the other side or say on the piston and this is the work done by the system using the stirrer mode on the stirrer or whatever is managing the stirrer when we write this such an expression it is always the work done by the system all components would also be work done by the system different in fact your question is essentially the first question in some different phrasing anymore now I will stop here but let me discuss for you some other problems which I will leave to you as homework the first problem is algebraic it just as warm up to students so leave it like that the second one also is a straightforward third one we have done the fourth one is you know yesterday the discussion came up is when is the gravitational pressure gradient in a fluid significant this is an illustration where we have to determine a is actually hydrostatics integration of the gravitational pressure gradient to get the net force on the piston b calculate the work done by the water on the piston that is first part of b so this f will have to be multiplied by the dx the movement of the piston and integrated from x1 to x2 so this is where you cannot although there is a p0 and hence there is a p here on the piston you cannot just calculate as pdv we will have to calculate as the pressure converted into force because the force is varying and will keep on varying because as x changes the height will also change but once you do this if you consider the relation between h and x as the piston moves h reduces as x increases it will have to be integrated out so this is essentially pdv but in a much more complicated way now 1.5 okay I think I will spend some time on 1.5 you are given 250 gram block of a metal and the pressure on that is increased quasi statically and isothermally from 0 to 1000 bar assume that the density and isothermal bulk modulus of the metal remain almost constant at 20 gram per centimeter cube and 10 raise to 10 2 into 10 raise to 12 9 per centimeter square respectively determine the work done in June isothermal bulk modulus is defined as b equals minus v dp by dv at constant now here our system happens to be some block of metal and it is under pressure it is getting compressed because of that pressure and the pressure goes up from 0 to 1000 bar now if it were a fluid or a gas at 0 bar that means at very low pressure it will have a very large volume at 1000 bar it will have a much much smaller volume it will get very significantly compressed so if it were a gas at 0 bar it will have a very large volume at 1000 bar it will have a small volume so the process would be I have sketched it in the wrong direction but the process would be something like this it is mentioned that it is quasi static this will be for a fluid or for a gas but this is a metal which is a solid so the volume will hardly change it says that the density remains almost constant why does it does it say that the density remain constant or exactly constant then what would be the answer if the density remains constant mass is fixed that means volume is constant the volume does not change there is no work done so if there is some work done that means there has to be some change in volume however small and mind you were going from 0 bar up to 1000 bar which is quite a high pressure so even a small amount of change in volume would lead to some work interaction so but depiction of the process diagram would be something like this now I will show volume on the small scale and pressure on a large scale let us say this is 1000 and this is 0 in bar and this is say the initial volume I will draw a line indicating the initial volume and say if this is the initial volume state 1 state 2 will be only a small volume below state 1 this is state 2 v2 will be less than v1 but v2 minus v1 a negative number compression will be a very small number now we know that for a metal or for any fluid the volume will be a function of pressure acting on it and its temperature if it is compressible as you increase the pressure the volume will change we know also that metals expand so as you heat up volume will change so if you want to determine dv this will be there used to partial derivative this will be partial of v with respect to p at constant t dp plus partial of v with respect to t at constant p dp this is the variation because of change in pressure this is the variation because of change in temperature so we know that dw in this case there is only pressure so this will be dw expansion which will be equal to p dv which will be equal to p dv and for dv we substitute this this will be I will put p first p into partial of v with respect to p partial of p dv by dp at constant t dp plus p dv by dt at constant p dp okay I have just substituted that here now what are we given we are given that pressure changes from 0 to 1000 bar that is okay density remains almost constants at 20 gram per centimeter cube and the isothermal bulk modulus which is given as minus v dp by dv at constant t remains constant at this value so let us use this information and it is given that it is compressed quasi statically and isothermal quasi static let me call this as equation 1 quasi static means equation 1 can be written as equation 1 can be written as equation 1 can be integrated during the process at every stage we know what will be the pressure we know what will be the volume and we know what will be the temperature so integration is possible if it were non quasi static we would say the path is not known so we cannot integrate second one given is it is isothermal means during the process dt is 0 hence the last term in equation 1 is 0 dt is 0 now this implies that your dw it simply p dv by dt dv by dp at constant t dp and now we are given that the pressure changes from 0 to 1000 bar given that b which is equal to the definition you will find here minus v dp by dv at constant t is given and also we are told that it is almost constant so that means we want dv by dp at constant t so this implies that we know the value of dv by dp at constant t will be that will be equal to minus v by b so substitute that here you will get this to be minus v by b into p dp and hence w will be minus integral 1 to 2 1 is our 0 bar pressure and 2 is our 1000 bar pressure v by b p dp and now here remember that density is mass m per unit volume so this implies that v equals m by m by m by m by m by m by m by m by m by rho so this becomes minus 1 to 2 m by rho b p dp and now mass anyway is constant we are told that density and bulk modulus are almost constant so this factor can be taken out so this will become minus 1 to 2 sorry minus m by rho b integral 1 to 2 p dp which can straight away be integrated and then take care of the units and you will get here the importance is to emphasize the this almost constant part of b and almost constant part of rho that word in the bracket almost constant is important although it is in the brackets that is the most important word in this because if we emphasize that density is constant that means it is incompressible and if it is incompressible then volume does not change there will be no work done you compress it no work done for compression the pressure will just try volume does not change the remaining problems 6 and 7 are of a similar kind except that the variation of force with displacement is given. Now after this we come to the real main part of thermodynamics that is first law.