DON'T MISS THE 4 PRACTICE PROBLEMS AT THE END. HOW TO IDENTIFY AND TACKLE IRREVERSIBLE IDEAL GAS PROCESSES FROM A GIVEN INFORMATION IN THE PROBLEM. THERE ARE MULTIPLE METHODS TO SOLVE SUCH PROBLEMS ON THE CONCEPT OF "IRREVERSIBLE IDEAL GAS PROCESSES , BUT IN THIS VIDEO WE PRESENT AN EXAM ORIENTED APPROACH FOR OBEJECTIVE TYPE QUESTIONS - 15,24 , PASSAGE 30,31 OF CHAPTER - 8 "THERMAL PHYSICS " SOLVED FROM THE BOOK "PATHFINDER PHYSICS"👇 CHECK OUT A PREVIOUS VIDEO (PART-1) FROM CHALLENGE YOUR UNDERSTANDING ON IRREVERSIBLE PROCESS 👇 th-cam.com/video/xZww1effRzc/w-d-xo.html
sir i know u have a very kind heart and your aim is not only likes and all those stuff and u will not make us sad u will continue making videos becoz u know we all love u from our bottom of our heart :)
For practice problem 1, I am in some doubts For practice problem 2 , at 25:15 , I am getting ABCD , For practice problem 3, at 25:55 , I am getting ABC For practice problem 4, at 26:25, I am getting u^2/7g
@@PHYSICSSIRJEE Yes sir, I told that in an already posted comment on the ambiguity of that problem where you have given a heart :) ... Basically about the gas being not mentioned at the right side...
sir in 14:41 you said work done by gas on piston=delta U.....but we know in first law the W stands for work done by surr and not work dne by system so how did you take work done by gas= delta U
I am a big fan of urs...First of all my huge respect and greetings to you...🙏🙏.. In the 3rd problem which you solved with friction u mentioned it as reversible process... Unfortunately it won't be...This process is better to be named as" Quasi-Static process with Friction", not Reversible process...Since this is a Quasi-Static process one can use work done equal to integral pdv and all the state properties will also be defined... So no issue in applying formulas...Still we cannot say it to be Reversible process because of the friction...In this process heat will be generated due to friction and once we reverse the process and again heat will be generated due to friction which will certainly affect the surrounding... Reversible means which when reversed both the system and surrounding should be in same previous state at all the times...Since there is non zero entropy change due to friction it won't be Reversible... I am not explaining these things to you because you yourself are a stalwart...I can't even think of this dare... Whatever little explanation I have written is only for students...I keep watching all your lectures with enthusiasm and gets affected when u say u won't upload further videos due to insufficient likes or subscribers... Those students are definitely lucky which get to learn Physics from you personally...I myself have shared your videos to many students... Numerous concept such as surface tension, rotating frame videos, resolved series videos, etc...There is so much to learn whenever u upload any video... Kindly keep making such videos...You are a teacher to me.. Happy Guru Purnima...Charan Sparsh...Kuch galti hui to maafi...🙏🙏🙏🙏...
24:45 Change in PE of ball = kinetic energy of piston And applying work energy theorem on piston Work done by gas + ext work(O as net displacement of piston is zero)= KE(-mgh) And 5/2V∆P= mgh ∆p=mg/s and V=h(s)
Sir I have two queries 1. what is basic definition of adiabatic process ∆ Q=0 or it should follow the equation pv^gamma =constant 2. You said we can use ∆Q=∆U +W for initial and final condition then why can't be use equation pv^gamma for the initial and final condition if it is a non quasi-static process and heat exchange is zero
Sir. In many of these problems, we get different answers for assuming irreversible and quasi static. But in the irreversible case, we get the ans, by doing Delta Q = Delta W + Delta U, and PV =NRT For the inital and final states. But in the quasistatic case, we use dQ = dw + dU. for the intermediate states and get a relation for the state varibales. But Doesnt dQ = dw + du imply delta q + delta w + delta U? So why are we getting diffferent answers for quasistaic and irreversible, if the equations we use can be the same
Sir plzzzz solve this question of olympiad 👉👉👉👉Let us consider a model of the Atwood machine, which consists of a square prism, ABCD, whose central axis is horizontal and perpendicular to this sheet of paper and whose face AB makes an angle of π 4 with the vertical line. On the upper faces AB and BC of the prism, there are massless bodies P and Q, respectively. As shown in Fig. 2.37, the two bodies P and Q are connected by an inextensible massless string, S1, and similar strings S0 and S2 are hung down from P and Q, respectively. Note that the string S1 is horizontal and strings S0 and S2 are vertical. Here, we assume there is static friction between body P and face AB as well as between body Q and face BC; the frictional coefficients for both cases are µ0. Suppose bodies P and Q are always in contact with faces AB and BC, respectively.
For the process to be isothermal dT=0 (T is not changing throughout the process )we can only comment on the Tf and Ti of the process,but in between the extreme cases there is surely change in temperature.
25:11 For Aits problem I m getting ABCD 26:25 sir I am getting u2/( 7g + PoA/m) where m is mass of piston and A is area of base. I think here massive word is used for the piston so is there any approximation required or I need to work it again.....
@@amiyancandol4499 the container is given conducting , so after long time , the gas will again acquire temperature of surroundings. That's what I wrote in orange at 7:47 regarding heat exchange
Sir I am unable to understand how will the equilibrium be reestablished in practice problem 1 at 24:20 , as the collision is elastic, so won't the ball keep moving up and down, and there is no energy loss as well? and the mass of the piston is taken to be light, so does that mean, piston will rapidly fall down? This is a bit confusing...
@@PHYSICSSIRJEE Ok sir, but I think this should have been mentioned in the question. Also, I think it would take infinite time to reach the state of equlibrium in theory...and after that is achieved, then we are supposed to solve that problem
@@PHYSICSSIRJEE Yes sir, I also thought about this, but we have to assume the gas there in order to maintain the initial condition of equilibrium given....
Thanks a lot sir. I thought that you will stop making videos. I thought that I should download all your videos. Please don't make my fear come true. Don't worry about likes and sharing sir, we are doing it but I don't know why the likes are not increasing. Please keep us helping 🙏 May god bless you ❤
Sir how did you come to the conclusion at 7:30 that the left piston will be at the bottom of the left tube and not anywhere in between the initial height and the bottom of the tube?
I don't remember about chemistry terminology . Isothermal means dT = 0 at every step in the process. ∆T = 0 is not a sufficient condition for isothermal process. Also , the process could have taken place in finite number of irreversible steps ( like we have in chemistry) during which temperature could have changed . So it's better , we don't mark option D here at 19:50 . PS: I request you to post time stamps for your doubts . It's encouraging for teachers to navigate easily to answer your doubt
Sir,at 11:13 , we are assuming that the piston would stop at eqm position but wouldn't it go till the other extreme as it has gained some momentum (just lyk SHM)would it really stop at equilibrium position?
It will oscillate back and forth before reaching the final state. We don't know and don't want to know about intermediate states for the solution taken
Sir for the 3rd practice problem even i had a doubt, that acc to the answer they considered pressure constant in the right chamber but actually , since it's a closed chamber, P must not be constant right ? Edit: I just checked the keys, multiple ans keys are given, one considering P constant on right side and another without that 😅
Yeah but it was given closed, so that's what we can get confused if giving the paper in the exam hall, there's no way we can think of assuming open container if given closed, i guess just things not in our control
Sir or Can Anyone give me the info about those discord server where pathfinder solns are discussed....What are they? Where can I find them?? I solved the second and last problem myself without soln😁....Thank u so much sir for these lectures!!
@@PHYSICSSIRJEE tq for ur advice sir. I am a jee 2022 aspirant. Sir I have not done my class 11th topics properly. But my preparation in 12th is going good . I feel more 1 year will fetch me a good rank Pls comment on this sir And thanks sir . U r channel has been a integral part of my preparation 🙏. I will share in my groups
@@physicsislove991 lots of time to prepare for 2022. Keep your hardwork going. No need to think of dropping a year if you work smartly towards your goal
DON'T MISS THE 4 PRACTICE PROBLEMS AT THE END. HOW TO IDENTIFY AND TACKLE IRREVERSIBLE IDEAL GAS PROCESSES FROM A GIVEN INFORMATION IN THE PROBLEM.
THERE ARE MULTIPLE METHODS TO SOLVE SUCH PROBLEMS ON THE CONCEPT OF "IRREVERSIBLE IDEAL GAS PROCESSES , BUT IN THIS VIDEO WE PRESENT AN EXAM ORIENTED APPROACH FOR OBEJECTIVE TYPE QUESTIONS - 15,24 , PASSAGE 30,31 OF CHAPTER - 8 "THERMAL PHYSICS " SOLVED FROM THE BOOK "PATHFINDER PHYSICS"👇
CHECK OUT A PREVIOUS VIDEO (PART-1) FROM CHALLENGE YOUR UNDERSTANDING ON IRREVERSIBLE PROCESS 👇
th-cam.com/video/xZww1effRzc/w-d-xo.html
sir i know u have a very kind heart and your aim is not only likes and all those stuff and u will not make us sad u will continue making videos becoz u know we all love u from our bottom of our heart :)
absolutely, well said !!!
That AITS question came in my coaching test
least discussed and the most important topic of thermal physics !! Thanks a lot sir !!
An absolute gem 💎💫. Sir, your concepts are absolutely crystal clear. I wish I can develop such clarity when I grow up. Love you sir.
Thank you 🙂🙏
Sir I have literally no words to express my sincere gratitude to you.
26:20 u^2/7g upwards
How??
Thank you sirji😊❤
Practice problem 2 : A,B,C,D (24:58)
Correct 👍🏼
@@PHYSICSSIRJEE Sir where can i find solutions to other practice problems
hi sir amazing explanation.I'm from sri chaitanya Meghan hills campus sir....happy with nice experienced teachers like you sir.
You are the bestt🔥🔥
For practice problem 1, I am in some doubts
For practice problem 2 , at 25:15 , I am getting ABCD ,
For practice problem 3, at 25:55 , I am getting ABC
For practice problem 4, at 26:25, I am getting u^2/7g
All three , you attempted are correct 👍🏼🙂. Any issue you found with JEE 2015 problem at 25:55 ?
@@PHYSICSSIRJEE Yes sir, I told that in an already posted comment on the ambiguity of that problem where you have given a heart :) ... Basically about the gas being not mentioned at the right side...
Tnx alot sir
Waiting sir..😍 this was always my weak link..these pistons,gas ,adiabatic and irreversible process etc thanks for bringing it
I love your videos and subscribed your channel.
Thanks sir 🙏🙏
Thanks a lot sir ☺️
Pathfinder & your explanation both are really mesmerising 🤩
Thank you very much sir
24:30 height is 7Hi/2, work done by gas is -5mgHi/2
m needs to be managed ig it's not given in question
Sir.. Ye section mein bhut problem aate the par aajse woh nhi ayge😃😃...
Thank you sir for these amazing content...
Thanks sir for this awesome video...cleared many concepts
Thank you very much sir 😇
After long time, again back to SIRJEE.😁
sir in 14:41 you said work done by gas on piston=delta U.....but we know in first law the W stands for work done by surr and not work dne by system so how did you take work done by gas= delta U
Thanks a lot sirjee..I was always confused with this things 🙏🙏
No doubt Sir You are the GOD of this realm!!!
Kind words 🙂🙏
Thanks sir we need this..
thank you very much for this sir
🙏🙏 thanks for the beautiful explanation sir ji
I am a big fan of urs...First of all my huge respect and greetings to you...🙏🙏..
In the 3rd problem which you solved with friction u mentioned it as reversible process... Unfortunately it won't be...This process is better to be named as" Quasi-Static process with Friction", not Reversible process...Since this is a Quasi-Static process one can use work done equal to integral pdv and all the state properties will also be defined... So no issue in applying formulas...Still we cannot say it to be Reversible process because of the friction...In this process heat will be generated due to friction and once we reverse the process and again heat will be generated due to friction which will certainly affect the surrounding... Reversible means which when reversed both the system and surrounding should be in same previous state at all the times...Since there is non zero entropy change due to friction it won't be Reversible...
I am not explaining these things to you because you yourself are a stalwart...I can't even think of this dare... Whatever little explanation I have written is only for students...I keep watching all your lectures with enthusiasm and gets affected when u say u won't upload further videos due to insufficient likes or subscribers... Those students are definitely lucky which get to learn Physics from you personally...I myself have shared your videos to many students... Numerous concept such as surface tension, rotating frame videos, resolved series videos, etc...There is so much to learn whenever u upload any video... Kindly keep making such videos...You are a teacher to me.. Happy Guru Purnima...Charan Sparsh...Kuch galti hui to maafi...🙏🙏🙏🙏...
24:45 Change in PE of ball = kinetic energy of piston
And applying work energy theorem on piston
Work done by gas + ext work(O as net displacement of piston is zero)= KE(-mgh)
And 5/2V∆P= mgh
∆p=mg/s and V=h(s)
Time stamp ?
Nice compilation of problems 🙏
Practice problem-2 , 25:30
Sir, Is the answer A,B,C,D ?
Waiting😍
Sir can you take problem 29 build pathfinder thermal physics
Irreversible process ....it's complete 🔥💥💥all my doubts busted
Sir I have two queries
1. what is basic definition of adiabatic process ∆ Q=0 or it should follow the equation pv^gamma =constant
2. You said we can use ∆Q=∆U +W for initial and final condition then why can't be use equation pv^gamma for the initial and final condition if it is a non quasi-static process and heat exchange is zero
1. ∆Q = 0
2. PV^gamma = C is a curve on which final state of irreversible process need not lie
Sir. In many of these problems, we get different answers for assuming irreversible and quasi static. But in the irreversible case, we get the ans, by doing Delta Q = Delta W + Delta U, and PV =NRT For the inital and final states. But in the quasistatic case, we use dQ = dw + dU. for the intermediate states and get a relation for the state varibales. But Doesnt dQ = dw + du imply delta q + delta w + delta U? So why are we getting diffferent answers for quasistaic and irreversible, if the equations we use can be the same
Sir plzzzz solve this question of olympiad 👉👉👉👉Let us consider a model of the Atwood machine, which consists
of a square prism, ABCD, whose central axis is horizontal and
perpendicular to this sheet of paper and whose face AB makes an
angle of π
4 with the vertical line. On the upper faces AB and BC
of the prism, there are massless bodies P and Q, respectively. As
shown in Fig. 2.37, the two bodies P and Q are connected by an
inextensible massless string, S1, and similar strings S0 and S2 are
hung down from P and Q, respectively. Note that the string S1
is horizontal and strings S0 and S2 are vertical. Here, we assume
there is static friction between body P and face AB as well as
between body Q and face BC; the frictional coefficients for both
cases are µ0. Suppose bodies P and Q are always in contact with
faces AB and BC, respectively.
Question OP
Janardhan Sir OP
20:25 Sir, can’t we say process is irreversible isothermal ?
For the process to be isothermal dT=0 (T is not changing throughout the process )we can only comment on the Tf and Ti of the process,but in between the extreme cases there is surely change in temperature.
Where can I get the solutions for the last practice questions?
Sir where is the video, where you suggested books for Olympiads, I can't find it?
Sir so u are back
I never left , till now 🙂
25:11 For Aits problem I m getting ABCD
26:25 sir I am getting u2/( 7g + PoA/m) where m is mass of piston and A is area of base. I think here massive word is used for the piston so is there any approximation required or I need to work it again.....
I think you will get exact answer
Ok sir
Sir in the first problem, why is it that initial and final temperature is same ?
Timestamp of your doubt ?
7:45
@@amiyancandol4499 the container is given conducting , so after long time , the gas will again acquire temperature of surroundings. That's what I wrote in orange at 7:47 regarding heat exchange
Okay thank you sir, I didn't read the conducting thing, sorry !
Sir I am unable to understand how will the equilibrium be reestablished in practice problem 1 at 24:20 , as the collision is elastic, so won't the ball keep moving up and down, and there is no energy loss as well? and the mass of the piston is taken to be light, so does that mean, piston will rapidly fall down? This is a bit confusing...
The orderly energy of ball gets transferred to the disorderly energy of gas via piston upon multiple collisions
@@PHYSICSSIRJEE Ok sir, but I think this should have been mentioned in the question. Also, I think it would take infinite time to reach the state of equlibrium in theory...and after that is achieved, then we are supposed to solve that problem
sir are u still in meghan towers , We got ray optics dpp prepared by You
26:14 is the ambiguity is the pressure on the right side of piston(spring side) and its work ?
Yes , JEE assumed there is another gas on right side but the diagram on right doesn't show dots like on left side
@@PHYSICSSIRJEE Yes sir, I also thought about this, but we have to assume the gas there in order to maintain the initial condition of equilibrium given....
@@vilakshangupta380 yes absolutely
Just curious , has irreversible processes come in jee earlier? Fantastic video as usual
Didn't see anything except free expansion . But never say never , I guess 😉
Thanks a lot sir. I thought that you will stop making videos. I thought that I should download all your videos. Please don't make my fear come true. Don't worry about likes and sharing sir, we are doing it but I don't know why the likes are not increasing. Please keep us helping 🙏 May god bless you ❤
Even if I stop, I won't delete past videos 🙂
Jee Adv 2021 is so near and you are thinking of stopping 😭
Sir how did you come to the conclusion at 7:30 that the left piston will be at the bottom of the left tube and not anywhere in between the initial height and the bottom of the tube?
Unbalanced downward force on piston as explained clearly is the reason
Sir how to solve the questions if they the amplitude or frequency or time period of the oscillatory motion in case of irreversible process?
It would be difficult to keep it oscillating. But any problem can be solved , if it's authentic , by reading the statement carefully word by word
Nice explanation sir 👍
Will try to solve the practice problems and comment ans.😀
Op 🔥💥👌
Sir for passage qn 30th one, in chemistry we say irreversible isothermal process in which initial and final temp are same, so shouldn't D be wrong ?
I don't remember about chemistry terminology . Isothermal means dT = 0 at every step in the process. ∆T = 0 is not a sufficient condition for isothermal process. Also , the process could have taken place in finite number of irreversible steps ( like we have in chemistry) during which temperature could have changed . So it's better , we don't mark option D here at 19:50 .
PS: I request you to post time stamps for your doubts . It's encouraging for teachers to navigate easily to answer your doubt
@@PHYSICSSIRJEE okay sir thank you and sorry I'll post the time stamp from next time
Sir,at 11:13 , we are assuming that the piston would stop at eqm position but wouldn't it go till the other extreme as it has gained some momentum (just lyk SHM)would it really stop at equilibrium position?
It will oscillate back and forth before reaching the final state. We don't know and don't want to know about intermediate states for the solution taken
@@PHYSICSSIRJEE ok sir ,I understand now ,thanks for giving the reply👌
Noice 🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
_/\_ thank you sir
Sir for the 3rd practice problem even i had a doubt, that acc to the answer they considered pressure constant in the right chamber but actually , since it's a closed chamber, P must not be constant right ?
Edit: I just checked the keys, multiple ans keys are given, one considering P constant on right side and another without that 😅
Did they give multiple keys ? I need to check
@@PHYSICSSIRJEE no sir I'm sorry 😅 i checked fiitjee keys , in the official one they haven't
Probably they considered an open vessel
Yeah but it was given closed, so that's what we can get confused if giving the paper in the exam hall, there's no way we can think of assuming open container if given closed, i guess just things not in our control
But we can't consider the pressure of right container gas constant
🥰
Sir or Can Anyone give me the info about those discord server where pathfinder solns are discussed....What are they? Where can I find them??
I solved the second and last problem myself without soln😁....Thank u so much sir for these lectures!!
Check the video on discord in description
25:00 ABCD
Sir is dropping for jee advanced a good idea . I believe I deserve more marks . I was not working hard
Pls enlighten me 🙏
First prepare and write the exam as if it's your last. Then you can decide after the result
@@PHYSICSSIRJEE tq for ur advice sir. I am a jee 2022 aspirant.
Sir I have not done my class 11th topics properly. But my preparation in 12th is going good . I feel more 1 year will fetch me a good rank
Pls comment on this sir
And thanks sir . U r channel has been a integral part of my preparation 🙏. I will share in my groups
@@physicsislove991 lots of time to prepare for 2022. Keep your hardwork going. No need to think of dropping a year if you work smartly towards your goal
@@PHYSICSSIRJEE thank you very much sir for enlightening me 🙏
25:30 ABCD
Sir, please pin your Discord Link
25:20 ABCD
Correct 🙂👍🏼