Sir I had heard a lot about you thru quora and I am blessed that I am able to study under you. Sir your initiative is very helpful to non coaching students and I have tonnes of respect for you sir
Sir before lecture u told change in state function cannot be statefunction the how both int energy(u) and change in int enegy (∆u) can be state functions
Let z be a state function. So in cyclic process ∆z = 0 But ∆z cannot be a state function because if it is a state function then in cyclic process ∆(∆z) = 0 which doesn't mean anything. U is always a state function but ∆U is not.
@@Sneha_Yadav16 bhai generally apply hota hai , the sign of compression is always +ve and for expansion is always -ve , so his point was right , really
What I think- sudden compression implies that the system is not in equilibrium and hence PV=nRT is not applicable and hence we will consider it to be a irreversible adiabatic process. People are free to improve upon my understanding as I might have made a mistake in understanding.
since, there is sudden change in the compression, that implies that there must have been a finite, measurable, difference between the pressure of the system and the pressure of the surroundings. Thus, the process is irreversible adiabatic.
As work done is zero or constant in both the processes, so quantities are defined for initial and final positions only. So no need to study about "process"
Isothermal me ye temperature wala lafda kyu nahi hua. . ho jata. Ho jata waha par bhi, par waha to T constant hai.matlab nRT = c ,,,,, Thus P.V ki initial and final value irev aur rev me same hogi. But as we are taking assumptipns here ,, aur irrev to bas area dikhane ke liye hai. Therefore hamne p aur v ki value bhi same hi assume kar li
Sir 30:03 diagram is wrong.....Sir this is conceptual mistake from your side.. So many mistakes you make sir I wasted so much time in 30:03 work done in expansion is negative so from initial state Tf more in the sense greater the value, if initial temp is 100 then if it will drop to 70 then Delta T is -30 and if final temp is 50 then 50-100= -50 and if tf 70>50 meaning work done should be more for irreversible but -30>-50 so the lower one should be irreversible
1. In regard to the diagram sir has drawn at 30:03, sir has clearly explained and written that we are speaking in context of the magnitude of the work done (i.e |W|) in both processes (rev and irr). Now, when we are concerned with the magnitude, we are not bothered about the sign (which of course is -ve for expansion). We are only and only concerned with the magnitude (numerical value of W). 2. Now, the temperature drop/temperature difference/temperture variation (T2-T1) would be higher for a reversible process as |W| ∝ (T2-T1) and |W| is higher for reversible adiabatic process. 3. Using your stated example only, if Ti=100 and Tf=70, Tf-Ti=-30 (this means, the temperature has dropped by 30 units) now, if Tf=50, Tf-Ti=-50 (this means, the temperature has dropped by 50 units). Here, the minus sign indicates the drop. Now, temperature drop is more in the second case, when Tf=50. (In terms of magnitude or units dropped.) Hence the magnitude of the work done |W| would be more in the second case (where the temperature drop is more) and hence the second case would be the case of reversible, adiabatic expansion. Basically, while taking or considering modulus, we have to take it and consider on both sides. (that helps us understand properly what minus or plus signs represent on both the sides and we are free from any confusion.) Hope you understood the points and explanation. Namaste.
To clarify doubts , I am a Chemistry faculty and Whatever Mala Narvekar has written is absolutely perfect . Roopa Sharma has misunderstood the concept so please ignore her comment .
Sir I had heard a lot about you thru quora and I am blessed that I am able to study under you.
Sir your initiative is very helpful to non coaching students and I have tonnes of respect for you sir
Best wishes
@@alokkumaralksir5963 hi sir
Shat shat naman sir.. charan sparsh.
awesome sir
Thank you sir ❤🙏
22:42 cant enthalpy be found using the formulae in the previous lectures?
it can be, you can use del(H) = del(U) + nR(del(T)) which gives the same result.. (it's too late but helpful note for me too..)
25:54 , 31:37 observed facts of Adiabatic process
SUPER SIR !!!!
God bless you my song where are you bow😂
Now*
@@eforeverything6482 damnn he's in iit madras (googled his name)
Sir at 12:29 how did you derive the formula TP^1-gamma/gamma= constant ?
He just raised the whole power by 1/gamma
Thnx
28:06
Sir how to know if work is delivered or used up during epansion/compression?
Great explanation
Sir your smile in the beginning is everything for me to get away from the dark side of jee prep…☺️☺️☺️
bhai iit nikla?
Sir u r great! #✌🙏
🎉26:00
1:50 3:22 4:41 6:12 7:44 9:47 14:20 16:17 18:22 19:20 21:50 22:48 24:43 25:42 28:16 28:29 to 30:08 30:08
Sir before lecture u told change in state function cannot be statefunction the how both int energy(u) and change in int enegy (∆u) can be state functions
where has sir said that change in internal energy is a state function?
Let z be a state function. So in cyclic process ∆z = 0
But ∆z cannot be a state function because if it is a state function then in cyclic process ∆(∆z) = 0 which doesn't mean anything.
U is always a state function but ∆U is not.
Use euler's equation of state functions .
Thank you sir😀😀
thank you sir
Watching on 06/10/22 and have to complete this playlist till 15/10/22 aspiring for 2024 .
Bro I am also aspiring for jee 2024 tumhara maths pura syllabus khatam ho gya ?
@viralmeme bro u taking about 11 th or 12 th complete
How was your jee
But in case of adiabatic expansion t final reversible is greater than t final irreversible 🤔🤔
Tf for irreversible is 230 and reversible is 150 so ofcourse irreversible is more
no vro,, t of irrev. > t of rev in adiabatic process
20:01
Can't understand please some body help😢😢
kya nahi samjha bhai
question me likha hai oheka part rev. adi. ka isliye liya spelling mistake tha
Sir how is Pext equals to p gas in isobaric process at 03:15
In a reversible process p ext is always equal to p gas bcos system and surrounding are in thermodynamic equilibrium
@@ishaan3441 but its equal only during reversible so sir should have specified ig
Because it's isobaric😂😂
Wonderful 🙏🙏 enjoyed by heart!!
17:50
bhai tum abhi se taiyaari kar rahe ho >?
17:30 Sir.ideal do bar ho gya😁😁😬
19:20
31:58
So damn underrated!
W rev(comp)
W rev
you havent used mod so W(comp)>W(exp) irrespective of rev or irr in my opinion, as work done for comp is +ve and work done for expansion is -ve
@@11thWasteHogyaYaar Cannot it applies for diff conditions, Can't say that this is generally happens
@@Sneha_Yadav16 bhai generally apply hota hai , the sign of compression is always +ve and for expansion is always -ve , so his point was right , really
sudden compression is reversible adiabatic or irreversible adiabatic ?
which Adiabatic formula we use
What I think- sudden compression implies that the system is not in equilibrium and hence PV=nRT is not applicable and hence we will consider it to be a irreversible adiabatic process.
People are free to improve upon my understanding as I might have made a mistake in understanding.
@@hyperplane69 you are right and as for reasoning for adiabaticity i think the process happens too quick for heat transfer.
@@hyperplane69 to apply PV = nRT do we need to have equilibrium?
since, there is sudden change in the compression, that implies that there must have been a finite, measurable, difference between the pressure of the system and the pressure of the surroundings. Thus, the process is irreversible adiabatic.
sir why didn't we talk about reversible and irreversible nature for isochoric and isobaric processes?
btw these lectures are gems!
Bhai bachelor's degree ke baad PhD krte hain.. seedha 12th ke bad hi nhi
i guess cuz Isochoric me volume change nhi ho rha and isobaric me pressure is contant throughout
Bro u got iit?
As work done is zero or constant in both the processes, so quantities are defined for initial and final positions only. So no need to study about "process"
@@vishwajeetchoudhary8264Yah
9:25 sirji aapka gamma lambda ke zyada close lagta hai hahaha
dH=q at const pressure
H=nc(t2-t1)
But this formula we used only at constant pressure na. Then whywe used it in adiabatic process.?? 😕🤔
See Your Lectures Properly
Sir Has Told For Any Process!!
@@VandanaThakur-uq6jc that too many times lol
This formula is applicable for any process if the gas is ideal and in this lecture everything was assumed to be ideal only
why is pressure ext constant for an irreversible adiabatic process but not constant for a reversible adiabatic process?
it's not always constant. but we are studying only for constant pext.
❤
❤️❤️❤️
Isothermal me ye temperature wala lafda kyu nahi hua.
. ho jata. Ho jata waha par bhi, par waha to T constant hai.matlab nRT = c ,,,,,
Thus P.V ki initial and final value irev aur rev me same hogi.
But as we are taking assumptipns here ,, aur irrev to bas area dikhane ke liye hai.
Therefore hamne p aur v ki value bhi same hi assume kar li
Sir 30:03 diagram is wrong.....Sir this is conceptual mistake from your side.. So many mistakes you make sir I wasted so much time in 30:03 work done in expansion is negative so from initial state Tf more in the sense greater the value, if initial temp is 100 then if it will drop to 70 then Delta T is -30 and if final temp is 50 then 50-100= -50 and if tf 70>50 meaning work done should be more for irreversible but -30>-50 so the lower one should be irreversible
Can you plz elaborate i didnt find any mistake
@@prabhjotsingh6697 did u got
thanks
1. In regard to the diagram sir has drawn at 30:03, sir has clearly explained and written that we are speaking in context of the magnitude of the work done (i.e |W|) in both processes (rev and irr).
Now, when we are concerned with the magnitude, we are not bothered about the sign (which of course is -ve for expansion). We are only and only concerned with the magnitude (numerical value of W).
2. Now, the temperature drop/temperature difference/temperture variation (T2-T1) would be higher for a reversible process as |W| ∝ (T2-T1) and |W| is higher for reversible adiabatic process.
3. Using your stated example only, if Ti=100 and Tf=70, Tf-Ti=-30 (this means, the temperature has dropped by 30 units) now, if Tf=50, Tf-Ti=-50 (this means, the temperature has dropped by 50 units). Here, the minus sign indicates the drop. Now, temperature drop is more in the second case, when Tf=50. (In terms of magnitude or units dropped.) Hence the magnitude of the work done |W| would be more in the second case (where the temperature drop is more) and hence the second case would be the case of reversible, adiabatic expansion.
Basically, while taking or considering modulus, we have to take it and consider on both sides. (that helps us understand properly what minus or plus signs represent on both the sides and we are free from any confusion.)
Hope you understood the points and explanation.
Namaste.
To clarify doubts , I am a Chemistry faculty and Whatever Mala Narvekar has written is absolutely perfect . Roopa Sharma has misunderstood the concept so please ignore her comment .