Thermodynamics | Advanced problem | Maximum speed of piston

[@dhruvsinghchoudhery2657]

Really good Q Nicely Explained...

[@yatharthgupta6468]

Sir why we took pvgamma = constant..the process is not reversible

[@abhirupiitkgp139]

Can we do it by another method? F=(p1-p2)A=mvdv/dx . Now for any displacement x of the piston we can find p1 and p2 by adiabatic formula and then substitute. For v to be maximum dv/dx=0 . hence we get value of x for maximum velocity . Then we can find v in terms of x by integrating the force equation and substitute earlier value of x obtained as limits?

[@Fighter_Believer_Achiever]

1:43,sir since temperature is not constant, I think we cannot directly say the Boyle law statement

[@Fighter_Believer_Achiever]

Thank you very much sir for such wonderful question🥳🥳🥳🥳

[@Fighter_Believer_Achiever]

Sir can you please reply - Will shm occur here?? Intuition says that it should shm but in thermal physics pathfinder mcq 9,it says that shm will stop after finite time. I cannot understand the reason

[@Deadpool63894]

Because there is no external heat supplied to the system , so after equilibrium both side pressure became same and the piston stays at rest👍

[@vipergaming3494]

@@Deadpool63894 bro woke up and chose to solve a 2 year old doubt💀

[@dr.vinitajain2649]

Sir should'nt the net work done by gases be W1-W2 instead of W!+W2 ?? like in previous example pls reply..

[@omgupta2590]

no because we are adding all the vaues of work .if work is negative then it will show in the value

[@Ramesh-ey8fe]

thank you sir

[@saiyamjain8729]

Nice problem

[@Diego_129]

Respected sir.... While writing net work w2=-ve so should nt it be Wnet= +W1 + (-W2)..... As in previous question u wrote this way........ Please correct me sir.🙏

[@neerajchoprarealid3935]

Bro here was total work=-internal energy Because in adiabtic process U=-w And w=w1+w2

[@yoyobro607]

Acc to me here formula gives work done by gas and will automatically include -ve sign.

[@sohamadak6811]

Amazing

[@tonystark8857]

Sir why speed is maximum when pressure on both sides becomes equal

[@vedbrahmbhattiit-j4497]

for equilibrium

[@arjunlad8557]

@Target IPhO thanks

[@lakshitasharma4709]

thank u sir jee

[@vishram9595]

Sir how can we use PV^gama =const here, the process is irreversible

[@PhysicswithAkashGoyal]

It would have been irreversible if one side is vacuum. In fact in reality no process can be reversible... But we don't have any other choice... Otherwise Pressure , temperature etc can't be defined at each instant... For reversible process pressure at both sides should differ by an infinitesimal amount (dP). That's definitely Not the case here

[@vishram9595]

@@PhysicswithAkashGoyal got it sir ur videos are really helpful hope that they will continue

[@NAYAN-t3e]

@@PhysicswithAkashGoyal But sir, it is given that the system is adiabatic & frictionless between moving parts. Are these assumptions sufficient to consider the problem formulation in reversible manner ?

[@abhirupiitkgp139]

Can we do it by another method? F=(p1-p2)A=mvdv/dx . Now for any displacement x of the piston we can find p1 and p2 by adiabatic formula and then substitute. For v to be maximum dv/dx=0 . hence we get value of x for maximum velocity . Then we can find v in terms of x by integrating the force equation and substitute earlier value of x obtained as limits?

[@ishan__001]

@@abhirupiitkgp139 sounds good I also tried with same at first but eqn are more complex usually this kind of procedural things involves lot of calculation in my experience with good ques and hence we will not be able to do this in exam