You are simply amazing. the concept seems so hard when you read it in the textbooks,but you made it look like the easiest thing in the world. thanks a ton!:D
This is the kind of lesson where you understand everything but you're like 'ok so what's the point?'. Ok so a system is in equilibrium when it has the macrostates well defined, the macrostates are well defined when process is really slow and we call that 'cvasistatic' and if we do this really slow process backwards, and we presume that there is no energy loss we have a 'reversible process'. Great, now how do we relate that with the real life pistons and the next chapters in Thermodynamics? Are engineers trying to create pistons that can keep as much as possible states of equilibrium between movements ? Having states of Equilibrium helps us finding out the work that a system does more easily? If cvasistatics processes are not real life relatable what's the point that we're making? etc Great lesson and a huge thank you for having this matterial for free for everyone that's willing to learn !
If macro state variables are not defined when u blew out half the load, then on what bases we use the lower of the pressure to calculate irreversible work done by the system, p2x(v2-v1). Thanks
It's like loss of heat is loss of energy and loss of energy is loss of information, if we lose energy due to heat we will never be able to go back or reverse to one macrostate because we don't have the same energy as in the beginning.
So why would volume not be well defined between equilebria? Are you saying the volume is kind of independent from the dimensions of the cylinder when there is no equilibrium?
Little bit late but may help others: What I understand is that pressure, volume and temperature are connected - if you compress the gas (decrease the volume) either pressure or temperature has to rise. Other way if you decompress the gas by pulling the piston up and making the volume bigger then pressure or temperature will decrease. So after instantaneous removing the rock the pressure causes a fast decompression, so fast that you can't measure the temperature and pressure inside. Depending on what was used in the experiment the pressure will overshoot the equilibrium same way the spring does. Then the gas will want to compress because there is still half of the rock pushing the piston down. Back and forth, back and forth and after some time it gets to the equilibrium and stops. This is the moment you can measure pressure, volume and temperature. Tl;dr: You could define volume between equilibrium, but it would make no sense as there are other factors - pressure and temperature - that you need to take into account.
Can we say Quasi static process leads to equilibrium Equilibrium means temperature & pressure is homogeneous Reversibility in s process exists due to the quasi static process equilibrium. One thing leads to the other. Reversible ensures the initial state of system & surrounding is establish able or restorable at every point in time during the process. And Drop or rise In the amount of heat, pressure, temperature, volume, are in included in this continual ability of the process to restore. Thanks
The hypothetical reversible process can be controlled to take any path. Example: it could constructed from several infinitesimal reversible isobaric and isochoric processes
For example, we have something on the piston, and this pigeon is mixed down and the piston is pushed because of the pressure of the pressure. If this mass is uniformly increased gas is expanded and the work performed by gas at this time will be really equal to the work that is calculated by quasi-static assumption? This means that quasi static process does not proceed with this process, which means that in any fixed time frame of the process there will be no different pressure and temperature. But then this quasi static process theoretical model is simply and such processes do not actually exist. Is not it?
Why do we need to consider reversible processed? Is it to compare thermodynamic efficiency or COP of various engines, heat pumps & refrigerator cycles with Carnot engine efficiency? Because the Carnot efficacy gives u the maximum possible efficiency & COP?
Please correct me: Reversibility ensures heat rejected to the surrounding is equal to the heat added to the system in the forward & backward isothermal expansion & compression. It ensures the change of entropy of the universes remains at 0?
Just like this process from 1 to 2 has to be extremely slow to make it a reversible process, adiabatic expansion process should go fast to keep it reversible? If so, why?
if we will take all balls at the same time at this time gas will complete a work and this work will be equal to work which we calculate cvazistatistical acces?
In quasi static, equilibrium & reversible process the the macro states of the system are continually going up or down without changing the equilibrium of the system, conversely not changing the equilibrium of the micro states of the surroundings, plus the net change of heat energy added to the system is equal to the heat energy rejected to the surrounding? Could u please verify this understanding? Thanks
For example, we have something on the piston, and this pigeon is mixed down and the piston is pushed because of the pressure of the pressure. If this mass is uniformly increased gas is expanded and the work performed by gas at this time will be really equal to the work that is calculated by quasi-static assumption? This means that quasi static process does not proceed with this process, which means that in any fixed time frame of the process there will be no different pressure and temperature. But then this quasi static process theoretical model is simply and such processes do not actually exist. Is not it?
Bc as it expands, work is flowing outside the system which means it's losing internal energy (since U=W+Q), the consequence being the system (or gas in this case) will cool down. You can easily prove this with spray cans, the more you discharge it the colder it'll get.
@@anguyen1920 Its is just ideal, but he didnt got into the reason on why it happens. I think i has somthing to do with the fact that when you compress a gas slowly the tempeture stays constant but when you do it fast it heats up. Cant yet understand why tho
The slides are ugly. what do you intend to do by showing such a lousy slide. it looks like a fish market. even though your content is proper and nice but the slides make it bad.
Better call Sal
Khan academy is the greatest thing to ever happen to education. 🙌🏾🙌🏾🙌🏾
You are simply amazing.
the concept seems so hard when you read it in the textbooks,but you made it look like the easiest thing in the world.
thanks a ton!:D
Please share me..what you understand here...what is quasi static process and what is reversible ? I cannot understand for this poor slides.
When I think infinitesimally small, I think calculus.
Same but i think this idea is very intertwined with calculus
that's why we have calculus, we need instantaneous values..
Why is my textbook so bad at explaining things? Khan Academy should write textbooks. :p
Please share me..what you understand here...what is quasi static process and what is reversible ? I cannot understand for this poor slides.
Khan Academy has its own way of explaining concepts
TH-cam is a way better teacher than any school teacher
Khan academy has revolutionized education
But I still feel it's underrated
A few less "little-littles" and "super-doupers" would've been enough, but very well explained!
Reversible processes are quasistatic and most quasistatic processes are reversible. Got it!
Now thats what I was looking for!!!!!!!!!!
This is unbelievably helpful. Thank you!!
Truth is when you face a problem you have a better solution for it to prevent others from facing it like you did in understanding reversible process
Completely Understood. Thank You.
2:25 all hell broke loose!!!! Lol
Awesome explanation searching the answer for almost 4 years but here I got it... Thankyou so much
Indeed! superbly well explained. lots of thanks
This man is a hero
just wow!! khan academy should write textbooks :)
10:58 Example of irreversible quasi static process?
thank. this really helped me understand these terms
This is the kind of lesson where you understand everything but you're like 'ok so what's the point?'. Ok so a system is in equilibrium when it has the macrostates well defined, the macrostates are well defined when process is really slow and we call that 'cvasistatic' and if we do this really slow process backwards, and we presume that there is no energy loss we have a 'reversible process'. Great, now how do we relate that with the real life pistons and the next chapters in Thermodynamics? Are engineers trying to create pistons that can keep as much as possible states of equilibrium between movements ? Having states of Equilibrium helps us finding out the work that a system does more easily? If cvasistatics processes are not real life relatable what's the point that we're making? etc
Great lesson and a huge thank you for having this matterial for free for everyone that's willing to learn !
If macro state variables are not defined when u blew out half the load, then on what bases we use the lower of the pressure to calculate irreversible work done by the system, p2x(v2-v1).
Thanks
Thank you so much for these videos.
It's like loss of heat is loss of energy and loss of energy is loss of information, if we lose energy due to heat we will never be able to go back or reverse to one macrostate because we don't have the same energy as in the beginning.
Could u explain reversibility in case of adiabatic compression & expansion?
Thanks
So why would volume not be well defined between equilebria? Are you saying the volume is kind of independent from the dimensions of the cylinder when there is no equilibrium?
Oscillation of the piston which only died down because there will be friction in reality
Little bit late but may help others: What I understand is that pressure, volume and temperature are connected - if you compress the gas (decrease the volume) either pressure or temperature has to rise. Other way if you decompress the gas by pulling the piston up and making the volume bigger then pressure or temperature will decrease. So after instantaneous removing the rock the pressure causes a fast decompression, so fast that you can't measure the temperature and pressure inside. Depending on what was used in the experiment the pressure will overshoot the equilibrium same way the spring does. Then the gas will want to compress because there is still half of the rock pushing the piston down. Back and forth, back and forth and after some time it gets to the equilibrium and stops. This is the moment you can measure pressure, volume and temperature.
Tl;dr: You could define volume between equilibrium, but it would make no sense as there are other factors - pressure and temperature - that you need to take into account.
The best teacher
Can we say
Quasi static process leads to equilibrium
Equilibrium means temperature & pressure is homogeneous
Reversibility in s process exists due to the quasi static process equilibrium.
One thing leads to the other.
Reversible ensures the initial state of system & surrounding is establish able or restorable at every point in time during the process.
And
Drop or rise In the amount of heat, pressure, temperature, volume, are in included in this continual ability of the process to restore.
Thanks
Side note: the graph should be a pv=constant graph as opposed to linear graph.
The hypothetical reversible process can be controlled to take any path. Example: it could constructed from several infinitesimal reversible isobaric and isochoric processes
Thanks Sal!
For example, we have something on the piston, and this pigeon is mixed down and the piston is pushed because of the pressure of the pressure. If this mass is uniformly increased gas is expanded and the work performed by gas at this time will be really equal to the work that is calculated by quasi-static assumption? This means that quasi static process does not proceed with this process, which means that in any fixed time frame of the process there will be no different pressure and temperature. But then this quasi static process theoretical model is simply and such processes do not actually exist. Is not it?
Thank you sir!
Thank you sir, may Allah bless you
Thank you!
Why do we need to consider reversible processed?
Is it to compare thermodynamic efficiency or COP of various engines, heat pumps & refrigerator cycles with Carnot engine efficiency?
Because the Carnot efficacy gives u the maximum possible efficiency & COP?
Good 😊
Please correct me:
Reversibility ensures heat rejected to the surrounding is equal to the heat added to the system in the forward & backward isothermal expansion & compression.
It ensures the change of entropy of the universes remains at 0?
Love the explanation
Excellent
I LOVE YOU!!! YOUR A LIFE SAVER!!!
Is mangenta colour your favorite?..
thank you soooooo much
Just like this process from 1 to 2 has to be extremely slow to make it a reversible process, adiabatic expansion process should go fast to keep it reversible?
If so, why?
Thank U a lot.....
if we will take all balls at the same time at this time gas will complete a work and this work will be equal to work which we calculate cvazistatistical acces?
wow!
thank you :)
In quasi static, equilibrium & reversible process the the macro states of the system are continually going up or down without changing the equilibrium of the system, conversely not changing the equilibrium of the micro states of the surroundings, plus the net change of heat energy added to the system is equal to the heat energy rejected to the surrounding?
Could u please verify this understanding?
Thanks
For example, we have something on the piston, and this pigeon is mixed down and the piston is pushed because of the pressure of the pressure. If this mass is uniformly increased gas is expanded and the work performed by gas at this time will be really equal to the work that is calculated by quasi-static assumption? This means that quasi static process does not proceed with this process, which means that in any fixed time frame of the process there will be no different pressure and temperature. But then this quasi static process theoretical model is simply and such processes do not actually exist. Is not it?
What happens if sand remove at once????
Gracias.
Why every reversible process is mandatory to be quasi-static??
what is quasi-static process?
What is equilibrium here? I mean, between what things , we are talking about equilibrium
That the stuff inside the container is homogenous and thus the properties be described by a single number, P,V,T.
I would have chosen the job of sal's car driver if he allowed me for 1-2 yrs..
HOW WOULD THE GAS HAVE A LOWER TEMPERATURE AFTER THE WEIGHT IS BLOWN OFF
Bc as it expands, work is flowing outside the system which means it's losing internal energy (since U=W+Q), the consequence being the system (or gas in this case) will cool down. You can easily prove this with spray cans, the more you discharge it the colder it'll get.
You didnt explaine WHY there isnt a lost of energy.
I think this is just a concept in an ideal situation (not achievable yet in real life) where there's no lost in energy
@@anguyen1920
Its is just ideal, but he didnt got into the reason on why it happens.
I think i has somthing to do with the fact that when you compress a gas slowly the tempeture stays constant but when you do it fast it heats up. Cant yet understand why tho
all hell broke loose!
similes of youth vera the eld
about information
Entropy irreversible
Simply waste
The slides are ugly. what do you intend to do by showing such a lousy slide. it looks like a fish market. even though your content is proper and nice but the slides make it bad.
bro it was made in 2009 back then only MS paint existed to work with.
Thanks a lot