Don't discount the fact that by now you're hearing this information for at least the second time. Try doing things the other way next time. Watch Khan first, then go to lecture. I promise you will enjoy lecture more than you do now. You will be prepared to work with this information on a new level in class and you will likely be able to ask better questions, which will only help you understand this stuff. Perhaps that may be the source of your trouble, too, that you don't ask questions. I don't know you, but I do know that most students remain quiet and confused, and thus frustrated at their professors or the content. It seems like most don't want to look stupid in front of peers. It's okay to look stupid when learning things you've never learned before. (You're all stupid! Why wouldn't you be? You've never done this before.) Regardless of the expectations you have for yourself, I promise you your instructor is not expecting you to completely understand this stuff after hearing it _one_ time. No one expects that. For the love of who or whatever you believe in, ask questions. If you're afraid to speak up in class, send an email, go to office hours, hell do your homework at office hours. "That's so awkward!" Oh well. This is why you pay tuition, not to sit in a class and hear something once, but to actually learn the material; learning requires this work on your part though, it's an active process. That's how to use your professors, and get the most out of a mortgage's worth of student loans you're going to have. Khan is great, but realize that you had to take time out of your day to participate in your learning. Your professor and Khan are both available, but it's up to you to actually utilize the resource.
You're animations are just fine man.. I've never taken formal physics courses, but for 10+ years now I've done my own research via books,magazines etc... Your videos by far are the most clear and to the point..your analogies are tough sometimes..but het..that's what pause and repay are for! Keep it up brotha! This is much appreciated!!
The day after tomorrow is my physics exam. While revising thermodynamics , i was facing some difficulties to understand this topic. Thanks to Sal for explaining the concept in a easy way. Love from Bangladesh❤️❤️
Thank you sir. Your video is worth a million dollar instead of reading the boring physical textbook. As long the concepts are grasp properly, it will stay with you forever.
All of these videos are really just incredible. I'm a sophomore engineer and I still find these videos really helpful! I wish there was some way to help you put out these videos. Really, really admirable--thank you
I can't thank you enough!!! SAL!!! MA MAN!!!! seriously, I was really troubled by this... I'm so grateful that youtube exists and there's a channel called khan academy
all lectures deliverd by this HONOURABLE professor are soooooooooo nice. i like all the lectures and enjoyed. i realy appreciate wow. keep it up sir. May God give u long life to educate us
shitty microsoft paint-esque program + 20 minutes = lucid explanation of the Carnot cycle....amazing!!! Thank you so much!!! I feel more comfortable with thermo than ever!
adiabatic process means isolated, so no heat transfer can ocurr,the system travels trough an isoterm and if the temperature doesn`t change for our system, neither do the Internal energy is going to change, so is constant. But in and adiabatic process the temperature of the system is not constant is not travelling along the isoterm anymore.
chemistry textbooks u = q + w physics textbooks u = q - w chemistry defines w as work done on the system physics defines w as work done by the system same equation MIND YOUR SIGNS
Shouldn't delta-U have a specific value considering for an adiabatic process Q = 0 always? In that example, W(AB) = 0; no work has been done. But the gas did do work considering the change in Volume... idk maybe I'm not seeing it.
I thought you couldn't draw the line from A to B because we can't measure the system while it's in flux...or do we always assume it's a quasistatic system...or can we draw the path because the system is kept at a constant temperature?
How do we know that Q2 < Q1? That is, how do we know that a smaller amount of heat is transferred from the system, between C and D, than to it, between A and B?
From B to C where is the heat flow? The temperature is changing from T1 to T2 therefore its not isothermal. Once the reservoir is gone the system is completely isolated so the heat exchange would stop.
So A to B is adiabatic and so is B to C? I'm sorry but isn't T2 going into the system? Or is this a system definition issue? Thanks for the video, though. Very good!
Our internal energy did not change? I thought pressure was kinetic energy, and if the pressure went down the internal kinetic energy went down? Just like if you removed pressure from a Co2 cartridge by firing a pellet gun, the internal energy changes and each successive pellet will fire at a slower velocity.
If the internal energy of the ideal gas from 1-2 was zero ( the work done (1-2) was to increase the volume and decrease the pressure ) , then what is the internal energy of the ideal gas during adiabatic expansion ( process 2-3 )
Wouldn't that also make it redundant? I mean Carnot cycle is utilised in an engine right.. you made an engine because you wanted to do work on something.. but if you have to do some outside work to make that happen.. then you're basically doing work to get work. Doesn't make sense to me. But ofcourse I'm obviously missing something. I just don't know what that is.
They were actually adapted from the rotation of a train wheel, when as the rocks decrease, it indicates the axes start to have less push on the wheel (or backward resistance), closing in on 90 degrees to the forward motion of the train. Keep on going, and the wheel’s spin pushes the piston in the opposite direction, compressing the gas and then there’s the next 2 stages, completing the cycle.
What I don't get is that you can easily devise a system where the piston does work when kt moves both ways. So in that case both increasing and decreasing volumes would mean the piston is doing work. So you can just add and remove the same qty of heat and have the system do work, whicb would violate the first law.
How is adiabatic process digrammed with isothermic process in 2D and in isothermic process T is const while in adiabatic process T is variable so there must be 3D diagram For 3 variables P&V&T
I think something's missing in the explanation: In the end, the system is at the exact same state as in the beginning: the same amount of pebbles at the same height. So how is there a net work done by the system?
This video was released when I was in the kindergarten
Music Forever ! Same
Same, I was 5!
same!
I was learning how to talk than
I was 8 xD
Haven't started watching this explanation yet, but I already know its going to be a shitload better then my lecturers
Don't discount the fact that by now you're hearing this information for at least the second time. Try doing things the other way next time. Watch Khan first, then go to lecture. I promise you will enjoy lecture more than you do now. You will be prepared to work with this information on a new level in class and you will likely be able to ask better questions, which will only help you understand this stuff.
Perhaps that may be the source of your trouble, too, that you don't ask questions. I don't know you, but I do know that most students remain quiet and confused, and thus frustrated at their professors or the content. It seems like most don't want to look stupid in front of peers. It's okay to look stupid when learning things you've never learned before. (You're all stupid! Why wouldn't you be? You've never done this before.) Regardless of the expectations you have for yourself, I promise you your instructor is not expecting you to completely understand this stuff after hearing it _one_ time. No one expects that. For the love of who or whatever you believe in, ask questions.
If you're afraid to speak up in class, send an email, go to office hours, hell do your homework at office hours. "That's so awkward!" Oh well. This is why you pay tuition, not to sit in a class and hear something once, but to actually learn the material; learning requires this work on your part though, it's an active process. That's how to use your professors, and get the most out of a mortgage's worth of student loans you're going to have. Khan is great, but realize that you had to take time out of your day to participate in your learning. Your professor and Khan are both available, but it's up to you to actually utilize the resource.
@@TurdFurgeson571 Some teachers really are just bad at teaching.
You're animations are just fine man.. I've never taken formal physics courses, but for 10+ years now I've done my own research via books,magazines etc... Your videos by far are the most clear and to the point..your analogies are tough sometimes..but het..that's what pause and repay are for! Keep it up brotha! This is much appreciated!!
10 more years have passed since your comment, how is your research going ?
The day after tomorrow is my physics exam. While revising thermodynamics , i was facing some difficulties to understand this topic. Thanks to Sal for explaining the concept in a easy way.
Love from Bangladesh❤️❤️
Thank you sir. Your video is worth a million dollar instead of reading the boring physical textbook. As long the concepts are grasp properly, it will stay with you forever.
You are so much better than my thermo professor....
All of these videos are really just incredible. I'm a sophomore engineer and I still find these videos really helpful! I wish there was some way to help you put out these videos.
Really, really admirable--thank you
amassad127 donate khanacademy.org
I can't thank you enough!!! SAL!!! MA MAN!!!!
seriously, I was really troubled by this... I'm so grateful that youtube exists and there's a channel called khan academy
all lectures deliverd by this HONOURABLE professor are soooooooooo nice.
i like all the lectures and enjoyed. i realy appreciate wow. keep it up sir. May God give u long life to educate us
For some reason I kept getting confused about how Q1 was coming into the system and Q2 was going out. This video helped so much!
im taking this right now with a prof that graduated from MIT as well ;)
thanks for all the help sal
I love captions so much...especially when I find that I can open a Chinese captions.感觉世界充满关怀
I was struggling to understand this. Thank you for this!
Fell much more comfortable about my Therm exam on Monday now, thanks!
It’s kinda crazy when I fully understood the mechanism behind it. It gave me that”aha!” moment. Again, I am thoroughly impressed by Khan academy
Such a great explanation. I'd forgotten just how incredible Khan Academy was.
wow i have to say this is awesome. I didnt think i would find a video that would explain carnot cycle to a point that i would understand. Thanks man !
I want to know what drugs carnot was doing, they seem pretty good!
Lmfao
Existential crisis.
why do i pay tuition
You summed up college for me.
Paagal
@@audacioustux Jesus. Poor guy just asked a question man.
but I feel why do I pay even to college?
For that piece of paper called a degree
these lectures should be put into number order
10:13 I think as pressure is decreased, the volume is increased. You said volume goes down. But awesome work here.
Teach forever sir
Very comprehensive video on Thermal cycles, nice work!
not finished repeating it but already IMMENSLY helpful
Simply made too simple by khan
i would like to see such a talented professer!!!!!!!
Thank you for the video! You made it all so simple to understand
also, at 7:00 : im taking pchem I right now, and we do define deltaU=q+w. so qab=-wab.
shitty microsoft paint-esque program + 20 minutes = lucid explanation of the Carnot cycle....amazing!!!
Thank you so much!!! I feel more comfortable with thermo than ever!
Thanks for helping me in learning it.
From B to C (adiabatic) process, why does Volume keep increasing since there's no change in heat transfer and there's no source / reservoir?
Fabrice Vieillesse when adiabatic expansion is taking place, the piston is just allowed move out. Therefore the gas expands increasing the volume.
Thankyou sir
at 10:11 Salman Could have meaned "my pressure would have kept going down and my volume would have kept going up"
Great Video.... I would have failed my physic xm if I wasn't there....
+Abhisek Upadhyaya no shit sherlock
+Ali Javed
Give him some ice for that burn
Janmay Patel lol :P i had forgotten about this
in both process the volume is changing going up during expansions and going down in a compresion
adiabatic process means isolated, so no heat transfer can ocurr,the system travels trough an isoterm and if the temperature doesn`t change for our system, neither do the Internal energy is going to change, so is constant.
But in and adiabatic process the temperature of the system is not constant is not travelling along the isoterm anymore.
TYSM 🥰
chemistry textbooks u = q + w
physics textbooks u = q - w
chemistry defines w as work done on the system
physics defines w as work done by the system
same equation
MIND YOUR SIGNS
THANK YOU!
8:43 "I should be talking about thermodinamics not drawing" lol
Thank you ❤️
Thank you dear!
useful engine ... .. childhood comes to mind D,:
I tripped on your shit. I understood it all. thanks.
Thank u so much! U r doing a GREAT work there...
Haaa if i had access to internet when i was a student....
you deserve a Nobel prize..nuf sayed
Shouldn't delta-U have a specific value considering for an adiabatic process Q = 0 always? In that example, W(AB) = 0; no work has been done. But the gas did do work considering the change in Volume... idk maybe I'm not seeing it.
Thanks a million 🙏❤❤❤❤❤
I understood, work done by the system is transferred the heat from hot reservoir to cold reservoir.
I thought you couldn't draw the line from A to B because we can't measure the system while it's in flux...or do we always assume it's a quasistatic system...or can we draw the path because the system is kept at a constant temperature?
Yeah you assume quasistatic
hey why the slop in the graph is more steaper at BC
How do we know that Q2 < Q1? That is, how do we know that a smaller amount of heat is transferred from the system, between C and D, than to it, between A and B?
all your videos are very helpful. thank you!
Thank you so much for your helpful tutorial! :D
From B to C where is the heat flow? The temperature is changing from T1 to T2 therefore its not isothermal. Once the reservoir is gone the system is completely isolated so the heat exchange would stop.
I just love your videos vai,love from Bangladesh.
first 2 mins ----> instant subscribe
THANK YOU, i understand now
its awesome i can learn phy very easily
Thanks khan academy👌👌All ur lactrs r v Funnt and amazing
ı wish ı could like it for billion times hahhah
So A to B is adiabatic and so is B to C? I'm sorry but isn't T2 going into the system? Or is this a system definition issue? Thanks for the video, though. Very good!
great video! it help me a lot
there is no reservoir from D to A as no heat is exchanged in a adiabatic process.
During the isothermal expansion process, does the system simultaneously absorb heat while doing work?
I understand. Work done went
wow!!! Thanks Sal.
bcos the volume expanded that's why the temperature decreases.
man you're really cranking these out, it's hard to keep up
11 years,
How those videos affected you?
Very good thanks Very much!!
thAnks
Brilliant video, you have helped me so much! Thank you :)
Great explanation!!keep it up
Thank's from Germany!
Thanks
Our internal energy did not change? I thought pressure was kinetic energy, and if the pressure went down the internal kinetic energy went down? Just like if you removed pressure from a Co2 cartridge by firing a pellet gun, the internal energy changes and each successive pellet will fire at a slower velocity.
If the internal energy of the ideal gas from 1-2 was zero ( the work done (1-2) was to increase the volume and decrease the pressure ) , then what is the internal energy of the ideal gas during adiabatic expansion ( process 2-3 )
Appreciate ur work!
Thanx so much!
One Question: By removing the Rocks, you have to use Work from the outside - would'nt that make the process irreversible?
Wouldn't that also make it redundant?
I mean Carnot cycle is utilised in an engine right.. you made an engine because you wanted to do work on something.. but if you have to do some outside work to make that happen.. then you're basically doing work to get work.
Doesn't make sense to me.
But ofcourse I'm obviously missing something.
I just don't know what that is.
They were actually adapted from the rotation of a train wheel, when as the rocks decrease, it indicates the axes start to have less push on the wheel (or backward resistance), closing in on 90 degrees to the forward motion of the train. Keep on going, and the wheel’s spin pushes the piston in the opposite direction, compressing the gas and then there’s the next 2 stages, completing the cycle.
Hi,
Love your videos, is there a way to watch just thermodynamics ones? Thank you
May allah (God) give u long life to educate us
Why temperature should goes down in 1 st and 2nd step if pressure is decreased and volume incresed ( if reservoir wasn't there)
Wow😍
He sounds like Tom Hanks at 1.25x speed lol
Yeah lol
Can we say that the work done to the system from A to C is equal to mgh of the stones where h is the distance traveled by the piston?
@TheMIkex24
He accidentally took the wrong word. The pressure is going down and the volume is going up, as you can can see in the figure.
What I don't get is that you can easily devise a system where the piston does work when kt moves both ways. So in that case both increasing and decreasing volumes would mean the piston is doing work. So you can just add and remove the same qty of heat and have the system do work, whicb would violate the first law.
THANK YOU SO MUCH!!!!
Gracias.
Good work.
How is adiabatic process digrammed with isothermic process in 2D and in isothermic process T is const while in adiabatic process T is variable so there must be 3D diagram For 3 variables P&V&T
It's a projection. The third variable is just mentioned.
it's a bit messy to do 3d and 2d is understandable
A graph is always drawn between independent variables
P,V&T altogether are not independent i.e. if two are known ,we can find the value of the third.
I can’t thank you enough :)
Do you use the mouse or a tablet? love your work! so awesome!
of course tablet
what do you do if it isn't a mono atomic ideal gas and if we are dealing with steam...?
👌👌explaination
from 2021
Same thing. Thermodynamics is one of the subsubjects of chemistry, even physics.
I think something's missing in the explanation: In the end, the system is at the exact same state as in the beginning: the same amount of pebbles at the same height. So how is there a net work done by the system?
because work is a path function, not a state function.