This managed to clear up my confusion with simple differential math AND thermodynamic derivations in 7 minutes flat. I've been trying to understand this stuff for well over a year now... Thanks for the help!
Ok while i do really appreciate what u have done I do want to point out that this isnt necessarily a Legendre Transform. What u have done is combined the second law in differential form with the definition of enthalpy and u have done the right mathematical manipulations to end up at dH.However, what the legendre transform does is to change the dependence of a fonction F(x,y) on its variables. So suppose we wanted to change the dependence of F from (x,y) to (X,Y) where X and Y are the coefficients of dx and dy in the expression of dF. That is, Im saying dF = Xdx + Ydy. But yeah ur working is legit tho.
Appreciate the feedback. However, this does count as a Legendre Transform. (From Wikipedia) A Legendre transform is used to convert functions of one quantity (such as position, pressure, or temperature) into functions of the conjugate quantity (momentum, volume, and entropy, respectively). In thermodynamics, we are using this to change from one set of variables to another set of variables. It may not be immediately obvious, but you’re shifting the variables of use here.
You did a good job uploading the video but Oscar is mostly right. Just because you changed the variables does not mean it is legendre transform. Changing velocity to momentum existed before Adrian Legendre, but he proposed a process that could perform transformation on non-obvious functions or variables, mainly by taking derivatives/ partial derivatives. The essence lies on the process, outcome can be gained through other means as well.
Jorge Mercent thanks for tuning in! And yes, this is basically an application of the distributive property! Without it, the Legendre Transform wouldn’t work
I dont understand. If right now we know H = U + pV and dU= TdS - pdV, and our goal is to get dH. First, I would get dH = dU + pdV + Vdp. Second, I would plug dU = TdS - pdV in the above equation and get dH = TdS + Vdp. So, mission complete, why bother to introduce d(pV), and what is this Legendre Transform for?
This is for the enthalpy transformation. You’re free to do it like that as well. It’s essentially the same thing but I decided to get through each step to visualize how to get there!
Parsa Rahimi basically, the rules of derivatives (and I guess distributive property) allow you to combine dU + d(PV) into d(U+PV)! Does that sorta make sense?
@@parsarahimi335 My Thermo professor do this kind of thing all the time, so I think it is ok. And a lot of thermo book also do that (like Van Ness - Introduction to Chemical Engineering Thermodynamics). I think this happens because this operation is properly called the "total differential", so this isn't the derivative in the strict way, but this is just a mathematical nuance, and I think this is thing we don't have to worry so much
Really nice casual style of lesson, really intuitive! Thanks man!
Isak Hallberg thank you for watching!!!
This managed to clear up my confusion with simple differential math AND thermodynamic derivations in 7 minutes flat. I've been trying to understand this stuff for well over a year now... Thanks for the help!
Ahhh I’m glad to have been of help!!!
You're the man. Way better explanation that many other locations.
Appreciate you watching!
i am currently binging your videos for a midterm and you are so helpful! i really appreciate you uploading these :)
Thanks for watching!!!
@@BaahirJinadu and rewatching over and over again :)) !! saving my life
@@trinityc5072 I am honored!!! I know the engineering struggle so I am glad to be of help! Thanks so much!
Besides teaching, your voice is wonderful too. Perfect for recitation!! 😀
Thanks!
Super helpful, was able to finish my homework because of this video :)
I'm glad!!! Thanks for watching!
Explained very well! Thanks so much!
SuperSneakysneaker thanks for tuning in!
Ok while i do really appreciate what u have done I do want to point out that this isnt necessarily a Legendre Transform. What u have done is combined the second law in differential form with the definition of enthalpy and u have done the right mathematical manipulations to end up at dH.However, what the legendre transform does is to change the dependence of a fonction F(x,y) on its variables. So suppose we wanted to change the dependence of F from (x,y) to (X,Y) where X and Y are the coefficients of dx and dy in the expression of dF. That is, Im saying dF = Xdx + Ydy. But yeah ur working is legit tho.
Appreciate the feedback. However, this does count as a Legendre Transform. (From Wikipedia) A Legendre transform is used to convert functions of one quantity (such as position, pressure, or temperature) into functions of the conjugate quantity (momentum, volume, and entropy, respectively). In thermodynamics, we are using this to change from one set of variables to another set of variables. It may not be immediately obvious, but you’re shifting the variables of use here.
You did a good job uploading the video but Oscar is mostly right. Just because you changed the variables does not mean it is legendre transform. Changing velocity to momentum existed before Adrian Legendre, but he proposed a process that could perform transformation on non-obvious functions or variables, mainly by taking derivatives/ partial derivatives. The essence lies on the process, outcome can be gained through other means as well.
Thank you for the great explaination! @5:40, you pulled out d as if it is a common term, is that even legal?
Jorge Mercent thanks for tuning in! And yes, this is basically an application of the distributive property! Without it, the Legendre Transform wouldn’t work
Erik Anderson appreciate it!
Thanks man
Any time
Thank you a lot!
No worries! Thanks for watching!
thanks bro
Any time
Super useful dude. Thanks
tripp thanks for watching!!!
prepare well next time please
thank you for your lesson still
I dont understand.
If right now we know H = U + pV and dU= TdS - pdV, and our goal is to get dH.
First, I would get dH = dU + pdV + Vdp.
Second, I would plug dU = TdS - pdV in the above equation and get dH = TdS + Vdp.
So, mission complete, why bother to introduce d(pV), and what is this Legendre Transform for?
This is for the enthalpy transformation. You’re free to do it like that as well. It’s essentially the same thing but I decided to get through each step to visualize how to get there!
good job
Christian Penister thank you! :)
5:35 lol, please explain
Parsa Rahimi basically, the rules of derivatives (and I guess distributive property) allow you to combine dU + d(PV) into d(U+PV)! Does that sorta make sense?
@@BaahirJinadu Never seen something like that before so no
Parsa Rahimi take a look at the distributive property online. If you still don’t get it, let me know and I can make a video on it!
@@parsarahimi335 My Thermo professor do this kind of thing all the time, so I think it is ok. And a lot of thermo book also do that (like Van Ness - Introduction to Chemical Engineering Thermodynamics). I think this happens because this operation is properly called the "total differential", so this isn't the derivative in the strict way, but this is just a mathematical nuance, and I think this is thing we don't have to worry so much