Enthalpy | Thermodynamics | Chemistry | Khan Academy
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- เผยแพร่เมื่อ 23 ก.ย. 2009
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Understanding why enthalpy can be viewed as "heat content" in a constant pressure system. Created by Sal Khan.
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"Let me do this in a different color" = Sal's catchphrase.
There should be a compilation of sal saying it
Another one: " let me use a Magenta colour!" I love it when he says that. 😆
Khan "Half of the video is me changing colours" Academy. lol.
"It's almost kinda too obvious for me to explain"
Oh Sal. You overestimate me 🥲
TH-cam asked me to give feedback about your comment lol
40 pages of textbook and five hours of lectures can't make me understand enthalpy, Khan did it in just 15 mins.
Yep, my thermo book tends to not explain concepts very well, rather it focuses almost entirely on the mathematics.
Tag me
Include me
Those books were probably explaining other stuff around enthalpy assuming you already knew what enthalpy is.
You have no idea how much appreciation I have for your videos. I'm practically self-teaching myself by watching your videos and I swear, I'm learning more than I'm lectured in class.
Truth
i would appreciate it if you could lend me your head for a day for my exams, will return to you ASAP. thanks
12 years later and you're videos are teaching us better than anything !
"if I'm sitting at the beach and I have my chemistry set" lol
13:35
Thats too nerdy even for a nerd like me
Understanding state function is vital important in studying Thermodynamic. This are also the key to understand entropy and enthropy.
Sal takes his chemistry set to the beach. What a nerd
Thank You very much. I'm studying for my biophys exam and I was getting pretty helpless until I watched your tutorial..:)
what I don't understand is why we need to know all of this. like why do we need to know HOW we get what equals to delta H.
You don't have to, if you don't want.
I think the biggest misconception about school is what we learn is important it is. But what is the most important thing about all these subjects is how you learn, not what you learn. Through learning and going through the processes you learn how to think differently and abstractly rather than the old way of 2+2=4
Hey, let's hope you are not a chemist for NASA now, since if you don't know why it only applies at constant pressure, your spaceshuttles will explode
Honestly same... I just want to get into nursing school. 😂
Its less about learning the specific lesson, and more about teaching the learning process. And putting you in situations where you have to learn and think for yourself.
Thank you for helping me not fail chemistry. Much love. -Julianne
He may sound funny, but he can explain things in such clear and simple manner only because he understood the subject very well and not just text book learning. When you think he teaches almost every subject with such mastery, he is AMAZING..!!
You made this understandable (amazing)
Thank you!!! This makes so much more sense to me now!
That's what i ever wanted! Someone who DOES know how to teach Thermodynamic's law... Thank's God a lot!
"So I´m sitting at the beach and I have my chemistry set". A perfectly reasonable scenario haha. Love these videos.
at 10:30 to 11:45 -- If pressure is constant, then delta P is zero. if you have a dP*V, you can't "factor" it to P*dV. dP means "change in pressure", so if you posit that pressure is constant, then dP= P2 - P1 = 0, because P2 = P1. Then you have dP/P = 0 = dV/V, which means that volume is also constant. What am I missing here?
It's d(PV)= dP*V + P*dV
Since dP =0, then d(PV) can be reduced to only P*dV
Cmiiw
already feel better about my exam, thanks!
Wow The best video explanation of not just enthalpy but why we need it and why it's helpful, you won't get this quality education if you paid for it.
Make a video on bond energy
way better than college explanation
thank you so much
MOTHER OF GOD I understand it! And just in time for tomorrow's chem test!! Thank you!
Awesome video, thank you for the effort
Your 10 minute videos teach me more stuff than a 3 hour lesson with my chemistry teacher
hi
You can rewrite dpv to: dpv = vdp + pdv. Which gives dH = Q + vdp. More compact and you can see immediately what happens when p is constant.
Thanks, don't know what I'd do academically without the internet!
you are literally the best teacher ever...
Thank you very much for explaining this! I needed this to solve one problem. I've been looking for this for a while. You made my day!
I wonder how awesome the teacher was who taught Sal.
EXCELLENT! Thank you!
Why don't you get a separate branch for the Indian curriculum for Chemistry like you have for Physics and Mathematics?? I would be really glad if you take this into consideration.
"good old P-V diagram" nice :)
Thank you so, so much! I finally understand this after many years))
does this work for constant volume as well?
you are a gift by god!!! he will bless u sal!!
I tried reading the P. W. Atkins book both in English and in my native language and after 50 pages I was so lost I had no frickin idea what it was talking about. thanks for explaining these things!
How about the volume is constant instead of constant pressure?
Still get same calculation or other results?
I am from India I am just 12 and I need to learn all these but khan bro done this very easy : )
holy shit this was a great explanation. totally explained my thermo class on enthalpy thanks! The whole reasoning for state properties was huge
I am still confused , if the enthalpy is state function then why it should increase after completion of cycle( as said in video that heat content is first 4 and then 6 ).
it was explained in the vedio that after each cycle heat content gonna increase in the system, it is right???
then how can we get work from system??
is change in internal energy equal to the heat added when volume is constant?
Thank you, i thought i wouldn't sleep tonight. This doubt of enthalpy definition was really bothering me.
hello, I have a question! >>>>> so enthalpy is a state variable that defines/expresses the heat content of that state at a constant pressure? which means that at p1 water at some point in the liquid state will have a heat content or enthalpy with a certain value no mater the route it got there>> what are the different possible routes that we can get there? that's what I don't understand, any materials scientists out there???????
I am a little confused here,
deltaPV=PdeltaV
say that Pressure is constant at 5 atm and volume is going to change from 6 to 7 liters.
wouldnt delta Pressure equal zero if there is no change? then you would end up with zero on the left since deltaP=0 and you would end up with 5x1 on the right because deltaV=1 and P= 5
?
deltaPV is delta(PV) not delta(P)*V
I think this will clear things up for you! :)
wow...THANKS SO MUCH
compressing something while keeping the internal temperature in the system the same at all times
i spent all night trying to get this. it took this video for me to actually get it
So is enthalpy only defined for isobaric processes? I dont see how it would be useful variable to define with any other process, as the work and P(delta V) terms wont cancel out nicely.
"good ol' PV diagram" lol this guy cracks me up, not to mention knows how to explain chemistry topics
In the textbook I have.. the first law of thermodynamics is U = q + w not q-w
I was about to ask this question
@Prachi Sharma what's the difference between work done on vs by the system?
@@spectrum792 (considering pressure-volume work)
Imagine there is a cylinder filled with gas with a piston.
If you apply pressure on the piston, the gas compresses which implies work is done on the system
If the gas expands moving the piston upwards it is said that work is done by the system
This is because in your book they consider every change in energy from te systems point of view. It is de convention in chemistry. I personaly think its less confusing. A negative q means there is heat taken from the system. A negative w means the work is done by the system. So it loses energy. This energy is now on the surroundings. That is why a negative Delta H is a exotherm reaction. Heath is coming out of the system so the energy content of the system decreases.
It changes btween chimestry and phy .the point of view is not the same(by/on)
Thank you very much..
Thank you kind sir, although it's a little difficult for me to understand since I'm only in high school chem, this helped a lot, nonetheless.
you say when that heat content would need to be incremented by 2 everytime you go around the path. but wouldn't the equivalent amount of energy be used up by work? as work is equal to heat in closed path? so the heat content should remain the same ryt? someone please answer this!!
excellent. thank you
he makes it seem so easy...if there is anyone out there that has time can you tutor me..I have in exam next week and if i dont pass it i fail the class. this is the only class that i dont understand.
what happens to enthalpy when pressure of a gas is reduced (joule thomson)
Sir of Khan Academy, congratulation, you solved a problem of Energy between to conditions of matter, but zero explanation about Enthalpy: how it got there in the first place?
Or if you want X, Y, Z didattical introduction and explanation to the the word Enthalpy as defined by the Physics. Then an example defintetly will help.
FINALLY I KNOW WHAT IS THE ENTHALPY!
But how is internal energy a state variable when it is the addition of a state variale and a non-state variable?
you help me soooo much thanks
wow which software is that to sketch?
i love these videos. thanks
If /del/ H = Q, and /del/ PV = P/del/V, doesn't that make Q a valid state variable, and consequently imply all paths are constrained to a single curve? Because I feel like the solution space for constant P sounds to restrictive, since that's only a line instead of a curve.
I am extremely confused at 10:10
So when you write "delta P V" you actually mean "delta (P*V)" ?
Because otherwise a constant P would result in "delta P=0" and then you would have "0*V = P*delta(V)" , which would not be true, right?
Am I missing something ? Does the first delta apply to both P and V or just to P?
delta (pV) = p delta V + V delta p
There's a bit of sloppy notation in the video.
how can we just assume that pressure is constant? it clearly can't be if the volume is changing. see pv=nrt.
The Volume increase is proportional to the increase in temperature. So, overall The pressure stays the same.
P=nRT/V
is that true for both an open and a closed beaker? is the open beaker the nice case where the formulas work just fine?
Why pressure is constant ? As P=1/V
14 years!!!! thats huge
DeltaP = 0 if constant P. You switch from the cycle to a general state. It is, in any case, too fast.
you have been my savior for my chem test tomorrow
Hi I need help with defining enthalpy. In chemistry we learnt that enthalpy = (q / mol of limiting reactants ) x coefficient
My question is that why enthalpy based on limiting reagent? Why is it not based on the TOTAL reacted reactant?
For example 2A+B -> C
Enthalpy = q/ (mole A+B)
What's the definition for enthalpy
fun fact: in india their is different sign conventions in physics and chemistry of work done by the system .
Can someone help me through this dilemma?
Work is defined as the (W = F * s) when a body moves a straight distance under a constant force.
However, since for a curved path of varying force, we break down the curve to infinitesimal segments where the Force is locally constant and the distance (ds) is essentially straight. Thus,
for this tiny line segment, we apply the Work concept and we have dW = Fds. Summing up all the dW would give us the total Work.
This is why dW = Fds (and never dW = Fds + sdF).
I have the following problems:
Under what circumstances is Enthalpy explicitly defined for a system?
How can we say that dH = du + PdV + VdP when we never say dW = Fds + sdF?
I know it's pretty late and you might've left these roads a lot behind in your life.
But, if you still wanna know (of course if don't already), it's actually d(PV) and not P.dV or V.dP as it is in the work done formula dW= F.ds
d(PV) expands in accordance with the chain rule of differentiation which gives us PdV+VdP. Either of these terms can get zero in an isobaric or isochoric process.
@@varunarora1256 Its been a long journey. I graduated since and currently a prospective doctoral student. And yes, I did figure it out eventually...
But thank you for taking the time to reply.
😆🥰
@@ishaaqahamed3888 ohh it's crazy how much one does in just three years 😌
Btw, a doctorate in what subject if I may ask? Is it chemistry related 😁
@louyclare wow that was a super good explanation.... couldn't have done it better myself. SO does this mean that Work in an adiabatic system is - or +?
Mathematically, why wouldn't it work if V=const.? Isn't it the same thing?
Why the Area between the two curves is the net work done? Don't it should be the sum of the individual areas under the curve?
I didn't understand how constant pressure made all the difference until now. It was an epiphany moment for me.
So, to make enthalpy equals to the amount of heat added we make the pressure constant so that in the calculation we can cancel out W and PV, right? Can you give me an example In real life, in which this condition really applies? Let's say in Energy Conversion thing, like the Heat Pump Cycle or Refrigeration... Really appreciate it, Sir!
what is isothermal compressibility?
You need to put PV in parenthesis. delta(PV)
if the entropy is close to zero, what does this say about the enthalpy- would it be exothermic, endothermic or something totally different?
even if delta H is = to -.00000000000001 it is exothermic. lol if it is plain out zero, you're doing something wrong. All reactions either release or absorb energy.
When you say deltaPV it would be nice if you would add brackets around delta(PV). Otherwise it reads like (deltaP)*(V)
meanwhile 49 other people do not believe that enthalpy is the heat content in a constant pressure system
11:05 , why isn't it:
∆PV = (Pf - Pi)V
Say P is *constant* , and P = 5
(5 - 5)V = 0
Or another way, why is there an xf and xi if there is no ∆ associated with them:
∆5x = 5x - 5x = 0
or
∆PV = ∆5V = 5(f)V - 5(i)V = 0
I assume then, that the ∆ refers to the entire "group" of variables that are multiplied together, and not just the one variable...
hahaha, he's such a good teacher. what a good thing.
it can be both, the work done 'w' is negative if the environment does work on the system, and positive if the system does work on the environment.
if heat is added to the system and system is doing equal work on surrounding then heat is converted to work and it's heat content will be constant after many cycles
This video did absolutely nothing to tell us what enthalpy is. but only what enthalpy is during constant pressure
correct!
My textbook shows deltaU = q + w. You're saying deltaU = q - w. Which is correct?
Yes, that's right. It's just a terminological thing, so some books assume the opposite
Great video love you sal
Best thing avialable on u tube
Just Perfect :)
If the system does net work, where does the extra energy to do the net work come from? FOr example, if a work is done on the system from 2-1 and then the system does work from 1-2 with a different path. We have a net work done by the system, but the system should have energy equal to the work done on it. So how can the system do extra work that is more than work applied to it?
So I understand the explanation that the change in enthalpy is equal to the head added in a constant pressure system. But mathematically, shouldn't that also be true in a constant volume system? It seems like the mathematical proof he does should apply just the same to volume in that equation as it does to pressure. Can someone explain to me what I'm missing here?
But there isnt any work done by Or on the system in case of constant volume so the way he explained in constant pressure case that heat at initial point would always increase by 2 after completing the cycle would not be there in case of constant volume... I hope it makes sense but that's what is my perspective
Lol Sal's idea of a vacation at the beach is so different from mine.
very clear
thank you :)
This was awesome
I know it's an old video, but aren't you missing a very important assumption? I mean, enthalpy is equals to heat at constant pressure IF the energy you give to the system is all heat, because if there's work involved that's also included in enthalpy (except of course the expansion work). As a more general rule I'd say that at constant pressure enthalpy = heat + work other than expansion.
if work is involved then for the pressure to remain constant there must also be some heat transfer, so the constant pressure assumption is sufficient