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.
@@benjaminrickdonaldsonhaha true they even my teacher did the same. And i asked him several times.he told , consider it as heat and I was like 🤷♂️ if it was that simple I wouldn't stuggle for hours at home
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..!!
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?
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
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.
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!
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.
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?
@@enanxahmed (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 is simpler to me if you consider Δ(PV), in parenthesis. Then Δ(PV) = (ΔP)V + (ΔV)P. The second term cancels out with W, and ΔH=Q in constant pressure conditions
@@Knipper2000 The non-delta values are the initial values (V1 and p1). Here (0:18), your case of variable pressure and constant temperature is explained. After a full cycle, d(p.V) = 0.
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...
@bagrasweetheart No, it will not work if V is constant. That is because work is equal to pressure times the change in volume. If the volume is held constant, then a change in pressure will not do any work. Remember, the mechanical definition of work is force times displacement. If you push against a brick wall you technically do no work.
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.
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!!
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???????
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.
@ccrunnerx92x W(by system) = -W(on system) which is Newton's Third Law of Motion (you are assuming the system obeys the assumption s of Ideal Gases, so particle collisions are elastic). He is using the work done BY the system, therefore it is negative. You can substitute in the work done ON the gas and hence it would be... dU=Q+W :)
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!
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?
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 ?
@so12what - I had the same impression, but the writings were too "readable" for paint... So there are two options, he either has some pen-like devide or is a heck of good mouse user!!! :)
Ah, midway thru the equation has delta p on one side and assumes constant pressure. Doesn't that make delta p 0 and then the whole equation falls apart?
@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 +?
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 ).
Sal, The Thermo concepts where deveolped (1800s) to extract the max work from a steam engine. They can be applied today when looking at alternate energy. Personal I don't believe in this hype. Energy at a high Temperature is not free and that's whats required to keep this PC and my lifesytle running on demand.
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.
"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 🥲
RUclips asked me to give feedback about your comment lol
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
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.
@@benjaminrickdonaldsonhaha true they even my teacher did the same. And i asked him several times.he told , consider it as heat and I was like 🤷♂️ if it was that simple I wouldn't stuggle for hours at home
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
"So I´m sitting at the beach and I have my chemistry set". A perfectly reasonable scenario haha. Love these videos.
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.
Sal takes his chemistry set to the beach. What a nerd
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..!!
Thank you for helping me not fail chemistry. Much love. -Julianne
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
way better than college explanation
thank you so much
That's what i ever wanted! Someone who DOES know how to teach Thermodynamic's law... Thank's God a lot!
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.
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!
MOTHER OF GOD I understand it! And just in time for tomorrow's chem test!! Thank you!
you are a gift by god!!! he will bless u sal!!
Your 10 minute videos teach me more stuff than a 3 hour lesson with my chemistry teacher
hi
Understanding state function is vital important in studying Thermodynamic. This are also the key to understand entropy and enthropy.
Thanks, don't know what I'd do academically without the internet!
you are literally the best teacher ever...
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.
You made this understandable (amazing)
already feel better about my exam, thanks!
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.
I wonder how awesome the teacher was who taught Sal.
i spent all night trying to get this. it took this video for me to actually get it
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.
Great video love you sal
I am from India I am just 12 and I need to learn all these but khan bro done this very easy : )
Thank You very much. I'm studying for my biophys exam and I was getting pretty helpless until I watched your tutorial..:)
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.
compressing something while keeping the internal temperature in the system the same at all times
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?
@@enanxahmed (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)
It is simpler to me if you consider Δ(PV), in parenthesis. Then Δ(PV) = (ΔP)V + (ΔV)P.
The second term cancels out with W, and ΔH=Q in constant pressure conditions
@ 10:07 d(p.V) = dp . V + p . dV; at constant pressure (dp = 0), thus d(p.V) = p . dV
@@Knipper2000 The non-delta values are the initial values (V1 and p1). Here (0:18), your case of variable pressure and constant temperature is explained. After a full cycle, d(p.V) = 0.
wow...THANKS SO MUCH
When you say deltaPV it would be nice if you would add brackets around delta(PV). Otherwise it reads like (deltaP)*(V)
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...
EXCELLENT! Thank you!
you have been my savior for my chem test tomorrow
Why is there need of Enthalpy when there are already functions like Q and U to represent energy
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.
ı am so happy because ı didnt need to listen the lecture from indian engilish. :D
"good old P-V diagram" nice :)
14 years!!!! thats huge
@bagrasweetheart
No, it will not work if V is constant. That is because work is equal to pressure times the change in volume. If the volume is held constant, then a change in pressure will not do any work. Remember, the mechanical definition of work is force times displacement. If you push against a brick wall you technically do no work.
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
Make a video on bond energy
"good ol' PV diagram" lol this guy cracks me up, not to mention knows how to explain chemistry topics
Thank you!!! This makes so much more sense to me now!
Delta p = change in pressure. since there is no change in pressure, as it is constant through out, delta p = 0. hope it helps.
Its ∆(PV). IF Pressure is constant then P∆V
meanwhile 49 other people do not believe that enthalpy is the heat content in a constant pressure system
Best thing avialable on u tube
Awesome video, thank you for the effort
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.
Thank you so, so much! I finally understand this after many years))
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!!
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???????
This was awesome
FINALLY I KNOW WHAT IS THE ENTHALPY!
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.
does this work for constant volume as well?
@ccrunnerx92x W(by system) = -W(on system) which is Newton's Third Law of Motion (you are assuming the system obeys the assumption s of Ideal Gases, so particle collisions are elastic).
He is using the work done BY the system, therefore it is negative. You can substitute in the work done ON the gas and hence it would be...
dU=Q+W :)
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!
you help me soooo much thanks
You need to put PV in parenthesis. delta(PV)
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!
Thank you, i thought i wouldn't sleep tonight. This doubt of enthalpy definition was really bothering me.
Lol Sal's idea of a vacation at the beach is so different from mine.
Great teacher
6:09 Oh you know he aint gonna "play around"
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?
fun fact: in india their is different sign conventions in physics and chemistry of work done by the system .
why does it look like we are counting pressure volume work twice ? Internal energy includes PV work. Why do you add it again when defining enthalpy.
The H is for the Greek capital Eta, not capital Aitch. It's H because Enthalpy begins with an Eta sound.
This video did absolutely nothing to tell us what enthalpy is. but only what enthalpy is during constant pressure
correct!
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! :)
@so12what - I had the same impression, but the writings were too "readable" for paint... So there are two options, he either has some pen-like devide or is a heck of good mouse user!!! :)
Ah, midway thru the equation has delta p on one side and assumes constant pressure. Doesn't that make delta p 0 and then the whole equation falls apart?
@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 +?
Did anyone hear the sound of a wave at 13:52 when he said sea level?
is change in internal energy equal to the heat added when volume is constant?
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 ).
DeltaP = 0 if constant P. You switch from the cycle to a general state. It is, in any case, too fast.
Thank you very much..
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?
Sal,
The Thermo concepts where deveolped (1800s) to extract the max work from a steam engine. They can be applied today when looking at alternate energy. Personal I don't believe in this hype. Energy at a high Temperature is not free and that's whats required to keep this PC and my lifesytle running on demand.
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.
Examkrackers type humor lol. I love it.
i love these videos. thanks
Ive just learned from this video
Delta pv=P delta V
Thank you so much
lmfao, "If i had my chemistry set at the beach and have me beaker of something and started throwing stuff into it"
This guy is so funny :D
wait but doesn't delta V and V also have to be the same? Because if they're different that doesnt cancel out the P(deltaV) with (deltaP)V?
@rscottalex I think he should have written the expression using parenthesis: ΔH = ΔU + Δ(PV)
hahaha, he's such a good teacher. what a good thing.
holy shit this was a great explanation. totally explained my thermo class on enthalpy thanks! The whole reasoning for state properties was huge