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Manipulate formula

  1. Apr 2, 2016 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    begin with ##H = U+PV##. Manipulate it into the form: ##(\frac{∂H}{∂U})_P##
    hint: you might find the resulting expression from part a useful)

    the resulting expression that I found was ##(\frac{∂U}{∂V})_P = \frac{C_V}{V*a} + π_T##


    2. Relevant equations
    ##π_T## = ##(\frac{∂U}{∂V})_T = T(\frac{∂P}{∂T})_V - P##

    ##a = \frac{1}{V}(\frac{∂V}{∂T})_P##

    equation of state
    ##dU = TdS - PdV##
    ##dH = TdS + VdP##
    3. The attempt at a solution

    I was able to solve part a. what is throwing me off is that the terms are not partials, but I need to manipulate it so.
    What I tried was to say that ##(\frac{∂H}{∂U})_P = dU##
    but... I don't think that's right, and if it is, I don't know where to go from there...

    you start with dU in the first equation. ##dU = C_VdT + π_TdV## and the resulting equation is what I wrote above... Maybe I plug that in for dU? but then what about the hint saying to do something with the resulting equation.

    maybe plug in the dU and dH... but then again, I keep coming back to that hint...

    edit: I forgot to write what a is equal to. adding that now
    edit2: ok it's there. I am pretty sure I have everything now. If I am missing something, just ask. I'll try my best.
     
    Last edited by a moderator: Apr 3, 2016
  2. jcsd
  3. Apr 3, 2016 #2

    grandpa2390

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    anyways. my problem is that I have to turn ##H = U + PV## and manipulate it into the form ##(\frac{∂U}{∂V})_P##

    I have to take that first equation and somehow turn it into ##1 + \frac{P}{\frac{C_V}{V}_A+π_T} = \frac{C_v + (P + π_T)V*a}{C_V + V*a*π_T}##
     
  4. Apr 3, 2016 #3
    I do not understand what is your problem. From my side i will like to clarify about what differentials you are writing from the content of thermodynamics. First some clarification about δ, ∇, ∂ and d. Suppose there is a function f(x) of only one variable, then ∇, d and δ are enough
    ∇x means (x2 - x1) and ∇f(x) = f(x2) - f(x1) with no conditions on how close x2 is going to x1. When we use δ, like δf and or δx, it is understood that at some stage it will tend to zero. But df or dx means surely it is tending to zero. ∇ means difference, δ means infinitesimal difference and d means differential. So we have y = f(x) or

    dy/dx = f'(x) = lim (δx tending to 0)[δy/δx] so we have another function. Just to revise
    ∇f = f(x2) - f(x1), δf = f'(x)δx is approximately true for any interval δx and tends to be true as δx tends to zero
    But df = f'(x)dx is an exact equation of differentials.

    Now coming to function of two independent variables say F(x, y), The exact differential dF can be written as
    dF = {(∂F/ ∂x)suffix, y}dx + {(∂F/ ∂y)suffix, x}dy
    This is all mathematical. We need not write the suffix when the two variables are fixed and always remain the same, But when the same function can be considered as a function of a large number of pairs of independent variables as happens in Thermodynamics. So in the per-view of thermodynamics there does not exist dF/dx at all dF exists but not dF/dx, it is always (∂F/ ∂x)
    This mathematical insight you must have.
    Coming to Physics,
    dU = TdS - PdV, is the mathematical formulation of the first two laws of thermodynamics for a closed system in equlibrium, You cannot get it from mathematics.
    The mathematical conclusions which you can draw from this equation of differentials are the following:
    1. U can be considered as a function of two independent variables S and V
    2. T and P are also functions of S and V and in fact
    3. T = {(∂U/ ∂S)suffix, V} and P = - (∂U/ ∂V)suffix, S}
     
  5. Apr 3, 2016 #4

    NascentOxygen

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    Perhaps you intend Δ here?
     
  6. Apr 3, 2016 #5
    Yes nascent oxygen I made that mistake there.
     
  7. Apr 3, 2016 #6
    Kindly read Δ(delta) this for ∇(grad) in my post no. 3 at 9.35 pm
     
  8. Apr 3, 2016 #7

    grandpa2390

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    never mind I will figure it out myself. I wrote the problem exactly the way he did. What I don't know is how to solve this problem.

    maybe I need to write out the solution to the first part in order for you to see what I am supposed to be doing, but writing all of that in LaTeX is a royal pain for zero gain. I would have to break forum rules and just post a picture of the work. I'll just keep working at it. thanks for you attempt :)
     
  9. Apr 3, 2016 #8

    grandpa2390

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    @Let'sthink
    current progress

    I am reverse engineering the problem. Trying to go backwards from the solution to the problem. I have made a little progress.

    ##H=U+PV## after a little manipulation becomes ##(\frac{∂H}{∂U})_P = 1 + (\frac{P}{\frac{∂U}{∂V}})##
    or maybe you just say ##∂H = ∂U + PV## and ##\frac{∂H}{∂U} = 1 + \frac{PV}{∂U}##
    but it would have to be ##1 + \frac{P∂V}{∂U}##

    How to get a ∂V ....

    I'm still working on it, but feel free to jump in with advice if you can.

    is it because P is the constant. if P is constant you automatically make V a ∂V as well?
     
    Last edited: Apr 3, 2016
  10. Apr 3, 2016 #9

    grandpa2390

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    or maybe since we are dividing PV by ∂U
    since U is a function of S and V, then automatically it can be assumed that it is ∂U/∂V since there is no S ?
     
  11. Apr 3, 2016 #10
    That is what i am trying to tell you grandpa2390, expressions like, "∂H=∂U+PV∂H=∂U+PV" are meaningless in mathematics which is alanguage of Physics. The possible expression is: dH=dU+d(PV). Now I write some correct mathematical steps:
    H = U + PV is function defining relation. The very fact you want (∂H/∂U)P(suffix) means you are considering H as a function of two independent variables U and P
    (∂H/∂U)P(suffix) = (∂U/∂U)P(suffix) + (∂(PV)/∂U)P(suffix) = 1 + {P*((∂V/∂U)P(suffix)}. You cannot find fault with these expressions, you msut note that here V is considered as a function of U and P.

    I do not understand your problem and also what is π in quoted expression,dU=CVdT+πTdV, Is T suffix or multiplicand?
    Tell me what you wishto prove and leave it for us whether we prove it forward or reverse. State your problem clearly and unambiguously. Whatever work you are doing each step in it must be a correct step mathematically and you should be able to explain it unambiguously.
     
  12. Apr 3, 2016 #11

    grandpa2390

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    That is what I needed. :)
    dH = dU + d(PV)
    P is constant so dH = dU + PdV
    dH/dU = 1 + P(dV/dU) I feel so stupid. Such a stupid simple mistake to take the derivative of H and U but not PV.

    thanks!
     
  13. Apr 3, 2016 #12

    grandpa2390

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    I just needed my simple mistake corrected. I am familiar with Multivariable calculus. I just made a dumb error.

    As for Pi sub T, I gave the value above in the related formulas. But it is not really relevant because dV/dU has the Pi sub T already in it and simplified. So no need to worry about that. it is all of the partials that need to be converted using the Thermodynamic identities and Maxwell Relations.

    Anyways thanks for pointing out where my mistake was. After you pointed out that I had made a mistake by taking a partial of just one term rather than the entire thing, I can take it from there :)

    Thanks!
     
  14. Apr 3, 2016 #13

    grandpa2390

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    ##C_VdT + π_TdV##
    the V and T are suffixes. ##C_V## is the Heat capacity at constant volume. ##π_T## is the Internal Pressure.

    The problem is exactly what I wrote. If you think it is ambiguous... blame my professor. but trust me. He is just going to say. if nobody understands it the way I wrote it, it is not my fault. Sorry you and the rest of the world is stupid. So I am not going to bother. In this case I understand it though.

    I'll walk you through it. now that, thanks to you, I know where my error was. ;)

    I love this stuff. I should have majored in Mathematics instead of Physics. It is not the Physics that I enjoy, it is the manipulation of all of these formulas and the math. When I understand it, it is fun. Like a puzzle :)
     
  15. Apr 4, 2016 #14

    grandpa2390

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    so we begin with ##H = U + PV## and we want to manipulate it into the form of ##(\frac{∂U}{∂V})_P##

    Now I know you don't like the way the notation is being used. But that is what the Physics professors do here. and they always comment that this will make the Math majors squirm... But they are Physicists not Mathematicians so they beg pardon. I don't know. Maybe my professors aren't very good. I thought it was a Physics thing in general. but maybe at MIT the instructor use the proper notation... oh well.

    so here are the steps:
    1. ##dH = dU + d(PV)## (take the derivative of the entire thing)

    2. ##(\frac{∂H}{∂U})_P = (\frac{∂U}{∂U}) + (\frac{∂(PV)}{∂U})_P## (divide by ∂U)

    3. ##(\frac{∂H}{∂U})_P = 1 + (\frac{P∂V}{∂U})_P## (P is constant so we can pull it out)

    4. ##(\frac{∂H}{∂U})_P = 1 + (P\frac{1}{\frac{∂U}{∂V}})_P## (apply the Partial Inverter to ##\frac{∂V}{∂U}##)

    5. ##(\frac{∂H}{∂U})_P = 1 + \frac{P}{\frac{C_V}{V*a}+π_T}## (Remember that ##(\frac{∂U}{∂V})_P = {\frac{C_V}{V*a}+π_T}##)

    simplify

    6. ##(\frac{∂H}{∂U})_P = {\frac{\frac{C_V}{V*a}+π_T}{\frac{C_V}{V*a}+π_T}} + \frac{P}{\frac{C_V}{V*a}+π_T}##

    7. ##(\frac{∂H}{∂U})_P = {\frac{\frac{C_V}{V*a}+π_T + P}{\frac{C_V}{V*a}+π_T}} ##

    8. ##(\frac{∂H}{∂U})_P = {\frac{\frac{C_V}{V*a}+ \frac{(π_T + P)*V*a}{V*a}}{\frac{C_V + V*a*π_T}{V*a}}} ##

    9. ##(\frac{∂H}{∂U})_P = \frac{C_V + (π_T + P)*V*a}{C_V + V*a*π_T} ##

    And that is the solution
    :)

    now on to part 2 of this question. I'll be back if I need anymore help. this should be the easy part though
     
  16. Apr 4, 2016 #15
    It suits me if you like mathematics more. But i think you are muddling with maths only. Now I write each step of yours and tell you what is wrong with it one by one.
    1.dH=dU+d(PV) (take the derivative of the entire thing)
    dH=dU+d(PV), This is correct, but the explanation in the brackets is wrong, dF is called the differential of F for any function whether of one variable or two variables.
    For one variable say x
    dF = F'(x)dx and for two independent variable, say X and Y dF = {(∂F/ ∂x)suffix, y}dx + {(∂F/ ∂y)suffix, x}dy
    So the rule used is differential of LHS = differentials of RHS and Differential of sum = sum of the differentials. shall post other things later
     
  17. Apr 4, 2016 #16

    grandpa2390

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    You keep wanting to get pedantic about it. Let me save you some trouble. It is too late. I don't even understand have of the things you are writing. LHS, RHS... I don't know what any of that means. using 50 different symbols. I am just doing it using the method/notation that my instructors use.

    I got the right answer. The way I did it, while it obviously causes mathematicians to cringe, got me the right answer. and in this class, it is the best I can hope for.

    I appreciate your effort. You want me to learn the right terminology and so forth so that I don't sound like an idiot when explaining my work. or so that my work is easier to follow. but you are trying to teach an old dog, who has only had 3 semesters of calculus, new tricks. The only thing I can understand at the moment is the way I was taught and demonstrated, and all of my energy is about how to solve these problems.

    you're just going to get frustrated with me; I'm simply not that smart. This is something I am going to have to spend a semester getting straight. I should take more math classes. But right now, I just need to be able to solve these problems. I need to just get through these last few classes so I can start teaching High School Algebra. the one thing I am qualified for ;)
     
    Last edited: Apr 4, 2016
  18. Apr 4, 2016 #17
    For the sake of completeness and that you would try to learn the correct terminology to use when and if you like I am completing my post may be it is the last. May be you will be less distraught after reading the whole of it. Your all other statements seem to be correct except the second statement is wrong and not necessary. You can get the third statement directly from 1. I had done this earlier and you had understood. Of course I do not understand π(suffix T) I will try to understand it next time you seek help.So finally my defense and your prosecution rests.
     
  19. Apr 5, 2016 #18

    grandpa2390

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    I'm not prosecuting... I'm sorry if you thought I was. I appreciate your help in pointing out my error in differentiating the one term. I was just showing you what the problem was looking for. I also appreciate you wanting to help me. if you just want to correct that one statement, I can handle that I think. It is just that the way you have been talking, you were planning on writing a textbook for me. ;)

    the pi_T is just the accepted notation for internal pressure.
    kind of like omega for ohms (I think it is omega...) a for acceleration, etc. That's all. I mean maybe I am not understanding what it is that you don't understand. and so my trying to explain it to you as if it is so simple, and I'm not even explaining the part of it that you don't understand is insulting or attacking you. in which case I apologize again.

    :) thanks for your help.

    I mean technically, I could have skipped the 2nd step. but when I am writing out my work, I like to be thorough. not because I believe the reader is a moron, or because I can't handle skipping steps. I don't I just like being thorough and clear and concise when presenting my work to someone else.... so if you want to tell me what step 2 should have said. I'm all for that.
     
  20. Apr 5, 2016 #19
    Sorry, this time it was my mistake of expression in English, I meant prosecution by me of yours and defense by me of myself again. I still maintain, although even textbooks may also write like that, it is not algebraic division by dU ((divide by ∂U, as you write), As in dy/dx, it is not dy divided by dx. It involves division of some quantities and then taking its limit. And writing dividing by ∂U is not at all acceptable because it does not have any meaning which at least du and δu have as i had explained earlier. i wish you all the best. Let us meet on a ground where we both understand each other even if it is school algebra.
     
  21. Apr 6, 2016 #20
    Your three steps:
    1.dH=dU+d(PV)dH=dU+d(PV)(take the derivative of the entire thing)

    2.(∂H∂U)P=(∂U∂U)+(∂(PV)∂U)P(∂H∂U)P=(∂U∂U)+(∂(PV)∂U)P(divide by ∂U)

    3.(∂H∂U)P=1+(P∂V∂U)P(∂H∂U)P=1+(P∂V∂U)P(P is constant so we can pull it out)
    could be replaced by just 2 steps:
    1. H = U+ PV, differentiating throughout with respect to U with P held as constant,{ This means we are considering very term as a function of U and P}, and 2
    2. (∂H∂U)Psuffix = 1+P(∂V∂U)suffix
     
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