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Obtaining an equation of state from compressibility and expansivity (integration)

  1. Sep 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Obtaining an equation of state from compressibility and expansivity.

    States of superheated steam are observed to have an isothermal compressibility k=(rNT) / (VP^2)
    and a volume expansivity B=(N/V)((r/P)+(am / T^m+1)).

    r,m and a are constants.

    a) Find dv in terms of dP and dT
    b) Deduce the equation of state for superheated steam up to an undetermined constant


    2. Relevant equations

    B= (1/V)(dV/dT) P const

    k= (-1/V)(dV/dP) T const



    3. The attempt at a solution

    a) I have arrived at:

    (rTN^2 / P^2) ((r/P)+(am / T^m+1)) = -(dV)^2 / dPdT

    I can simplify and rearange that for dV = -sqrt((rTN^2 / P^2) ((r/P)+(am / T^m+1)) dP dT


    If a) is correct, then I need help with b); how would I go about integrating the expression.

    Do I start with (dV)^2 = ?

    Thank you very much!
    Any help would be sincerely appreciated (and needed).
     
  2. jcsd
  3. Sep 25, 2014 #2

    BiGyElLoWhAt

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    I guess I'm not sure what exactly you're trying to do. Are you looking for 1 equation that relates all 3 differentials? Or are you looking for 2 equations i.e. partials? It appears as though you did some form of substitution to get the 2 equations together, but I'm not seeing what you did. Could you post your work?
     
  4. Sep 25, 2014 #3

    BiGyElLoWhAt

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    Also, I don't think your notation is quite right:

    ##\frac{(dV)^2}{dPdT}## is not the same as ##\frac{\partial^2V}{\partial P\partial T}##
     
  5. Sep 25, 2014 #4
    Hi!

    Are you referring to disparity between d and the other thing, or d^2 V and (dV)^2?

    I came by it by multiplying two equations together (dV)(dV)
     
  6. Sep 25, 2014 #5

    I did (1/V)(dV/dT) = (N/V)((r/P)+(am / T^m+1))

    (-1/V)(dV/dP) =(rNT) / (VP^2)

    then muliplied them together.

    For a) I a trying to find dV=.

    For b), I think that I am trying trying to find V= (or P= or T=)

    Thanks!
     
  7. Sep 25, 2014 #6

    BiGyElLoWhAt

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    Both actually. Im assuming that youre not looking for a partials solution.

    Also, all of temperature, pressure, and volume are changing?

    ##\frac{1}{V}\frac{dV}{dT} = B## is only true when P is constant, no? If they're changing then that relationship isn't true.
    The same for the latter equation for constant T.

    Is this for thermo or something?
     
  8. Sep 25, 2014 #7
    Tis indeed for thremo
     
  9. Sep 25, 2014 #8

    BiGyElLoWhAt

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    Also, what made you decide to multiply the equations, rather than adding or subtracting?
     
  10. Sep 25, 2014 #9
    It looked like the easier way to go about combining the equations (at the time).
    Would you suggest the latter?
     
  11. Sep 25, 2014 #10

    BiGyElLoWhAt

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    Yea I would if that's what you're intending to do. Did you read my post about the legitimacy of the equations? Double check that what you're doing is correct before proceeding (don't wanna do a lot of unnecessary work XD)
     
  12. Sep 25, 2014 #11
    Ok. I've come to:

    dV = [Nam / T^(m+1)] [1+ TdP / PdT]

    Could someone help with the integration?
     
  13. Sep 25, 2014 #12

    BiGyElLoWhAt

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    That definitely looks more reasonable. (I didn't work through it, so I'm taking what you did as correct)

    As far as the integration goes, try distrubiting the first bracketed stuff through to the second
    bracketed stuff, get your exponent for T in line, and then it should be fairly simple, as most of what you will have left is constants.

    Just to clarify, is this what you have?
    ##dV = [\frac{N*a*m}{T^{m+1}}][1+\frac{T}{P}\frac{dP}{dT}]##
     
  14. Sep 25, 2014 #13
    Just to clarify, is this what you have?
    ##dV = [\frac{N*a*m}{T^{m+1}}][1+\frac{T}{P}\frac{dP}{dT}]##[/QUOTE]

    Ya
     
  15. Sep 25, 2014 #14

    BiGyElLoWhAt

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    Actually maybe that won't be as easy to get into total differential form as I thought... :nb)
     
  16. Sep 25, 2014 #15
    Still doable?
     
  17. Sep 25, 2014 #16

    BiGyElLoWhAt

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    Haha, that's what I'm working on. It's not jumping out at me like it normally does, but I'm not necessarily a math wizard so that doesn't mean much. With this type of equation I think it should be setup as such:
    http://tutorial.math.lamar.edu/Classes/CalcIII/Differentials.aspx
    Once you get it into a form like that you can simply integrate each term, but I'm kind of struggling. I'm looking for a clever substitution for that pesky dT right now to at least get it out of the denominator in a useful fashion.

    Edit
    This function might not be nice enough for that though. Anyone else have any pointers? Otherwise I'm inclined to believe that you made some arithmetic error somewhere :(
     
    Last edited: Sep 25, 2014
  18. Sep 25, 2014 #17
    Edit
    This function might not be nice enough for that though. Anyone else have any pointers? Otherwise I'm inclined to believe that you made some arithmetic error somewhere :([/QUOTE]

    Oh dear, I may have stuffed it up (sorry!)

    I'm going through it again.
     
  19. Sep 25, 2014 #18
    There should be a dT in the brackets with the Nam.

    Does that help anything?
     
  20. Sep 25, 2014 #19
    If you ever get around to it, could you most what you get so that I can verify what I have?

    Thanks so much for taking the time to help!
     
  21. Sep 25, 2014 #20

    BiGyElLoWhAt

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    Yea no problem man, I'll try to get to it here soon... I'm trying to fix some stupid computer problems right now...
     
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