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Rearranging Equations

  1. Oct 18, 2005 #1
    Alright I'm going absolutely nuts here. It's a simple matter of rearranging an equation and for the life of me I can't do it. I've been working at it for like an hour now to no avail. Here's the equation.


    I'm trying to solve for v. I've manipulated this equation every way I can think of and still can't get it. I'm sure it's something simple but I'm stuck. Please help.
  2. jcsd
  3. Oct 18, 2005 #2
    you'll get a quadratic in V. You are aware of this, right?
  4. Oct 18, 2005 #3
    Nope. Completely over my head. How in the world would you get that in the form of ax^2+bx+c=0?
  5. Oct 18, 2005 #4
    Well here's the real problem. I'm trying to get the partial of (dv/dT) with constant pressure from that equation. What I was trying to do was to find v from the equation and take the partial of that side with respect to v and the partial of the other side with respect to T. Is this what I should be doing?
  6. Oct 19, 2005 #5


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    Actually, you don't get a quadratic, you get a cubic! The least common denominator of the fractions, if I am reading this properly,
    [tex]P= \frac{RT}{v-b}-\frac{a}{v(v+b)T^{0.5}}[/tex]
    (Is that what you intended? Your fractions are ambiguous.)
    is v(v-b)(v+b)T0.5, a cubic in v.
    Cubics in general are hard to solve. Since you say you really want the derivative of y wrt T, P being constant, I would recommend "implicit differentiation". Writing the equation as
    [tex]P= RT(v-b)^{-1}- av^{-1}(v-b)^{-1}T^{-1/2}[/tex]
    differentiate both sides: using product and chain rules, wrt T:
    [tex]0= R(v-b)^{-1}- RT(v-b)^{-2}v'+av^{-2}(v-b)^{-1}T^{-1/2}v'+ av^{-1}{v-b}^{-2}T^{-1/2}v'+ (a/2)v^{-1}(v-b)^{-1}T^{-3/2}[/tex]
    and solve for v', the derivative of v with respect to T.
  7. Oct 19, 2005 #6
    I think my biggest problem is I'm not really seeing how to solve partials. Anyone have any good literature about the subject. Thanks.
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