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Simple Integral Help

  1. Feb 2, 2007 #1
    I'm trying to integrate the Van der Waals equation of state for an isothermal problem, but based on my results I think I'm doing a bit of simple calculus wrong and hope someone here can help.

    [tex]P = \int \frac{RT}{v-b} dv[/tex]
    where R, b and T are known constants.

    I tried to do a u-substitution for 1/(v-b) with u=(v-b)
    so, du = 1.

    Thus. [tex] P = \int \frac{RT}{u} du[/tex]

    So, P = RT ln(u) = RT ln(v/b)

    Any ideas where I went wrong? Thanks.
     
  2. jcsd
  3. Feb 3, 2007 #2

    quasar987

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    ln(v-b) does not equal ln(v/b) to the best of my knowledge. The rule is ln(A/B)=ln(A)-ln(B)

    The result makes physical sense too. b is a measure of the intermolecular repulsion force. As v-->b, P-->oo.
     
  4. Feb 3, 2007 #3

    Gib Z

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    Ahh sorry I didn't help earlier lol, as soon as I read the first sentence, I thought I didn't know the physics involved so i couldn't help. As soon as you told me they were all constants it made it much easier.

    [tex]P=RT \int \frac{1}{v-b} dv= RT \ln (v-b)[/tex], which can not be simplified. I thinking quasar987 also pointed that out, just clarifying or sumthin..
     
  5. Feb 3, 2007 #4
    Ah, I see. Thanks for the help!

    (of course, since you are familiar with the physics, Gib z, you may see I'm trying use the relation of your namesake to fund du, and I could do the attraction potion which is the ultimately simple a/v^2.

    Again, I greatly appreciate the help!
     
  6. Feb 3, 2007 #5

    ssd

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    I do not have good idea about Physics, but mathematically we shall add the constant of integration with what GibZ has shown.
     
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