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How is the b term used in the van der Waals equation?

  1. Jun 22, 2015 #1
    What does the following statement below mean?

    "The excluded volume 92eb5ffee6ae2fec3ad71c777531578f.png is not just equal to the volume occupied by the solid, finite-sized particles, but actually four times that volume. To see this, we must realize that a particle is surrounded by a sphere of radius 2r (two times the original radius) that is forbidden for the centers of the other particles. If the distance between two particle centers were to be smaller than 2r, it would mean that the two particles penetrate each other, which, by definition, hard spheres are unable to do."

    if there is n molecules won't b just be 4/3r^3 where r is the radius of the gas molecule?
     
  2. jcsd
  3. Jun 22, 2015 #2

    mfb

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    Following the logic of the quoted part it would have to be 8 times the volume. Atoms and molecules are not solid spheres, however, so both numeric values don't make much sense.
     
  4. Jun 26, 2015 #3
    hmm so what would the b term be based on?
     
  5. Jun 26, 2015 #4

    It says within the 2r distance from the centre of a particle another particle cannot remain. So, volume is taken considering this 2r
     
  6. Jun 26, 2015 #5

    How come? For r, area is 4πr2,
    And for 2r it is 4π(2r)2=16πr2

    So, it is 4 times, not 8

    EDIT: sorry, the post is wrong
     
    Last edited: Jun 26, 2015
  7. Jun 26, 2015 #6
    but why do we take the volume considering 2r? Why not just r?
     
  8. Jun 26, 2015 #7
    It is something like that, a particle has radius r and another particle has also radius r. So when the
    first particle comes nearere to second particle or vice versa then the two
    CENTRES will remain at least 2r distance
    apart. Otherwise the particles will penetrate each other. It is what your post statement says.


    Though I am not sure why centre distance is considered.

    N.B. Edit is done.
     
    Last edited: Jun 26, 2015
  9. Jun 26, 2015 #8

    TeethWhitener

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    But this is volume, so [itex]V=\frac{4}{3}\pi (2r)^3=8\times \frac{4}{3}\pi r^3[/itex]. The reason you have a factor of 4 and not 8 is because this is the excluded volume for 2 particles, so you have to divide by 2: [itex]V=\frac{1}{2}\times \frac{4}{3}\pi (2r)^3=4\times \frac{4}{3}\pi r^3[/itex]
    It made enough sense to win van der Waals a Nobel prize!
     
  10. Jun 26, 2015 #9
    I've found out an answer.

    Can any body tell me how can I share a pdf file?
     
  11. Jun 26, 2015 #10
    Now I am getting confused
     
  12. Jun 26, 2015 #11

    TeethWhitener

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    Since 2 particles can't come closer than 2r to each other (for hard spheres), we get a factor of 8. But since there are two particles involved in this interaction, the factor of 8 is the excluded volume per two particles. In order to get the volume excluded per one particle, we have to divide by 2, giving us 8/2=4
     
  13. Jun 26, 2015 #12
    Hmm can you explain why we have to use 2r instead of just r to calculate the volume to exclude per molecule?
     
  14. Jun 26, 2015 #13

    Attached Files:

  15. Jun 26, 2015 #14
    But still a question arises, here it says that in these area the two particles cannot move freely, but I think a slight part of the volume is open for some other particles/molecules
     
  16. Jun 26, 2015 #15
    But even if 2 molecules can't fit into the same 'centre' area they can still be side-by-side so how can that area be excluded?
     
  17. Jun 28, 2015 #16

    mfb

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    He didn't get a Nobel Prize for thinking atoms would be solid spheres. And they are not.
    The additional volume term makes sense, but replacing atoms by solid spheres is a bad approximation, and using their radius is certainly over-stretching that model.
     
  18. Jul 1, 2015 #17

    TeethWhitener

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    Actually, he did.
    Remember, he won the prize in 1910, when there was still some disagreement as to whether molecules and atoms even existed (Einstein's explanation of Brownian motion was only 5 years old, and experiments supporting Einstein were only about a year or two old, and somewhat contradictory to boot). The van der Waals equation is theoretically important because it predicts a phase transition (which the ideal gas law does not), and it's actually a significant quantitative improvement over the ideal gas law in certain P-T regimes. In practice, the b term of the vdW equation can be fitted and the relation described in this thread gives a parameter that is known as the van der Waals radius, which is still incredibly important in things like MD simulations of biomolecules.
     
  19. Jul 1, 2015 #18

    mfb

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    He got the prize for the equation (and some other work), not for the assumption atoms would be solid spheres.
    That is of purely historic interest, however, we know much more about atoms now.
     
  20. Jul 1, 2015 #19

    TeethWhitener

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    Van der Waals radii are still used extensively in large molecular dynamics simulations. Another bit of trivia: Pauling (ref. "The Nature of the Chemical Bond") was one of the first to point out that the vdW radii of nonmetals were really close to certain features observed in their x-ray diffraction patterns (namely, they represent the distance between non-bonded structures in the crystal--e.g., the distance between layers in 2D layered compounds such as graphite or MoS2). It turns out that the hard-sphere approximation isn't a terrible one for a lot of different applications.
     
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