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I am having trouble understanding why the excluded volume of a particle is four times its actual volume.

According to Wikipedia, "The excluded volume b is not just equal to the volume occupied by the solid, finite-sized particles, but actually four times the total molecular volume for one mole of a Van der waals' gas. 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."

enter image description hereI do understand that the centre of one particle will not be able to enter the sphere of radius 2r surrounding the other particle. But shouldn't atleast half the volume of the particle be able to enter the sphere?

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2 balls and nothing else, taking 1/4 of volume in the double size free sphere, is a model of a minimal free space for free molecule moving.

Anything less and balls get blocked or locked by each other. For distances L > 2r, molecules pass each other. For L < 2r, they block each other.

For the pressure-aware gas state equation, it effectively means like if the own molecular volume were like 4 times more than it really is.

Note that this has much more free space than the tight ball arrangement. It is obvious that much more free space is needed for balls freely passing around neighbor balls.

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  • $\begingroup$ Aren't the balls actually blocking each other? Both of them were travelling on their own paths when they collided or blocked each other right? $\endgroup$
    – Obinna
    Aug 1 at 8:17
  • $\begingroup$ Yes, they are, typically 10E10 times per second for air. The point is, not the neighbour ones on regular bases. The mean free flight path in air is 70 nm, several hundred times the molecule size and several dozen times the mean molecule distance.// Take the tightest equal ball arrangement and calculate how much you have to expand it if balls are to be just able to pass their neighbours. It could be a nice homework. $\endgroup$
    – Poutnik
    Aug 1 at 8:19
  • $\begingroup$ So like if we take those two balls and label the point where they touch as P, then we assume that both of the balls could be spinning in any direction around P and thus they form a sphere of radius 2r. Is this the case? $\endgroup$
    – Obinna
    Aug 1 at 8:34
  • $\begingroup$ What spinning ? Imaging 2 tightly passing balls. $\endgroup$
    – Poutnik
    Aug 1 at 8:36
  • $\begingroup$ Alright, thanks a lot! $\endgroup$
    – Obinna
    Aug 1 at 9:35

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