1
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • $\begingroup$ This is from the perspective of a single particle, bearing in mind that on average, the other particles are not "entering each other's spheres" because it is a gas. Should the gas condense, the packing will be tighter, i.e. each particle will use up less space than a sphere with radius 2r (compare densest packing of spheres). $\endgroup$
    – Karsten ♦
    Aug 31, 2022 at 9:43

2 Answers 2

Reset to default
1
$\begingroup$

Imagine the centre molecule to be stationary. The collision partner approaches from where you are viewing and goes straight into the image. The collision will occur if the partner is within a disc of radius $2R$ as shown in the image, that is within the hemisphere of radius $2R$ (i.e. hemisphere facing you) which is $4$ times the volume of the spherical molecule with radius $r$.

Improve this answer
$\endgroup$
0
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$
    – Pumpkin_Star
    Aug 1, 2022 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, 2022 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? $\endgroup$
    – Pumpkin_Star
    Aug 1, 2022 at 8:34
  • $\begingroup$ What spinning ? Imaging 2 tightly passing balls. $\endgroup$
    – Poutnik
    Aug 1, 2022 at 8:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.