The excluded volume of molecule in motion is how many times the actual volume of a molecule in rest
Answers
Consider one mole of gas composed of non-interacting point particles that satisfy the ideal gas law:
p=RTVm=RTvp=RTVm=RTv
Next assume that all particles are hard spheres of the same finite radius r (the van der Waals radius). The effect of the finite volume of the particles is to decrease the available void space in which the particles are free to move. We must replace VV by V−bV−b, where bb is called the excluded volume or "co-volume". The corrected equation becomes:
p=RTVm−bp=RTVm−b
The excluded volume bb 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 2r2r (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 2r2r, it would mean that the two particles penetrate each other, which, by definition, hard spheres are unable to do.
The excluded volume for the two particles (of average diameter dd or radius rr) is:
b′2=4πd33=8⋅(4πr3/3)b2′=4πd33=8⋅(4πr3/3)
which divided by two (the number of colliding particles) gives the excluded volume per particle:
b′=b′2/2→
Answer:
excluded volume is 4 times the proper volume
Explanation:
The excluded volume for 2 articles (of average radius r) is,
b
′
=
3
4π(2r)
3
=8(
3
4πr
3
)
Whiche divided by 2 gives excluded volume per particle =4(
3
4πr
3
)
Hence, this is 4 times the proper volume of the particle at rest.