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I was just reading over the EOS topic when where I stumbled upon excluded volume. Although the pictorial representation of the hard sphere model clearly shows that the excluded volume is 4 times the volume occupied by molecules, but what i am not unable to understand is that why do we take the sphere with diameter equal to one of the molecules in the first place.

Why do we take the dotted circle as excluded volume

So, my question is that why do we have to take the whole spehere..

For example like this

I guess this has something to do with viewing the molecules in 3-dimensions, but i am unable to visualize.

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    $\begingroup$ Draw the exclusion volume centered on one of the particles, rather than encompassing the two. In your picture, take the lower right particle as A and the upper left particle as B. Draw your circle concentrically around A. Then it becomes clearer that the center of B can not penetrate into that circle. The center of particle B can never enter that volume, one considers this as particle B can never occupy that volume. $\endgroup$ – Jon Custer Aug 23 '18 at 19:41
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We assume that the molecules are identical hard spheres with diameter $d$. The excluded volume is emphatically not the volume of the sphere, but rather the volume excluded to the other particles by the presence of that sphere.

When two particles collide, they touch each other at some point on their surface, and the distance between the center of one molecule to that of the other is $d$. Indeed, two particles cannot get closer than a distance $d$ to each other, and hence the excluded volume of each particle is the volume of a sphere of diameter $d$—-we must take into account the radii of both colliding spheres.

Edit: This suggests that the excluded volume should be eight times as large as the original volume of a sphere, not four times. However, a collision event involves two particles, and this excluded volume is shared, in a sense, by both particles. We must therefore divide by a factor of two, and that gives the resulting multiplier of four.

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  • $\begingroup$ I knew that part, whats confusing me is that why do we take a whole sphere? In the second picture, cant a third molecule (shown in red) occupy that space? shouldn't the area enclosed with solid black line equal excluded volume? Thanks $\endgroup$ – Shah M Hasan Aug 23 '18 at 12:19
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    $\begingroup$ @ShahMHasan, from my earlier comment, because you should think of the excluded volume as a sphere of radius d centered about one of the particles. $\endgroup$ – a-cyclohexane-molecule Aug 23 '18 at 13:44
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    $\begingroup$ A sphere with twice the diameter of a molecule should have 8 times the volume, shouldn't it? $\endgroup$ – Ivan Neretin Aug 23 '18 at 15:20
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    $\begingroup$ @IvanNeretin, yes, sorry, I've neglected to address that point. A collision event involves two particles, and this excluded volume is shared, in a sense, by both particles. We must therefore divide by a factor of two, and that gives the resulting multiplier of four. $\endgroup$ – a-cyclohexane-molecule Aug 23 '18 at 15:50
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    $\begingroup$ @ShahMHasan - Just two particles colliding is considered because it is much more probable than three particles colliding. However three particles can collide, so the simplification isn't entirely accurate. $\endgroup$ – MaxW Aug 23 '18 at 20:51
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If the probability that a pair of balls overlaps is $8v/V$ where $v$ is the ball volume, and $V$ is the container volume, the probability that they don't overlap is

$$1 - 8v/V$$

Given $N$ balls, there are $N(N-1)/2$ combinations of pairs. The probability that none of them overlap is approximately

$$(1 - 8v/V)^{N(N - 1)/2} \sim (1 - 4vN/V)^N$$

Multiplying by $V^N$ as in calculation of partition functions, one gets

$$(V-4vN)^N,$$

which looks like an excluded volume of $4v,$ not $8v.$

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