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Old Aug 24th 2010, 08:40 PM   #1
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Number Density !!!

Hi All

Consider that I have a unit volume. Now there are two species in this volume, say, molecules. Dia of one type is d1 and the other type is d2. Say the number density of these molecules is n1 and n2 respectively. The volume fraction occupied by these two species is known (say v1 and 1-v1). Thus the number density for unit volume can be written as
n1=v1/(pi*d1*d1*d1/6) and n2=(1-v1)/(pi*d2*d2*d2/6)
Will the sum of the number density of these species be equal to 1 i-e. n1+n2 = 1 ? If not, then what will it be equal to?

Thanks.
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Old Aug 25th 2010, 03:12 PM   #2
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A few issues here:

1. You are assuming that the number density can be calculated by knowing the volume of an individual item - in this case a molecule - like this:

N = Unit volume/volume per object

But you are using an expression for the volume of an individual molecule that assumes it is spherical, like a tennis ball, and spheres can not be packed perfectly together. Thus the number of tennis balls that can fit into a volume v is not n = v/(pi*d^3/6), but is actually less than that due to the interstitial holes between adjacent balls.

2. Further, the values of n1 and n2 are the number of molecules per unit volume. That number is certainly much greater than 1. So no way n1 + n2 = 1. As a check, suppose d1 = d2: in that case your formula gives you n1+n2 = 1/(pi d^3/8), which is not 1 unless the volume of a single molecule is equal to a the unit volume.

What may be of use here is that if you know the values for n1 and n2, then if you mix these two types of objects together in a unit size container (v1+v2=1), then the total number density is:

N = v1*n1+(1-v1)*n2

This is basically the weighted average of n1 and n2.
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Old Aug 25th 2010, 08:35 PM   #3
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Hi Chip
Thanks for the detailed answer. But I may I know what is the difference between N and n1 & n2? If posible, plz state an example.
Thanx.
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Old Aug 26th 2010, 02:17 PM   #4
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Here's an example:

Suppose n1 = 10^6, n2 = 10^7, v1=40%, and v2 = 1-v1 = 60%. Then

N = v1*n1 + v2*n2 = 0.4 x 10^6 + 0.6 x 10^7 = 6.4*10^6

So note that N is a weighted average of n1 and n2; as such its value will be somewhere between the two.
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