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Topic: Comparing the average distance of two atoms with the size of the atom alone  (Read 2856 times)

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Offline remeday86

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Hello,

I am trying to figure out the average distance between two Neon atoms. First of all, consider that one mole of gas of Ne is under SATP conditions and assume that it behaves as a perfect gas.

I have to find the volume occupied by one Ne atom so i used where Vm = 24.789 dm^3 mol^-1 @ SATP.
If the space is equally divided into tiny cubes containing one atom, then the volume would be (24.489 dm^3 mol^-1)x(6.02x10^23 molecules^-1) = 4.1178x10^-23 dm^3 per atom such that it's large enough to contain a single atom.

So, if the volume occupied by one Neon atom is 4.1178x10^-23 dm^3, then the length of the cube is the volume containing one Ne atom. I used the volume of a cube formula, V= L^3 ==> L = 3.453x10^-8 dm . That is the length of the cube which would be the diameter of the atom enclosed in it.

Therefore, if the diameter of the molecule is the length of the cube, the molecule's radius should be half of the cube's length. Since the space between the two molecules is equally divided into tiny cubes each containing one Ne atom, the 2r or diameter of the molecule would be the average distance between the two atoms which is also 3.453x10^-8 dm .
 

However, i have to compare that length with the atomic radius of an Ne atom and resources claim that the atomic radius of Ne is 38 pm. Since the length is the diameter, I divided 3.453x10^-8 dm by 2 to get the radius and converted dm into pm. However, I got ~1725 pm.  :-[

I know somewhere along the way I must've missed something. So please tell me where I did wrong. Thanks in advance.

Offline Borek

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Re: Comparing the average distance of two atoms with the size of the atom alone
« Reply #1 on: September 07, 2010, 02:47:51 AM »
Atoms in gas are not touching their neighbors. In solids and liquids you may assume they do, but not in gases.

Note that your approach will give size highly dependent on the temperature and pressure, that's not the case.
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