Calculate Spacing b/w Atoms of Ideal Gas@273K,1atm

[hhhmortal]

Homework Statement

Hi, I've got the following problem:

Calculate the spacing between atoms of an ideal gas at a temperature of 273K and a pressure of 1atm.

Homework Equations

Mean free path equation:

ʎ = 1 / 4(pi)(n)(σ²)

The Attempt at a Solution

I first used the ideal gas equation PV =nRT to get n/V and thus I got 2.67 x 10^25 particles/m^3.

But I don't know how I can find the radius of the atom. Am I using the wrong equation to get the spacing between atoms?

Thanks!

 

[mgb_phys]

If you consider the spacing between the centres of the atoms - you don't really need the size of the atom.
Take a look at the answer and you will see that the size of an atom (0.1nm) is tiny compared to the spacing.

 

[hhhmortal]

If you consider the spacing between the centres of the atoms - you don't really need the size of the atom.
Take a look at the answer and you will see that the size of an atom (0.1nm) is tiny compared to the spacing.

I don't quite understand you. Do you mean just assuming the radius of the atom is 0.1nm and then using the mean free path equation to get the spacing between atoms? If so wouldn't that give me a spacing between atoms very similar to its radius.

 

[fluidistic]

I don't quite understand you. Do you mean just assuming the radius of the atom is 0.1nm and then using the mean free path equation to get the spacing between atoms? If so wouldn't that give me a spacing between atoms very similar to its radius.

I don't think so. mgb_phys wanted you to consider the atoms as points without any length and see the distance between them. You will find this distance rather large compared to the radius of an atom. (around 0.1 nm).
That's what I understand from mgb_phys.

 

[hhhmortal]

I don't think so. mgb_phys wanted you to consider the atoms as points without any length and see the distance between them. You will find this distance rather large compared to the radius of an atom. (around 0.1 nm).
That's what I understand from mgb_phys.

Does this mean (n)^(1/3) which equals 3x10^8 and hence spacing between atoms is 3.3nm?

So what really is the difference between the mean free path and spacing between atoms?

Thanks.

 

[mgb_phys]

A mole at stp is 22.4litres.
So you have 6 10^23 atoms/22.4litres = 2x10^19 atoms/cc
Or roughly 3million atoms per cm = 3nm apart.

A typical atom is roughly 0.1nm in diameter so the size of the atom doesn't really matter compared to the spacing.

 

[sailormoon]

A mole at stp is 22.4litres.
So you have 6 10^23 atoms/22.4litres = 2x10^19 atoms/cc
Or roughly 3million atoms per cm = 3nm apart.

A typical atom is roughly 0.1nm in diameter so the size of the atom doesn't really matter compared to the spacing.

may i know does all these apply to solid?
quest: est the atomic spacing of the iron atoms.atomic mass 55.9u,density of iron found in quest is 6.585g/cm^3.