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

  • Thread starter Thread starter hhhmortal
  • Start date Start date
  • Tags Tags
    Atoms
Click For Summary
SUMMARY

The discussion focuses on calculating the spacing between atoms of an ideal gas at a temperature of 273K and a pressure of 1 atm using the mean free path equation. Participants clarified that the mean free path equation, ʎ = 1 / 4(pi)(n)(σ²), can be utilized without needing the exact radius of the atom, which is approximately 0.1 nm. The calculated atomic spacing is around 3 nm, significantly larger than the atomic radius. The conversation also touches on the distinction between mean free path and atomic spacing, emphasizing that atomic size is negligible compared to the spacing.

PREREQUISITES
  • Understanding of the ideal gas law (PV = nRT)
  • Familiarity with the mean free path equation (ʎ = 1 / 4(pi)(n)(σ²))
  • Basic knowledge of atomic structure and dimensions
  • Concept of standard temperature and pressure (STP)
NEXT STEPS
  • Explore the derivation and applications of the mean free path equation in different gas conditions.
  • Study the implications of atomic spacing in various states of matter, including solids and liquids.
  • Investigate the relationship between atomic density and spacing in different materials.
  • Learn about the kinetic theory of gases and its relevance to molecular spacing and behavior.
USEFUL FOR

Students studying chemistry or physics, particularly those focusing on gas laws, atomic theory, and molecular interactions. This discussion is beneficial for anyone seeking to understand atomic spacing in ideal gases and its implications in various scientific contexts.

hhhmortal
Messages
175
Reaction score
0

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!
 
Physics news on Phys.org
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.
 
mgb_phys said:
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.
 
hhhmortal said:
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.
 
fluidistic said:
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.
 
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.
 
mgb_phys said:
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.
 

Similar threads

Replies
1
Views
3K
Replies
24
Views
3K
  • · Replies 69 ·
3
Replies
69
Views
7K
Replies
3
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
3K
Replies
1
Views
1K