Solid property of inert gas- LJ potential

Click For Summary

Homework Help Overview

The discussion revolves around the properties of inert gases and the Lennard-Jones potential, specifically focusing on calculating equilibrium separation and molar volume in a solid-state context. Participants are attempting to solve a problem related to interatomic distances and molar volume calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of equilibrium separation and the interpretation of molar volume. There are questions about unit conversions and the application of formulas related to interatomic distances.

Discussion Status

Some participants have provided calculations and attempted to clarify the molar volume concept. There is ongoing exploration of different interpretations of the formulas, and some guidance has been offered regarding unit conversions and assumptions about the structure of the solid.

Contextual Notes

There are indications of confusion regarding unit conversions between cubic centimeters and cubic meters, as well as the assumptions made about the atomic structure in the calculations. Participants are also navigating the constraints of homework guidelines.

ibysaiyan
Messages
441
Reaction score
0

Homework Statement


Hi,
We meet again PF, this time I am little stumped on the following question(link posted down).

Homework Equations



Link to the question

"Question"

The Attempt at a Solution

I think I get the gist of it for part (i) my calculated value for equlibirum separation is about 0.32nm which seems sensible... now carrying on to part two where basically they want me to use the given equation for : The term molar volume is a little vague to me..the value i get on part (i) is about 19.9 cm^3... is that right ? so do I just plug that into the equation or do I convert it into m^3 or simply meter? (this is something i cam confused about).

Thanks
-ibysaiyan
 
Last edited by a moderator:
Physics news on Phys.org
Anyone please ?

Here's the question :
http://pastebin.com/HazHGdZh
 
Last edited by a moderator:
ibysaiyan said:
I think I get the gist of it for part (i) my calculated value for equlibirum separation is about 0.32nm which seems sensible... now carrying on to part two where basically they want me to use the given equation for : The term molar volume is a little vague to me..the value i get on part (i) is about 19.9 cm^3... is that right ? so do I just plug that into the equation or do I convert it into m^3 or simply meter? (this is something i cam confused about).

How did you get 19.9 cm^3 for the molar volume?
If you plug in the molar volume in cm^3 into the equation, you get the interatomic distance in cm.

ehild
 
ehild said:
How did you get 19.9 cm^3 for the molar volume?
If you plug in the molar volume in cm^3 into the equation, you get the interatomic distance in cm.

ehild

On wiki it said : Molar volume = Molecular mass/ density...
so I simply plugged in the values : 39.95 / 1.8 to give me 22.2 cm^3..
but to get the separation distance at equlibirum using : Vm = Na * r0^3 / √2
I converted Molar volume into m^3... i.e (22.2/100)^3 = 10.92*10^-3 (m^3)

There are basically two answers for the separation distance.. one i get is 0.32nm and other using above equation is 2.9nm
 
So the molar volume is 22.2 cm^3, that is correct. The first question asks you to determine the interatomic distance assuming simple cubic structure. In that case the atoms sit on the vertexes of the unit cells. The unit cells are cubes and there are Na such cubes in the molar volume and the edge of such cub is the cubic root of its volume.
You need to use the other formula when answering the second question.

ehild
 
ehild said:
So the molar volume is 22.2 cm^3, that is correct. The first question asks you to determine the interatomic distance assuming simple cubic structure. In that case the atoms sit on the vertexes of the unit cells. The unit cells are cubes and there are Na such cubes in the molar volume and the edge of such cub is the cubic root of its volume.
You need to use the other formula when answering the second question.

ehild

Certainly , that is exactly what I have done . For party two using the formula given I get a value of about 8 times larger than part one, is that reasonable ?
 
ibysaiyan said:
Certainly , that is exactly what I have done . For party two using the formula given I get a value of about 8 times larger than part one, is that reasonable ?

No, it is wrong. Show your calculation in detail.

ehild
 
ehild said:
No, it is wrong. Show your calculation in detail.

ehild

Sure.

[itex]V_{m}[/itex] = 22.2 [itex]cm^{3}[/itex] which I have converted into m^3 .. (22.2/100)^3 = 10.92*10^-3 m^3

Formula given is:
[itex]V_{m}[/itex] = Na * r0^3 / √2
which gives me about 2.9nm
 
ibysaiyan said:
[itex]V_{m}[/itex] = 22.2 [itex]cm^{3}[/itex] which I have converted into m^3 .. (22.2/100)^3 = 10.92*10^-3 m^3

That is totally wrong... 1 cm =0.01 m, 1 cm^3=10-6 m3, so 22.2 cm3=22.2x10-6 m.

The correct procedure for the transformation would be

22.2 cm3= 22.2( cm*m/(100 cm))3

ehild
 
  • #10
ehild said:
That is totally wrong... 1 cm =0.01 m, 1 cm^3=10-6 m3, so 22.2 cm3=22.2x10-6 m.

The correct procedure for the transformation would be

22.2 cm3= 22.2( cm*m/(100 cm))3

ehild

What was I thinking, the above conversion which I was doing doesn't make any sense.. had I took cube root of that then it would have given me a value in meters.
Thanks for all your help.

Happy new year to you and to the rest of the users.
 
  • #11
Happy New Year to you, too.

ehild
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
7
Views
3K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K