# Solid property of inert gas- LJ potential

1. Dec 27, 2011

### ibysaiyan

1. The problem statement, all variables and given/known data
Hi,
We meet again PF, this time I am little stumped on the following question(link posted down).

2. Relevant equations

"Question" [Broken]

3. 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: May 5, 2017
2. Dec 27, 2011

### ibysaiyan

Here's the question :
http://pastebin.com/HazHGdZh [Broken]

Last edited by a moderator: May 5, 2017
3. Dec 28, 2011

### ehild

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

4. Dec 28, 2011

### ibysaiyan

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 seperation 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

5. Dec 28, 2011

### ehild

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

6. Dec 28, 2011

### ibysaiyan

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 ?

7. Dec 28, 2011

### ehild

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

ehild

8. Dec 28, 2011

### ibysaiyan

Sure.

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

Formula given is:
$V_{m}$ = Na * r0^3 / √2

9. Dec 29, 2011

### ehild

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. Dec 31, 2011

### ibysaiyan

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.