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Solid property of inert gas- LJ potential

  1. Dec 27, 2011 #1
    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

    Link to the question

    "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. jcsd
  3. Dec 27, 2011 #2
    Anyone please ?

    Here's the question :
    http://pastebin.com/HazHGdZh [Broken]
     
    Last edited by a moderator: May 5, 2017
  4. Dec 28, 2011 #3

    ehild

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    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
     
  5. Dec 28, 2011 #4
    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
     
  6. Dec 28, 2011 #5

    ehild

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    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
     
  7. Dec 28, 2011 #6
    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 ?
     
  8. Dec 28, 2011 #7

    ehild

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

    ehild
     
  9. Dec 28, 2011 #8
    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
     
  10. Dec 29, 2011 #9

    ehild

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    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
     
  11. Dec 31, 2011 #10
    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.
     
  12. Dec 31, 2011 #11

    ehild

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    Happy New Year to you, too.

    ehild
     
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