--%>
Assignment Help - homework help

Molecular Diameters

The excluded volume b, introduced by vander Wall's as an empirical correction term, can be related to the size gas molecules. To do so, we assume the excluded volume is the result of the pairwise coming together of molecules. This assumption is justified when b values are obtained from second viral coefficient data. Fitting values for the empirical constants are derived from van der Waal's equation. Some b values obtained in this way are given in table.


So that we need to deal with a single molecular size parameter, we treat molecules as spherical particles. The diameter of a molecule is d. the volume of a molecule is v

The volume in which a pair of molecules cannot move because of each other's presence is indicated by the lightly shaded region. The radius of this excluded volume sphere is equal to the molecule diameter d. the volume excluded to the pair of molecules is 4/3πd3. We thus obtain,

= 4[4/3π (d/2)]3

The expression in brackets is the volume of a molecule.vander Waal's b term is the excluded volume per mole of the molecules. Thus we have, with N representing Avogadro's number,

B= 4n [4/3π (d/2)3] = 4N (vol. of molecule)

Molecular size and Lennard Jones intermolecular Attraction term based on second virial coefficient data:

Gas Excluded volume B, L mol-1 Molecular diam. D, pm ELJ, J × 10-21
He 0.021 255 0.14
Ne 0.026 274 0.49
Ar 0.050 341 1.68
Kr 0.058 358 2.49
Xe 0.084 405 3.11
H2 0.031 291 0.52
N2 0.061 364 1.28
O2 0.058 358 1.59
CH4 0.069 380 1.96
C(CH3)4 0.510 739 3.22


Van der Waal's equation and the Boyle temperature:

Gas Tboyle, K Tboyle/TC
H2 110 3.5
He 23 4.5
CH4 510 2.7
NH3 860 2.1
N2 330 2.6
O2 410 2.7


Example: calculate the radius of the molecule from the value of 0.069 L mol-1 for the excluded volume b that is obtained from the second virial coefficient data.

Solution: the volume of 1 mol of methane molecules is obtained by dividing the b value of 0.069 L mol-1 = 69 × 10-6 m3 mol-1 value by 4. Then division by Avogadro's number gives the volume per molecule. We have:

Volume of methane molecule = 69 × 10-6 m3/4 × 6.022 × 1023 

= 2.86 × 10-29 m3

The volume is equal to 4/3∏r3 and on this basis we calculate:

r = 1.90 × 10-10 m and d = 3.80 × 10-10 m = 380 pm

   Related Questions in Chemistry

  • Q : Colligative property problem Which is

    Which is not a colligative property: (a) Refractive index (b) Lowering of vapour pressure (c) Depression of freezing point (d) Elevation of boiling point    

    		•
  • Q : Explain Vapour Pressure Composition A

    A pressure composition diagram for a liquid vapor system can be used to show the composition of the liquid and equilibrium vapor.Vapor equilibrium data are useful in the study of distillations. It is of value to have diagrams showing not only the vapor pre

    		•
  • Q : Atmospheric pressure Give me answer of

    Give me answer of this question. The atmospheric pressure is sum of the: (a) Pressure of the biomolecules (b) Vapour pressure of atmospheric constituents (c) Vapour pressure of chemicals and vapour pressure of volatile (d) Pressure created on to atmospheric molecules

    		•
  • Q : Describe Point Groups. For any

    For any symmetric object there is a set of symmetry operations that, together, constitute a mathematical group, called a point group.It is clear from the examples that most molecules have several elements of symmetry. The H2O

    		•
  • Q : State octet rule in chemistry Explain

    Explain what is octet rule in chemistry?

    		•
  • Q : Solutions The relative lowering of

    The relative lowering of vapour pressure of 0.2 molal solution in which solvent is benzene

    		•
  • Q : Laws of Chemical Combination Laws of

    Laws of Chemical Combination- In order to understand the composition of the compounds, it is necessary to have a theory which accounts for both qualitative and quantitative observations during chem

    		•
  • Q : Chemistry brief discription of relative

    brief discription of relative lowering of vapour pressure

    		•
  • Q : Hydroxide is highly insoluble in

     : 1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to

    		•
  • Q : Finding strength of HCL solution Can

    Can someone please help me in getting through this problem. 1.0 gm of pure calcium carbonate was found to require 50 ml of dilute  HCL for complete reaction. The strength of the HCL  solution is given by: (a) 4 N  (b) 2 N  (c) 0.4 N  (d) 0.2 N

    		•