--%>
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 : Explain Photoelectron Spectroscopy. The

    The energies of both the outer and inner orbitals of atoms and molecules can be determined by photoelectron spectroscopy.Energy changes of the outermost or highest energy electron of molecules were dealt with here in a different passion. The energies of ot

    		•
  • Q : Volume hydrogen peroxide Choose the

    Choose the right answer from following. The normality of 10 lit. volume hydrogen peroxide is: (a) 0.176 (b) 3.52 (c) 1.78 (d) 0.88 (e)17.8

    		•
  • Q : What are Vander Waal's Radii? Vander

    Vander Waal's radii can be assigned to the atoms of molecules on the basis of the closeness of approach of these atoms in crystalline substances. Diffraction studies of crystals give information about hoe molecules can approach each other and can pack

    		•
  • Q : Structure of a DNA molecule Elaborate

    Elaborate the structure of a DNA molecule?

    		•
  • Q : Decision about dipole moment is present

    How can you decide if there is a dipole moment or not?

    		•
  • Q : Maximum vapour pressure Provide

    Provide solution of this question. Which solution will show the maximum vapour pressure at 300 K: (a)1MC12H22O11 (b)1M CH3 COOH (c) 1MNacl2 (d)1MNACl

    		•
  • Q : Problem on thermodynamic equilibrium In

    In the manufacture of sulphuric acid by the contact process, S02 is oxidized to SO3 over a vanadium catalyst: The reactor is adiabatic and operates at atmospheric pressure. The gases enter the reactor at 410&d

    		•
  • Q : Preparation of normal solution Give me

    Give me answer of this question. What weight of ferrous ammonium sulphate is requiored to prepare 100 ml of 0.1 normal solution (mol. wt. 392): (a) 39.2 gm (b) 3.92 gm (c)1.96 gm (d)19.6 gm

    		•
  • Q : Strength of Nacl in solution To 5.85gm

    To 5.85gm of Nacl one kg of water is added to prepare of solution. What is the strength of Nacl in this solution (mol. wt. of nacl = 58.5)? (a) 0.1 Normal (b) 0.1 Molal (c) 0.1 Molar (d) 0.1 FormalAnswer:

    		•
  • Q : Surface Tension Vapour Pressure The

    The vapor pressure of small liquid drops depends on the drop size. Although the surface properties of a liquid are different from those of the bulk liquid, the special surface properties can be ignored except in a few situations. One is the case in which a liquid is dispersed into fine dr

    		•