--%>
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 : What is Spectroscopy? This is a very

    This is a very important aspect of Physical Chemistry in which knowledge of the size, shape, rigidity and electronic structure of molecules deduced from the experimental methods treated here goes hand in hand with the theoretical approaches of chemical reactions. Spec

    		•
  • Q : Chemists have not created a periodic

    Explain the reason behind that the chemists have not created a periodic table of compounds?

    		•
  • 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 : Problem associated to vapour pressure

    Provide solution of this question. 60 gm of Urea (Mol. wt 60) was dissolved in 9.9 moles, of water. If the vapour pressure of pure water is P0 , the vapour pressure of solution is:(a) 0.10P0 (b) 1.10P0 (c) 0.90P0 (d) 0.99P0

    		•
  • Q : How to test a gas to see if it was

    Write a short note to describe how to test a gas to see if it was hydrogen or not?

    		•
  • Q : What is adsorption and its examples. In

    In a liquid a solid substance a molecule present within the bulk of the substance is being attracted infirmly from all sides by the neighbouring molecules. Hence there is no bet force acting on the molecule or there are no unbalanced forces of the molecule. On the oth

    		•
  • Q : P- block why pentahalids are more

    why pentahalids are more covalent than tetrahalids

    		•
  • Q : BASIC CHARACTER OF AMINES IN GAS PHASE,

    IN GAS PHASE, BASICITIES OF THE AMINES IS JUST OPPOSITE TO BASICITY OF AMINES IN AQEUOUS PHASE .. EXPLAIN

    		•
  • Q : Partial vapour pressure of volatile

    Choose the right answer from following. For a solution of volatile liquids the partial vapour pressure of each component in solution is directly proportional to: (a) Molarity (b) Mole fraction (c) Molality (d) Normality

    		•
  • Q : Basic concepts Determination of correct

    Determination of correct mol. Mass from Roult's law is applicable to :

    		•