--%>
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 : Short note on the function of

    Write down a short note on the function of mitochondria?

    		•
  • Q : Explosions produce carbon dioxide

    Illustrate all the explosions produce carbon dioxide?

    		•
  • Q : Solutions The relative lowering of

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

    		•
  • Q : Colligative property associated question

    Give me answer of this question. Which of the following is not a colligative property : (a)Optical activity (b)Elevation in boiling point (c)Osmotic pressure (d)Lowering of vapour pressure

    		•
  • Q : Difference in Mendeleevs table and

    Briefly describe the difference in the Mendeleev’s table and modern periodic table?

    		•
  • Q : Describe Enzyme Catalyzed reactions

    Many enzyme catalyzed reactions obeys a complex rate equation that can be written as the total quantity of enzyme and the whole amount of substrate in the reaction system. Many rate equations that are more complex than first and se

    		•
  • Q : Molarity of sodium hydroxide Can

    Can someone please help me in getting through this problem. Determine the molarity of a solution having 5g of sodium hydroxide in 250ml  solution is: (i) 0.5  (ii) 1.0  (iii) 2.0   (d) 0.1Answer: The right answer i

    		•
  • Q : Question on Raoults law Give me answer

    Give me answer of this question. For a dilute solution, Raoult's law states that: (a) The lowering of vapour pressure is equal to mole fraction of solute (b) The relative lowering of vapour pressure is equal to mole fraction of solute (c) The relative lowering of v

    		•
  • Q : Schrodinger equation with particle in a

    Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functi

    		•
  • Q : Explain methods for industrial

    The important methods for the preparation of alcohol on large-scale are given below:    

    		•