--%>

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 m
3

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 : Problem related to molality Help me to

    Help me to solve this problem. What is the molality of a solution which contains 18 g of glucose (C6,H12, O6) in 250 g of water:  (a) 4.0 m (b) 0.4 m (c) 4.2 m (d) 0.8 m

  • Q : Molar concentration Choose the right

    Choose the right answer from following. Molar concentration (M) of any solution : a) No. of moles of solute/Volume of solution in litre (b) No. of gram equivalent of solute / volume of solution in litre (c) No. of moles os solute/ Mass of solvent in kg  (

  • 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 : How can enzymes act as catalyst?

    Enzymes are complex proteinous substances, produced by living bodies, such as act as catalysis in the physiological reactions. The enzymes are, also called biochemical catalysts and the phenomenon is known as bio-chemical catalysis because numerous reactions that occur the bodies of animals and p

  • Q : Explain solid in liquid solutions. The

    The French chemist Francois Marie Raoult (1886) carried out a series of experiments to study the vapour pressure of a number of binary solutions. On the basis of the results of the experiments, he proposed a generalization called Raoult's law which states that, <

  • Q : What are biodegradable polymers?

      These are polymers that can be broken into small segments by enzyme-catalysed reactions. The required enzymes are produced by microorganism. It is a known fact that the carbon-carbon bonds of chain growth polymers are inert to enzyme-catalysed reactions, and hence they are non biod

  • Q : Define tripod and its use Illustrate a

    Illustrate a tripod? And how it’s used?

  • Q : Molarity of Sodium hydroxide Select the

    Select the right answer of the question. Molarity of 4% NaOH solution is : (a) 0.1M (b) 0.5M (c) 0.01M (d) 0.05M

  • Q : Structure of a DNA molecule Elaborate

    Elaborate the structure of a DNA molecule?

  • Q : Alkaline medium The amount of KMnO 4

    The amount of KMnO4 required to prepare 100 ml of 0.1N solution in alkaline medium is: (a) 1.58 gm (b) 3.16 gm (c) 0.52 gm (d) 0.31 gmAnswer: (a) In alkaline medium KMnO4 act as ox