--%>
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 : Molecular weight of substance The

    The boiling point of a solution of 0.11 gm of a substance in 15 gm of ether was found to be 0.1oC higher than that of the pure ether. The molecular weight of the substance will be (Kb = 2.16)       (a) 148 &nbs

    		•
  • Q : Colligative property related question

    Select the right answer of the question. Which of the following is not a colligative property : (a) Osmotic pressure (b) Elevation in B.P (c) Vapour pressure (d) Depression in freezing point

    		•
  • Q : Microwave Adsorption The absorption of

    The absorption of microwave radiation increases the rotational energy of molecules and gives information about the moment of inertia of the molecules.Now we can begin the study of the spectroscopy that explores the different ways in which the energy of the

    		•
  • Q : What are the chemical properties of

    Haloalkanes are extremely reactive category of aliphatic compounds. Their reactivity is due to the presence of polar carbon-halogen bond in their mole

    		•
  • 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

    		•
  • Q : Explain Ionic Bond with examples. The

    The bonding in ionic molecules can be described with a coulombic attractive term.For some diatomic molecules we take quite a different approach from that used in preceding sections to describe the bonding. Ionic bonds are interpreted in terms of the coulom

    		•
  • Q : Film Mass Transport Sulfur trioxide

    Sulfur trioxide (SO3) is manufactured by the gas-phase oxidation of SO2 over a platinum catalyst: SO2 + ½ O2 à SO3 The catalyst is a non-porous ext

    		•
  • Q : Question related to colligative

    The colligative properties of a solution depend on: (a) Nature of solute particles present in it (b) Nature of solvent used (c) Number of solute particles present in it (d) Number of moles of solvent only

    		•
  • 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 : 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

    		•