--%>
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 are electromotive force in

    The main objective of this particular aspect of Physical Chemistry is to examine the relation between free energies and the mechanical energy of electromotive force of electrochemical cells. The ionic components of aqueous solutions can be treated on the basis of the

    		•
  • Q : Problem on preparing of a solution Give

    Give me answer of this question. How many grams of CH3OH should be added to water to prepare 150 solution of@M CH3 OH: (a) 9.6 (b) 2.4 (c) 9.6x 103 (d) 2.4 x103

    		•
  • Q : Adiabatic compression A lean natural

    A lean natural gas is available at 18oC and 65 bars and must be compressed for economical pipeline transportation. The gas is first adiabatically compressed to 200 bars and then isobarically (i.e. at constant pressure) cooled to 25°C. The gas, which is

    		•
  • Q : Calculate PH value for a acetic acid 1.

    1. A solution of 0.100 M acetic acid is prepared. a) What is its pH value? b) If 20% of the initial acetic acid is converted to the acetate form by titration with NaOH, what is the resultant pH?

    		•
  • Q : Molar mass lculwhat is the equation for

    lculwhat is the equation for caating molar mass of non volatile solute

    		•
  • Q : Importance of organic chemistry

    Describe the importance of organic chemistry?

    		•
  • 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 : Define Bond Energies - Bond Charges

    Energy changes in some chemical reactions can be used to deduce the energies of chemical bonds. Our understanding of the molecular basis of thermodynamic properties is extended when we ask why the enthalpy change for a reaction is what it is. We deduce,

    		•
  • 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 : Non-ideal Gases Fugacity The fugacity

    The fugacity is a pressure like quantity that is used to treat the free energy of nonideal gases.Now we begin the steps that allow us to relate free energy changes to the equilibrium constant of real, nonideal gases. The thermodynamic reaction 

    		•