--%>
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 : Strength of dilute acid of Sulfuric acid

    Select the right answer of the question.10ml of conc.H2SO4 (18 molar) is diluted to 1 litre. The approximate strength of dilute acid could be: (a)0.18 N (b)0.09 N (c) 0.36 N (d)1800 N

    		•
  • Q : What is Ideal Mixtures Ideal mixing

    Ideal mixing properties can be recognized in the formation of an ideal gas mixture from ideal gases. Consider the formation of a mixture of gases i.e. a gaseous solution, from two mixtures of pure gases. A useful characterization of an ideal mixture, or soluti

    		•
  • Q : Dipole moment Elaborate a dipole moment

    Elaborate a dipole moment?

    		•
  • Q : Mole 2.0gram of dolomite is heated to a

    2.0gram of dolomite is heated to a constant weight of 1.0g. Calculate the total volume of CO2 produced at STP by this reation

    		•
  • Q : PH of an Alkyl Halide Briefly state the

    Briefly state the pH of an Alkyl Halide?

    		•
  • Q : Examples of reversible reaction

    Describe some examples of a reversible reaction?

    		•
  • Q : Molecular weight of solute Select right

    Select right answer of the question. A dry air is passed through the solution, containing the 10 gm of solute and 90 gm of water and then it pass through pure water. There is the depression in weight of solution wt by 2.5 gm and in weight of pure solvent by 0.05 gm. C

    		•
  • Q : Molar mass of compound The freezing

    The freezing point of a solution having 4.8 g of a compound in 60 g of benzene is 4.48. Determine the molar mass of the compound (Kf = 5.1 Km-1) , (freezing point of  benzene = 5.5oC)          &n

    		•
  • Q : Modes of concentration Which of the

    Which of the given modes of expressing concentration is fully independent of temperature: (1) Molarity (2) Molality (3) Formality (4) Normality Choose the right answer from above.

    		•
  • Q : Anti-aromatic and the non-aromatic

    What is main difference among anti-aromatic and the non-aromatic compounds?

    		•