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 :Coordination number of a cation The

The coordination number of a cation engaging a tetrahedral hole is: (a) 6  (b) 8  (c) 12  (d) 4 Answer: (d) The co-ordination number of a cation occupying a tetrahedral hole is 4.

• ##### Q :Describe the properties of the

Briefly describe the properties of the carbohydrates?

• ##### Q :Concentration of Sodium chloride

Provide solution of this question. If 25 ml of 0.25 M NaCl solution is diluted with water to a volume of 500ml the new concentration of the solution is : (a) 0.167 M (b) 0.0125 M (c) 0.833 M (d) 0.0167 M

• ##### Q :Preparation of normal solution Give me

Give me answer of this question. What weight of ferrous ammonium sulphate is requiored to prepare 100 ml of 0.1 normal solution (mol. wt. 392): (a) 39.2 gm (b) 3.92 gm (c)1.96 gm (d)19.6 gm

• ##### Q :Mole fraction of solute The mole

The mole fraction of the solute in 1 molal aqueous solution is: (a) 0.027 (b) 0.036 (c) 0.018 (d) 0.009What is the correct answer.

• ##### Q :Entropy on molecular basis. The

The equation S = k in W relates entropy to W, a measure of the number of different molecular level arrangements of the system.In the preceding developments it was unnecessary to attempt to reach any "explana

• ##### Q :Number of moles present in water

Provide solution of this question. How many moles of water are present in 180 of water: (a)1 mole (b)18 mole (c)10 mole (d)100 mole

• ##### 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 :Molarity Give me answer of this

Give me answer of this question. If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is:(a) 0.1M (b) 0.2M (c)0.3M (d)0.4M