--%>
Assignment Help - homework help

Diffusion Molecular View

When the diffusion process is treated as the movement of particles through a solvent the diffusion coefficient can be related to the effective size of diffusing particles and the viscosity of the medium.

To see how the experimental coefficients can be treated to properties of the system and particularly of the solute macromolecules we take a molecular view of the diffusion process. Consider across a distance interval dx over which the concentration changes from c to c-dc. The force that drives the molecules to the ore dilute region can be related to the difference in the, molar free energy of the solute at concentration c and at concentration c-dc. If deal behaviour is assumed, the free energy differences per molecule is

Gc - dc - Gc = RT/N In (c -dc)/c 

Or

dG = RT/N In (1 - dc/c) - RT/N dc/c  where the relation In (1 - y) = -y for small y has been used.

This free energy difference corresponds to the mechanical energy needed to transfer one macromolecule across the distance dx. This energy can therefore be written as a force times the distance dx. Thud dG = driving force × dx, or

Driving force = dG/dx = RT/N 1/c dc/dx

A frictional force sets in and balances this diffusion force when some constant velocity is reached. The frictional force exerted by a viscous solvent fluid of viscosity η has been derived for a macroscopic sphere of radius r by G. G strokes as 

Frictional force = 6∏rη dx/dt

It appears suitable to apply this expression to the motion of reasonably spherical macromolecules. The diffusion velocity increases, therefore, until the force balances that equation. Then

6∏rη dx/dt = - RT/N 1/c dc/dx 

Or

cdx/dt = - RT/(6∏rη) dc/dx

Since c implies a mass per unit volume measure of concentrations, the product c dx/dt can be interrupted as the rate with which the diffusing substance moves through a unit cross section at x. this follows suggests, from the fact that dx/dt, the average diffusion velocity in the x direction, is the distance the diffusing molecules travel per unit time. Thus all the molecules within a distance dx/dt of a cross section will pass cross section in unit time. These molecules are in a volume equal to dx/dt times the cross section area. The mass of these molecules is the product of this volume and the concentration expressed as mass per unit volume. Thus c dx/dt is the amount per unit time, i.e. the rate with which the solute passes through the cross section. We can write now

D ∂c/∂x = - RT/(6∏rη) ∂c/∂x

This leads to the identification

D = RT/(6∏rη) 

And 6∏rη = RT/DN

Measurements of D and η could therefore lead to a value of the radius r for the macromolecule. Such a procedure is a little unsatisfactory. Molecules do not necessarily obey Strokes' law, even if they are spherical. Furthermore, macromolecules will generally be solvated and in moving through the solution will to some extent vary along this salvation layer. Equation is important however, in that it provides a way of determining the effective value of the group of terms 6∏rη for a solute characterized by molecules with radius r and a solvent characterized by viscosity η. 

   Related Questions in Chemistry

  • Q : Thermodynamics I) Sulphur dioxide (SO2)

    I) Sulphur dioxide (SO2) with a volumetric flow rate 5000cm3/s at 1 bar and 1000C is mixed with a second SO2 stream flowing at 2500cm3/s at 2 bar and 200C. The process occurs at steady state. You may assume ideal gas behaviour. For SO2 take the heat capacity at constant pressure to be CP/R = 3.267

    		•
  • Q : Haloalkene with the help of polarity of

    with the help of polarity of c-x bond show that aryl halides are less reactive than alkyl halides

    		•
  • 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 : Relative lowering of vapour pressure

    explain the process of relative lowering of vapour pressure

    		•
  • Q : Explain the molecular mass with respect

    During the formation of polymers, different macromolecules have different degree of polymerisation i.e. they have varied chain lengths. Thus, the molecular masses of the individual macromolecules in a particular sample of the polymer are different. Hence, an average value of the molecular mass is

    		•
  • Q : Molecular Diameters The excluded volume

    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

    		•
  • Q : What are aliphatic amines and its

    In common system, the aliphatic amines are named by using prefix for alkyl group followed by the word amine.In case of mixed amines, the name of alkyl groups are arranged in alphabetical order. This is followed by the word amine. However, for simple secondary or tertiary amines anothe

    		•
  • Q : Symmetry Elements The symmetry of the

    The symmetry of the molecules can be described in terms of electrons of symmetry and the corresponding symmetry operations.Clearly some molecules, like H2O and CH4, are symmetric. Now w

    		•
  • Q : Neutralisation of phosphorous acids

    Provide solution of this question. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3 PO3) the volume of 0.1 M aqueous KOH solution required is: (a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL

    		•
  • Q : Volume of solution containing solute

    What volume of solution contains 0.1 mole of the solute: (a) 100ml (b) 125ml  (c) 500ml (d) 62.5ml Choose the right answer from above.

    		•