Transcript Presentation453.08

Lecture 8 – Viscosity of Macromolecular Solutions
Ch 24
pages 643-650
Summary of lecture 7
 We have expanded the concept of diffusion coefficient from its
simple relationship to size and viscosity provided by Stoke’s law
f  6R
D  k bT / f
Summary of lecture 7
 Hydration affects diffusion so that the frictional coefficient f that
we measure is related to the frictional coefficient of an unhydrated
molecule by the relationship:
3
 f  V 2   1V1
  
V2
 f0 
 The denominator is the volume of an un-hydrated molecule, and
the term above is the total hydrodynamic volume, including
hydration. If a molecule is approximately spherical:
3
 f 
r
    
 f0 
 r0 
3
Summary of lecture 7
 The diffusional properties of non-spherical objects can be
calculated (for simple shapes analytically) and used to provide
frictional coefficients
 For a rod-like particle of length 2a and radius b, for example
f  f0
2 / 31 / 3 P 2 / 3
ln( 2 P )  0.30
 For a prolate ellipsoid where the lengths a and b are called the
major and minor semi-axis lengths
f  f0
P 1 / 3 P 2  1

ln P  P 2  1

Viscosity
We have introduced the concept of friction by expressing the
counterforce acting on a particle moving in a viscous medium,
F=fu. Although Newtons’ law is exercised on each particle (e.g.
electrophoresys or sedimentation), and therefore one would
expect the particle to accelerate under the influence of a certain
force F (an electric or centrifugal field, for example), the viscous
drag defines a certain velocity as the steady state speed at which
the particles move under the influence of an acting external
force and of the viscosity of the medium. Friction increases with
speed, so that the speed of the particle will only increase up to a
point, until it will reach a steady state value u
Friction
Consider a molecule in solution. If an external force F is applied to
the particle, the particle obviously accelerates according to F=ma
The particle will not accelerate for long. After a brief period, the
velocity becomes constant as a result of resistance from the
surrounding fluid. This velocity is called the steady state velocity
vT and fulfills the condition:
fvT  F
Friction
fvT  F
fv is the frictional force exerted by the surrounding fluid on the
particle and f is the frictional coefficient of the particle
The fictional coefficient depends on the size and shape of the
particle but not on its mass. For a spherical particle with radius R
f  6R
Where  is the viscosity of the fluid (Stoke’s Law)
Viscosity
Viscosity measures the resistance of fluids to flow
Consider a flowing liquid constrained between two plates (see
Figure). The plates are large and the lower plate stationary. The
upper plate moves in a plane parallel to the lower plate
Viscosity
As the upper plane moves, an infinitesimally thin layer of liquid
sticks to each plate: a layer of liquid adheres to the lower plate
and thus does not move, while a layer of liquid translates with
the upper plate. Intervening layers move with velocities that
vary as a function of the distance from the lower stationary plate
The velocity gradient in the direction perpendicular to the
planes may be thought of as a deformation of the liquid Dx/Dy
(after all:
u x  dx / dt
The velocity gradient is:
du x / dy
Viscosity
The velocity gradient
du x / dy
can be related to the ‘deformation’ in the fluid) and is called
shear.
Viscosity
A way to look at viscosity is by considering that there will be a
momentum change in different layers in the liquid; as the
velocity increases creating a velocity gradient, there is a net
momentum transfer in the opposite direction of the gradient
The change in velocity will induce a change in momentum p
perpendicular to the motion of the planes (p=mu). We can
express the change in velocity in terms of a flow of momentum
perpendicular to the plane, and therefore we can introduce a
’current’ describing the flux of momentum
Viscosity
We can relate this flux to the momentum gradient by an
equation analogous to Fick’s diffusion equation:
du
J p  
dy
 is a phenomenological proportionality constant called
viscosity
The viscosity  has units of gm cm-1 sec-1=1 poise. For example,
the viscosity of pure water at room temperature is
approximately 0.01 poise (1 centipoise)
Viscosity
Another way to look at viscosity is by introducing the shear
strain, which is defined as:
lim Dy 0
Dx dx

Dy dy
The shear stress is defined as the force F pushing the upper plate
in the x direction divided by the area A of the upper plate:
  shear stress 
F
A
Viscosity
The shear stress is defined as the force F pushing the upper plate
in the x direction divided by the area A of the upper plate:
  shear stress 
F
A
For many liquids, called Newtonian liquids, the shear stress and
strain are related by a simple proportionality. The constant of
proportionality that relates the shear stress to the velocity
gradient is called the viscosity 
 
d dx d dx

dy dt dt dy
Viscosity of Macromolecular Solutions
Viscosity and friction are measures of the rate of energy
dissipation in a flowing fluid. The addition of a macromolecule to
a solution greatly increases the viscosity and thus increases the
rate of energy dissipation
The rate of energy dissipation in a macromolecular solution is
greater than the rate of dissipation in pure solvent:
 E 
 E 
 
 
 t  so ln  t  solvent
Einstein showed that the rate of energy dissipation in a dilute
macromolecular solution is defined by the equation:
 E 
 E 
1  v 

 
 
 t  so ln  t  solvent
Viscosity of Macromolecular Solutions
 E 
 E 
1  v 
 
 

t

t
  so ln   solvent
 represent the fraction of the solution volume occupied by
macromolecules and n is a numerical factor related to the shape of
the macromolecule. For spherical particles n=5/2, while for nonspherical molecules n>5/2. For ellipsoids of rotation, for example,
the asymmetry factor n is dependent on the axial ratio a/b:
v
a / b2
15ln( 2a / b)  3 / 2 

Oblate Ellipsoid of Rotation:
v
16 a / b 
15 tan 1 a / b 
a / b2
5ln( 2a / b)  1 / 2) 

14
15
Intrinsic Viscosity
Because the viscosity is proportional to the rate of energy
dissipation in macromolecular solutions, i.e:
dE

dt
we may write the ratio of the viscosity of a dilute macromolecular
solution solution to the viscosity of pure solvent 0 as :
 solution
 1  n
0
for more concentrated solutions, higher order terms must be
considered, but this approximation is valid at relatively low
concentration of solute
Intrinsic Viscosity
 solution
 1  n
0
The ratio:
The quantity:
 solution
 r
0
is called relative viscosity
 solution
r  1 
 1  n  nV C
0
is called specific viscosity
C is the concentration of the solute in gm/mL and V is the
partial specific volume of the hydrated macromolecule
Intrinsic Viscosity
The intrinsic viscosity is defined as
  Lim
C0
 sp
C
 nV
It directly reflects the shape of the solute
Experimentally, the intrinsic viscosity is measured by
measuring the specific viscosity divided by C, i.e. sp/C, and
extrapolating to C=0:
Intrinsic Viscosity
The difficulty with interpreting intrinsic viscosity data is that
both the hydration and shape (asymmetry parameter)
contribute to it. That is, if two intrinsic viscosities differ, it
may be a result of variations in the asymmetry factor and/or
the specific volume, which reflects hydration. In the case of
viscosity calculations, the specific volume corresponds to the
hydrated molecule (e.g. protein or DNA). As stated in the
previous lecture, the specific volume of a hydrated molecule
can be expressed from its partial specific volume and from
the volume associated with hydration molecules:
V  V2   1V1
Intrinsic Viscosity
The difficulty with interpreting intrinsic viscosity data is that
both the hydration and shape (asymmetry parameter)
contribute to it. That is, if two intrinsic viscosities differ, it
may be a result of variations in the asymmetry factor and/or
the specific volume, which reflects hydration. In the case of
viscosity calculations, the specific volume corresponds to the
hydrated molecule (e.g. protein or DNA). As stated in the
previous lecture, the specific volume of a hydrated molecule
can be expressed from its partial specific volume and from
the volume associated with hydration molecules:
V  V2   1V1
V2 is the partial specific volume of the unhydrated molecule,
V1 is the partial specific volume of water (essentially the
inverse of the density of water), 1 is the number of grams of
water hydrating a gram of protein