7.3. Beyond equilibrium: Osmotic flow[[Student version, January 17, 2003]] 225
bonuswewill also learn about nonequilibrium flow, which will be useful when we study ion transport
in Chapters 11–12. More generally, our discussion will lay the groundwork for understanding many
kinds of free-energy transducers. For example, Chapter 10 will use such ideas to explain force
generation in molecular machines.
7.3.1 Osmotic forces arise from the rectification of Brownian motion
Osmotic pressure gives a force pushing the pistons in Figure 1.3 on page 11 relative to the cylinder.
Ultimately this force has got to come from themembraneseparating the two chambers, since only
the membrane is fixed relative to the cylinder. Experimentally one sees this membrane bow as it
pushes the fluid, which in turn pushes against the piston. So what we really want to understand is
how, and why, the membrane exerts force on (transmits momentum to) the fluid.
Tomake the discussion concrete we’ll need a number of simplifying assumptions. Some are
approximations, while others can be arranged to be literally true in carefully controlled experiments.
Forexample, we will assume that our membrane is totally impermeable to solute particles. Such
amembrane is calledsemipermeable;the “semi” reminds us that waterdoespass through such a
membrane. We will also take the fluid to be essentially incompressible, like water. Finally, as usual
westudy a flat surface, where everything is constant in thexandydirections.
Imagine a fluid with an external force acting directly on it, like that of gravity. For example,
the pressure in a swimming pool increases with depth, because in equilibrium each fluid element
must push upward to balance the weight of the column of fluid above it:
p(z)=p 0 +ρm,wg×(z 0 −z). (7.11)Herep 0 is atmospheric pressure,z 0 −zis the depth, andρm,wgis the weight (force) per unit volume
(a similar expression appears in Equation 3.22 on page 74). More generally, the force acting on a
fluid may not be a constant. LetF(z)beanexternal force per volume acting in the +zˆdirection at
positionzand, consider a small cube of fluid centered at (x, y, z). Balancing the forces on the cube
again shows that in equilibrium the pressure cannot be constant, but instead must vary (Figure 7.5):
[
−p(z+^12 dz)+p(z−^12 dz)
]
dxdy+F(z)dxdydz=0. (7.12)Taking dzto be small and using the definition of the derivative gives dp/dz=F(z), the condition
for mechanical equilibrium (in this case calledhydrostatic equilibrium). Taking the force density
Fto be the constant−ρm,wgand solving recovers Equation 7.11 as a special case.
Next imagine a suspension of colloidal particles in a fluid with number densityc(z). Suppose
aforcef(z)acts along +ˆzoneach particle,depending on the particle’s position. (For a literal
example of such a situation, imagine two perforated parallel plates in the fluid with a battery
connected across them; then a charged particle will feel afwhen it’s between the plates, but zero
force elsewhere.)
In the low Reynolds-number regime inertial effects are negligibly small (see Chapter 5), so the
applied force on each particle is just balanced by a viscous drag from the fluid. The particles in turn
push back on the fluid, transmitting the applied force to it. Thus even though the force does not
act directly on the fluid, it creates an average force density ofF(z)=c(z)f(z), and a corresponding
pressure gradient:
dp
dz
=c(z)f(z). (7.13)