Biological Physics: Energy, Information, Life

226 Chapter 7. Entropic forces at work[[Student version, January 17, 2003]]

r

F(r)dxdydz

p(z)dxdz p(z)dxdz

p(z−^12 dz)dxdy

p(z+^12 dz)dxdy

x

y

z

Figure 7.5: (Schematic.) Forces on a small element of fluid. An external force densityF(r)can be regarded as
acting on the element’s center of mass, atr=(x, y, z). The fluid’s pressure pushes inward on all six sides of the box.
It is assumed to be constant inxandy,but not inz.Thuswhile the net pressure forces in thexandydirections
cancel, there is a nontrivial requirement for force balance in theˆzdirection.

The force on each particle reflects the gradient of that particle’s potential energy:f(z)=−dU/dz.
Forexample, an impenetrable solid wall creates a zone where the potential energy goes to infinity;
the force increases without limit near the wall, pushing any particle away. We’ll make the convention
thatU→ 0 far from the membrane (see Figure 7.6b).
Equation 7.13 presents us with an apparent roadblock: We have just one equation but two
unknown functions,c(z)andp(z). Luckily we know something else aboutc:Inequilibrium, the
Boltzmann distribution gives it as a constant times e−U(z)/kBT,and the constant is just the con-
centration( c 0 in the force-free region, far from the membrane. Then the force density alongˆzis
c 0 e−U/kBT

)(

−dU/dz

)

,which we rewrite asc 0 kBTddz

[

e−U(z)/kBT

]

.According to Equation 7.13,
this expression equals dp/dz:
dp
dz
=kBT
dc
dz

.

Integrating the equation across the membrane channel and out into the force-free region then gives
that ∆p=c 0 kBT,ormore generally that:

The equilibrium pressure difference across a semipermeable membrane equals

kBTtimes the difference in concentration between the two force-free regions on

either side of the membrane.

(7.14)

Wehave just recovered the van ’t Hoff relation, Equation 7.7 on page 220. Compared to the
discussion of Section 7.2.1, however, this time we have a greater level of detail.
Gilbert says: Now I can see the actual mechanism of force generation. When a membrane is
impermeable to solute particles, for example because it contains channels which are physically too
small for particles to pass, then the particles entrain some water as they bounce off the membrane,
due to viscous friction. This entrainment pulls water through the channel; a backward pressure
gradient is needed to stop the flow.
Sullivan: But wait. Even when the particles arefree(no membrane) their Brownian motion
disturbs the surrounding fluid! What’s your argument got to do with the osmotic case?