228 Chapter 7. Entropic forces at work[[Student version, January 17, 2003]]
the membrane is obliged to supply a kick to the right. Each kick delivers some momentum; these
kicks don’t average to zero. Instead they pull fluid through the channel until equilibrium is reached.
Your Turn 7c
Now suppose that there are particles onbothsides of the membrane, with concentrationsc 1 and
c 2 .Supposec 1 >c 2 .Redraw Figure 7.6 and find the simple form taken by the van ’t Hoff relation
in this case.Our discussion makes clear how misleading it can be to refer to “the osmotic pressure.” Suppose
wethrow a lump of sugar into a beaker. Soon we have a very nonuniform concentrationc(r)ofsugar.
And yet the pressurep(r)iseverywhere constant, not equal tokBTc(r)aswemight have expected
from a na ̈ıve application of the van ’t Hoff relation. After all, we know that osmotic pressures can
behuge; the fluid would be thrown into violent motion if it suddenly developed such big pressure
variations. Instead it sits there quietly, and the concentration spreads only by diffusion.
The flaw in the na ̈ıve reasoning is the assumption that concentration gradients themselves
somehow cause pressure gradients. But pressure can only change if aforceacts (Equation 7.12).
Thus osmotic pressure can only arise if there is a physical object, the semipermeable membrane,
present to apply force to the solute particles. In the absence of such an object, for instance if we
just throw a lump of sugar into the water, there is no force and no pressure gradient. Similarly,
in the experiment sketched in Figure 1.3, initially there will be no osmotic force at all. Only when
solute molecules have had a chance to diffuse from the initial lump of sugar to the membrane will
the latter begin to rectify their Brownian motion and so transmit force to them, and thence to the
fluid.
7.3.2 Osmotic flow is quantitatively related to forced permeation
The preceding subsection argued that the membrane repels particles, which in turn drag fluid away
from the membrane, creating a low-pressure layer there. This layer is the depletion zone; see the
solid curve in Figure 7.6c.
Now suppose that we applynoforce to the pistons in Figure 1.3a. Then there will be no net
pressure difference between the sides. After all, pressure is force per area, namely zero on each side.
(More realistically, it’s likely to be atmospheric pressure on each side, but still there’s no jump.)
Doesn’t this contradict the van ’t Hoff relation? No, the van ’t Hoff relation doesn’t give the actual
pressure, but rather that pressure whichwouldbeneeded tostoposmotic flow, that is, the pressure
dropifthe systemwerebrought to equilibrium. We can certainly maintain a smaller pressure drop
thanc 0 kBT;then the osmotic effect will actually pull water through the pores from thec=0side
to thec=c 0 side. This process is osmotic flow.
The solid curve in Figure 7.6d summarizes the situation. In equilibrium, the fluid pressure was
constant throughout the pore (Figure 7.6c), but now it cannot be. The discussion leading to the
Hagen–Poiseuille relation (Equation 5.17 on page 160) then gives the flow rateQneeded to create
auniform pressure drop per unit lengthp/L.The system simply chooses that flow rate which gives
the required pressure drop. These observations apply to reverse osmosis as well (see Section 1.2.2
on page 10): If we push against the natural osmotic flow with a force per area even greater than
c 0 kBT,then the flow needed to accommodate the imposed pressure drop goes backward. This
situation is shown as the dashed curve in Figure 7.6d.
Wecan summarize the entire discussion in a single master formula. First we note that even if we