220 Chapter 7. Entropic forces at work[[Student version, January 17, 2003]]
∫ Thus for dilute solutions we can do all the integrals over the solute particle locations
d^3 r 1 ···d^3 rN first: Just as in the derivation of Equation 7.2, we again getVN. This timeV
is the volume of just that part of the chamber accessible to solute (right-hand side of Figure 1.3).
Since the membrane is essentially invisible to water molecules,nothing else in the partition func-
tion depends onV.Hence the sum over the positions and momenta of all the water molecules just
contributes a constant factor toZ,and such factors cancel from formulas like Equation 7.2.
The equilibrium osmotic pressurepequilin Figure 1.3 is thus given by the ideal gas law:
pequil=ckBT. van ’t Hoff relation (7.7)Herec=N/Vis the number density of solute molecules.pequilis the force per area that we must
apply to the solute side of the apparatus in order to get equilibrium.
The discussion above was appropriate to the situation shown in Figure 1.3 on page 11, where
wesomewhat artificially assumed there was no air, and hence no atmospheric pressure, outside the
apparatus. In the more common situation shown in Figure 7.1, we again get a relation of the form
Equation 7.7, but this time for thedifferencein pressure between the two sides of the membrane.
Thus ∆p=zfρm,wg,wherezf is the final height of the column of fluid,ρm,wis the mass density
of water, andgis the acceleration of gravity. Thus in this case we conclude thatthe equilibrium
height of the fluid column is proportional to the solute concentration in the cup.
The van ’t Hoff relation lets us understand a mysterious empirical fact from Chapter 1. Consider
again Figure 1.3b on page 11. Suppose that the solvent flows until the volume of the left side (with
solute) has doubled. Throughout the flow, the piston has been harnessed to a load. To extract the
maximum possible work from the system, we continuously adjust the load to be almost, but not
quite, big enough to stop the flow.
Your Turn 7b
Integrate Equation 7.7 to find the total work done by the piston against the load. Compare your
answer to Equation 1.7 on page 12 and find the actual value of the constant of proportionalityγ.Estimates Weneed some estimates to see if osmotic pressure is really significant in the world
of the cell. Suppose a cell contains globular proteins, roughly spheres of radius 10nm,atacon-
centration such that 30% of the cell’s volume is occupied with protein (we say that the “volume
fraction”φequals 0.3). This is not an unreasonable picture of red blood cells, which are stuffed
with hemoglobin. To find the concentrationcin Equation 7.7, we set 0.3 equal to the number of
proteins per volume times the volume of one protein:
0 .3=c×
4 π
3
(10−^8 m)^3. (7.8)Thusc≈ 7 · 1022 m−^3 .Tophrase this in more familiar units, remember that one mole per liter
corresponds to a concentration ofNmole/(10−^3 m^3 ). We’ll call a solution of 1 mole/Laonemolar
solution, defining the symbolM=mole/L,sowehaveasolution withc=1. 2 · 10 −^4 M,ora“0. 12 mM
solution.”^3
(^3) In this book we will pretend that dilute-solution formulas are always applicable, and so will not distinguish
between molar and molal concentration.