Biological Physics: Energy, Information, Life

7.1. Microscopic view of entropic forces[[Student version, January 17, 2003]] 217

entropic forces, notably the electrostatic, hydrophobic, and depletion effects. As always, we will
look for quantitative confirmation of our formal derivations in controlled experiments.
The Focus Questions for this chapter are:
Biological question: What keeps cells full of fluid? How can a membrane push fluid against a
pressure gradient?
Physical idea:Osmotic pressure is a simple example of an entropic force.

7.1 Microscopic view of entropic forces

Before proceeding to new ideas, let’s take two last looks at the ideal gas law. We already understand
the formula for pressure,p=kBTN/V,from a mechanistic point of view (Section 3.2.1 on page
73). Let’s now recover this result using the partition function—a useful warmup for our study of
electrostatic forces later in this chapter, and for our study of single-molecule stretching in Chapter 9.

7.1.1 Fixed-volume approach

Suppose a chamber with gas is in thermal contact with a large body at fixed temperatureT.In
an ideal gas,Nparticles move independently of each other, constrained only by the walls of their
vessel, a cube with sides of lengthL.The total energy is then just the sum of the molecules’ kinetic
energies, plus a constant for their unchanging internal energies.
The Example on page 198 gives us a convenient way to calculate the free energy of this system,
using its relation to the partition function. Indeed, in this situation the general formula for the
partition function (Equation 6.33 on page 198) becomes very simple. To specify a state of the
system, we must give the positions{ri}and momenta{pi}of every particle. To sum over all
possible states, we therefore must sum over every such collection ofr 1 ,...,pN.Since position and
momentum are continuous variables, we write the sums as integrals:

Z(L)=C

∫L

0

d^3 r 1

∫∞

−∞

d^3 p 1 ···

∫L

0

d^3 rN

∫∞

−∞

d^3 pNe−(p^1

(^2) +···+pN (^2) )/(2mkBT)

. (7.1)

Don’t confuse the vectorspi(momentum) with the scalarp(pressure)! The limits on the integrals
mean that each of the three components ofrirun from 0 toL. The factorCincludes a factor
of e−i/kBTfor each particle, whereiis the internal energy of moleculei.For an ideal gas these
factors are all fixed, and soCis a constant; we won’t need its numerical value. (In Chapter 8 we
willlet the internal energies change, when we study chemical reactions.) The Example on page 198
then gives the free energy as−kBTlnZ(L).
Equation 7.1 looks awful, but all we really want is thechangein free energy as we change
the volume of the box. That’s because an entropic force is a derivative of the free energy (see
Equation 6.17 on page 188); to get the dimensions of pressure (force/area, or energy/volume), we
need−dF(L)/d(L^3 ). Butmost of the integrals in Equation 7.1 are just constants,aswesee by
rearranging them:

Z(L)=C

(∫L

0

d^3 r 1 ···

∫L

0

d^3 rN

)(∫∞

−∞

d^3 p 1 e−p^1

(^2) /(2mkBT)
···

∫∞

−∞

d^3 pNe−pN

(^2) /(2mkBT)

)

.

The only things that depend onLare thelimitsof the first 3Nintegrals, and each of these integrals
just equalsL.ThusZis a constant timesL^3 N,and soF(L)isaconstant plus−kBTNlnL^3.