Biological Physics: Energy, Information, Life

8.1. Chemical potential[[Student version, January 17, 2003]] 261

point (see the Example on page 178), but we need to remember thatEappearing there was only
the kinetic energyEkin.

Your Turn 8a

As a first step to evaluating Equation 8.1, calculate the derivative ofSwith respect toNfor an

ideal gas, holding fixed the kinetic energy. TakeNto be very large and find

dS

dN

∣∣

∣∣

Ekin

=kB

3

2 ln

(

1

3 π

m


^2

Ekin

N

(

V

N

) 2 / 3 )

.

Tofinish the derivation ofμ,weneed to convert the formula you just found to give the derivative
at fixed total energyE,not fixedEkin.Ifweinject a molecule into the system holdingEkinfixed,
thenextractan amount of kinetic energy equal to the internal energyof that molecule, this has
the net effect of holding the total energyEfixed while changing the particle number by dN=1.
Thus we need to subtract a correction term from the result in Your Turn 8a above.

Example Carry out the step just described in words, and show that the chemical potential of

an ideal gas can be written as

μ=kBTln(c/c 0 )+μ^0 (T). ideal gas or dilute solution (8.3)

In this formulac=N/V is the number density,c 0 is a constant (the “reference

concentration”), and

μ^0 (T)=−^32 kBTln

mkBT

2 π
^2 c 02 /^3

. ideal gas (8.4)

Solution: Translating the words into math, we need to subtractdEdSkin

∣∣

∣Nfrom the

result in Your Turn 8a. Combining the resulting formula with Equation 8.1, and

using the fact that the average kinetic energyEkin/Nequals^32 kBT,then gives

μ=kBTlnc−

3

2 kBTln

[

4 π

3

m

(2π
)^2

3

2 kBT

]

.

The above formula appears to involve the logarithms of dimensional quantities. To

make each term separately well-defined, we add and subtractkBTlnc 0 ,obtaining

Equations 8.3–8.4.

Wecallμ^0 thestandard chemical potentialat temperatureTdefined with respect to the chosen
reference concentration. The choice of the reference value is a convention; the derivation above
makes it clear that its value drops out of the right-hand side of Equation 8.3. Chemists refer to
the dimensionless quantity e(μ−μ

(^0) )/kBT
as theactivity.ThusEquation 8.3 states that the activity
equals approximatelyc/c 0 for an ideal gas.
Equation 8.3 also holds for dilutesolutionsas well as for low-density gases. As argued in our
discussion of osmotic pressure (Section 7.2 on page 219), the entropic term is the same in either
case. For a solute in a liquid, however, the value ofμ^0 (T)will no longer be given by Equation 8.4.
Instead,μ^0 (T)will now reflect the fact that the solvent (water) molecules themselves arenot
dilute, so the attractions of solvent molecules to each other and to the solute are not negligible.
Nevertheless, Equation 8.3 will still hold with some measurable standard chemical potentialμ^0 (T)