Biological Physics: Energy, Information, Life

178 Chapter 6. Entropy, temperature, and free energy[[Student version, January 17, 2003]]

denotedS:
S≡kB
K
I=kBln Ω. (6.5)
Before saying another word about these abstractions, let’s pause to evaluate the entropy explic-
itly, for a system we know intimately.

Example Find the entropy for an ideal gas.

Solution: Wewantto count all states allowed by the conservation of energy. We

express the energy in terms of the momentum of each particle:

E=

∑N

i=1

m

2 vi

(^2) =^1
2 m

∑N

i=1

pi^2 =

1

2 m

∑N

i=1

∑^3

J=1

(

(pi)J

) 2

.

Here (pi)J is the component of particlei’s momentum along theJaxis. This

formula resembles the Pythagorean formula: In fact, ifN=1then it says precisely

that

√

2 mEis the distance of the pointp from the origin, or in other words that

the allowed momentum vectors lie on the surface of an ordinary (two-dimensional)

sphere (recall Figure 3.4 on page 72).

When there are lots of molecules, the locus of allowed values of{(pi)J}is the

surface of a sphere of radiusr =

√

2 mEin 3N-dimensional space. The number

of states available forNmolecules in volumeV at fixedEis then proportional to

the surface area of this hyper-sphere. Certainly that area must be proportional to

the radius raised to the power 3N− 1 (think about the case of an ordinary sphere,

N=1,whose area is 4πr^2 =4πr^3 N−^1 ). SinceNis much larger than one, we can

replace 3N−1byjust 3N.

Tospecify the microstate we must give not only the momenta but also the locations

of each particle. Since each may be located anywhere in the box, this means that

the number of available states also contains a factor ofVN.SoΩisaconstant times

(2mE)^3 N/^2 VN,andS=NkBln[(E)^3 /^2 V]+const.

The complete version of the last result is called theSakur–Tetrode formula:

S=kBln

[(

2 π^3 N/^2

(3N/ 2 −1)!

)

(2mE)^3 N/^2 VN

1

N!

(2π
)−^3 N

1

2

]

. (6.6)

This is a complex formula, but we can understand it by considering each of the factors in turn.
The first factor in round parentheses is the area of a sphere of radius 1, in 3Ndimensions. It can
beregarded as a fact from geometry, to be looked up in a book (or see Section 6.2.2′on page 206
below). Certainly it equals 2πwhen 3N=2,and that’s the right answer: The circumference of a
unit circle in the plane really is 2π.The next two factors are just what we found in the Example
above. The factor of (N!)−^1 reflects the fact that gas molecules are indistinguishable; if we exchange
r 1 ,p 1 withx 2 ,p 2 ,weget a different list ofri’s andpi’s, but not a physically different state of the
system.
is the Planck constant, a constant of Nature with the dimensionsML^2 T−^1. Its origin
lies in quantum mechanics, but for our purposes it’s enough to note thatsomeconstant with these
dimensions is needed to make the dimensions in Equation 6.6 work out properly. The actual value
of
won’t enter any of our physical predictions.
Equation 6.6 looks scary, but many of the factors enter in non-essential ways. For instance,
the first “2” in the numerator gets totally overwhelmed by the other factors whenNis big, so we
can drop it, or equivalently put in an extra factor of 1/2, as was done at the end of Equation 6.6.