218 Chapter 7. Entropic forces at work[[Student version, January 17, 2003]]
This result is reasonable: In an ideal gas the total potential energy is a constant (the particles
don’t change, and they don’t interact with each other) and so is the total kinetic energy (it’s just
3
2 kBTN). Hence the free energyF=E−TSis a constant minusTS.Sowehaverecovered a fact
wealready knew, that the entropy is a constant plusNkBlnL^3 (see the Example on page 178).
Differentiating the free energy recovers the ideal gas law:
p=−
dF
d(L^3 )=
kBTN
V. (7.2)
7.1.2 Fixed-pressure approach
Aslight rephrasing of the argument just given will prove useful for our discussion of macromolecular
stretching in Chapter 9. Instead of fixingV and findingp,let’s fix thepressureand find the
equilibriumvolume. That is, instead of a box of fixed volume, we now imagine a cylinder with a
sliding piston. The displacementLof the piston is variable; its areaAis fixed. A forcefpushes
the piston inward; thus, the potential energy of the piston isfL. The available volume for the
gas molecules is nowAL.Thuswe’d like to compute the average value〈L〉,given the externally
supplied force. As always, this average is given by a sum over all states ofLtimes the probability
to be in the given state.
The Boltzmann distribution gives the probability to have a specified set of positions and mo-
menta:
P(r 1 ,...,pN,L,ppiston)=C 1 exp[
−
(
p 12 +···+pN^2
2 m+
(ppiston)^2
2 M
+fL)
/kBT]
. (7.3)
In this formulaMis the mass of the piston andppistonis its momentum, so (ppiston)^2 /(2M)isits
kinetic energy.
Wewish to calculate
∫
L×P(L,...). It’s convenient to use the fact thatP,like any probability
distribution, is normalized^1 ;thus, its integral equals one, and we can rewrite our desired quantity
as
〈L〉=∫
L∫×P(L,r 1 ,...)d^3 r 1 ···d^3 pNdppistondL
P(L,r 1 ,...)d^3 r 1 ···d^3 pNdppistondL. (7.4)
This time therintegrals give (AL)N.The reason it was convenient to introduce the denominator
in Equation 7.4 is that now the remaining integrals simply cancel between the numerator and
denominator, as does the constantC 1 ,leaving
〈L〉=∫∞
0 dLe−fL/kBTLN×L
∫∞
0 dLe−fL/kBTLN. (7.5)
Your Turn 7a
a. Check that this expression equals (N+1)kBT/f.[Hint: Use integration by parts to make
the numerator of Equation 7.5 look more like the denominator.]
b. Show that we have once again derived the ideal gas law. [Hint: Remember thatNis so big
thatN+1≈N.]Here is one last reformulation. The trick of differentiating under the integral sign^2 shows that
Equation 7.5 can be compactly written as〈L〉=d(−kBTlnZ(f))/df.ReplacingfbypAandL
(^1) See Section 3.1.1 on page 66.
(^2) See Equation 3.14 on page 70.