TEST BANK FOR Statistical Physics of Particles By Mehran Kardar (Solution manual)

I. Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
II. Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
III. Kinetic Theory of Gases . . . . . . . . . . . . . . . . . . . . . . . . 38
IV. Classical Statistical Mechanics . . . . . . . . . . . . . . . . . . . . . 72
V. Interacting Particles . . . . . . . . . . . . . . . . . . . . . . . . . . 93
VI. Quantum Statistical Mechanics . . . . . . . . . . . . . . . . . . . . 121
VII. Ideal Quantum Gases . . . . . . . . . . . . . . . . . . . . . . . . 138
Problems for Chapter I - Thermodynamics
1. Surface tension: Thermodynamic properties of the interface between two phases are
described by a state function called the surface tension S. It is defined in terms of the
work required to increase the surface area by an amount dA through ¯dW = SdA.
(a) By considering the work done against surface tension in an infinitesimal change in
radius, show that the pressure inside a spherical drop of water of radius R is larger than
outside pressure by 2S/R. What is the air pressure inside a soap bubble of radius R?
• The work done by a water droplet on the outside world, needed to increase the radius
from R to R + R is
W = (P − Po) · 4πR2 · R,
where P is the pressure inside the drop and Po is the atmospheric pressure. In equilibrium,
this should be equal to the increase in the surface energy SA = S · 8πR · R, where S
is the surface tension, and
Wtotal = 0, =⇒ Wpressure = −Wsurface ,
resulting in
(P − Po) · 4πR2 · R = S · 8πR · R, =⇒ (P − Po) =
2S
R
.
In a soap bubble, there are two air-soap surfaces with almost equal radii of curvatures,
and
Pfilm − Po = Pinterior − Pfilm =
2S
R
,
leading to
Pinterior − Po =
4S
R
.
Hence, the air pressure inside the bubble is larger than atmospheric pressure by 4S/R.
(b) A water droplet condenses on a solid surface. There are three surface tensions involved
Saw, Ssw, and Ssa, where a, s, and w refer to air, solid and water respectively. Calculate
the angle of contact, and find the condition for the appearance of a water film (complete
wetting).
• When steam condenses on a solid surface, water either forms a droplet, or spreads on
the surface. There are two ways to consider this problem:
Method 1: Energy associated with the interfaces
1
In equilibrium, the total energy associated with the three interfaces should be minimum,
and therefore
dE = SawdAaw + SasdAas + SwsdAws = 0.
Since the total surface area of the solid is constant,
dAas + dAws = 0.
From geometrical considerations (see proof below), we obtain
dAws cos θ = dAaw.
From these equations, we obtain
dE = (Saw cos θ − Sas + Sws) dAws = 0, =⇒ cos θ =
Sas − Sws
Saw
.
Proof of dAws cos θ = dAaw: Consider a droplet which is part of a sphere of radius R,
which is cut by the substrate at an angle θ. The areas of the involved surfaces are
Aws = π(Rsin θ)2, and Aaw = 2πR2(1 − cos θ).
Let us consider a small change in shape, accompanied by changes in R and θ. These
variations should preserve the volume of water, i.e. constrained by
V =
πR3
3
􀀀
cos3 θ − 3 cos θ + 2

.
Introducing x = cos θ, we can re-write the above results as


Aws = πR2 􀀀
1 − x2
,
Aaw = 2πR2 (1 − x) ,
V =
πR3
3
􀀀
x3 − 3x + 2

.
The variations of these quantities are then obtained from


dAws = 2πR

dR
dx
(1 − x2) − Rx

dx,
dAaw = 2πR

2
dR
dx
(1 − x) − R

dx,
dV = πR2

dR
dx
(x3 − 3x + 2) + R(x2 − x)

dx = 0.
2
From the last equation, we conclude
1
R
dR
dx
= −
x2 − 1
x3 − 3x + 2
= −
x + 1
(x − 1)(x + 2)
.
Substituting for dR/dx gives,
dAws = 2πR2 dx
x + 2
, and dAaw = 2πR2 x · dx
x + 2
,
resulting in the required result of
dAaw = x · dAws = dAws cos θ.
Method 2: Balancing forces on the contact line
Another way to interpret the result is to consider the force balance of the equilibrium
surface tension on the contact line. There are four forces acting on the line: (1) the surface
tension at the water–gas interface, (2) the surface tension at the solid–water interface, (3)
the surface tension at the gas–solid interface, and (4) the force downward by solid–contact
line interaction. The last force ensures that the contact line stays on the solid surface, and
is downward since the contact line is allowed to move only horizontally without friction.
These forces should cancel along both the y–direction x–directions. The latter gives the
condition for the contact angle known as Young’s equation,
Sas = Saw · cos θ + Sws , =⇒ cos θ = Sas − Sws
Saw
.
The critical condition for the complete wetting occurs when θ = 0, or cos θ = 1, i.e. for
cos θC = Sas − Sws
Saw
= 1.
Complete wetting of the substrate thus occurs whenever
Saw ≤ Sas − Sws.
(c) In the realm of “large” bodies gravity is the dominant force, while at “small” distances
surface tension effects are all important. At room temperature, the surface tension of
water is So ≈ 7×10−2Nm−1. Estimate the typical length-scale that separates “large” and
“small” behaviors. Give a couple of examples for where this length-scale is important.
3
• Typical length scales at which the surface tension effects become significant are given
by the condition that the forces exerted by surface tension and relevant pressures become
comparable, or by the condition that the surface energy is comparable to the other energy
changes of interest.
Example 1: Size of water drops not much deformed on a non-wetting surface. This is given
by equalizing the surface energy and the gravitational energy,
S · 4πR2 ≈ mgR = ρV gR =
4π
3
R4g,
leading to
R ≈
s
3S
ρg ≈
s
3 · 7 × 10−2N/m
103kg/m3 × 10m/s2 ≈ 1.5 × 10−3m = 1.5mm.
Example 2: Swelling of spherical gels in a saturated vapor: Osmotic pressure of the gel
(about 1 atm) = surface tension of water, gives
πgel ≈
N
V
kBT ≈
2S
R
,
where N is the number of counter ions within the gel. Thus,
R ≈

2 × 7 × 10−2N/m
105N/m2

≈ 10−6m.
********
2. Surfactants: Surfactant molecules such as those in soap or shampoo prefer to spread
on the air-water surface rather than dissolve in water. To see this, float a hair on the
surface of water and gently touch the water in its vicinity with a piece of soap. (This is
also why a piece of soap can power a toy paper boat.)
(a) The air-water surface tension So (assumed to be temperature independent) is reduced
roughly by NkBT/A, where N is the number of surfactant particles, and A is the area.
Explain this result qualitatively.
• Typical surfactant molecules have a hydrophilic head and a hydrophobic tail, and prefer
to go to the interface between water and air, or water and oil. Some examples are,
CH3 − (CH2)11 − SO−3 · Na+,
4
CH3 − (CH2)11 − N+(CH3)3 · Cl−,
CH3 − (CH2)11 − O − (CH2 − CH2 − O)12 − H.
The surfactant molecules spread over the surface of water and behave as a two dimensional
gas. The gas has a pressure proportional to the density and the absolute temperature,
which comes from the two dimensional degrees of freedom of the molecules. Thus the
surfactants lower the free energy of the surface when the surface area is increased.
Fsurfactant =
N
A
kBT · A = (S − So) · A, =⇒ S = So −
N
A
kBT.
(Note that surface tension is defined with a sign opposite to that of hydrostatic pressure.)
(b) Place a drop of water on a clean surface. Observe what happens to the air-watersurface
contact angle as you gently touch the droplet surface with a small piece of soap,
and explain the observation.
• As shown in the previous problem, the contact angle satisfies
cos θ = Sas − Sws
Saw
.
Touching the surface of the droplet with a small piece of soap reduces Saw, hence cos θ
increases, or equivalently, the angle θ decreases.
(c) More careful observations show that at higher surfactant densities
∂S
∂A

T
=
NkBT
(A − Nb)2 −
2a
A

N
A
2
, and
∂T
∂S

A
= −
A − Nb
NkB
;
where a and b are constants. Obtain the expression for S(A, T) and explain qualitatively
the origin of the corrections described by a and b.
• When the surfactant molecules are dense their interaction becomes important, resulting
in
∂S
∂A

T
=
NkBT
(A − Nb)2 −
2a
A

N
A
2
,
and
∂T
∂S

A
= −
A − Nb
NkB
.
Integrating the first equation, gives
S(A, t) = f(T) −
NkBT
A − Nb
+ a

N
A
2
,
5
where f(T) is a function of only T, while integrating the second equation, yields
S(A, T) = g(A) −
NkBT
A − Nb
,
with g(A) a function of only A. By comparing these two equations we get
S(A, T) = So −
NkBT
A − Nb
+ a

N
A
2
,
where So represents the surface tension in the absence of surfactants and is independent
of A and T. The equation resembles the van der Waals equation of state for gas-liquid
systems. The factor Nb in the second term represents the excluded volume effect due to
the finite size of the surfactant molecules. The last term represents the binary interaction
between two surfactant molecules. If surfactant molecules attract each other the coefficient
a is positive the surface tension increases.
(d) Find an expression for CS − CA in terms of ∂E
∂A

T , S, ∂S ∂A

T
, and ∂T
∂S

A
, for ∂E
∂T

A =
∂E
∂T

S
.
• Taking A and T as independent variables, we obtain
δQ = dE − S · dA, =⇒ δQ =
∂E
∂A

T
dA +
∂E
∂T

A
dT − S · dA,
and
δQ =

∂E
∂A

T − S

dA +
∂E
∂T

A
dT.
From the above result, the heat capacities are obtained as


CA ≡
δQ
δT

A
=
∂E
∂T

A
CS ≡
δQ
δT

S
=

∂E
∂A

T − S

∂A
∂T

S
+
∂E
∂T

S
,
resulting in
CS − CA =

∂E
∂A

T − S

∂A
∂T

S
.
Using the chain rule relation
∂T
∂S

A ·
∂S
∂A

T ·
∂A
∂T

S
= −1,
6
we obtain
CS − CA =

∂E
∂A

T − S

·

 −1
∂T
∂S

A · ∂S ∂A

T

.
********
3. Temperature scales: Prove the equivalence of the ideal gas temperature scale , and
the thermodynamic scale T, by performing a Carnot cycle on an ideal gas. The ideal gas
satisfies PV = NkB, and its internal energy E is a function of only. However, you
may not assume that E ∝ . You may wish to proceed as follows:
(a) Calculate the heat exchanges QH and QC as a function of H, C, and the volume
expansion factors.
• The ideal gas temperature is defined through the equation of state
θ =
PV
NkB
.
The thermodynamic temperature is defined for a reversible Carnot cycle by
Thot
Tcold
=
Qhot
Qcold
.
For an ideal gas, the internal energy is a function only of θ, i.e. E = E(θ), and
¯dQ = dE − ¯dW =
dE
dθ · dθ + PdV.
adiabatics (DQ = 0)
1
pressure P
volume V
2
3
4
Q hot
Q cold
q
q