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Problems & Solutions

for

Statistical Physics of Particles

Updated July 2008

by

Mehran Kardar

Department of Physics

Massachusetts Institute of Technology

Cambridge, Massachusetts 02139, USA

Table of Contents

• I. Thermodynamics
• II. Probability
• III. Kinetic Theory of Gases
• IV. Classical Statistical Mechanics
• V. Interacting Particles
• VI. Quantum Statistical Mechanics
• VII. Ideal Quantum Gases

In equilibrium, the total energy associated with the three interfaces should be mini-

mum, and therefore

dE = Saw dAaw + 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(R sin θ)

2 , and Aaw = 2πR

2 (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 =

πR

3

cos

3 θ − 3 cos θ + 2

Introducing x = cos θ, we can re-write the above results as

Aws = πR

2

1 − x

2

Aaw = 2πR

2 (1 − x) ,

V =

πR

3

x

3 − 3 x + 2

The variations of these quantities are then obtained from

dAws = 2πR

[

dR

dx

(1 − x

2 ) − Rx

]

dx,

dAaw = 2πR

[

dR

dx

(1 − x) − R

]

dx,

dV = πR

2

[

dR

dx

(x

3 − 3 x + 2) + R(x

2 − x)

]

dx = 0.

From the last equation, we conclude

R

dR

dx

x

2 − 1

x^3 − 3 x + 2

x + 1

(x − 1)(x + 2)

Substituting for dR/dx gives,

dAws = 2πR

2 dx

x + 2

, and dAaw = 2πR

2 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

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

− 2 N m

− 1

. Estimate the typical length-scale that separates “large” and

“small” behaviors. Give a couple of examples for where this length-scale is important.

CH 3 − (CH 2 ) 11 − N

(CH 3 ) 3 · Cl

− ,

CH 3 − (CH 2 ) 11 − O − (CH 2 − CH 2 − 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

kB T · ∆A = (S − So) · ∆A, =⇒ S = So −

N

A

kB T.

(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-water-

surface 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

N kB T

(A − N b)^2

2 a

A

N

A

, and

∂T

∂S

A

A − N b

N kB

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

N kB T

(A − N b)

2

2 a

A

N

A

and

∂T

∂S

A

A − N b

N kB

Integrating the first equation, gives

S(A, t) = f (T ) −

N kB T

A − N b

• a

N

A

where f (T ) is a function of only T , while integrating the second equation, yields

S(A, T ) = g(A) −

N kB T

A − N b

with g(A) a function of only A. By comparing these two equations we get

S(A, T ) = So −

N kB T

A − N b

• a

N

A

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 N b 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 C S

− 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

Consider the Carnot cycle indicated in the figure. For the segment 1 to 2, which undergoes

an isothermal expansion, we have

dθ = 0, =⇒ ¯dQhot = P dV, and P =

N kB θhot

V

Hence, the heat input of the cycle is related to the expansion factor by

Qhot =

∫ V

2

V 1

N kB θhot

dV

V

= N kB θhot ln

V 2

V 1

A similar calculation along the low temperature isotherm yields

Qcold =

∫ V

3

V 4

N kB θcold

dV

V

= N kB θcold ln

V 3

V 4

and thus

Qhot

Qcold

θhot

θcold

ln (V 2 /V 1 )

ln (V 3 /V 4 )

(b) Calculate the volume expansion factor in an adiabatic process as a function of Θ.

• Next, we calculate the volume expansion/compression ratios in the adiabatic processes.

Along an adiabatic segment

¯dQ^ = 0,^ =⇒^ 0 =^

dE

dθ

· dθ +

N kB θ

V

· dV, =⇒

dV

V

N kB θ

dE

dθ

· dθ.

Integrating the above between the two temperatures, we obtain

ln

V 3

V 2

N kB

∫ (^) θ hot

θcold

θ

dE

dθ

· dθ, and

ln

V 4

V 1

N kB

∫ (^) θ hot

θcold

θ

dE

dθ

· dθ.

While we cannot explicitly evaluate the integral (since E(θ) is arbitrary), we can nonethe-

less conclude that

V 1

V 4

V 2

V 3

(c) Show that QH /QC = ΘH /ΘC.

• Combining the results of parts (a) and (b), we observe that

Qhot

Qcold

θhot

θcold

Since the thermodynamic temperature scale is defined by

Qhot

Qcold

Thot

Tcold

we conclude that θ and T are proportional. If we further define θ(triple pointH 20 ) =

T (triple pointH 20 ) = 273. 16 , θ and T become identical.

4. Equations of State: The equation of state constrains the form of internal energy as

in the following examples.

(a) Starting from dE = T dS − P dV , show that the equation of state P V = N kB T , in fact

implies that E can only depend on T.

• Since there is only one form of work, we can choose any two parameters as independent

variables. For example, selecting T and V , such that E = E(T, V ), and S = S(T, V ), we

obtain

dE = T dS − P dV = T

∂S

∂T

V

dT + T

∂S

∂V

T

dV − P dV,

resulting in

∂E

∂V

T

= T

∂S

∂V

T

− P.

Using the Maxwell’s relation

†

∂S

∂V

T

∂P

∂T

V

we obtain

∂E

∂V

T

= T

∂P

∂T

V

− P.

Since T

∂P ∂T

V

= T

NkB V

= P , for an ideal gas,

∂E ∂V

T

= 0. Thus E depends only on T , i.e.

E = E(T ).

(b) What is the most general equation of state consistent with an internal energy that

depends only on temperature?

• If E = E(T ),

∂E

∂V

T

= 0, =⇒ T

∂P

∂T

V

= P.

The solution for this equation is P = f (V )T, where f (V ) is any function of only V.

† dL = Xdx + Y dy + · · · , =⇒

∂X ∂y

x

∂Y ∂x

y

∂

2 L ∂x·∂y

where s and v are molar entropy and volume, respectively. Thus, the coexistence line

satisfies the condition

dP

dT

coX

Sg − Sl

Vg − Vl

sg − sl

vg − vl

For an alternative derivation, consider a Carnot engine using one mole of water. At the

source (P, T ) the latent heat L is supplied converting water to steam. There is a volume

increase V associated with this process. The pressure is adiabatically decreased to P − dP.

At the sink (P − dP, T − dT ) steam is condensed back to water.

(a) Show that the work output of the engine is W = V dP + O(dP

2 ). Hence obtain the

Clausius–Clapeyron equation

dP

dT

boiling

L

T V

• If we approximate the adiabatic processes as taking place at constant volume V (vertical

lines in the P − V diagram), we find

W =

P dV = P V − (P − dP )V = V dP.

pressure

volume

1 2

(^4 )

Q cold

Q hot

liq. gas T

T-dT

P

P-dP

V

Here, we have neglected the volume of liquid state, which is much smaller than that

of the gas state. As the error is of the order of

∂V

∂P

S

dP · dP = O(dP

2 ),

we have

W = V dP + O(dP

2 ).

The efficiency of any Carnot cycle is given by

η =

W

QH

TC

TH

and in the present case,

QH = L, W = V dP, TH = T, TC = T − dT.

Substituting these values in the universal formula for efficiency, we obtain the Clausius-

Clapeyron equation

V dP

L

dT

T

, or

dP

dT

coX

L

T · V

(b) What is wrong with the following argument: “The heat QH supplied at the source to

convert one mole of water to steam is L(T ). At the sink L(T − dT ) is supplied to condense

one mole of steam to water. The difference dT dL/dT must equal the work W = V dP ,

equal to LdT /T from eq.(1). Hence dL/dT = L/T implying that L is proportional to T !”

• The statement “At the sink L(T − dT ) is supplied to condense one mole of water” is

incorrect. In the P − V diagram shown, the state at “1” corresponds to pure water, “2”

corresponds to pure vapor, but the states “3” and “4” have two phases coexisting. In

going from the state 3 to 4 less than one mole of steam is converted to water. Part of the

steam has already been converted into water during the adiabatic expansion 2 → 3, and

the remaining portion is converted in the adiabatic compression 4 → 1. Thus the actual

latent heat should be less than the contribution by one mole of water.

(c) Assume that L is approximately temperature independent, and that the volume change

is dominated by the volume of steam treated as an ideal gas, i.e. V = N kB T /P. Integrate

equation (1) to obtain P (T ).

• Consider a horizontal slab of area A between heights h and h + dh. The gravitational

force due to mass of particles in the slab is

dFgravity = mg

N

V

Adh = mg

P

kB T

Adh,

where we have used the ideal gas law to relate the density (N/V ) to the pressure. The

gravitational force is balanced in equilibrium with the force due to pressure

dFpressure = A [P (h) − P (h + dh)] = −

∂P

∂h

Adh.

Equating the two forces gives

∂P

∂h

= −mg

P

kB T

, =⇒ P (h) = p 0 exp

mgh

kB T

assuming that temperature does not change with height.

(f) Use the above results to estimate the boiling temperature of water on top of Mount

Everest (h ≈ 9 km). The latent heat of vaporization of water is about 2. 3 × 10

6 Jkg

− 1 .

• Using the results from parts (c) and (e), we conclude that

PEverest

Psea

≈ exp

mg

kB T

(hEverest − hsea)

≈ exp

[

L

kB

TEverest(boil)

Tsea(boil)

)]

Using the numbers provided, we find TEverest(boil) ≈ 346

o K (

o C≈ 163

o F).

6. Glass: Liquid quartz, if cooled slowly, crystallizes at a temperature Tm, and releases

latent heat L. Under more rapid cooling conditions, the liquid is supercooled and becomes

glassy.

(a) As both phases of quartz are almost incompressible, there is no work input, and changes

in internal energy satisfy dE = T dS + μdN. Use the extensivity condition to obtain the

expression for μ in terms of E, T , S, and N.

• Since in the present context we are considering only chemical work, we can regard

entropy as a function of two independent variables, e.g. E, and N , which appear naturally

from dS = dE/T − μdN/T. Since entropy is an extensive variable, λS = S(λE, λN ).

Differentiating this with respect to λ and evaluating the resulting expression at λ = 1,

gives

S(E, N ) =

∂S

∂E

N

E +

∂S

∂N

E

N =

E

T

N μ

T

leading to

μ =

E − T S

N

(b) The heat capacity of crystalline quartz is approximately CX = αT

3 , while that of

glassy quartz is roughly CG = βT , where α and β are constants.

Assuming that the third law of thermodynamics applies to both crystalline and glass

phases, calculate the entropies of the two phases at temperatures T ≤ Tm.

• Finite temperature entropies can be obtained by integrating ¯dQ/T , starting from S(T =

1. = 0. Using the heat capacities to obtain the heat inputs, we find

Ccrystal = αT

3

T

N

dScrystal

dT

, =⇒ Scrystal =

N αT

3

Cglass = βT =

T

N

dSglass

dT

, =⇒ Sglass = βN T.

(c) At zero temperature the local bonding structure is similar in glass and crystalline

quartz, so that they have approximately the same internal energy E 0. Calculate the

internal energies of both phases at temperatures T ≤ Tm.

• Since dE = T dS + μdN , for dN = 0, we have

dE = T dS = αN T

3 dT (crystal),

dE = T dS = βN T dT (glass).

Integrating these expressions, starting with the same internal energy Eo at T = 0, yields

E = Eo +

αN

T

4 (crystal),

E = Eo +

βN

T

2 (glass).

(d) Use the condition of thermal equilibrium between two phases to compute the equilib-

rium melting temperature Tm in terms of α and β.

• From the condition of chemical equilibrium between the two phases, μcrystal = μglass,

we obtain ( 1

· αT

4

· βT

2 , =⇒

αT

4

βT

2

• We have Cx = T

∂S ∂T

= A(x)T , where S = S(T, x). Thus

∂A

∂x

∂x

∂S

∂T

∂T

∂S

∂x

from part (a), implying that A is independent of x.

(c) Give the expression for S(T, x) assuming S(0, 0) = S 0.

• S(x, T ) can be calculated as

S(x, T ) = S(0, 0) +

∫ T ′=T

T ′=

∂S(T

′ , x = 0)

∂T ′^

dT

′

∫ (^) x′=x

x′=

∂S(T, x

′ )

∂x′^

dx

= S 0 +

∫ T

0

AdT

′

∫ (^) x

0

(b − cx

′ )dx

′

= S 0 + AT + (b −

c

x)x.

(d) Calculate the heat capacity at constant tension, i.e. CJ = T ∂S/∂T |J as a function of

T and J.

• Writing the entropy as S(T, x) = S(T, x(T, J)), leads to

∂S

∂T

J

∂S

∂T

x

∂S

∂x

T

∂x

∂T

J

From parts (a) and (b),

∂S ∂x

T

= b − cx and

∂S ∂T

∣ (^) x = A. Furthermore, ∂x ∂T

J

is given by

a

∂x ∂T

− b + cx + cT

∂x ∂T

= 0, i.e.

∂x

∂T

b − cx

a + cT

Thus

CJ = T

[

A +

(b − cx)

2

(a + cT )

]

Since x =

J+bT a+cT

, we can rewrite the heat capacity as a function of T and J, as

CJ = T

[

A +

(b − c

J+bT a+cT

2

(a + cT )

]

= T

[

A +

(ab − cJ)

2

(a + cT )

3

]

8. Hard core gas: A gas obeys the equation of state P (V − N b) = N kB T , and has a

heat capacity CV independent of temperature. (N is kept fixed in the following.)

(a) Find the Maxwell relation involving ∂S/∂V | T,N

• For dN = 0,

d(E − T S) = −SdT − P dV, =⇒

∂S

∂V

T,N

∂P

∂T

V,N

(b) By calculating dE(T, V ), show that E is a function of T (and N ) only.

• Writing dS in terms of dT and dV ,

dE = T dS − P dV = T

∂S

∂T

V,N

dT +

∂S

∂V

T,N

dV

− P dV.

Using the Maxwell relation from part (a), we find

dE(T, V ) = T

∂S

∂T

V,N

dT +

T

∂P

∂T

V,N

− P

dV.

But from the equation of state, we get

P =

N kB T

(V − N b)

∂P

∂T

V,N

P

T

, =⇒ dE(T, V ) = T

∂S

∂T

V,N

dT,

i.e. E(T, N, V ) = E(T, N ) does not depend on V.

(c) Show that γ ≡ CP /CV = 1 + N kB /CV (independent of T and V ).

• The hear capacity is

CP =

∂Q

∂T

P

∂E + P V

∂T

P

∂E

∂T

P

+ P

∂V

∂T

P

But, since E = E(T ) only,

∂E

∂T

P

∂E

∂T

V

= CV ,

and from the equation of state we get

∂V

∂T

P

N kB

P

, =⇒ CP = CV + N kB , =⇒ γ = 1 +

N kB

CV

which is independent of T , since CV is independent of temperature. The independence of

CV from V also follows from part (a).