CHAPTER 2.

THERMODYNAMICS

66

where m = M/ is the magnetization per mole of substance, and



M
1

isothermal susceptibility:
=
T =
H T



M
1

adiabatic susceptibility:
=
S =
H S

2G
H 2

(2.307)

2H
.
H 2

(2.308)

Here the enthalpy and Gibbs free energy are

H = E HM
G = E T S HM

dH = T dS M dH
dG = S dT M dH .

(2.309)
(2.310)

Remark: The previous discussion has assumed an isotropic magnetic system where M and H are collinear, hence
H M = HM .


2G
1
M

(2.311)
T =
H T
H H


M
2H
1

.
(2.312)
S =
H S
H H
In this case, the enthalpy and Gibbs free energy are
H = E H M
G = E T S H M

dH = T dS M dH
dG = S dT M dH .

(2.313)
(2.314)

2.11.5 Joule effect: free expansion of a gas

Previously we considered the adiabatic free expansion of an ideal gas. We found that Q = W = 0 hence E = 0,
which means the process is isothermal, since E = (T ) is volume-independent. The entropy changes, however,
since S(E, V, N ) = N kB ln(V /N ) + 21 f N kB ln(E/N ) + N s0 . Thus,
 
V
(2.315)
Sf = Si + N kB ln f .
Vi
What happens if the gas is nonideal?
We integrate along a fictitious thermodynamic path connecting initial and final states, where dE = 0 along the
path. We have




E
E
0 = dE =
dV +
dT
(2.316)
V T
T V
hence

We also have

Thus,

T
V

E
V

=T

T
V

(E/V )T
1
=
(E/T )V
CV

S
V

p=T

p
T

E
V

(2.317)

p.

(2.318)

"

 #
p
1
pT
.
=
CV
T V

(2.319)

2.11. APPLICATIONS OF THERMODYNAMICS

gas
Acetone
Argon
Carbon dioxide
Ethanol
Freon
Helium
Hydrogen
Mercury
Methane
Nitrogen
Oxygen
Water

L2 bar
mol2

14.09
1.363
3.640
12.18
10.78
0.03457
0.2476
8.200
2.283
1.408
1.378
5.536

67

L
mol

0.0994
0.03219
0.04267
0.08407
0.0998
0.0237
0.02661
0.01696
0.04278
0.03913
0.03183
0.03049

pc (bar)

Tc (K)

vc (L/mol)

52.82
48.72
7404
63.83
40.09
2.279
12.95
1055
46.20
34.06
50.37
220.6

505.1
150.9
304.0
516.3
384.9
5.198
33.16
1723
190.2
128.2
154.3
647.0

0.2982
0.0966
0.1280
0.2522
0.2994
0.0711
0.0798
0.0509
0.1283
0.1174
0.0955
0.0915

Table 2.3: Van der Waals parameters for some common gases. (Source: Wikipedia.)

Note that the term in square brackets vanishes for any system obeying the ideal gas law. For a nonideal gas,

ZVf 
T
,
T = dV
V E

(2.320)

Vi

which is in general nonzero.

Now consider a van der Waals gas, for which


a
p + 2 (v b) = RT .
v
We then have
pT

p
T

a
a 2
=

.
v2
V2

(2.321)

In 2.11.3 we concluded that CV = 12 f R for the van der Waals gas, hence
2a
T =
fR

ZVf

Vi



2a 1
1
dV
.
=

V2
f R vf
vi

(2.322)

Thus, if Vf > Vi , we have Tf < Ti and the gas cools upon expansion.
Consider O2 gas with an initial specific volume of vi = 22.4 L/mol, which is the STP value for an ideal gas, freely
expanding to a volume vf = for maximum cooling. According to table 2.3, a = 1.378 L2 bar/mol2 , and we
have T = 2a/f Rvi = 0.296 K, which is a pitifully small amount of cooling. Adiabatic free expansion is a
very inefficient way to cool a gas.

CHAPTER 2. THERMODYNAMICS

68

Figure 2.22: In a throttle, a gas is pushed through a porous plug separating regions of different pressure. The
change in energy is the work done, hence enthalpy is conserved during the throttling process.

2.11.6 Throttling: the Joule-Thompson effect

In a throttle, depicted in Fig. 2.22, a gas is forced through a porous plug which separates regions of different
pressures. According to the figure, the work done on a given element of gas is
ZVi
ZVf
W = dV pf dV pi = pf Vf pi Vi .

(2.323)

Now we assume that the system is thermally isolated so that the gas exchanges no heat with its environment, nor
with the plug. Then Q = 0 so E = W , and
Ei + pi Vi = Ef + pf Vf

(2.324)

Hi = Hf ,

(2.325)

where H is enthalpy. Thus, the throttling process is isenthalpic. We can therefore study it by defining a fictitious
thermodynamic path along which dH = 0. The, choosing T and p as state variables,




H
H
dp +
dT
(2.326)
0 = dH =
T p
p T
hence

T
p

(H/p)T
.
(H/T )p

The numerator on the RHS is computed by writing dH = T dS + V dp and then dividing by dp, to obtain

 



H
S
V
=V +T
=V T
.
p T
p T
T p
The denominator is

H
T

(2.327)

(2.328)



S
H
=
S p T p


S
= Cp .
=T
T p

(2.329)

" 
#

1
v
T
=
v
cp
T p


v
=
T p 1 ,
cp

(2.330)

Thus,


T
p

2.11. APPLICATIONS OF THERMODYNAMICS

69

Figure 2.23: Inversion temperature T (p) for the van der Waals gas. Pressure and temperature are given in terms
of pc = a/27b2 and Tc = 8a/27bR, respectively.
where p =

1
V

V
T p

is the volume expansion coefficient.

From the van der Waals equation of state, we obtain, from eqn. 2.285,


T v
RT /v
vb
T p =
=
=

2 .
a
2ab
2a
v T p
p v2 + v3
v RT vb
v

Assuming v

a
RT

, b, we have

Thus, for T > T =

2a
Rb ,

we have

T
p H

T
p

1
cp


2a
b .
RT

(2.331)

(2.332)

< 0 and the gas heats up upon an isenthalpic pressure decrease. For

T < T , the gas cools under such conditions.

In fact, there are two inversion temperatures T1,2

for the van der Waals gas. To see this, we set T p = 1, which is
the criterion for inversion. From eqn. 2.331 it is easy to derive
r
bRT
b
=1
.
(2.333)
v
2a

We insert this into the van der Waals equation of state to derive a relationship T = T (p) at which T p = 1 holds.
After a little work, we find
r
a
3RT
8aRT
2 .
(2.334)
p=
+
2b
b3
b
This is a quadratic equation for T , the solution of which is
!2
r
2p
2a
3b
2 1
.
(2.335)
T (p) =
9 bR
a

CHAPTER 2. THERMODYNAMICS

70

In Fig. 2.23 we plot pressure versus temperature in scaled units, showing the curve along which
volume, pressure, and temperature scales defined are
vc = 3b

pc =

a
27 b2

Tc =

8a
.
27 bR

T
p H

= 0. The

(2.336)

Values for pc , Tc , and vc are provided in table 2.3. If we define v = v/vc , p = p/pc , and T = T /Tc , then the van der
Waals equation of state may be written in dimensionless form:


3
p + 2 3v 1) = 8T .
(2.337)
v
In terms of the scaled parameters, the equation for the inversion curve
q
2

p = 9 36 1 13 T

T
p

= 0 becomes

q

T =3 1 1

1
9

2

(2.338)

Thus, there is no inversion for p > 9 pc . We are usually interested in the upper inversion temperature, T2 ,
corresponding to the upper sign in eqn. 2.335. The maximum inversion temperature occurs for p = 0, where
2a

= 27
Tmax
= bR
4 Tc . For H2 , from the data in table 2.3, we find Tmax (H2 ) = 224 K, which is within 10% of the
experimentally measured value of 205 K.
 
< 0 and p < 0, we have T > 0.
What happens when H2 gas leaks from a container with T > T2 ? Since T
p
H

The gas warms up, and the heat facilitates the reaction 2 H2 + O2 2 H2 O, which releases energy, and we have
a nice explosion.

2.12 Phase Transitions and Phase Equilibria

A typical phase diagram of a p-V -T system is shown in the Fig. 2.24(a). The solid lines delineate boundaries
between distinct thermodynamic phases. These lines are called coexistence curves . Along these curves, we can
have coexistence of two phases, and the thermodynamic potentials are singular. The order of the singularity is
often taken as a classification of the phase transition. I.e. if the thermodynamic potentials E, F , G, and H have
discontinuous or divergent mth derivatives, the transition between the respective phases is said to be mth order.
Modern theories of phase transitions generally only recognize two possibilities: first order transitions, where the
order parameter changes discontinuously through the transition, and second order transitions, where the order parameter vanishes continuously at the boundary from ordered to disordered phases11 . Well discuss order parameters
during Physics 140B.
For a more interesting phase diagram, see Fig. 2.24(b,c), which displays the phase diagrams for 3 He and 4 He. The
only difference between these two atoms is that the former has one fewer neutron: (2p + 1n + 2e) in 3 He versus (2p
+ 2n + 2e) in 4 He. As we shall learn when we study quantum statistics, this extra neutron makes all the difference,
because 3 He is a fermion while 4 He is a boson.

2.12.1 p-v-T surfaces

The equation of state for a single component system may be written as
f (p, v, T ) = 0 .
11 Some

(2.339)

exotic phase transitions in quantum matter, which do not quite fit the usual classification schemes, have recently been proposed.

2.12. PHASE TRANSITIONS AND PHASE EQUILIBRIA

generic
substance

(b)

(c)

pressure

(a)

71

temperature

3He

4He

Figure 2.24: (a) Typical thermodynamic phase diagram of a single component p-V -T system, showing triple point
(three phase coexistence) and critical point. (Source: Univ. of Helsinki.) Also shown: phase diagrams for 3 He (b)
and 4 He (c). What a difference a neutron makes! (Source: Brittanica.)
This may in principle be inverted to yield p = p(v, T ) or v = v(T, p) or T = T (p, v). The single constraint f (p, v, T )
on the three state variables defines a surface in {p, v, T } space. An example of such a surface is shown in Fig. 2.25,
for the ideal gas.
Real p-v-T surfaces are much richer than that for the ideal gas, because real systems undergo phase transitions in
which thermodynamic properties are singular or discontinuous along certain curves on the p-v-T surface. An
example is shown in Fig. 2.26. The high temperature isotherms resemble those of the ideal gas, but as one cools
below the critical temperature Tc , the isotherms become singular. Precisely at T = Tc , the isotherm p = p(v, Tc )
becomes perfectly horizontal at v = vc , which is the critical molar volume. This means that the isothermal com

pressibility, T = v1 v
p T diverges at T = Tc . Below Tc , the isotherms have a flat portion, as shown in Fig. 2.28,
corresponding to a two-phase region where liquid and vapor coexist. In the (p, T ) plane, sketched for H2 O in Fig. 2.4
and shown for CO2 in Fig. 2.29, this liquid-vapor phase coexistence occurs along a curve, called the vaporization
(or boiling) curve. The density changes discontinuously across this curve; for H2 O, the liquid is approximately
1000 times denser than the vapor at atmospheric pressure. The density discontinuity vanishes at the critical point.
Note that one can continuously transform between liquid and vapor phases, without encountering any phase
transitions, by going around the critical point and avoiding the two-phase region.
In addition to liquid-vapor coexistence, solid-liquid and solid-vapor coexistence also occur, as shown in Fig. 2.26.
The triple point (Tt , pt ) lies at the confluence of these three coexistence regions. For H2 O, the location of the triple
point and critical point are given by
Tt = 273.16 K
pt = 611.7 Pa = 6.037 10

Tc = 647 K
3

atm

pc = 22.06 MPa = 217.7 atm

2.12.2 The Clausius-Clapeyron relation

Recall that the homogeneity of E(S, V, N ) guaranteed E = T S pV +N , from Eulers theorem. It also guarantees
a relation between the intensive variables T , p, and , according to eqn. 2.148. Let us define g G/ = NA , the
Gibbs free energy per mole. Then
dg = s dT + v dp ,
(2.340)

CHAPTER 2. THERMODYNAMICS

72

Figure 2.25: The surface p(v, T ) = RT /v corresponding to the ideal gas equation of state, and its projections onto
the (p, T ), (p, v), and (T, v) planes.
where s = S/ and v = V / are the molar entropy and molar volume, respectively. Along a coexistence curve
between phase #1 and phase #2, we must have g1 = g2 , since the phases are free to exchange energy and particle
number, i.e. they are in thermal and chemical equilibrium. This means
dg1 = s1 dT + v1 dp = s2 dT + v2 dp = dg2 .
Therefore, along the coexistence curve we must have
 
dp
s s1

= 2
=
,
dT coex v2 v1
T v

(2.341)

(2.342)

where
T s = T (s2 s1 )

(2.343)

is the molar latent heat of transition. A heat must be supplied in order to change from phase #1 to phase #2, even
without changing p or T . If is the latent heat per mole, then we write as the latent heat per gram: = /M ,
where M is the molar mass.
Along the liquid-gas coexistence curve, we typically have vgas vliquid , and assuming the vapor is ideal, we may
write v vgas RT /p. Thus,
 

p
dp
=
.
(2.344)

dT liqgas
T v
RT 2
If remains constant throughout a section of the liquid-gas coexistence curve, we may integrate the above equation
to get
dT
dp
=
p(T ) = p(T0 ) e/RT0 e/RT .
(2.345)
=
p
R T2

2.12. PHASE TRANSITIONS AND PHASE EQUILIBRIA

73

Figure 2.26: A p-v-T surface for a substance which contracts upon freezing. The red dot is the critical point and the
red dashed line is the critical isotherm. The yellow dot is the triple point at which there is three phase coexistence of
solid, liquid, and vapor.

2.12.3 Liquid-solid line in H2 O

Life on planet earth owes much of its existence to a peculiar property of water: the solid is less dense than the
liquid along the coexistence curve. For example at T = 273.1 K and p = 1 atm,
vwater = 1.00013 cm3 /g

vice = 1.0907 cm3 /g .

(2.346)

The latent heat of the transition is = 333 J/g = 79.5 cal/g. Thus,


dp
dT

liqsol

333 J/g
=
T
v
(273.1 K) (9.05 102 cm3 /g)

dyn
atm
= 1.35 10
= 134 .
2
C
cm K

(2.347)

The negative slope of the melting curve is invoked to explain the movement of glaciers: as glaciers slide down
a rocky slope, they generate enormous pressure at obstacles12 Due to this pressure, the story goes, the melting
temperature decreases, and the glacier melts around the obstacle, so it can flow past it, after which it refreezes.
But it is not the case that the bottom of the glacier melts under the pressure, for consider a glacier of height
h = 1 km. The pressure at the bottom is p gh/
v 107 Pa, which is only about 100 atmospheres. Such a pressure
can produce only a small shift in the melting temperature of about Tmelt = 0.75 C.
Does the Clausius-Clapeyron relation explain how we can skate on ice? My seven year old daughter has a mass
of about M = 20 kg. Her ice skates have blades of width about 5 mm and length about 10 cm. Thus, even on one
12 The

melting curve has a negative slope at relatively low pressures, where the solid has the so-called Ih hexagonal crystal structure. At
pressures above about 2500 atmospheres, the crystal structure changes, and the slope of the melting curve becomes positive.

CHAPTER 2. THERMODYNAMICS

74

Figure 2.27: Equation of state for a substance which expands upon freezing, projected to the (v, T ) and (v, p) and
(T, p) planes.
foot, she only imparts an additional pressure of
p =

20 kg 9.8 m/s2
Mg

= 3.9 105 Pa = 3.9 atm .

A
(5 103 m) (101 m)

(2.348)

The change in the melting temperature is thus minuscule: Tmelt 0.03 C.

So why can my daughter skate so nicely? The answer isnt so clear!13 There seem to be two relevant issues in play.
First, friction generates heat which can locally melt the surface of the ice. Second, the surface of ice, and of many
solids, is naturally slippery. Indeed, this is the case for ice even if one is standing still, generating no frictional
forces. Why is this so? It turns out that the Gibbs free energy of the ice-air interface is larger than the sum of free
energies of ice-water and water-air interfaces. That is to say, ice, as well as many simple solids, prefers to have a
thin layer of liquid on its surface, even at temperatures well below its bulk melting point. If the intermolecular
interactions are not short-ranged14 , theory predicts a surface melt thickness d (Tm T )1/3 . In Fig. 2.30 we
show measurements by Gilpin (1980) of the surface melt on ice, down to about 50 C. Near 0 C the melt layer
thickness is about 40 nm, but this decreases to 1 nm at T = 35 C. At very low temperatures, skates stick
rather than glide. Of course, the skate material is also important, since that will affect the energetics of the second
interface. The 19th century novel, Hans Brinker, or The Silver Skates by Mary Mapes Dodge tells the story of the
poor but stereotypically decent and hardworking Dutch boy Hans Brinker, who dreams of winning an upcoming
ice skating race, along with the top prize: a pair of silver skates. All he has are some lousy wooden skates, which
wont do him any good in the race. He has money saved to buy steel skates, but of course his father desperately
needs an operation because I am not making this up he fell off a dike and lost his mind. The family has no
other way to pay for the doctor. What a story! At this point, I imagine the suspense must be too much for you to
bear, but this isnt an American Literature class, so you can use Google to find out what happens (or rent the 1958
13 For
14 For

a recent discussion, see R. Rosenberg, Physics Today 58, 50 (2005).

example, they could be of the van der Waals form, due to virtual dipole fluctuations, with an attractive 1/r 6 tail.

2.12. PHASE TRANSITIONS AND PHASE EQUILIBRIA

75

Figure 2.28: Projection of the p-v-T surface of Fig. 2.26 onto the (v, p) plane.
movie, directed by Sidney Lumet). My point here is that Hans crappy wooden skates cant compare to the metal
ones, even though the surface melt between the ice and the air is the same. The skate blade material also makes a
difference, both for the interface energy and, perhaps more importantly, for the generation of friction as well.

2.12.4 Slow melting of ice : a quasistatic but irreversible process

Suppose we have an ice cube initially at temperature T0 < 273.15 K (i.e. = 0 C) and we toss it into a pond
of water. We regard the pond as a heat bath at some temperature T1 > . Let the mass of the ice be M . How much
heat Q is absorbed by the ice in order to raise its temperature to T1 ? Clearly
Q = M cS ( T0 ) + M + M cL (T1 ) ,

(2.349)

where cS and cL are the specific heats of ice (solid) and water (liquid), respectively15 , and is the latent heat
of melting per unit mass. The pond must give up this much heat to the ice, hence the entropy of the pond,
discounting the new water which will come from the melted ice, must decrease:
Spond =

Q
.
T1

(2.350)

Now we ask what is the entropy change of the H2 O in the ice. We have
Sice =

dQ

=
T

ZT1
Z
M cS
M
M cL
dT
+
+ dT
T

T0

 
 
M
T

+
+ M cL ln 1 .
= M cS ln
T0

15 We

assume cS (T ) and cL (T ) have no appreciable temperature dependence, and we regard them both as constants.

(2.351)