Download Chapter 1 Thermodynamics

Chapter 1
Thermodynamics
1.1
Preliminaries
Degree of freedom (d.o.f) is a number of parameters necessary to
formulate the initial value problem divided by two.
For a harmonic oscillator the initial value problem is specified with only
two parameters: position coordinate (or angle) and velocity coordinate (or
angular velocity) which together correspond to only a single d.o.f.
In Hamiltonian mechanics the position q and momentum p coordinates
come in conjugate pairs (q, p). Thus the number of d.o.f. is the number of
3
CHAPTER 1. THERMODYNAMICS
4
such pairs or one half of the total dimensionality of the phase space.
Problem: How many d.o.f. in a problem of binary collision of
protons in the LHC (Large Hadron Collider)?
Solution: Binary collisions involve collisions of only two particles. Each particle is specified by position and velocity vector
coordinates. This makes the number of d.o.f. (2 · 3 + 2 · 3)/2 = 6.
Our main objective in this course is to understand the behavior of systems
with a very large number of d.o.f. N ∫ 1 (e.g. NA = 6.02214 ◊ 1023
molecules in a box). The task is (in some sense) much more ambitious than
the problems we are used to in physics courses where the number of d.o.f. is
usually small. For N . 10 analytical methods may be useful; for N . 1010
computer may work; for N ≥ 1googol = 10100 statistical physics may be the
only tool.
There are two standard ways to study the large N limit:
• phenomenological (e.g. thermodynamics) and
• fundamental (e.g. statistical mechanics).
We start with phenomenological approach and later will use a kinetic theory
to justify a more fundamental (?) approach.
Thermodynamics is a phenomenological theory of systems with
many d.o.f.
The main idea is that only a small number of measurable parameters (e.g.
volume V , pressure P , temperature T , etc.) should be sufficient for describing
the so-called equilibrium states.
Equilibrium state is the state whose thermodynamic parameters
do not change with time.
It is an important experimental fact that in equilibrium all of the parameters are either extensive or intensive. For example, volume is extensive, but
pressure is intensive.
Extensive parameter is proportional to the amount of substance
and intensive parameter is independent on the amount of substance.
Problem: Give an example of neither extensive nor intensive
quantity. Explain.
CHAPTER 1. THERMODYNAMICS
5
Solution: When two wires with electrical resistances R1 and
R2 are connected in series than the total resistance is an extensive quantity Rtotal = R1 + R2 , but when the wires are connected in parallel the total resistance is not extensive 1/Rtotal =
1/R1 + 1/R2 . Therefore, electrical resistance is not an extensive
nor intensive quantity.
Not all of the thermodynamic parameters are independent of each other.
Equation of state,
f (P, V, T ) = 0,
(1.1)
describes the dependence and reduces the number of independent parameters.
For example, the equation of state for an ideal gas (sufficiently diluted gas)
is given by,
P V = N kT,
(1.2)
where N is the number of molecules and k is the Boltzmann’s constant.
Because of a universal character of ideal gases (1.2) can be used to set a
relative temperature scale (but to set an absolute scale and to give a meaning
to T = 0 it is necessary to postulate the Third Law of thermodynamics to be
discussed later in the course). For example, we can measure PNVk when water
boils and denote this temperature by T = 0, and then we can measure PNVk
when water freezes and denote this temperature by T = 100. Then we can
use linear interpolation and extrapolation to assign temperature to arbitrary
values of PNVk . This is the Celsius scale.
CHAPTER 1. THERMODYNAMICS
6
It is convenient to think of f (P, V, T ) as a 3D function which describes
a given thermodynamical system and of (P, V, T ) as a point in 3D which
describes a given state of the system. The projection of the surface of the
equation of state on P ≠ V plane is known as the P ≠ V diagram.
Reversible transformation of different types such as
• 1-2. isothermal (i.e. constant temperature)
• 2-3. isobaric (i.e. constant pressure)
• 3-1. isochoric (i.e. constant volume)
are described by different paths on the P ≠ V diagram.
1.2
Zeroth Law
Every theory is based on a number of statements which cannot be proved
(not that we can prove anything in physics). In mathematics such statements
are called axioms, but in physics they go by different names: principles, laws,
assumptions, etc. With this respect thermodynamics is not an exception as
it is based on four laws of thermodynamics numbered from 0 to 3 (for purely
historical reasons).
Zeroth Law in words: If systems A and C are each
in thermal equilibrium with system B, then A is in
thermal equilibrium with system C.
Zeroth Law in symbols:
A ≥ B and C ≥ B ∆ A ≥ C
(1.3)
where ≥ is a relation between systems such that “A ≥ B” reads
as “A is in a thermal equilibrium with B”.
Problem: Under assumption that every system is in equilibrium
with itself, prove that ≥ is an equivalence relation.
Solution: Equivalence relation must satisfy three properties:
1) Reflexivity: By assumption, A ≥ A.
2) Symmetry: Let A ≥ B. From Reflexivity B ≥ B. Using
Zeroth Law, B ≥ B and A ≥ B ∆ B ≥ A.
3) Transitivity: Let A ≥ B and B ≥ C. From Symmetry C ≥ B.
From Zeroth Law, A ≥ B and C ≥ B ∆ A ≥ C.
CHAPTER 1. THERMODYNAMICS
7
An important consequence of the equivalence relation “~” is that it divides all
of the system into equivalence classes which can be labeled by temperature.
Of course, this does not tell us why the temperature should be a real number
as opposed to, for example, integers, complex or even more exotic p-adic
number.
1.3
First Law
The Zeroth Law does not tell us why temperature is a real number. Thus
additional laws must be postulated before this fact can be established not
only experimentally (e.g. using equation of state for ideal gases), but also
more theoretically. The next assumption is the First Law which should be
viewed as a statement about conservation of energy.
First Law in words: The increment in the internal energy U of a system is equal to the difference between
the increment of heat Q accumulated by the system
and the increment of work W done by it.
First Law in symbols:
dU = dQ ≠ dW,
where dU is an exact differential.
(1.4)
s
Exact differential dX is a differential whose integral dX depends
only on the limits of integration, but not on the path.
The First law is used to define the state function U which is
an extensive quantity: doubling the mass doubles the internal
energy. For quasi-static (or sufficiently slowly varying) processes
the work done by the system dW = P dV such that
dU = dQ ≠ P dV.
(1.5)
Problem: Are the differentials dQ and dW exact or not?
s
Solution: W = P dV is not a full derivative and therefore depends not only on the initial and final points but also on the path.
This can be shown explicitly by integrating along different path
from (P1 , V1 ) to (P2 , V2 ) assuming that P1 < P2 and V1 < V2 :
⁄ (P1 ,V2 )
(P1 ,V1 )
⁄ (P2 ,V2 )
P dV +
(P1 ,V2 )
P dV = P1 (V2 ≠V1 ) ”= P2 (V2 ≠V1 ) =
⁄ (P2 ,V1 )
(P1 ,V1 )
⁄ (P2 ,V2 )
P dV +
(P2, V1 )
P dV.
CHAPTER 1. THERMODYNAMICS
8
Therefore, dW is not an exact differential and since dU is exact
due to the First Law dQ = dU ≠ dW is also not exact.
It is convenient to think of the conjugate pair P and V as a generalized
force (intensive quantity) and generalized displacement (extensive quantity)
respectively. Other commonly used conjugate pairs include temperature T
and entropy S, chemical potential µi and number of particles Ni of type i,
etc. Then the total change in internal energy can be expressed as the sum
of the products of generalized forces and generalized displacements:
dU = T dS ≠ P dV +
ÿ
µi dNi .
(1.6)
i
In other words, when a given generalized force is not balanced it causes a
generalized displacement whose product equals to the energy transfer in or
out of the system.
1.4
Second Law
Second Law in Kelvin words: No process is possible in which the
sole result is the absorption of heat from a reservoir and its
complete conversion into work.
Second Law in Clausius words: No process is possible whose sole
result is the transfer of heat from a body of lower temperature to a body of higher temperature.
To show the equality of the two statements it will be useful to define a heat
engine. Heat engine is a system which undergoes a cyclic transformation that
takes heat Q1 from a warmer reservoir, converts some of it to work W and
rejects the rest Q2 = Q1 ≠ W to a colder reservoir. In contrast, refrigerator,
is a heat engine running backwards in time: use work to extract heat from a
colder reservoir and to reject it to a warmer reservoir.
Problem: Prove that the Kelvin and the Clausius statements of
the Second Law are equivalent.
Solution: Assume that Kelvin statement is false ∆ Extract Q
of heat from a reservoir at temperature T2 and convert it entirely
to work W = Q. Then convert this work back into heat Q = W
and transfer it to reservoir at temperature T1 > T2 ∆Clausius
statement is false.
Assume that Clausius statement is false ∆ Extract Q of heat
from a reservoir at temperature T2 and transfer it to reservoir at
temperature T1 > T2 ∆ Operate an engine between temperatures
T1 and T2 designed such that W = Q ∆Kelvin statement is false.
CHAPTER 1. THERMODYNAMICS
1.5
9
Carnot engine
Carnot engine is an engine consisting of two isothermal paths (i.e. T = 0)
operating at two different temperatures and two adiabatic paths (i.e. Q =
0) connected in a cyclic transformation.
The efficiency of engines is defined as
÷=
W
,
Q1
(1.7)
and for a Carnot engine it is given by
Q2
.
(1.8)
Q1
Carnot’s theorem: a) All irreversible engines operating between temperatures T1 and T2 < T1 are less efficient than a Carnot engine operating
between the same temperatures.
b) All reversible engines operating between temperatures T1 and T2 < T1
are equally efficient as a Carnot engine operating between the same temperatures.
÷ =1≠
CHAPTER 1. THERMODYNAMICS
10
Proof: a) Combine an arbitrary heat engine whose efficiency is ÷ with a
reversed Carnot engine whose efficiency is ÷ Õ so that there is no work done
by the combined system. If ÷ = ÷ Õ then the process does nothing in conflict
with irreversibility assumption. If ÷ > ÷ Õ then the process is in the conflict
with Clausius statement of the Second Law, i.e.
W = Wc ∆ ÷Q = ÷ Õ QÕ ∆
Q=
÷
Q ≠ Q > 0.
÷Õ
Thus, ÷ Æ ÷c . Together ÷ Æ ÷c and ÷ ”= ÷c give us ÷ < ÷c .
b) We have already proved that ÷ Æ ÷ Õ regardless of reversibility. Now we
can reverse the process by combining a reversed heat engine with a Carnot
engine which leads to conclusion ÷ Õ Æ ÷. Together ÷ Æ ÷ Õ and ÷ Õ Æ ÷ give us
÷ = ÷Õ.
According to Carnot’s theorem the efficiency of reversible process between
any two temperatures is a universal number, i.e. ÷(T1 , T2 ). This allows us
to define not only relative, but also absolute temperature scale. Consider
three Carnot cycles 1-2, 2-3 and 1-3 operating between different temperatures
T1 and T2 , T2 and T3 , T1 and T3 respectively, where without loss of generality
we assume that T1 > T2 > T3 . We can now construct a combine cycle such
CHAPTER 1. THERMODYNAMICS
11
that the heat Q2 rejected by 1-2 is absorbed by 2-3 which is a reversible and
thus a cycle 1-3 by Carnot’s theorem. Then the heat absorbed by reservoir
at T3 must satisfy both
Q3 = Q1 ≠ W13 = Q1 (1 ≠ ÷(T1 , T3 ))
and
Q3 = Q2 ≠W23 = Q2 (1≠÷(T2 , T3 )) = (Q1 ≠W12 )(1≠÷(T2 , T3 )) = Q1 (1≠÷(T1 , T2 ))(1≠÷(T2 , T3 )).
Therefore
and
(1 ≠ ÷(T1 , T3 )) = (1 ≠ ÷(T1 , T2 ))(1 ≠ ÷(T2 , T3 ))
1 ≠ ÷(T1 , T2 ) =
f (T2 )
f (T1 )
(1.9)
for an arbitrary function f (T ) which is by convention is set to be a linear
function, i.e.
T2
T1
(1.10)
Q2
T2
= .
Q1
T1
(1.11)
÷ =1≠
and from the definition of efficiency
1.6
Entropy
The Second Law suggests a new thermodynamical quantity, called entropy
and usually denoted by S. It is conveniently introduced using Clausius’s
theorem.
Clausius’s Theorem: For any cyclic transformation
j
dQ
Æ 0.
T
(1.12)
The equality holds for reversible transformations.
Proof: Subdivide the cycle into infinitesimal transformations
where the temperature T remains roughly constant. During each
transformation the system receives dQ of heat and does dWS
of work. Arrange a series of Carnot cycles operating between
CHAPTER 1. THERMODYNAMICS
12
temperatures T and TR where is the temperature of an arbitrary
reservoir. The whole purpose of each Carnot cycle is to take dQR
of heat from reservoir and deliver to the system dQ of heat and
to do dWC of work. Then according to the absolute definition of
temperature
dQR
dQ + dW
TR
=
=
.
(1.13)
dQ
dQ
T
By the Kelvin’s statement
of the Second Law the
total work must
i
i
s
be non-positive dW = (dWC + dWS ) = (dQ + dWC ) Æ 0
and thus,
j
(dQ + dWC ) = TR
j
j
dQ
dQ
Æ0 ∆
Æ0
T
T
(1.14)
since TR > 0. For a reversible cycle
we can run the process in
i dQ
opposite direction to show that T Ø 0 which together with
i dQ
i
Æ 0 implies dQ
= 0.
T
T
An immediate consequence of the Clausius’s Theorem is sthat for reversible
transformations between two states A and B the integral AB dQ
does not deT
pend on the path. Indeed if we take two distinct reversible paths parametrizes
by dQ and dQÕ then together they would form a reversible cycle, i.e.
⁄ B
A
and thus,
dQ ⁄ A dQÕ
+
=0
T
T
B
(1.15)
⁄ B
dQ ⁄ B dQÕ
=
.
(1.16)
T
T
A
A
This suggests that for a reversible transformation we can define and exact
differential
dQ
.
(1.17)
T
whose integral defines relative entropy up to an arbitrary constant of integration. This also produces a new pair of conjugate thermodynamic variables
generalized force T and generalized displacement S, i.e.
dS ©
dU = T dS ≠ P dV.
(1.18)
Although the entropy was defined using only reversible processes (as a
reference), the Clausius’s Theorem implies that for all systems
CHAPTER 1. THERMODYNAMICS
13
⁄ B
dQ
.
(1.19)
T
(This can be seen by connecting the irreversible
process
from A to sB with
i
s B dQ s A dQ
B dQ
any reversible process from B to A such that dQ
=
+
A T
B T = A T +
T
S(B) ≠ S(A) Æ 0.) Moreover, for thermally isolated systems (i.e. dQ = 0)
S(B) ≠ S(A) Ø
A
S(B) ≠ S(A) Ø 0
(1.20)
where the equalities hold for only reversible transformations. This shows that
in a thermal equilibrium (when the state of the system does not change) the
system must be in a state of maximal entropy.
Second Law in symbols:
dS
Ø0
(1.21)
dt
Not all of the processes we observe in nature are reversible. Many processes
are irreversible, which is a rather surprising fact given that the fundamental
laws of physics (as we know them) are usually time symmetric. Then, why
does the nature always picks some initial conditions and not other? Why
don’t we see a thermal state? Why is there an asymmetry between past and
future? Why do we see arrow of time?
1.7
dQ equations
Consider a system specified by three parameters P, V, T any two of which
are independent variables due to equation of state. Then, the differential of
internal energy can be expressed in three possible forms
dU (P, V ) =
dU (T, V ) =
dU (P, T ) =
A
ˆU
ˆP
B
A
ˆU
ˆT
B
ˆU
ˆP
B
A
V
A
B
dV
(1.22)
A
B
dV
(1.23)
A
B
dT.
(1.24)
ˆU
dP +
ˆV
V
ˆU
dT +
ˆV
T
ˆU
dP +
ˆT
P
T
P
Using the First Law given by Eq. (1.5) we obtain can the so-called ”Q
equations:
dQ(P, V ) = dU (P, V ) + P dV =
A
ˆU
ˆP
B
V
dP +
AA
ˆU
ˆV
B
P
B
+ P dV
(1.25)
CHAPTER 1. THERMODYNAMICS
A
dQ(T, V ) = dU (T, V ) + P dV =
AA
dQ(P, T ) = dU (P, T )+P dV =
B
ˆU
ˆT
ˆU
ˆP
14
AA
.dT +
V
B
+P
T
A
ˆU
ˆV
B B
ˆV
ˆP
B
B
+ P dV
T
AA
dP +
T
ˆU
ˆT
(1.26)
B
+P
P
A
(1.27)
where the heat capacities at constant pressure and volume are defined by
CP ©
A
ˆQ
ˆT
B
=
P
CV ©
A
ˆQ
ˆT
A
ˆU
ˆT
B
B
+P
P
=
V
A
A
ˆU
ˆT
B
ˆV
ˆT
B
(1.28)
P
(1.29)
.
V
Roughly speaking the heat capacities describe how much heat can be stored
in a system as its temperature is increased.
Problem: Under assumption that U (T ) is a function of temperature alone and CV does not depend on temperature, prove that
for an ideal gas it is more efficient (requires less energy) to heat
up the system at constant volume than at constant pressure.
Solution: Since U is not a function of P nor V ,
A
ˆU
ˆT
B
=
P
A
ˆU
ˆT
B
=
V
dU
dT
and the difference between specific heat capacities (1.28) and
(1.29) is given by
CP ≠ CV = P
A
ˆV
ˆT
B
P
=
A
ˆ(P V )
ˆT
B
= N k.
P
where the last equality is obtained from the equation of state
for an ideal gas (1.2). Since both the number of molecules and
the Boltzmann’s constant are positive number we conclude that
CP ≠ CV > 0 or that it is easier heat up the system at constant
pressure rather than at constant volume.
The dQ equation in their present form are not very useful. Using the Second
Law we can rewrite equations (1.26) and (1.27)
dQ
1
= dS =
T
T
A
ˆU
ˆT
B
V
1
dT +
T
AA
ˆU
ˆV
B
T
B
+ P dV
(1.30)
ˆV
ˆT
B B
P
dT
CHAPTER 1. THERMODYNAMICS
A
dQ
= dS =
T
1
T
A
ˆU
ˆP
B
T
P
+
T
A
ˆV
ˆP
15
B B
A
A
1
dP +
T
T
ˆU
ˆT
B
P
+
T
A
ˆV
ˆT
B B
=
A
A
ˆS
ˆV
B B
P
(1.31)
and use the exactness of dS, i.e.
dS =
A
ˆS
ˆT
B
V
A
ˆS
dT +
ˆV
B
T
A
dV ∆
ˆ
ˆV
A
A
AA
ˆU
ˆV
A
1
T
A
ˆU
ˆP
B
=
A
ˆU
ˆP
B
ˆS
ˆT
B B
V
T
ˆ
ˆT
dT
P
T
V
to derive the following relations
ˆ
ˆP
A
1
T
A
ˆ
ˆV
A
ˆU
ˆT
B
1
T
A
ˆU
ˆT
P
P
+
T
A
ˆP
ˆT
B B
A
V
ˆV
ˆT
T
ˆ
=
ˆT
B B
P
T
1
T
ˆ
=
ˆT
B
+P
T
BB
(1.32)
V
T
P
+
T
A
ˆV
ˆP
B B
T
or
A
ˆU
ˆV
B
+P = T
T
B
V
and ≠T
A
ˆV
ˆT
B
P
T
+P
A
ˆV
ˆP
B
P
(1.33)
. (1.34)
T
Then one can use these expressions to simplify the dQ equations
T dS = CV dT + T
T dS = CP dT ≠ T
1.8
A
ˆP
ˆT
B
dV
(1.35)
A
ˆV
ˆT
B
dP.
(1.36)
V
P
Chemical potential
Chemical potential, denoted byµ, is defined as work required to increase the
number of particle by one. The name - chemical potential - is due to the
fact that µ, describes the tendency of particles to move from higher densities
to lower densities. Paired together with a number of particles N it forms
another conjugate pair which appears in the most fundamental equation of
thermodynamics:
dU = T dS ≠ P dV + µdN
(1.37)
CHAPTER 1. THERMODYNAMICS
16
which combines in a single equations the First and Second laws.
As was already noted T ,P and µ are intensive quantities and S, V and
N are extensive. This can be expressed as a scaling law
U (⁄S, ⁄V, ⁄N ) = ⁄U (S, V, N )
(1.38)
which can differentiated with respect to ⁄ at ⁄ = 1,
A
ˆU
ˆS
B
A
ˆU
S≠
ˆV
V,N
B
A
ˆU
V +
ˆN
S,N
B
N = U (S, V, N ).
(1.39)
S,V
By combined with (1.37) we obtain Euler equation
U = ST ≠ P V + µN
(1.40)
SdT ≠ V dP + µdN = 0.
(1.41)
as well as the Gibbs-Duhem equation
In a more general form it reads as
SdT ≠ V dP +
ÿ
µi dNi = 0,
(1.42)
i
where µi is a chemical potential and Ni is the number of particle of type i.
1.9
Perpetual motion machines
If the First Law could be violated then it would open an exciting possibility of
perpetual motion machine of the first kind: do work without input of energy.
Such machines would be in a conflict with conservation of energy, but there
is certainly no prove that they cannot exist. Moreover it is well known that
for gravitational (more precisely general relativistic) systems the mass and
energy is not always well defined.
Although the perpetual motion machines of the first kind are not allowed
by the First Law, there are perpetual motion machines of the second kind
which are not in conflict with conservation of energy, yet they have never
been constructed. It is an experimental fact that not all of the processes
allowed by the conservation of energy are observed in nature. For example,
we see ice cubes melting in warm water to cool it down, but we never see a
time-reversed process where cool water splits into warm water with ice cubes.
Such processes are called irreversible, in contrast to reversible processes which
can run backwards in time if its initial and final conditions are interchanged.
In thermodynamic such processes are prohibited by postulating the Second
Law. If the Second Law could be violated it would allow perpetual motion
machines of the second kind: do work using thermal energy.
CHAPTER 1. THERMODYNAMICS
1.10
17
Thermodynamic potentials
As we have seen the extrema (maximum) of entropy corresponds to an equilibrium state of an isolated system (i.e. dQ = dW = 0)
dS
= 0.
(1.43)
dt
Can this concept be generalized to open systems? For this purpose in
addition to U we define three new thermodynamic potentials: Enthalpy,
Helmholtz free energy, Gibbs free energy and Landau free energy.
• Enthalpy
H = U + PV
(1.44)
It is convenient to express H in differential form as a function of S and
P
dH = dU + P dV + V dP = T dS + V dP.
(1.45)
When there is no heat exchange (i.e. dQ = 0) and the external force is
constant (i.e. P = const),
dH = dU + P dV + V dP = dW + P dV Æ 0.
(1.46)
where the equality corresponds to quasi-static processes (i.e. dW = ≠P dV ).
• Helmholtz free energy
A = U ≠ TS
(1.47)
It is convenient to express A in differential from as a function of V and
T
dA = dU ≠ T dS ≠ SdT = ≠P dV ≠ SdT.
(1.48)
When there is no external work (i.e. dW = 0) and the temperature remains
constant (i.e. T = const),
dA = dU ≠ SdT ≠ T dS = dQ ≠ T dS Æ 0.
(1.49)
where the equality corresponds to reversible processes (i.e. dQ = T dS).
As an example of the variational principle consider gas at a constant
temperature T in a box of volume V divided by a sliding piston into V1 and
CHAPTER 1. THERMODYNAMICS
18
V2 . In the equilibrium state
0 = dA =
=
=
or
A
A
ˆA
ˆV1
ˆA
ˆV1
AA
A
B
T
ˆA
dV1 +
ˆV2
T
ˆA
dV1 +
ˆV2
B
B
ˆA
ˆV1
T
B
ˆA
ˆV1
A
≠
=
T
A
A
A
dV2
(1.50)
T
B
T
B B
ˆA
ˆV2
ˆA
ˆV2
B
d(V ≠ V1 )
dV1
T
B
T
and using (1.48) we can conclude that in the equilibrium state the pressures
on both sides of the piston must be equal.
• Gibbs free energy
G = U ≠ TS + PV
(1.51)
It is convenient to express G in differential form as a function of P and
T since
dG = dU ≠ SdT ≠ T dS + P dV + V dP = V dP ≠ SdT.
(1.52)
When the external force (i.e. P = const) and temperature remains constant
(i.e. T = const) ,
dG = dU ≠SdT ≠T dS+P dV +V dP = dU +P dV ≠T dS = dQ≠dW +P dV ≠T dS Æ 0.
(1.53)
where the equality corresponds to reversible (i.e. dQ = T dS) and quasi-static
(i.e. dW = ≠P dV ) processes.
Note that the exactness of differential dH, dA, and dG, implies
A
ˆT
ˆP
B
A
ˆP
ˆT
B
and
A
ˆV
ˆT
B
S
A
ˆV
ˆS
B
V
A
ˆS
ˆV
B
P
=
=
A
.
P
ˆS
=≠
ˆP
.
T
B
T
.
CHAPTER 1. THERMODYNAMICS
19
Other thermodynamic relations are conveniently summarized by the socalled Maxwell or thermodynamic square:
For a thermodynamic potential of interest (in bold) whose arguments are
placed in the neighboring corners (in italic) the derivative of the potential
with respect to one of its argument with other argument held fixed is determined by going along a diagonal line either with or against the direction
1 2of
ˆU
the arrow. Going against the arrow yields a minus sign (e.g. P = ≠ ˆV
,
S=≠
1.11
1
2
S
ˆG
).
ˆT P
Third Law
Entropy of a system is defined for all states using a reference state by connecting it with a reversible transformation. However, if the surface defined
by the equation of state is disconnected then the reversible transformations
might not exist. Moreover it is important to have definition of entropy for
distinct systems which may later come into contact with each other. The
role of the Third Law is to defines an absolute scale of entropy which defines
uniquely the entropy of an arbitrary equilibrium state of any system.
Third Law in words: The entropy of any system at
absolute zero is zero.
Third Law in symbols:
lim S(T ) = 0
T æ0
(1.54)
Consequences of the Third Law:
1. Partial derivative of S at T = 0 with respect to every thermodynamic
CHAPTER 1. THERMODYNAMICS
20
parameter X vanishes:
lim
T æ0
A
ˆS
ˆX
B
= 0.
(1.55)
T
2. Heat capacity CX at T = 0 with fixed thermodynamic parameter X
vanishes:
lim CX = lim
T æ0
T æ0
A
ˆQ
ˆT
B
X
= lim T
T æ0
A
ˆS
ˆT
B
= 0.
(1.56)
X
3. Absolute zero cannot be reached in a finite number of steps. This
statement is often used as an alternative definition of the third law.
We will come back to it at the end of the course when the quantum
statistical mechanics is introduced.
Physical theories are always based on assumptions which very often turn out
to be false (sooner or later). Now, that you have seen all of the theoretical assumption (laws of thermodynamics) which go into a phenomenological
theory of thermodynamics it is a good time to ask which of the assumption
is likely to be wrong?
Zeroth Law (Universality of Temperature):
A ≥ B and C ≥ B ∆ A ≥ C
(1.57)
First Law (Conservation of Energy):
dU = dQ ≠ dW.
(1.58)
Second Law (Arrow of Time):
dS
Ø0
dt
(1.59)
Third Law (Quantum Mechanics):
lim S(T ) = 0
T æ0
(1.60)