Lecture-Notes (Thermodynamics) - niser

Chapter 4
Entropy and second law of
thermodynamics
4.1 Carnot cycle
In a cyclic transformation, where the final state of a system is the same as the initial one,
∆U = 0
since the internal energy U is a state function. A reversible cyclic process can be represented
by a closed loop in the P − V diagram (Fig. 4.1).
Figure 4.1: Reversible cyclic process. The shaded area of the loop is the total work done
by the system in one cycle.
It follows from the first law of thermodynamics that, since ∆U = 0, the work done by the
system during a cycle is equal to the heat absorbed, i. e.,
−∆W = ∆Q = PdV = Area enclosed.
⋆ In a cyclic process, work is converted into heat and in the end of each cycle the
system returns to its original state. We will introduce the heat engine which is
based on the properties of a cyclic process.
25

26 CHAPTER 4. ENTROPY AND SECOND LAW OF THERMODYNAMICS
Let us consider the operation of a heat engine in more detail (Fig. 4.2).
Part of the heat that is transferred to the system
from a heat bath with temperature T1, Q1, is converted
into work, W, and the rest, Q2, is delivered
to a second bath with T2 < T1 (condenser). Following
the first law of thermodynamics,
Q1 − |Q2| = |W |.
Carnot process: This a reversible cycle process
bounded by two isotherms and two adiabatic lines.
The Carnot process can be realized with an arbitrary
working substance, but we shall consider here
an ideal gas (Fig. 4.3).
Figure 4.2: Heat engine.
Figure 4.3: Ideal gas as a working substance for the Carnot process.
In the P − V diagram the Carnot process can be represented as shown in Fig. 4.4.
Figure 4.4: The Carnot process in the P − V diagram.
One Carnot cycle consists of four consecutive thermodynamic processes, which are
1. A → B: isothermal expansion at T = T1; the volume changes as
VA → VB,
while heat Q1 is absorbed from a bath and the system performs work.

4.2. SECOND LAW OF THERMODYNAMICS 27
2. B → C: adiabatic expansion, during which
and
while δQ = 0.
T1 → T2
VB → VC,
3. C → D: isothermal compression at T = T2; the system ejects heat Q2 to the bath
(Q2 < 0 is a convention).
4. D → A: adiabatic compression, during which
T2 → T1,
VD → VA,
δQ = 0.
In one cycle operation, the system receives an amount of heat Q1 from a hot reservoir,
performs work, and rejects ”waste heat” Q2 to a cold reservoir.
From the first law of thermodynamics we have:
0 = dU = (δQ + δW) = Q + W = Q1 + Q2 + W,
where −W is the work performed by the system, equal to the area enclosed in the loop
(shaded area in Fig. 4.4).
The efficiency of the Carnot engine is defined as
η ≡
performed work
absorbed heat
= −W
Q1
= Q1 + Q2
Q1
= Q1 − |Q2|
Q1
η is 100% if there is no waste heat (Q2 = 0). However, we will see that this is impossible
due to the second law of thermodynamics.
4.2 Second law of thermodynamics
Definition by Clausius:
” There is no thermodynamic transformation whose sole effect is to deliver heat
from a reservoir of lower temperature to a reservoir of higher temperature.”
Summary: heat does not flow upwards.
Definition by Kelvin:
”There is no thermodynamic transformation whose sole effect is to extract heat
from a reservoir and convert it entirely to work”.
Summary: a perpetuum mobile of second type does not exist.
.

28 CHAPTER 4. ENTROPY AND SECOND LAW OF THERMODYNAMICS
In order to prove that both definition are equivalent, we will
show that the falsehood of one implies the falsehood of the
other. For that purpose, we consider two heat reservoirs with
temperatures T1 and T2 with T1 > T2.
(1) If Kelvin’s statement were false, we could extract heat
from T2 and convert it entirely to work. We could then
convert the work back to heat entirely and deliver it to T1
(there is no law against this) (Fig. 4.5). Thus, Clausius’
statement would be negated.
Figure 4.5: Q2 = W =
Q1 process.
(2) If Clausius’ statement were false, we could let an amount
of heat Q1 flow from T2 to T1 (T2 < T1). Then, we could
connect a Carnot engine between T1 and T2 such as to
extract Q1 from T1 and return an amount |Q2| < Q1 back to T2. The net work
output of such an engine would be |Q1| − |Q2| > 0, which would mean that an
amount of heat |Q1| − |Q2| is converted into work, without any other effect. This
would contradict Kelvin’s statement.
From the microscopic point of view
- heat transfer is an exchange of energy due to the random motion of atoms;
- work’s performance requires an organized action of atoms.
In these terms, heat being converted entirely into work means chaos changing spontaneously
to order, which is a very improbable process.
→ Usually,
- one configuration corresponds to order;
- many configurations correspond to chaos.

4.3. ABSOLUTE TEMPERATURE 29
4.3 Absolute temperature
In order to introduce the concept of absolute temperature, let us discuss the consequences
of the second law of thermodynamics.
i) The second law of thermodynamics implies that a Carnot engine cannot be 100%
efficient since otherwise all heat absorbed from a warm reservoir would be converted
into work in one cycle.
ii) No engine working between two given temperatures can be more efficient than a
Carnot engine.
A Carnot engine is a reversible engine between two baths. The previous statement means
as well that an irreversible engine cannot be more efficient than a reversible one. In order
to show this, we consider two engines C and X (with X not necessarily a Carnot one)
working between baths at T1 (warm) and T2 (cold). We run the Carnot engine C in
reverse, as a refrigerator ¯ C, and feed the work output of X to ¯ C (Fig. 4.6).
Figure 4.6: Linked engine X and Carnot refrigerator ¯ C.
The total work output of such a system is
Wtot = (|Q ′ 1| − |Q ′ 2|) − (|Q1| − |Q2|) .
If we adjust the two engines so that |Q ′ 1 | = |Q1|, no net heat is extracted from the heat
bath at T1. In this case, an amount of heat |Q2| − |Q ′ 2 | is extracted from the heat bath
at T2 and converted entirely to work, with no other effect. This would violate the second
law of thermodynamics unless
|Q2| ≤ |Q ′ 2 |.
We divide this inequality by |Q1| and, using the fact that |Q1| = |Q ′ 1 |, get
and
|Q2|
|Q1| ≤ |Q′ 2 |
|Q ′ 1|
1 − |Q2|
|Q1| ≥ 1 − |Q′ 2|
.
|
|Q ′ 1

30 CHAPTER 4. ENTROPY AND SECOND LAW OF THERMODYNAMICS
From this follows that the efficiencies of the engines compare as
ηC ≥ ηX .
Consequence: all Carnot engines have the same efficiency since X can be, as a special
case, a Carnot engine.
⇒ The Carnot engine is universal, i.e., it depends only on the temperatures involved
and not on the working substance.
Definition: absolute temperature θ of a heat reservoir (bath) is defined such that the ratio
of the absolute temperatures of two reservoirs is given by
θ2
θ1
≡ |Q2|
|Q1|
−Q2
= 1 − η = ,
where η is the efficiency of a Carnot engine operating between the two reservoirs. Since
|Q2| is strictly greater than zero, |Q2| > 0, according to the second law of thermodynamics,
the absolute temperature is bounded from below:
θ > 0 .
⋆ This means that the absolute zero θ = 0 is a limiting value that can never be reached
since this would violate the second law of thermodynamics.
If an ideal gas is considered as a working substance in a Carnot engine, the absolute
temperature coincides with the temperature of the ideal gas defined as
T = PV
NkB
= θ.
4.4 Temperature as integrating factor
The absolute temperature T can be considered as an integrating factor that converts the
inexact differential δQ into an exact differential δQ/T.
Let us consider the following theorem stated by Clausius:
⇒ ”In an arbitrary cyclic process P, the following inequality holds:
P
δQ
T
≤ 0;
where the equality holds for P reversible.“
Q1

4.4. TEMPERATURE AS INTEGRATING FACTOR 31
Proof. Let us divide the cycle P into n segments so that on each segment
its temperature Ti (i = 1, . . .,n) is constant. We consider now a reservoir
at temperature T0 > Ti(∀i) (Fig. 4.7) and introduce Carnot engines
between the reservoir at T0 and Ti.
Figure 4.7: Cycle P connected to the reservoir at T0 via Carnot engines.
For each Carnot engine,
Wi + Q (0)
i + QC i
For cycle P,
with
Q (0)
i
T0
+ QC i
Ti
= 0 (first law of thermodynamics) and
= 0 (definition of absolute temperature).
W +
n
Qi = 0 (first law of thermodynamics),
i=1
Qi = −Q C i .
Then, the total heat absorbed from the reservoir at T0 is
n
n Q
= −T0
C i
= T0
Q (0)
T =
i=1
Q (0)
i
and the work performed by the system is
n
WT = − W +
=
n
Qi +
i=1
i=1
n
i=1
Wi
i=1
Ti
Q (0)
i − Qi =
n
i=1
n
i=1
Q (0)
i
Qi
Ti
= Q(0)
T .

32 CHAPTER 4. ENTROPY AND SECOND LAW OF THERMODYNAMICS
> 0, the combined machine would convert completely the heat from reservoir at
T0 into mechanical work. This would violate Kelvin’s principle and therefore
If Q (0)
T
i.e.,
For n → ∞,
The theorem of Clausius is proved.
T0
Q (0)
T
n
i=1
P
δQ
T
≤ 0,
Qi
Ti
≤ 0.
≤ 0 .
For a reversible process, we can revert the energy, which gives us
δQ
≥ 0.
T
This implies that
for a reversible process.
P
P
δQ
T
= 0
Corollary: for a reversible open path P, the integral
dQ
T
depends only on the end points and not on the particular path.
4.5 Entropy
Since
dQ
T
= 0 for a reversible process, δQ
T
P
dS = δQ
T
Eq. (4.1) implies that there exists a state function S.
is an exact differential:
. (4.1)
Definition: The state function S (potential), whose differential is given as
is called entropy.
dS = δQ
T

4.5. ENTROPY 33
The entropy is defined up to an additive constant. The difference between entropies of
any two states A and B is
B
δQ
S(B) − S(A) ≡
T ,
where the integral extends along any reversible path connecting A and B, and the result
of the integration is independent of the path.
What happens when the integration is along an irreversible
path? Since I − R is a cycle (see Fig. 4.8), it
follows from Clausius’ theorem that
⇒
I
δQ
T ≤
Therefore, in general
B
A
I−R
δQ
T
R
δQ
T
δQ
T
≤ 0 ⇒
= S(B) − S(A) .
≤ S(B) − S(A) ,
and the equality holds for a reversible process.
A
Figure 4.8: I − R (irreversiblereversible)
cycle.
In particular, for an isolated system, which does not exchange heat with a reservoir,
δQ = 0 and therefore
∆S ≥ 0 .
This means that the entropy of an isolated system never decreases and remains constant
during a reversible transformation.
Note:
i) The joint system of a system and its environment is called ”universe”. Defined in this
way, the ”universe” is an isolated system and, therefore, its entropy never decreases.
However, the entropy of a non-isolated system may decrease at the expense of the
system’s environment.
ii) Since the entropy is a state function, S(B) − S(A) is independent of the path, regardless
whether it is reversible or irreversible. For an irreversible path, the entropy
of the environment changes, whereas for a reversible one it does not.
iii) Remember that the entropy difference
B
S(B) − S(A) =
only when the path is reversible; otherwise the difference is larger that the integral.
A
δQ
T

34 CHAPTER 4. ENTROPY AND SECOND LAW OF THERMODYNAMICS
4.6 The energy equation
We note that for a reversible process in a closed system the first law of thermodynamics
can be written as
TdS = dU − δW = dU + PdV.
Let us consider T and V as independent variables (f.i., in a gas):
S = S(T, V ),
U = U(T, V ) → caloric equation of state,
P = P(T, V ) → thermic equation of state.
Since usually the internal energy U is not directly measurable, it is more practical to
consider dS = δQ/T. We will make use of one of the heat equations derived in section
3.5:
Dividing both sides of eq. (4.2) by T we get
On the other hand,
∂U ∂U
δQ = dT + + P dV
∂T V ∂V T
∂U
= CV dT + + P dV (4.2)
∂V
δQ
T
= dS = CV
T
dS =
dT + 1
T
T
∂U
+ P dV . (4.3)
∂V T
∂S ∂S
dT + dV . (4.4)
∂T V ∂V T
Comparing eq. (4.3) and (4.4), we obtain the following expressions:
∂S
∂T V
∂S
∂V
T
= CV
T ,
= 1
T
∂U
+ P .
∂V T
We use the commutativity of differentiation operations,
to write
1
T
∂
∂V
∂U
∂T V T
∂
∂V
∂S
∂T
T
= ∂
∂T
∂S
∂V ,
∂ CV
=
∂V T
∂
1 ∂U
∂T T ∂V T
= − 1
T 2
∂U
+ P +
∂V
1
∂
T ∂T
+ P ,
∂U
∂V T V
+
∂P
;
∂T V

4.7. USEFUL COEFFICIENTS 35
we cancel identical terms:
and get
∂U
∂V T
= T
∂
∂V
∂P
∂T V
∂U
=
∂T
∂U
∂T∂V
− P ⇒ energy equation. (4.5)
In the energy equation, the derivative of the internal energy is written in terms of measurable
quantities.
Examples: for an ideal gas
∂U
∂V T
= T nR
V
− P = 0,
which means that the internal energy of the ideal gas is not dependent on its volume:
Gay-Lussac experiment:
4.7 Useful coefficients
U = U(T) = CV T + const.
U = 3
2 NkBT → gas of monoatomic molecules,
U = 5
2 NkBT → gas of diatomic molecules.
As
we
saw in the last section, the energy equation relates the experimentally inaccessible
∂U
∂V T to the experimentally accessible
∂P
∂T V . We will show now that
∂P
can be
∂T V
related to other thermodynamic coefficients as well. For that purpose, we will make use
of the chain rule
∂x ∂y ∂z
= −1 ,
∂y z ∂z x ∂x y
(4.6)
which can be derived by considering the differential of function f(x, y, z):
df = ∂f ∂f ∂f
dx + dy +
∂x ∂y ∂z dz,
∂x
∂y
∂y
∂z
z
x
= − ∂f/∂y
∂f/∂x ,
= − ∂f/∂z
∂f/∂y ,

36 CHAPTER 4. ENTROPY AND SECOND LAW OF THERMODYNAMICS
Using (4.6), we get
∂P
∂T V
where
= −
α = 1
∂V
V ∂T P
kT = − 1
∂V
V ∂P T
kS = − 1
∂V
V ∂P
∂z
= −
∂x y
∂f/∂x
∂f/∂z .
1
(∂T/∂V ) P (∂V/∂P) T
S
= − (∂V/∂T) P
(∂V/∂P) T
= α
kT
(coefficient of thermal expansion),
(isothermal compressibility),
(adiabatic compressibility).
We substitute in
∂U
δQ = TdS = CV dT + + P dV
∂V T
expressions (4.7) and (4.5) and get the following useful relations.
, (4.7)
1) The absorbed heat is expressed in terms of directly measurable coefficients as
where T and V are independent variables.
2) If T and P are used as independent variables, then
δQ = TdS = CV dT + α
TdV , (4.8)
kT
TdS = CPdT − αTV dP (4.9)
3) If V and P are used as independent variables, dT can be rewritten in terms of dV
and dP as
∂T ∂T
dT = dV + dP =
∂V P ∂P V
1 kT
+ dP .
αV α
The corresponding TdS equation is then
TdS = CP
dV +
αV
CPkT
α
− αTV dP . (4.10)
There is also an important connection between CP and CV (which follows from eq.
(4.8) and (4.9)):
CP − CV = α2
TV = −T
kT
∂V 2
∂T P
∂V
∂P T
> 0 . (4.11)

4.8. ENTROPY AND DISORDER 37
By introducing the isochore pressure coefficient
it is possible to rewrite CP − CV as
Indeed,
β = 1
P
α
kT
⇒ CP − CV = α2
β = 1
P
4.8 Entropy and disorder
kT
∂P
∂T V
β 2 TV kTP 2 .
⇒ α = βPkT ⇒
TV = β2 P 2 k 2 T
kT
,
TV = β 2 TV kTP 2 .
We want now to establish a connection between thermodynamics and statistical mechanics,
which we will be learning in the second half of the course.
1. We give a definition of the multiplicity of a macrostate as
⎛
number of microstates
multiplicity of
= ⎝ that correspond to the
a macrostate
macrostate
2. The entropy can be defined in terms of the multiplicity as
S ≡ k ln (multiplicity) .
We shall see in statistical mechanics that this definition is equivalent to the phenomenological
concept we have learnt previously in this section.
3. With the above definition of entropy at hand, we can formulate the second law of
thermodynamics in this way:
4.
”If an isolated system of many particles is allowed to change, then, with large probability,
it will evolve to the macrostate of largest entropy and will remain in that
macrostate.”
(energy input by heating)
∆Sany system ≥ ,
T
where T is the temperature of the reservoir.
5. If two macroscopic systems are in thermal equilibrium and in thermal contact, the
entropy of the composite system equals the sum of the two individual entropies.
⎞
⎠.

38 CHAPTER 4. ENTROPY AND SECOND LAW OF THERMODYNAMICS
6. Since entropy is proportional to the ”multiplicity of a macrostate”, it can be viewed
as a ”measure of disorder”:
Order Disorder
⇓ ⇓
strong correlation absence of correlation
⇓ ⇓
small multiplicity large multiplicity
As an illustration, let us consider a class in the school, where all kids sit always in
the same place. This situation can be defined as ”ordered” since there is a strong
correlation between the kids’ positions. For example, if Peter sits in the first row,
you know straight away who is sitting behind and besides him: there is only one
possibility (multiplicity is very low).
Looking now at a class in the kindergarten, we will find that the children there are
all mixed, sitting and moving all the time. Now, if Peter sits in the first row, you
have no idea where the other kids are sitting. There are no correlations between the
kids’ positions and there are many options to have the children in the class. Such a
”system” of kids is in a ”disordered state”, with very high multiplicity. The entropy
of the kindergarten class is high!
4.9 Third law of thermodynamics (Nernst law)
While with the second law of thermodynamics we can only define the entropy (up to a
constant), the third law of thermodynamics tells us how S(T) behaves at T → 0. Namely,
it states that the entropy of a thermodynamic system at T = 0 is a universal constant,
that we take to be 0:
lim S(T) = 0.
T →0
S(T → 0) is independent from the values that the rest of variables take.
Consequences:
1) The heat capacities of all substances disappear at T = 0:
lim
T →0 CV
∂S
= lim T = 0,
T →0 ∂T V
lim
T →0 CP
∂S
= lim T = 0.
T →0 ∂T P
Note that an ideal gas provides a contradiction to the previous statement since in
that case CV = const and CP = const:
CV = 3
2 NkB,
CP = CV + NkB = 5
2 NkB,
CP − CV = NkB = 0. (4.12)

4.9. THIRD LAW OF THERMODYNAMICS (NERNST LAW) 39
2)
But, for T → 0 the ideal gas is not anymore a realistic system because a real gas
undergoes at low temperatures a phase transition - condensation.
lim β = lim
T →0 T →0
3) The absolute T = 0 (zero point) is unattainable.
1
V
∂V
= 0
∂T
In order to show the validity of the last statement, let us analyze what happens when we
are trying to reach low temperatures by subsequently performing adiabatic and isothermal
transformations. If a gas is used as a working substance (Linde method), such a sequence
of transformations looks in the P − V diagram as shown in Fig. 4.9.
Figure 4.9: Linde method: sequence of adiabatic and isothermal transformations in a gas.
Let us consider now the processes involved.
A → B: isothermic compression.
Work is performed on the gas and an amount of heat Q1 < 0 is given to the reservoir
in a reversible process. As a result, ∆S = Q1
T diminishes.
B → C: adiabatic expansion.
The gas performs work. Since δQ = 0, the entropy remains constant and the
temperature diminishes.

40 CHAPTER 4. ENTROPY AND SECOND LAW OF THERMODYNAMICS
In principle, by repeating this process many times, one could reach the T = 0 point.
But, according to the third law of thermodynamics, all entropy curves must end at 0 for
T → 0, which means that only through an infinite number of steps is it possible to reach
T = 0.