Slideshow chapter 1 3 physical chemistry 1 dr ngo thanh an

Dr. Ngo Thanh An
PHYSICAL CHEMISTRY 1
1

• The principal goal of physical chemistry is to
understand the properties and behaviour of material
systems and to apply this understanding in useful
ways.
Introduction
2

Introduction
Thermodynamics:
phenomenological theory to
describe equilibrium
properties of macroscopic
systems based on few
macroscopically measureable
quantities.
Or: The study of heat and its
transformation into
mechanical energy is called
thermodynamics.
4

• Macroscopic system: A large system containing
many atoms or molecules
• Microscopic system: a system consisting of a single
atom or molecule
• Macroscopic properties (such as temperature and
pressure) apply only to a macroscopic system and
are properties of the whole system.
• Microscopic properties (such as kinetic energy and
momentum) are mechanical in nature.
Definitions
7

A body or system whose condition
is altered without gaining heat from
or losing heat to the surroundings
(energy is transferred only as work.)
Definitions
8

9
Thermodynamic definition of work: It is a kind of interaction that would occur at the
system boundaries. It can be positive or negative.
Heat: Heat is a mode of energy transfer that takes place between the system and the
surroundings solely due to the temperature difference. Thus, heat is a transient
phenomenon. It can be recognized only during a process.
Energy exists in many forms, such as mechanical energy, heat, light, chemical
energy, and electrical energy. Energy is the ability to bring about change or to do
work.
Definitions

Definitions
A thermodynamic process is a passage of a
thermodynamic system from an initial state
to a final state 10

When a gas is compressed or expanded so that no heat enters or
leaves a system, the process is said to be adiabatic.
Adiabatic changes of volume can be achieved by performing the
process rapidly so that heat has little time to enter or leave or by
thermally insulating a system from its surroundings.
Do work on a pump by pressing
down on the piston and the air is
warmed.
Definitions
11

When a gas adiabatically expands, it does work on its surroundings
and gives up internal energy, and thus becomes cooler.
Blow warm air onto your hand from your wide-open mouth. Now
reduce the opening between your lips so the air expands as you blow.
Adiabatic expansion—the air is cooled.
Definitions
12

13
A source supplies
energy in the
form of heat, and
a sink absorbs it.
• A hypothetical body with a relatively large thermal energy capacity (mass x specific
heat) that can supply or absorb finite amounts of heat without undergoing any
change in temperature is called a thermal energy reservoir, or just a reservoir.
• In practice, large bodies of water such as oceans, lakes, and rivers as well as the
atmospheric air can be modeled accurately as thermal energy reservoirs because of
their large thermal energy storage capabilities or thermal masses.
Bodies with relatively large thermal
masses can be modeled as thermal
energy reservoirs.
Definitions

• State Variables
• system quantity whose values are fixed at constant
temperature, pressure, composition
• State Function
• a system property whose values depends only on the
initial and final states of the system.
• Path Functions
• system quantity whose value is dependent on the manner
in which the transformation is carried out.
Definitions
14

An intensive property is a bulk property, meaning that it is a physical property of
a system that does not depend on the system size or the amount of material in
the system. Ex: Temperature, Pressure, density, specific heat, …
By contrast, an extensive property is additive for independent, non-interacting
subsystems.The property is proportional to the amount of material in the system.
Ex: volume, mole number, mass, length, entropy, enthalpy, Gibbs free energy …
Definitions
State 1
(x1,y1,z1)
(depending on state
variables)
State 2
(x2,y2,z2)
(depending on state
variables)
State
function
• State functions: Can we find the exact form of functions?
 We don’t need to know the functions because we just pay
attention to the initial and final state.
• State 1 or state 2 is completely determined by the value of
its state functions at a certain set of state variables. 15

State variables:
Describe equilibrium state of thermodynamic system uniquely.
Intensive: homogeneous of degree 0, independent of system size
Extensive: homogeneous of degree 1, proportional to system
size
Note: Intensive state variables serves as equilibrium parameters.
Ex: temperature (T), pressure (P) and chemical potential ().
Definitions
--------------------------------------------
Euler's Theorem: Let f(x1,…,xn) be a function such that
Then f is said to be a homogeneous function of degree n.
1 1( ,..., ) ( ,..., )n
n nf x x f x x  
16

Intensive and extensive variables
Definitions
17

• Gas - a substance that is characterised by widely
separated molecules in rapid motion
• Mixtures of gases are uniform. Gases will expand to
fill containers.
• Ex:
- Common gases include - O2 and N2, the major
components of "air"
- Other common gases - F2, Cl2, H2, He, and N2O
(laughing gas)
Definitions
19

• The pressure of a gas is best defined as the forces
exerted by gas on the walls of the container
• Define P = force/area
• The SI unit of pressure is the Pascal
• 1 Pa = N/m2 = (kg m/s2)/m2
Some state variables
Pr
Force
essure
Area

20

• How do we measure gas pressure?
• We use an instrument called the barometer -
invented by Torricelli
• Gas pressure conversion factors
• 1 atm = 760 mm Hg = 760 torr
• 1 atm = 101.325 kPa = 1.01325 bar
• 1 bar = 1 x 105 Pa (exactly)
22

Definition: Temperature measures the degree of hotness of a body (“how
hot”). It doesn’t depend on the mass or the material of an object. It can be
thought of as a measure of the average kinetic energy of the atoms or
molecules in a body.
23

If two thermodynamic systems are each in thermal equilibrium with a third,
then they are in thermal equilibrium with each other.
24

As thermal motion of atoms increases, temperature increases. There seems to
be no upper limit of temperature but there is a definite limit at the other end of
the temperature scale. If we continually decrease the thermal motion of atoms
in a substance, the temperature will drop.
Absolute zero is the temperature at which no more energy can be extracted
from a substance.
At absolute zero, no further lowering of its temperature is possible. This
temperature is 273 degrees below zero on the Celsius scale. Absolute zero
corresponds to zero degrees on the Kelvin, or thermodynamic, scale and is
written 0 K (short for “zero Kelvin”).
Unlike the Celsius scale, there are no negative numbers on the thermodynamic scale.
Degrees on the Kelvin scale are the same size as those on the Celsius scale. Ice melts at
0°C, or 273 K, and water boils at 100°C, or 373 K.
25

P versus T plots of the
experimental data obtained
from a constant-volume gas
thermometer using four
different gases at different (but
low) pressures.
0
lim
p
PV
T
nR

(for ideal gas)
26

Consider thermodynamic system described by state variables {Z1, Z2,…, Zn}
Subspace of equilibrium states: f(Z1, Z2,…, Zn) = 0  This is the equation of
state (EOS)
Ideal gas: {T, P, V}  thermodynamicsEOS: pV = nRT
Equation of state
28

• Experiments with a wide variety of gases revealed
that four variables were sufficient to fully describe
the state of a gas
• Pressure (P)
• Volume (V)
• Temperature (T)
• The amount of the gas in moles (n)
Ideal gas law
29

Ideal gas law
PV=nRT
Ideal gas law
30

• Combine these relationships into a single fundamental
equation of state - the ideal gas equation of state
moleK
atm
08206.0314.8
L
moleK
J
R
nRTPV


Ideal gas law
31

• The pressure exerted by gas #1: P1 = n1 RT / V
• The pressure exerted by gas #2: P2= n2 RT / V
• The total pressure of the gases: pT = nT RT / V
nT represents the total number of moles of gas present in the
mixture.
P1 and P2 are the partial pressures of gas 1 and gas 2, respectively.
• PT = P1 + P2 = nT (RT/V)
• PT = P1 + P2 + P3 = Pi
Ideal gas law
32

• Gaseous mixtures - gases exert the same pressure
as if they were alone and occupied the same
volume.
• The partial pressure of each gas, Pi, is related to the
total pressure by Pi = Xi PT
• Xj is the mole fraction of gas i.
• Xj= nj / nT
Ideal gas law
33

• An ideal gas is a gas that obeys totally the ideal gas
law over its entire P-V-T range
• Ideal gases – molecules have negligible
intermolecular attractive forces and they occupy a
negligible volume compared with the container
volume
Ideal gas
34

Real gas
PvT diagram for real gases
37

40
Several equations have been proposed to represent the P-v-T behavior of substances
accurately over a larger region with no limitations.
Critical isotherm
of a pure
substance has an
inflection point
at the critical
state.
This model includes two effects not considered in the
ideal-gas model: the intermolecular attraction forces and
the volume occupied by the molecules themselves. The
accuracy of the van der Waals equation of state is often
inadequate.
Real gas

critical point is defined as the point at which the saturated liquid and saturated vapor
states are identical.
At the critical point, only one phase exists. There is an inflection point in the constant-
temperature line (critical isotherm) on a PV diagram. This means that at the critical
point:
Real gas
42

46
Compressibility factor Z A factor
that accounts for the deviation
of real gases from ideal-gas
behavior at a given temperature
and pressure.
The farther away Z is from unity, the more the gas
deviates from ideal-gas behavior.
Gases behave as an ideal gas at low densities (i.e.,
low pressure, high temperature).
Question: What is the criteria for low pressure and
high temperature?
Answer: The pressure or temperature of a gas is high
or low relative to its critical temperature or pressure.
Real gas

All substances obey the same equation of state in terms of reduced variables
Real gas
47

48
Comparison of Z factors for various gases.
Reduced
temperature
Reduced
pressure
Pseudo-reduced specific volume Z can also be determined from a
knowledge of PR and vR.
Real gas

As the weights fall, they give up potential energy and warm the water
accordingly. This was first demonstrated by James Joule, for whom the unit of
energy is named.
Schematic diagram for Joule´s experiment.
Insulating walls to prevent heat transfer
enclose water.
As the weights fall at constant speed, they
turn a paddle wheel, which does work on
water.
If friction in mechanism is negligible, the work
done by the paddle wheel on the water equal
the change of potential energy of the weights.
Chapter 2 – First law of thermodynamics

Internal energy involves energy on the microscopic scale
For thermochemistry  the internal energy
is the sum of the kinetic energy of motion of
the molecules, and the potential energy
represented by the chemical bonds
between the atoms and any other
intermolecular forces that may be operative.

U = Q - A

U = Q1 – A1
= Q2 – A2
= Q3 – A3 1
2
Q1
A1
Q2
A2
A3
Q3
For a infinitesimal process
dU Q A  
Where:
“d” used for describing state function
“” used for describing path function
U: state function
5

Chapter 2 – First Law of Thermodynamics

Thermodynamic definition of work: It is a kind of interaction that would occur at the
system boundaries. It can be positive or negative.

Work
(A)
Heat (Q)
release > 0 < 0
absorb < 0 > 0
Endothermic exothermic
Adiabatic system
Isothermal system
12

System
+
Heat, Qin Work, Wout
Work, WinHeat, Qout System
-

In the figure, the gas absorbs 400 J of heat and at the
same time does 120 J of work on the piston. What is the
change in internal energy of the system?
Qin
400 J
Wby =120 J
U = Q + W
= 400 J - 120 J
= +280 J
Q is positive: +400 J (Heat IN)
W is negative: -120 J (Work OUT)

Distinguish between heat, temperature, and thermal energy
• Thermal energy is an energy of the system due to the motion
of its atoms and molecules. Thermal energy is a state variable,
it may change during a process. The system’s thermal energy
continues to exist even if the system is isolated and not
interacting thermally with its environment
• Heat is energy transferred between the system and the
environment as they interact. Heat is not a particular form of
energy, nor is it a state variable. Heat may cause the system’s
thermal energy to change, but that does not mean that heat
and thermal energy are the same thing.
• Temperature is a state variable, it is related to the thermal
energy per molecule. But not the same thing.

Def. The specific heat capacity (c) of a substance is the amount of energy needed to
raise the temperature of 1kg of the substance by 1K (or 1 oC)

Specific heat capacity: Q
C
dT


Specific average heat capacity:
2 1
Q Q
C
T T
 
 
CP = ao + a1T + a2T2
CP = ao + a1T + a-2T-2
CP = ai Ti (i = 0, 1, 2, –2)
Effect of temperature on specific heat capacity:
For 01 mole of ideal gas: CP – CV = R
2
1
T
T
Q CdT 

• For ideal gas:
Cp = Cv + R

• Constant-
pressure
specific
• heats for some
gases

.A p dV 
 .dU Q p dV 
2
1
.
V
V
U Q p dV   
23

a. Isochoric process (V = const or dV = 0)
2
1
0 . 0
V
v V
V
dV A p dV
Q U
   
  

24

b. Isobaric process (p = const or dp = 0)
 2
1
2
p
1
pQ = U + A = U + p V = U + (pV)
= (U + pV) = H
. .
V
p V
P
A p dV p V V p V
Q H
    
     
 
  

25
Enthalpy is the sum of internal energy and the product of pV

Ap = p V = (pV) = (nRT) = nRT

R: ideal gas constant, R = 1,987 cal/mol.K
= 8,314 J/mol.K
= 0,082 l.atm/mol.K
Up = Qp – nR T
Ideal gas equation
pV = nRT :
or Up = H – nR T
c. Isobaric process for ideal gas
26

Joule’s law: (for ideal gas)
Internal energy of ideal gas just depends on
temperature: U = f(T).
TU 0  
2 2
1 1
2 1
1 2
. ln ln
V V
T T V V
nRT V p
Q A p dV dV nRT nRT
V V p
     
d. Isothermal process for ideal gas
27

Áp dụngnguyênlý 1: 𝛿𝑞 − 𝛿𝑤 = du (1)
Đặttỷlệtruyềnnănglượng: 𝐾 =
𝛿𝑞
𝛿𝑤
(2)
Với𝑑𝑢 = 𝑐 𝑣 𝑑𝑇 (3)
Ta sẽcó: 𝐾 − 1 𝑃𝑑𝑣 = 𝑐 𝑣 𝑑𝑇 (4)
Biếtrằngphươngtrìnhkhíthực:𝑃 × 𝑣 = 𝑍 × 𝑅 × 𝑇 (5)
Ta có: 𝑃𝑑𝑣 + 𝑣𝑑𝑃 = 𝑍𝑅𝑑𝑇 (6)
Ta cũngcó: 𝑐 𝑝 − 𝑐 𝑣 = 𝑍𝑅 (7)
Thếphươngtrình (7) vàovếphảicủaphươngtrình (6), sẽcó:
𝑃𝑑𝑣 + 𝑣𝑑𝑃 = 𝑐 𝑝 − 𝑐 𝑣 𝑑𝑇 (8)
Đặthệsốγ =
𝑐 𝑝
𝑐 𝑣
. Đâychínhlàhệsốđoạnnhiệt.
Chia phươngtrình (8) chophươngtrình (4), sẽcó:
−
𝑣𝑑𝑃
𝑃𝑑𝑣
= 1 − 𝛾 𝐾 + 𝛾 (9)
Nếuđặtgiátrị𝑛 = 1 − 𝛾 𝐾 + 𝛾 (10)
Ta sẽcó: −
𝑣𝑑𝑃
𝑃𝑑𝑣
= 𝑛 (11)
Hay là: 𝑃𝑣 𝑛
= 𝑐𝑜𝑛𝑠𝑡 (12)
e. Polytropic process

𝑷 × 𝑽 𝒏= 𝒄𝒐𝒏𝒔𝒕
𝑃2
𝑃1
=
𝑣1
𝑣2
𝑛
𝑇2
𝑇1
=
𝑣1
𝑣2
𝑛−1

A biatomic ideal gas undergoes a cycle starting at point A
(2 atm, 1L). Process from A to B is an expansion at
constant pressure until the volume is 2.5 L, after which is
cooled at constant volume until its pressure is 1 atm. It is
then compressed at constant pressure until the volume is
again 1L, after which it is heated at constant volume until
it is back in its original state. Find (a) the work, heat and
change of internal energy in each process (b) the total
work done on the gas and the total heat added to it during
the cycle.
A system consisting of 0.32 mol of a monoatomic ideal gas
occupies a volume of 2.2 L, at a pressure of 2.4 atm.
The system is carried through a cycle consisting:
1. The gas is heated at constant pressure until its volume
is 4.4L.
2. The gas is cooled at constant volume until the pressure
decreased to 1.2 atm
3. The gas undergoes an isothermal compression back to
initial point.
(a) What is the temperature at points A, B and C
(b) Find W, Q and ΔU for each process and for the entire
cycle

Dr. Ngo Thanh An
PHYSICAL CHEMISTRY 1

2
A cup of hot coffee does not
get hotter in a cooler room.
Transferring heat to
a wire will not
generate electricity.
Transferring heat
to a paddle
wheel will not
cause it to rotate.
These processes cannot
occur even though they
are not in violation of the
first law.
Chapter 3 – Second law of thermodynamics

3
Processes occur in a certain direction, and not in the reverse
direction.
A process must satisfy both the first
and second laws of thermodynamics
to proceed.
MAJOR USES OF THE SECOND LAW
1. The second law may be used to identify the direction of processes.
2. The second law also asserts that energy has quality as well as quantity. The first law is
concerned with the quantity of energy and the transformations of energy from one form
to another with no regard to its quality. The second law provides the necessary means to
determine the quality as well as the degree of degradation of energy during a process.
3. The second law of thermodynamics is also used in determining the theoretical limits for
the performance of commonly used engineering systems, such as heat engines and
refrigerators, as well as predicting the degree of completion of chemical reactions.

4
A source supplies
energy in the
form of heat, and
a sink absorbs it.
• A hypothetical body with a relatively large thermal energy capacity (mass x specific
heat) that can supply or absorb finite amounts of heat without undergoing any
change in temperature is called a thermal energy reservoir, or just a reservoir.
• In practice, large bodies of water such as oceans, lakes, and rivers as well as the
atmospheric air can be modeled accurately as thermal energy reservoirs because of
their large thermal energy storage capabilities or thermal masses.
Bodies with relatively large thermal
masses can be modeled as thermal
energy reservoirs.

5
The devices that convert heat to work.
1. They receive heat from a high-
temperature source (solar energy, oil
furnace, nuclear reactor, etc.).
2. They convert part of this heat to
work (usually in the form of a
rotating shaft.)
3. They reject the remaining waste
heat to a low-temperature sink (the
atmosphere, rivers, etc.).
4. They operate on a cycle.
Heat engines and other cyclic devices
usually involve a fluid to and from
which heat is transferred while
undergoing a cycle. This fluid is
called the working fluid.
Part of the heat received
by a heat engine is
converted to work, while
the rest is rejected to a
sink.
Work can always be converted to heat
directly and completely, but the reverse
is not true.

6
Some heat engines perform better than
others (convert more of the heat they
receive to work).
Even the most
efficient heat engines
reject almost one-half
of the energy they
receive as waste heat.
Schematic of a
heat engine.

7
A heat-engine cycle cannot be completed without
rejecting some heat to a low-temperature sink.
In a steam power plant, the
condenser is the device where
large quantities of waste heat is
rejected to rivers, lakes, or the
atmosphere.
Can we not just take the
condenser out of the plant and
save all that waste energy?
The answer is, unfortunately, a
firm no for the simple reason
that without a heat rejection
process in a condenser, the
cycle cannot be completed.
Every heat engine must waste some energy by transferring it to a low-
temperature reservoir in order to complete the cycle, even under idealized
conditions.

8
A heat engine that violates the Kelvin–
Planck statement of the second law.
It is impossible for any device that
operates on a cycle to receive heat from
a single reservoir and produce a net
amount of work.
No heat engine can have a thermal
efficiency of 100 percent, or as for a power
plant to operate, the working fluid must
exchange heat with the environment as well
as the furnace.
The impossibility of having a 100% efficient
heat engine is not due to friction or other
dissipative effects. It is a limitation that
applies to both the idealized and the actual
heat engines.

9
• The transfer of heat from a low-
temperature medium to a high-temperature
one requires special devices called
refrigerators.
• Refrigerators, like heat engines, are cyclic
devices.
• The working fluid used in the refrigeration
cycle is called a refrigerant.
• The most frequently used refrigeration
cycle is the vapor-compression refrigeration
cycle.
Basic components of a refrigeration
system and typical operating conditions.
In a household refrigerator, the freezer
compartment where heat is absorbed by the
refrigerant serves as the evaporator, and the
coils usually behind the refrigerator where
heat is dissipated to the kitchen air serve as
the condenser.

10
The objective of a
refrigerator is to remove QL
from the cooled space.
The efficiency of a refrigerator is expressed in terms
of the coefficient of performance (COP).
The objective of a refrigerator is to remove heat (QL)
from the refrigerated space.
Can the value of COPR be greater than unity?

11
The objective of a
heat pump is to
supply heat QH
into the warmer
space.
Can the value of COPHP be
lower than unity?
What does COPHP=1
represent?
The work supplied
to a heat pump is
used to extract
energy from the
cold outdoors and
carry it into the
warm indoors.

12
It is impossible to construct a device that operates in a
cycle and produces no effect other than the transfer of
heat from a lower-temperature body to a higher-
temperature body.
It states that a refrigerator cannot operate unless its
compressor is driven by an external power source, such
as an electric motor.
This way, the net effect on the surroundings involves the
consumption of some energy in the form of work, in
addition to the transfer of heat from a colder body to a
warmer one.
To date, no experiment has been conducted that
contradicts the second law, and this should be taken as
sufficient proof of its validity.
A refrigerator that violates
the Clausius statement of
the second law.

13
Two familiar reversible
processes. Reversible processes deliver the most and consume the
least work.
Reversible process: A process that can be reversed without leaving any trace on the
surroundings.
Irreversible process: A process that is not reversible.
• All the processes occurring in nature are irreversible.
• Why are we interested in reversible processes?
• (1) they are easy to analyze and (2) they serve as idealized
models (theoretical limits) to which actual processes can be
compared.
• Some processes are more irreversible than others.
• We try to approximate reversible processes. Why?

Quá trình thuận nghịch:
Là quá trình mà khi đi từ A đến B và ngược lại từ B đến A thì hệ không
gây ra bất kỳ sự biến đổi nào trong hệ cũng như cho môi trường. Không
gây biến đổi  tức không tiêu hao năng lượng  tức không có entropy
nội sinh  không sinh ra entropy
Điểm chính của quá trình thuận nghịch, đó là quá trình không gây
ra biến đổi entropy!!!!
Quá trình Bất thuận nghịch: quá trình không thỏa mãn các điều kiện trên
--
Đối với quá trình thuận nghịch:
- Công hệ sinh đạt cực đại  Tại sao????? (liên quan hiệu suất
nhiệt cực đại)
- Công hệ nhận đạt cực tiểu  Tại sao???? (liên quan giá trị COP
cực đại)

16
Friction renders a
process irreversible.
Irreversible compression
and expansion processes.
(a) Heat transfer
through a temperature
difference is
irreversible, and (b) the
reverse process is
impossible.
• The factors that cause a process to be irreversible are called
irreversibilities.
• They include friction, unrestrained expansion, mixing of two
fluids, heat transfer across a finite temperature difference,
electric resistance, inelastic deformation of solids, and
chemical reactions.
• The presence of any of these effects renders a process
irreversible.

►The Carnot cycle provides an example of a
reversible cycle that operates between two
thermal reservoirs.
►In a Carnot cycle, the system executing the cycle
undergoes a series of four internally reversible
processes: two adiabatic processes alternated
with two isothermal processes.

18
Reversible Isothermal Expansion (process 1-2, TH = constant)
Reversible Adiabatic Expansion (process 2-3, temperature drops from TH to TL)
Reversible Isothermal Compression (process 3-4, TL = constant)
Reversible Adiabatic Compression (process 4-1, temperature rises from TL to TH)
Execution of the Carnot cycle in a closed system.

19
P-V diagram of the Carnot cycle. P-V diagram of the reversed Carnot
cycle.
The Carnot heat-engine cycle is a totally reversible cycle.
Therefore, all the processes that comprise it can be reversed, in which case it
becomes the Carnot refrigerationcycle.

20
1. The efficiencyof an irreversibleheat engine is always less than the efficiencyof a
reversible one operating between the same two reservoirs.
2. The efficiencies of all reversible heat engines operating between the same two
reservoirs are the same.
The Carnot principles.
Proof of the firstCarnot principle.

21

22
The arrangement of heat engines used to
develop the thermodynamic temperature scale.
A temperature scale that is independent of the
properties of the substances that are used to
measure temperature is called a thermodynamic
temperaturescale.
Such a temperature scale offers great
conveniences in thermodynamic calculations.

If we select (T) = T, then
For a reversible heat engine operating between two reservoirs at temperatures TH and TL,
the above equation can be written as

26
For reversible cycles, the heat
transfer ratio QH /QL can be
replaced by the absolute
temperature ratio TH /TL.
A conceptual experimental setup to
determine thermodynamic temperatures on
the Kelvin scale by measuring heat transfers
QH and QL.
This temperature scale is
called the Kelvin scale, and
the temperatures on this
scale are called absolute
temperatures.

27
The Carnot heat
engine is the
most efficient of
all heat engines
operating
between the
same high- and
low-temperature
reservoirs.
No heat engine can have a higher efficiency
than a reversible heat engine operating
between the same high- and low-
temperature reservoirs.
Any heat engine Carnot heat engine

28
The fraction of heat that can be
converted to work as a function
of source temperature.
The higher the temperature
of the thermal energy, the
higher its quality.
How do you increase the thermal
efficiency of a Carnot heat engine?
How about for actual heat engines?
Can we use C
unit for
temperature
here?

29
No refrigerator can have a higher COP than a
reversible refrigerator operating between the same
temperature limits.
How do you increase the COP
of a Carnot refrigerator or
heat pump? How about for
actual ones?
Carnot refrigerator or heat
pump
Any refrigerator or heat pump

30
The COP of a reversible refrigerator or heat pump is the maximum theoretical
value for the specified temperature limits.
Actual refrigerators or heat pumps may approach these values as their designs
are improved, but they can never reach them.
The COPs of both the refrigerators and the heat pumps decrease as TL decreases.
That is, it requires more work to absorb heat from lower-temperature media.

Ex 1: An automobile engine has an efficiency of 22.0% and produces 2510 J of
work. How much heat is rejected by the engine?
Ans: 8900 J
Ex 2: An ideal or Carnot heat pump is used to heat a house to a temperature of
TH = 294 K (21 °C). How much work must be done by the pump to deliver QH =
3350 J of heat into the house when the outdoor temperature TC is (a) 273 K (0
°C) and (b) 252 K (–21 °C)?
(a)
(b)

32
Ex 3: Each drawing represents a hypothetical heat engine or a hypothetical heat
pump and shows the corresponding heats and work. Only one is allowed in nature.
Whichis it?

Ex 4: The lowest possible temperature is absolute zero, at
a. 0 on the Kelvin scale and 0 degrees on the Celsius scale.
b. 0 on the Kelvin scale and -100 degrees on the Celsius scale.
c. 0 on the Kelvin scale and -273 degrees on the Celsius scale.
d. 373 on the Kelvin scale and -273 degrees on the Celsius scale.
Ex 5: The second law of thermodynamics tells us that heat cannot flow from
a. hot to cold ever.
b. cold to hot ever.
c. hot to cold without external energy.
d. cold to hot without external energy.
Ex 6: Heat engines such as jet engines are more efficient when run at
a. high temperatures.
b. constant temperatures.
c. low temperatures.
d. a constantrate.

Ex 7:

Ex 8:

For heat engine, we have: 𝜂 𝑡ℎ = 1 −
𝑄 𝐿
𝑄 𝐻
𝜂 𝑡ℎ,𝑟𝑒𝑣 = 1 −
𝑇𝐿
𝑇 𝐻
Carnot’s principle: 𝜂 𝑡ℎ ≤ 𝜂 𝑡ℎ,𝑟𝑒𝑣
Therefore: 1 −
𝑄 𝐿
𝑄 𝐻
≤ 1 −
𝑇𝐿
𝑇 𝐻
Or: 𝑄 𝐻
𝑇 𝐻
−
𝑄 𝐿
𝑇𝐿
≤ 0

37
• The cyclic integral indicates
that the integral should be
performed over the entire
cycle and over all parts of
the boundary.
2 3 4 1
1 2 3 4
Q Q Q Q
T T T T
   
      

38
0 0H L
H L
Q Q
T T
   
2 3 4 1
1 2 3 4
Q Q Q Q
T T T T
   
      
0 T
Q

39
• Clausius inequality results in two important concepts:
– Entropy (S)
– Generated entropy (Sg)

• A thermodynamic (energy) function that describes the
degree of randomness or probability of existence.
• The more disordered the system, the larger its entropy.
• As a state function – entropy change depends only on
the initial and final states, but not on how the change
occurs.

• Nature spontaneously proceeds toward the state that
has the highest probability of (energy) existence –
highest entropy
• Entropy is used to predict whether a given
process/reaction is thermodynamically possible;

42
All paths are arbitrary
2 2
1 1A C
Q Q
T T
    
   
   
 Subtracting gives
2 1
1 2
0
C B
Q Q
T T
    
     
   
 
For reversible cycle A-B
2 1
1 2
0
A B
Q Q
T T
    
     
   
 
For reversible cycle C-B
Q
the quantity is independent of the path and dependent on the end states only
T

 
Since paths A and C are arbitrary, it follows that the integral of Q/T has the
same value for ANY reversible process between the two sates.

43
 
is a thermodynamic property
we call it entropy S
δQ
T
 
Entropy (the unit)
S = entropy (kJ/K); s = specific entropy
(kJ/kg K)
 












2
1
12gintegratin
revrev T
Q
SS
T
Q
dS

S2 – S1 depends on the end states
only and not on the path,
 it is same for any path reversible
or irreversible

44
Giả sử quá trình AB là thuận nghịch, quá trình CB: BTN
2 2
1 1C
C
δQ
dS
T
 
   
 
 
ර
𝛿𝑄
𝑇
= න
1
2
𝛿𝑄
𝑇 𝐴
+ න
2
1
𝛿𝑄
𝑇 𝐵
= 0
ර
𝛿𝑄
𝑇
= න
1
2
𝛿𝑄
𝑇 𝐶
+ න
2
1
𝛿𝑄
𝑇 𝐵
≤ 0
Cho quá trình A-B (thuận nghịch)
Cho quá trình C-B (bất thuận nghịch) න
1
2 𝛿𝑄
𝑇 𝐴
≥ න
2
1 𝛿𝑄
𝑇 𝐶
Ngoài ra, ta cũng có:
න
1
2
𝛿𝑄
𝑇 𝐴
= න
1
2
𝑑𝑆𝐴 = න
1
2
𝑑𝑆 𝐶

45
2nd law of thermodynamics for a closed system
0 for irreversible process
entropy generation
0 for a reversible process
genS

 

In any irreversibleprocessalways entropy is generated (Sgen > 0) due to
irreversibilities occurringinside the system.
gen
Q
dS S
T

 
2
2 1 1 gen
Q
S S S
T

   gen
for any process,
with S 0



This can be written out in a common form as an equality
or
Entropy Balance Equation for a closed system
δQ
dS
T

2
2 1 1
or
δQ
S S
T
  
equality for reversible
inequality for irreversible

Example: Entropy change during an isothermal process

47
There is some entropy generated during an irreversible
process such that
Entropy transfer
with heat
Entropy generation
due to irreversibility
genS
T
Q
SS  
2
1
12

Entropy
change
This is the entropybalance for a closed system.

48
 The entropy change can be evaluated independently of the
process details.
 However, the entropy generation depends on the process, and
thus it is not a property of the system.
 The entropy generation is always a positive quantity or zero and
this generation is due to the presence of irreversibilities.
 The direction of entropy transfer is the same as the direction of
the heat transfer: a positive value means entropy is transferred
into the system and a negative value means entropy is transferred
out of the system.
genS
T
Q
SS  
2
1
12


• Number of particles: N
• Arrangement: 2N
• Each particle has only two possible quantum states  or .
• Distribution: m
• Total number of microstates (): 2N
• Total magnetic moment of all the particles is
M = (N-N).
• M just depends on the relative number of up and down moments and not
on the detail of which are up or down.
• The relative number of ups and downs is constrained by:
N + N = N

Microstate: a microscopic description would necessitate specifying
the state of each particle
Macrostate (state) including a lot of microstate (don’t care which is
up or down. Just focuses on how much is up and down)
 01 (State)  01 (m), which m equals N-N
Number m tell us the distribution of N particles among possible
states:
N = (N-m)/2; N = (N+m)/2
 t(m): total number of microstates responding to a (macro)state:
𝑡 𝑚 =
𝑁!
𝑁 ↑ ! 𝑁 ↓ !
𝑡 𝑚 =
𝑁!
𝑁 − 𝑚
2
!
𝑁 + 𝑚
2
!

53
• For an isolated (or simply an adiabatic closed system), the heat
transfer is zero, then
 This means that the entropy of an adiabatic system during a
process always increases or, In the limiting case of a
reversible process, remains constant.
 In other words, it never decreases.
 This is called Increase of entropy principle.
 This principle is a quantitative measure of the second law.
2
2 1 gen1
S
Q
S S S
T

     genSadiabaticS  

54
• Now suppose the system is not adiabatic.
• We can make it adiabatic by extending the
surrounding until no heat, mass, or work are
crossing the boundary of the surrounding.
• This way, the system and its surroundings can be
viewed again as an isolated system.
• The entropy change of an isolated system is the
sum of the entropy changes of its components (the
system and its surroundings), and is never less
than zero.
• Now, let us apply the entropy balance for an
isolated system:
0 surrsystotalgen SSSS

55








processimposible
processreversible
processleirreversib
Sgen
0
0
0
Lưu ý: Entropy chỉ được truyền vào hệ theo 2 hình thức: hoặc truyền nhiệt, hoặc
truyền khối. Truyền công vào hệ không gây ra sự truyền entropy!!!!!!

56
• Processes can occur in a certain direction only , not in any
direction. A process must proceed in the direction that
complies with the increase of entropy principle. A process that
violates this principle is impossible.
• Entropy is a non-conserved property. Entropy is conserved
during the idealized reversible process only and increases
during all actual processes.
• The performance of engineering systems is degraded by the
presence of irreversibilities, and the entropy generation is a
measure of the magnitude of the irreversibilities present
during a process.

TNQ
T

  S
a. Heating (cooling) process at constant pressure/constant volume:
Isochoric process:
Isobaric process: PC dT
T
 
TNQ
T

  S VC dT
T
 
rev
T
QS SS int
2
112  



 

b. Isothermal process:
TNQ
T

  S
Phase transfer
TNQ
T

Ideal gas expansion:
Chemical reaction
T 2 1
1 2
Q V
ln ln
T V
   
P
S nR nR
P
TQ
T

  S
T
ΔS°reaction = ΣnpS°products – ΣnrS°reactants

• States, S.
– The microscopic energy levels
available in a system.
• Microstates, W.
– The particular way in which particles are distributed
amongst the states. Number of microstates = W.
• The Boltzmann constant, k = 1.38x10-23 J/K
– Effectively the gas constant per molecule = R/NA.
S = k lnW
2
2 1
1
W
ln
W
S S S k   

T
0
limS 0o
T
S

 
S = ST – So = ST
3rd law of thermodynamics: The entropy of a perfect crystal
at 0 K is zero.
Because:
- all molecular motion stops
- all particles are in their place

62
Xét quá trình biến đổi của 1 chất như sau:
 

 

 
 
1 3
o
42
5 7
T64
raén 1 raén 2
noùng chaûychuyeån phaS S
S SS
loûng hôi
noùng chaûy hoùa hôiS S
SSS
0(K) T T
T T T(K)
Tính ST 
 
         
2
1
7 7
1 1
T
T o i i p
i i T
dT
S S S S C
T T
Ví duï: So
298 chính laø entropy tuyeät ñoái cuûa moät
chaát ôû 1 atm, 298K (ñieàu kieän chuaån ).
2/26/2018 607010 - Chương 2

63
Ví duï:
Tính bieán thieân entropy cuûa cuûa quaù trình
ñoâng ñaëc benzen döôùi aùp suaát 1 atm trong
caùc tröôøng hôïp sau:
a) Quaù trình ñoâng ñaëc thuaän nghòch ôû 5oC vôùi
nhieät ñoâng ñaëc laø ññ = – 2370 cal/mol
b) Quaù trình ñoâng ñaëc baát thuaän nghòch ôû –
5oC .
Bieát nhieät dung cuûa benzen loûng vaø benzen raén
laàn löôït laø 30,3 vaø 29,3 cal/mol.K
2/26/2018 607010 - Chương 2

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