Chapter 07 Thermodynamics 139 - Universiti Tunku …staff.utar.edu.my/limsk/Physics/07...
Transcript of Chapter 07 Thermodynamics 139 - Universiti Tunku …staff.utar.edu.my/limsk/Physics/07...
Contents
Page
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Chapter 07 Thermodynamics .......................................... 139
7.0 Introduction ............................................................................139
7.2 Review the Basis of Thermodynamic ...................................139
7.2.1 Heat ............................................................................................... 139
7.2.2 Internal Energy ............................................................................ 139
7.2.3 Temperature Measurement and Unit of Measurement ........... 141
7.2.4 Specific Heat ................................................................................. 143
7.3 Thermal Expansion ................................................................144
7.4 Latent Heat .............................................................................145
7.5 Heat Transfer Mechanism .....................................................147
7.5.1 Conduction ................................................................................... 147
7.5.2 Convection .................................................................................... 149
7.5.3 Radiation ...................................................................................... 150
7.6 Laws of Thermodynamics .....................................................152
7.6.1 Zeroth
Law of Thermodynamics ................................................. 152
7.6.2 First law of thermodynamics ...................................................... 153
7.6.2.1 Heat Capacities and First Law of Thermodynamics .................... 155
7.6.3 Second Law of Thermodynamics ............................................... 156
7.6.3.1 Heat Transfer from Hot Object to Cold Object ............................ 157
7.7 Thermodynamics Potentials ..................................................158
7.8 Thermodynamic Processes ....................................................160
7.8.1 Adiabatic process ......................................................................... 161
7.8.2 Isochoric Process .......................................................................... 162
7.8.3 Cyclical process ............................................................................ 162
7.8.4 Isothermal Process ....................................................................... 162
7.8.5 Isobaric process ............................................................................ 163
7.9 Heat Engines ...........................................................................164
7.9.1 Carnot Engine .............................................................................. 165
7.9.2 Refrigeration ................................................................................ 167
Tutorials ........................................................................................168
List of Figures
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Figure 7.1: Internal energy contribution of gas, liquid, and solid .................................... 140 Figure 7.2: Comparison of internal energy for copper and water at temperature zero
Celsius ............................................................................................................ 141 Figure 7.3: Constant-volume gas thermometer ................................................................ 142 Figure 7.4: Results of temperature-pressure plot of constant-volume thermometer filled
with three-gas types ....................................................................................... 143 Figure 7.5: Specific heats of some materials at room temperature .................................. 144 Figure 7.6: Heat of fusion and heat of vaporization of some materials ........................... 146 Figure 7.7: Thermal conduction between hot reservoir and cold reservoir via a slab ...... 148
Figure 7.8: An illustration of convection of thermal heat transfer ................................... 150 Figure 7.9: Illustration of Zero
th law of thermodynamic .................................................. 153
Figure 7.10: Interpretation of First law of thermodynamic ................................................ 154 Figure 7.11: The work done graph of an adiabatic process ................................................ 161
Figure 7.12: The work done graph of an isochoric process ............................................... 162 Figure 7.13: The work done graph of an isothermal process ............................................. 163 Figure 7.14: The work done graph of an isobaric process ................................................. 164 Figure 7.15: A schematic representation of a heat engine.................................................. 165
Figure 7.16: Pressure-volume diagram of a Carnot engine ................................................ 166 Figure 7.17: The schematic of a refrigerator ...................................................................... 167
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Chapter 07
Thermodynamics
_____________________________________________
7.0 Introduction
Thermodynamics is a branch of physics, which deals with energy and work of a
system. It was first studied in the 19th century as scientists were first discovering
how to build and operate steam engines. Thermodynamics deals with the large-
scale response of a system in microscopic change, which can be observed and
measured in experiments. Small-scale gas interactions are described by the
kinetic theory of gasses, which is a compliment to thermodynamics.
In this chapter, review of basic thermodynamic, heat transfer mechanism
the laws of thermodynamic and other associated topics shall be discussed in
details.
7.2 Review the Basis of Thermodynamic
Let’s review some basics of thermodynamic. They are covered in the following
three sub-sections.
7.2.1 Heat
Heat is the mechanism by which energy is transferred between system and its
environment because there is temperature difference between them. It is
transferred from hot region to cold region by various mechanisms which will be
dealt later.
7.2.2 Internal Energy
Internal energy U of the system is the energy associated with the microscopic
component of the system such as atom and molecule, when they are viewed
from reference frame of rest with respect to the system. Internal energy consists
of two components, which are kinetic energy and potential energy. Kinetic
energy is associated with the random translational, rotational, while potential
energy is associated with vibrational motion of the atoms or molecules and
intermolecular forces.
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For an ideal mono-atomic gas, the internal energy U is just the translational
kinetic energy of the linear motion of the atom. For polyatomic gases, there are
rotational and vibrational kinetic energies. In liquids and solids, there is
potential energy associated with the intermolecular attractive forces. A
simplified visualization of the contributions to internal energy for gas, liquid,
and solid is shown in Fig. 7.1.
Figure 7.1: Internal energy contribution of gas, liquid, and solid
If water is placed in a pot and heated by flame, the internal energy of water
increases due to the heat transferred from the flame to water. Comparison of
internal energy for copper and water is shown in Fig. 7.2. Copper has less
potential energy than water. Thus, it takes lesser energy to increase the
temperature by one degree Celsius for a fixed mass. The energy required to
increase the temperature of one kg of material by one degree Celsius is called
specific heat. Take note the specific heat of copper is 386J/kg-K, while the
specific heat of water is 4,186J/kg-K.
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Figure 7.2: Comparison of internal energy for copper and water at temperature zero Celsius
7.2.3 Temperature Measurement and Unit of Measurement
To set-up a temperature scale, one has to pick a reproducible thermal
phenomenon and arbitrary assigns a certain Kelvin temperature to its
environment. This standard fixed-point temperature is selected at the triple point
of water. The triple point temperature T3 of water is T3 = 273.16K. The triple
point temperature is the temperature where liquid water, solid ice, and water
vapor can coexist in thermal equilibrium at a fixed set value of pressure and
temperature. The triple point temperature can be measured using a constant
volume gas thermometer shown in Fig. 7.3 filled with different type of gas such
as hydrogen, helium, or nitrogen. At constant volume, the pressure P in the
thermometer is dependent on the temperature of the liquid where it is immersed.
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Figure 7.3: Constant-volume gas thermometer
Temperature is dependent on pressure, which can be expressed by equation
(7.1).
T = CP (7.1)
where C is a constant and P the pressure is equal to P = Po - hg, where is the
density of the liquid and g is earth acceleration, which is 9.8m/s2. Po is the
atmospheric pressure, which is 1.01x105 pascal Pa (Newton/m
2) or 760 torr or
14.7lb/in2. At triple temperature T3, the pressure is P3. An equation relating
temperature and pressure can be expressed as
3P
PK15.373T (7.2)
However, it was found that at boiling point of water, different gas gives
different result. Thus, it is necessary to measure the triple point temperature
when the volume in the thermometer approaches zero, where it is now
independent of the gas type. Equation (7.2) shall then be changed to
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3
0V P
PlimK15.373T . Figure 7.4 shows the results of the temperature-pressure
plot of constant-volume thermometer filled with helium, nitrogen, and hydrogen
gases.
Figure 7.4: Results of temperature-pressure plot of constant-volume thermometer filled with
three-gas types
There are two ways that the unit of heat is being defined. In British system, it is
the British thermal unit BTU, in which one BTU is defined as the heat required
to raise the temperature of 1.0lb of water from 630 F to 64
0 F. Alternatively, one
calorie is the energy required to raise the temperature of 1.0g water from 14.50C
and 15.50C.
In 1948, scientist linked the heat like work defined as the energy transfer
and defined one calorie c as equals to 4.186J, which is equal to 3.969x10-3
BTU.
This shall mean 1.0 BTU = 251.9 calorie = 1,954.67J.
7.2.4 Specific Heat
The specific heat cm of a material is defined as the amount of energy necessary
to raise the temperature of one kilogram of that material by one degree Celsius.
In mathematical expression, it is equal to
)TT(m
QCm
if (7.3)
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where m is the mass and Q is the heat or energy or heat capacity.
Figure 7.5 shows the specific heats of some materials at room temperature.
One will find that generally the specific heat of conductor is lower than the non-
conductor.
Substance Specific Heat Cm
cal/g-K J/kg-K
Lead 0.0305 128
Tungsten 0.0321 134
Silver 0.0564 236
Copper 0.0923 386
Aluminum 0.215 900
Brass 0.092 380
Granite 0.19 790
Glass 0.20 840
Ice (-100C) 0.530 2,220
Mercury 0.033 140
Ethyl alcohol 0.58 2,430
Sea water 0.93 3,900
Water 1.00 4,190
Figure 7.5: Specific heats of some materials at room temperature
7.3 Thermal Expansion
When the temperature of material increases, its internal energy increases. This
result its dimensions increase due to increase in inter-atomic or intermolecular
distance that can be caused by vibration, rotational moment, and etc.
If the temperature of metal rod of length L is raised by temperature T, its
length is increased by an amount L following equation (7.4).
L = LT (7.4)
where is the coefficient of linear expansion, which has unit per Kelvin or per
degree Celsius.
The volume of the material will also be increased with the increase of
temperature, whereby the increase of temperature follows equation (7.5).
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V = VT (7.5)
where is the coefficient of volume expansion. The relationship between
coefficient of linear expansion and coefficient of volume expansion is = 3.
Example 7.1
On a hot day in Las Vegas, an oil tanker loaded 37,000 liters of diesel fuel. The
driver unloaded the entire load of diesel in Payson, Utah, where the temperature
is 25.0K lower than in Las Vegas. How many liters did he deliver? In order to
compensate the loss, what is the percentage increase of the cost price per liter of
diesel? Given that the coefficient of volume expansion of diesel is 9.54x10-4
/0C.
Solution
The change in volume of diesel in Payson is 37,000x9.54x10-4
x(-25) = -882.45
liters. In Payson the driver delivers 37,000 – 882.45 = 36,117.55 liters of diesel.
The increase of the cost of diesel is (37,000/36,117.55 –1)100 % = 2.44 % per
liter.
7.4 Latent Heat
When matter absorbs energy, it is not necessary used to raise the temperature of
the material. The energy may be used to change from one phase or state to
another phase or state. All matters can exist in three common phases i.e. solid,
liquid, and gas states. Take for an example, the temperature of solid ice and ice
water is the same, and the temperature of boiling water and steam is the same.
The energy absorbed by the solid ice is used to break the intermolecular force so
that the molecules can become loosely bonded. This energy is termed as heat of
fusion. It has unit J/kg. Boiling water absorbs energy to break the weak
intermolecular bond into free molecular gas. The energy required is termed as
heat of vaporization. Figure 7.6 shows the heat of fusion and heat of
vaporization of the some substances.
Substance
Melting Boiling
Melting Point K Heat of
Fusion kJ/kg
Boling
Point K
Heat of
Vaporization kJ/kg
Hydrogen 14.0 58.0 20.3 455
Oxygen 54.8 13.9 90.2 213
Water 273 333 373 2,256
Mercury 234 11.4 630 296
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Lead 601 23.2 2,017 858
Copper 1,356 207 2,868 4,730
Silver 1,235 105 2,323 2,336
Figure 7.6: Heat of fusion and heat of vaporization of some materials
From Fig. 7.6, it shows that the heat of vaporization is higher than heat of fusion,
which is logically true because it requires less energy to break the solid bond
into loose bond than to free the atom or molecule from inter-atomic or
intermolecular bonds.
Example 7.2
How much heat must be absorbed by ice of mass m = 720g at temperature -100C
to take it to liquid state at 150C?
Solution
Three steps are to be considered which are shown in the diagram below.
The energy required to raise the temperature of solid ice from -10
0C to 0
0C is
10x0.720x2,200J/kg-K = 15.84x103J.
The energy required to break the intermolecular bond to loose bond is the heat
of fusion, which is equal to 0.72x333x103 = 239.8x10
3J.
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The energy required to heat ice water from 00C to 15
0C is 0.72x15x4,190 =
45.25x103J.
The total energy required is 3.008x105J.
7.5 Heat Transfer Mechanism
The study of the change in energy of a system by measuring the heat exchange
with its surrounding is called calorimetry. Heat can be transferred between a
system and its environment via three mechanisms, which are conduction,
convection, and radiation.
7.5.1 Conduction
Conduction is the transfer of kinetic energy and vibrational energy of atoms to
the atom of lower kinetic energy and vibrational energy. The rate of conduction
Pcond, which is also equal to rate of heat transferdt
dQ follows equation (7.6).
L
TTAP CH
cond
k (7.6)
where k is the thermal conductivity, L is the thickness of the transfer material, A
is the face contact area, and TH>TC. Figure 7.7 illustrates the conduction
mechanism from higher temperature region to lower temperature region via slab
of thickness L and face area of A.
Thermal resistance R is defined as
R = k
L (7.7)
Thus, the thickness the slab, the higher will be the thermal resistance R, which
implies that the lower will be rate of transfer of heat. i.e.
k/L
TTAP CH
cond
=
R
TTAP CH
cond
.
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Figure 7.7: Thermal conduction between hot reservoir and cold reservoir via a slab
If there are two insulating layers of different thermal conductivity k materials, it
can be shown that the rate of conduction is
ii
CH
21
CHcond
/L
TTA
RR
TTAP
k, where
R1 and R2 are respectively the thermal resistance of material 1 and material 2.
If two materials are connected in parallel with the ends contact the hot and
cold reservoirs then the rate of conductor Pcond(total) shall be equal to the sum of
rate of conduction Pcond(1) of material 1 and rate of conduction Pcond(2) of material
2. Thus, equation (7.8), the rate of conduction Pcond(total) is
Pcond(total) = 1
CH
1R
TTA
+
2
CH
2R
TTA
(7.8)
Example 7.3
The wall of house is made of white pine, unknown materials, and a brick wall as
illustrated from figure below.
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The thickness Ld of brick wall is two times the thickness of white pine La. The
thermal conductivity kd of the brick is five times the thermal conductivity ka of
white pine. The temperature of indoor T1 is = 250C. The temperature T2 = 20
0C,
T5 = - 100C. If steady state transfer of heat is attained, what is the temperature
T4 of the interface between brick and the unknown material?
Solution
Since the steady state of transfer of heat is attained, the conduction rate of white
pine is same as the brick wall, which isa
21cond
L
TTAP
ak and
d
54
condL
TTAP
dk .
Thus,a
21cond
L
TTAP
ak
a
54
d
54
L2
TTA5
L
TTA
ad kk . The equation is now contains
one unknown T4 and T4 is found to be – 80C.
7.5.2 Convection
Convection occurs when liquid or gases come in contact with object whose
temperature is higher than that of liquid or gas. When gas or liquid comes in
contact with hot object, its temperature increases and becomes less dense. It
buoyant force causes it to rise. The surround liquid or gas is then flowed in to
take the place of rising warm liquid or gas. The process of heat flow is termed
as convection. An illustration of thermal heat transfer by convection by a heater
is shown in Fig. 7.8. The heater heats the air surrounding it. Hot air arises and
cold air flows in to take the place. The process will continue. This is one of the
way that heater is used to warm the room during winter.
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Figure 7.8: An illustration of convection of thermal heat transfer
Basically there are two types of heat convection. They are free convection and
force convection. Heat transfer of both types is described by Newton's law of
cooling, which follows equation (7.9).
TAPConvection h (7.9)
The rate of heat transfer PConvection to the surrounding fluid is proportional to the
exposed area A of the object and the difference between the object temperature
and the fluid free-stream temperature. h is termed as convection heat-transfer
coefficient. Other terms describing h include film coefficient and film
conductance.
7.5.3 Radiation
The transfer of heat between a system and its environment via electromagnetic
waves is called thermal radiation. A very good example of thermal radiation is
visible light. When a person stands outside where there is Sun light, he/she feels
hot after a while.
The rate of thermal radiation Prad at which an object emits energy via
electromagnetic radiation depends on surface area A, temperature of the area,
and emissivity . The value of emissivity is a value between 0 and 1, which is
surface dependent. A surface that has maximum emissivity of 1 is said to be a
blackbody radiator. The rate of radiation Prad follows equation (7.10).
Prad = 4AT (7.10)
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where is the Stefan-Boltzmann constant that has value 5.67x10-8
J/s-m2-K
4.
Stefan-Boltzmann constant is a universal constant that applies to all bodies,
regardless of the nature of surface.
The rate of absorption Pabs of an object from its environment, which has
uniform temperature Tenv follows equation (7.11).
Pabs = 4
envAT (7.11)
Owing to the fact that the object radiates energy to the environment and absorbs
energy from the environment, the net rate of radiation Pnet follows equation
(7.12).
Pnet = 4
env
4 TTA (7.12)
Example 7.4
The supergiant star Betelgeuse in constellation Orion has a surface temperature
about 2,900K and emits a radiant power of approximately 4.0x1030
W. The
temperature is about half and the power is about 10,000 greater than the Sun.
Assuming that Betelgeuse and Sun have perfect emissivity and spherical shape,
calculate the radii of the supergiant and Sun.
Solution
From Stenfan-Boltzmann equation, Prad =4AT , the surface area A is A=
4
rad
T
P
.
The radius R of supergiant shall be R = 4
rad
T4
P
=
48
30
2900x10x67.5x4
10x0.4
=
2.82x1011
m, which is larger than the orbit of Mar, which is 2.28x1011
m.
The radius of Sun is R = 4
rad
T4
P
=
48
26
5800x10x67.5x4
10x0.4
= 7.04x108m.
Example 7.5
The temperature of an unused wood-burning stove in a room is 180C. A fire is
started inside the stove. Eventually the temperature of the stove surface reaches
a constant temperature of 1980C and the room warms to a constant temperature
of 290C. The stove has an emissivity of 0.9 and a surface area of 3.50m
2.
Determine the net radiant power generated by the stove when the stove (a) is
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unheated and has temperature equal to the room temperature (b) has temperature
of 1980C.
Solution
The power generated by unheated stove is Prad = 4AT = 5.67x10-
8x0.9x3.5x291
4 = 1,280.75W.
The power absorbed from the surrounding by the stove is Prad = 4AT =
5.67x10-8
x0.9x3.5x2914 = 1,280.75W. The net power generated by the stove is
zero.
The net power generated by the stove when it is heated is Pnet =
4
env
4 TTA = 448 30247110x67.5x5.3x9.0 = 7,304.1W.
7.6 Laws of Thermodynamics
There are three principal laws of thermodynamics, which are zeroth law – the
thermodynamic equilibrium law, first law - internal energy law, second law -
entropy law. Each law leads to the definition of thermodynamic properties,
which helps to understand and predicts the operation of a physical system.
There is another law, which is third law. This law states that it is impossible to
reach zero Kelvin temperature in finite number steps. This law shall not be
discussed.
7.6.1 Zeroth
Law of Thermodynamics
Zeroth
law of thermodynamic involves some simple definitions of
thermodynamic equilibrium and temperature. It is observed that some properties
of an object, like the pressure in a volume of gas, the length of a metal rod, or
the electrical conductivity of a wire, can change when the object is heated or
cooled. Some of these phenomena have been demonstrated in the earlier section.
Zeroth law states that if two systems are at the same time in thermal
equilibrium with a third system, they are in thermal equilibrium with each other.
If the two objects are initially at different temperatures into physical contact,
they will eventually achieve thermal equilibrium. During the process of
reaching thermal equilibrium, heat is transferred between the objects and there
is a change in the property of both objects such as their internal energy.
Thermodynamic equilibrium leads to the large-scale definition of temperature.
When two objects are in thermal equilibrium they have the same temperature.
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The details of the process of reaching thermal equilibrium are described in the
first and second laws of thermodynamics.
As an illustration in Fig. 7.9, object A and object B are in physical contact
and in thermal equilibrium. Object B is also in thermal equilibrium with object
C. There is initially no physical contact between object A and object C. If object
A and object C are brought into contact, it is observed that they are in thermal
equilibrium. This simple observation allows making of thermometers, in which
it can be calibrated to measure the change in a thermal property such as the
length of a column of mercury by putting the thermometer in thermal
equilibrium with a known physical system. If the thermometer is brought into
thermal equilibrium with any other system such as placing under tongue, the
temperature of the other system is known by noting the change in the thermal
property. Objects in thermodynamic equilibrium have the same temperature.
Figure 7.9: Illustration of Zeroth
law of thermodynamic
7.6.2 First law of thermodynamics
First law of thermodynamics relates the various forms of energy, kinetic and
potential, in a system to the work, which a system can perform and to the
transfer of heat. It is used extensively in the discussion of heat engines.
The first law of thermodynamics defines that change of internal energy U
is equal to the difference of the heat transfer Q into a system and the work W
done by the system.
U = U2 - U1 = Q – W (7.13)
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Heat Q is absorbed by the system uses to increase the internal energy (U2 – U1).
The internal energy is used to perform work done W, resulting decrease of
internal energy. This is the rationale how equation is formed.
Heat removed from a system would be assigned a negative sign in the
equation. Similarly work done on the system is assigned a negative sign. Figure
7.10 shows the interpretation of the law.
In chemistry textbook, first law is written as U = Q + W. It is the same
law as the thermodynamic expression of the conservation of energy principle. It
is just that W is defined as the work done on the system instead of work done by
the system.
In the context of physics, the common scenario is one adding heat to a
volume of gas and using the expansion of the gas to do work, as in case pushing
down of a piston in an internal combustion engine. In the context of chemical
reactions and process, it may be more common to deal with situations where
work is done on the system rather than by the system.
Figure 7.10: Interpretation of First law of thermodynamic
State 1 illustrates that heat is being absorbed in the system with increase of
volume. State 2 shows that there is decrease in volume, therefore work done is
negative. Work done W is also defined as 2
1
v
V
pdVW . It is clearly shown that if
V2 < V1, the work done is negative.
Example 7.6
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The temperature of three moles of a mono-atomic ideal gas is reduced from
temperature Ti = 540K to Tf = 350K by two different methods. In the first
method 5,500J of heat flows into the gas, while in the second method, 1,500J of
heat flows into it. In each case find the change of internal energy and work done
by the gas.
Solution
Since the gas is mono-atomic type, the internal energy of the gas is only the
translational kinetic energy, which follows equation nRT2
3.
The change in internal energy U of the gas from temperature 540K to 350K is
)TT(nR2
3if = )540350(314.8x3
2
3 = -7,105J. 8.314 is the gas constant R, which
has unit J/K/mol.
The change of internal energy for both cases is the same, which -7,105J.
The work done by first method yields W = Q - U = 5,500 J +7,105J = 12,605J,
while the work done by second method yields 1,500J + 7,105J = 8,605J.
7.6.2.1 Heat Capacities and First Law of Thermodynamics
Let’s understand the relationship of specific heat capacity and first law of
thermodynamic and also look at some special process cases that applying First
Law of Thermodynamics.
Heat capacity is defined as Q = CmmT. The heat capacity Q for a gas is Q
= CnnT. Instead of mass, it is now replaced by number of mole.
The internal energy U of an ideal mono-atomic gas is U = nRT2
3. The
change of internal energy U from temperature Ti to Tf shall be U =
)TT(nR2
3if .
For isobaric process – constant pressure case P, the work done is W = P(Vf
– Vi). However, for ideal gas PV = nRT. This shall mean that the work done is
also W = nR(Tf – Ti). For isochoric process – constant volume case, the work
done is equal to zero since there is no volume change.
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Using first law equation Q = U + W, for isobaric process is equal to Q =
)TT(nR2
3if + nR(Tf – Ti). This shall mean that the specific heat capacity is
R2
5. We shall use CP to denote specific heat for constant pressure.
Similarly the heat supplied for isochoric process, Q = U, i.e. Q =
)TT(nR2
3if . This implies that the specific heat Cn is equal to R
2
3, for n=1. We
shall use CV to denote specific heat for constant volume.
7.6.3 Second Law of Thermodynamics
Ice cream melts and a cold can of soda get warm-up when they are left in hot
day. They never get colder when left in hot day. This spontaneous flow of heat
is the focus of one of the most profound laws of science, which is the second
law of thermodynamics.
They are several ways that second law of thermodynamic can be defined.
Based on the above examples, the first statement of the second law of
thermodynamics states that heat flows spontaneously from a hot to a cold body.
An explanation for this form of the second law can be obtained from
Newton's laws and the microscopic description of how heat flow through
conduction occurred when fast atoms collide with slow atoms, transferring some
of their kinetic energies in the process. One might wonder why the fast atoms
don't collide with the cold atoms and subsequently speed up, thereby gaining
kinetic energy as the cold atoms lose kinetic energy. This would involve the
spontaneous transfer of heat from a cold object to a hot object, in which it is the
violation of the second law. From the laws of conservation of momentum and
energy, in a collision between two objects, the faster object slows down and the
slower object speeds up.
It is possible to make a cold object in a warm place colder like the
refrigerator but this involves the input of some external energy. As such, the
flow of heat is not spontaneous in this case.
The second form of the second law of thermodynamic states heat cannot be
completely converted into other forms of energy that places some practical
restrictions on the efficiency like internal combustion and steam powered
engines.
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The third statement of second law of thermodynamic states that there exists
an useful state variable called entropy, which is defined as the measure of
number of state variable for a system at a given time. The change in entropy S
is equal to the heat transfer Q divided by the temperature T.
S = Sf – Si = Q/T = f
iT
dQ
AvgT
Q (7.14)
The entropy of the system and the environment will remain a constant if the
process is reversible. If the initial and final states are denoted by Si and Sf
respectively then Sf = Si for reversible system. An example of a reversible
process would be ideally forcing a flow through a constricted pipe. Ideal means
no losses. As the flow moves through the constriction, the pressure, temperature,
and velocity would change, but these variables would return to their original
values downstream of the constriction. The state of the gas would return to its
original conditions and the change of entropy of the system would be zero.
The second law also states that if the physical process is irreversible, the
entropy of the system and the environment must increase. The final entropy
must be greater than the initial entropy. An example of an irreversible process is
a hot object put in contact with a cold object. Eventually, they both attain the
same equilibrium temperature. If objects are separated after attaining thermal
equilibrium, they do not naturally return to their original different temperature
states. The process of bringing them to the same temperature is irreversible.
7.6.3.1 Heat Transfer from Hot Object to Cold Object
Let’s look at how the second law describing why heat is transferred from the hot
object with temperature T1 to the cold object with temperature T2. If heat is
transferred from the hot object to the cold object, the amount of heat transferred
is Q and the final equilibrium temperature for both objects is Tf. The
temperature of the hot object changes as the heat is transferred away from the
object. The average temperature Th of the hot object during the process is the
average of T1 and Tf. Similarly, for the cold object, the final temperature of the
cold object is Tf. The average temperature Tc during the process is the average
of Tf and T2. The entropy change for the hot object will be (-Q/Th), with the
minus sign applied because the heat is transferred away from the object. For the
cold object, the entropy change is (Q/Tc), which is positive value because the
heat is transferred into the object. So according to equation (7.14), the total
entropy change for the whole system would be given by the equation.
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Sf - Si = -Q/Th + Q/Tc (7.15)
where Si and Sf being the initial and final values of the entropy.
Temperature Th is greater than temperature Tc, because T1 is greater than
T2. The term Q/Tc will always be greater than -Q/Th and therefore, Sf will be
greater than Si, as the second law predicts.
If heat is being transferred from the cold object to the hot object, then the
final equation would be Sf = Si + Q/Th -Q/Tc. The signs on the terms would be
changed because of the direction of the heat transfer. Th would still be greater
than Tc, and this would result in Sf being less than Si. The entropy of the system
would decrease, which would violate the second law of thermodynamics.
7.7 Thermodynamics Potentials
There are four thermodynamic potentials that are useful in the chemical
thermodynamics of reactions and non-cyclic processes. They are internal energy,
enthalpy, Helmholtz free energy, and Gibbs free energy.
Enthalpy H is defined as H = U + PV. It is analogue to first law of
thermodynamic Q = ∆U + P∆V. It is a useful quantity for tracking chemical
reactions. In an exothermic (heat release) reaction, energy is released to a
system. It has shown up in some measurable forms in terms of the state
variables. An increase in the enthalpy H = U + PV may be associated with an
increase in internal energy which can be measured by calorimeter, or with work
done by the system, or a combination of the two processes. If the process
changes the volume, as in a chemical reaction which produces gas, then work
must be done to produce the change in volume. For a constant pressure process
the work done to produce a volume change ∆V is P∆V.
Helmholtz free energy F states that F = U – TS, where T is the absolute
temperature and S is the final entropy. The system in an environment of
temperature T, energy can be obtained by spontaneous heat transfer to and fro
between environment and the system. Helmholtz free energy is a measure of the
amount of energy that put in to create a system once the spontaneous energy
transfer to the system from the environment is accounted.
Gibbs free energy G states that G = U – TS + PV, where P is the absolute
pressure and V is the final volume. As discussed in enthalpy, an additional
amount of work PV must be done if the system is created from a very small
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volume into large volume system. As discussed in Helmholtz free energy, an
environment at constant temperature T will contribute an amount TS to the
system, reducing the overall energy necessary for creating the system. This net
energy contribution for a system created in environment with temperature T
from an initial very small volume is Gibbs free energy.
The change in Gibbs free energy ∆G, in a reaction is a very useful
parameter. It can be treated as the maximum amount of work obtainable from a
reaction. For example, in the oxidation of glucose, the change in Gibbs free
energy is ∆G = 686kcal = 2,870kJ. This reaction is the main energy reaction in
living cells.
Example 7.7
Electrolysis of water into hydrogen and oxygen is a very good example for the
application of the above mentioned thermodynamic potentials. This process is
presumed is done at temperature 298K and one atmosphere pressure, and the
relevant values are taken from a table of thermodynamic properties shown
below.
The process must provide the energy for the dissociation and the energy to
expand the produced gases. Both of those are included in the change in enthalpy
which is shown in the table.
Quantity H2O H2 0.5 O2 Change
Enthalpy -285.83 kJ 0 0 ∆H = 285.83 kJ
Entropy 69.91J/K 130.68J/K 0.5 x 205.14J/K T∆S = 48.7 kJ
At temperature 298K and one atmosphere pressure, the system work is W =
P∆V = (101.3x103 Pa)(1.5 moles)(22.4x10
-3m
3/mol)(298K/273K) = 3,715.0J
Since the enthalpy H= U + PV, the change in internal energy U is then equal to
∆U = ∆H - P∆V = 285.83 kJ - 3.72 kJ = 282.1kJ.
This change in internal energy must be accompanied by the expansion of the
gases produced, so the change in enthalpy represents the necessary energy to
accomplish the electrolysis. However, it is not necessary to put in this whole
amount in the form of electrical energy. Since the entropy increases in the
process of dissociation, the amount T∆S can be provided from the environment
at temperature T. The amount must be supplied by the battery, which is actually
the change in the Gibbs free energy. i.e. ∆G = ∆H - T∆S = 285.83kJ - 48.7kJ =
237.1kJ.
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Since the electrolysis process results in an increase in entropy, the
environment contributes energy amounted to T∆S. Gibbs free energy tells the
amount of energy in other forms must be supplied to get the process to proceed.
7.8 Thermodynamic Processes
A thermodynamic process is the way that a system changes from one state of
thermal equilibrium to another state such as the gas may be in thermodynamic
equilibrium for specified temperature, pressure, and volume. Heat transfer and
work done will change this equilibrium state to another state.
Thermodynamic process can be divided into quasi-static (reversible)
processes and irreversible processes. If a thermodynamic system undergoes a
change from one state in thermal equilibrium to another state slowly enough
such that in any instant of time the system is in thermal equilibrium, it is said to
be reversible process. In reality this process does not exist. Any thermal process
which is not a quasi-static process is an irreversible process. In nature, all
thermodynamic processes are irreversible.
There are four special types of quasi-static processes that are useful in
equilibrium thermodynamics. The processes are characterized by the
thermodynamic parameter that being kept constant during process while other
parameters change. The types are: adiabatic process or isentropic process
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where no heat is transferred in or out of the system from or to the environment,
isothermal process where the temperature of the system is kept constant,
isobaric process where the pressure is kept constant, and isochoric process
where the volume is kept constant.
7.8.1 Adiabatic process
The adiabatic process is a process occurring so rapidly that there is no transfer
of heat between the system and its environment. From first law of
thermodynamics with Q = 0 shows that the change in internal energy is in the
form of work done. Therefore, equation (7.13) becomes U2 - U1 = – W.
Free expansion is an adiabatic process whereby there is no transfer of heat
between the system and its environment, no work done on and by the system.
Thus, from equation (7.13), heat transfer is Q = W = 0.
For adiabatic condition PV = constant K, where is the ratio of Cp/Cv heat
capacity at constant pressure and constant volume. This implies that P = K/V
and the work done shall be
2
1
v
V
dVV
KW =
1
)VV(K 1
i
1
f. The pressure-volume
diagram is shown in Fig.7.11.
Figure 7.11: The work done graph of an adiabatic process
Note that for an ideal monatomic gas, value is equal to 5/3.
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7.8.2 Isochoric Process
Isochoric process is a constant volume process that there is no change in volume.
Thus, the work done is zero. Therefore, equation (7.13) becomes U2 - U1 = Q,
which is the change of internal energy equals to CVnT. From the ideal heat
equation VP = nRT, this shall mean Q = )PP(VR
CV
if . The pressure-volume
diagram is shown in Fig.7.12.
Figure 7.12: The work done graph of an isochoric process
7.8.3 Cyclical process
Cyclical process is a process after certain exchange of heat and work done, the
system restores back to its initial state. Thus, there is no change in its internal
energy. This implies that equation (7.13) becomes Q = W.
7.8.4 Isothermal Process
The isothermal process is a process whereby the temperature is kept constant
that there is no change of internal energy. This implies that equation (7.13)
becomes Q = W. The work done is 2
1
v
V
pdVW . Substituting P = V
nRT into the
equation 2
1
v
V
dVV
nRTW = nRT
i
f
V
Vln . The pressure-volume diagram is shown in
Fig. 7.13.
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Figure 7.13: The work done graph of an isothermal process
Example 7.8
Two moles of the monatomic gas argon expand isothermally at temperature
298K from an initial volume of Vi = 0.025m3 to a final volume of Vf = 0.050m
3.
Assuming that argon is an ideal gas, find (a) the work done by the gas, (b) the
change in the internal energy of the gas, and (c) the heat supplied to the gas.
Solution
The work done by isothermal process is W = nRT
i
f
V
Vln = 2x8.31x298
025.0
050.0ln
= 3,432.9J.
The internal energy of U = nRT2
3. Since the temperature is constant, the change
of kinetic energy is zero.
Using first law equation, the heat supplied Q is 3,432.9J.
7.8.5 Isobaric process
Isobaric process is process where the pressure is kept constant. The work done
W is W = P∆V = P(Vf – Vi). T = PV/(nR), thus the change in temperature ∆T =
)VV(nR
Pif . The heat absorbed is Q = Cp )VV(
R
Pif . The PV diagram is shown
in Fig. 7.14.
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Figure 7.14: The work done graph of an isobaric process
7.9 Heat Engines
A heat engine is a device that uses heat to perform work. Essentially it has three
features. It has a hot reservoir that supplies heat. Part of the heat is used to
perform work by working substance of the engine like gasoline- air mixture in
the automobile engine. The remaining part of the heat is rejected at temperature
lower than the input temperature called cold reservoir. The schematic of a heat
engine is shown in Fig. 7.15. There is no heat transfer directly from heat
reservoir to cold reservoir.
For an engine to be efficient, it must produce a relatively large amount of
work from the input heat. The efficiency e of a heat engine is defined as the
ratio of the work done by the engine to the input heat QH. i.e. e = HQ
W. However,
QH is also equal to QH = QC + W, thus, the efficiency is
e = H
C
Q
Q1 (7.16)
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Figure 7.15: A schematic representation of a heat engine
Example 7.9
An automotive engine has an efficiency of 22.0% and produces 2,510J of work.
How much heat is rejected?
Solution
From equation (7.16), The heat rejected is QC = QH(1- e). But QH is equal to QC
+ W. Thus, QC = (QC + W)(1 - e). This implies that QC = W/e - W =
2,510(1/0.22 - 1). = 8,899.0J.
7.9.1 Carnot Engine
According to a French engineer Sadi Carnot, in order to get maximum
efficiency for an engine, the process within the engine must be reversible. This
shall mean that the process will return to its initial states with no wasteful
transfer of energy. Figure 7.16 shows the pressure-volume plot of the Carnot
engine.
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Figure 7.16: Pressure-volume diagram of a Carnot engine
During the process step a to b, which is an isothermal expansion, heat QH is
absorbed by the working substance from the hot reservoir. Since the process is
an isothermal expansion process, the work done by the system is also equal to
heat absorbed. During the process step c to d, the working substance is releasing
heat QC to the cold reservoir. Since this is an isothermal compression, work is
done by the environment on the system. From concept of heat engine, there is
no heat transfer between hot and cold reservoir directly. Thus, the process step
bc and de must be the adiabatic process. The work done by the Carnot process
shall be the orange color area of the graph shown in Fig. 7.16.
According to first law of thermodynamic, U = Q –W. Since the process is
reversible, therefore the change of internal energy U is zero. Also for
differential change dQ = dW, the work done W is W = QH – QC.
From process step a to b there is a positive change of entropy SH = QH/TH
and during the process step c to d, there is a negative change of entropy SL =
QL/TL. There is no change of entropy for process step bc and da because there is
not heat absorbed or released. The change of entropy shall be
S = QH/TH - QL/TL (7.17)
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Since Carnot engine is a reversible engine, the change of entropy shall be zero.
This implies that QH/TH = QL/TL. From the equation TH>TC, this shall mean
QH>QC. This analysis confirms the second statement of second law of
thermodynamic, which is heat cannot be completely converted into other form
of energy.
The efficiency of the Carnot engine shall be e = W/QH = (QH – QC)/QH = 1-
TL/TH.
7.9.2 Refrigeration
Refrigeration is a device that uses work to transfer energy from low temperature
reservoir to high temperature reservoir. Figure 7.17 shows the schematic of a
refrigerator.
Figure 7.17: The schematic of a refrigerator
A measure of the efficiency of a refrigerator is the coefficient of performance K,
which is defined as the ratio of heat extracted from the cold reservoir to the
work done. For an ideal Carnot refrigerator the coefficient of performance K
=LH
L
Q
. The change of entropy for ideal refrigerator shall be S =
H
H
L
L
T
Q
T
Q .
Notice that TH >TL, implying the S is a negative value, which is not allow for
second law of thermodynamic. Therefore, there is not a perfect refrigerator.
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Tutorials
7.1. An aluminum pole is 33m high. What is length of this pole if there is an
increase of temperature of 15oC? The linear expansion of aluminum is
22.2x10-6
/oC.
7.2. Prove that the area expansion coefficient is approximately equal to two
times the linear expansion coefficient.
Let the length and width of the area be a and b respectively. The area A is
ab.
The figure illustration is shown below.
7.3. How much heat must be absorbed by ice of mass m = 720g at temperature
-100C to take it to liquid state at 100
0C? And how much additional heat
required to convert 50% of 1000C hot water into hot steam?
7.4. A certain diet doctor encourages people to diet by drinking ice water. His
theory is that the body must burn off enough fat to raise the temperature
of water from 0.00C to the body temperature of 37.0
0C. How many liters
of ice water would have to be consumed to burn-off 454g of fat, assuming
that this much fat required 3,500kCal be transferred to ice water? Why
this is not advisable to follow this diet.
7.5. An aluminum sphere at temperature 100oC is placed on a copper ring at
temperature 0oC. At thermal equilibrium the diameter of the aluminum
sphere is same as the diameter of the copper ring. Find the thermal
equilibrium temperature with the assumption no heat is lost to the
environment. Given that the linear expansion coefficient of copper and
aluminum is 16.6x10-6
/oC and 22.2x10
-6/oC respectively.
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7.6. What is the rate of energy loss in watt per square meter through the glass
window 3.0mm thick if the outside temperature is -50C and the inside
temperature is 250C? Given that the thermal conductivity of glass window
is 1.0W/m-K. Thermal conductivity of air is 0.026W/m-K.
7.7. Referring to question 7.6, if a storm window having the same thickness of
the glass is installed parallel to the first window with an air gap of 7.5cm
between the two windows, what is rate of energy loss due to conduction
only?
7.8. A tank of water has been outdoor in cold weather and a slab of ice 4.0cm
thick has formed on its surface. The air above the ice is –100C. Calculate
the rate (centimeter per sec) of ice formation. Take the thermal
conductivity of the ice to be 0.0040cal/s-cm-K and density 0.92g/cm3.
Assume the energy is transferred through the wall or bottom of the tank is
negligible.
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7.9. A wall consists of four layers is shown in figure below. The thermal
conductivities are k1 = 0.060W/m-k, k3 = 0.040W/m-k, k4 = 0.12W/m-k.
The thickness of layers are L1 = 1.5cm, L3 = 2.8cm, and L4 = 3.5cm. Find
temperature T1 and T2 when the energy of transfer is in steady state.
7.10. Find the change of entropy when a 2.3kg block of ice melts slowly
(reversible) at 273K.
7.11. 1,200J of heat flows spontaneously through a copper rod from a hot
reservoir at 650K to a cold reservoir at 350K. Determine the amount the
change in entropy for this irreversible process.
7.12. A Carnot engine operates between temperature TH = 850 K and TL =
300K. The engine performs 1,200J of work for each cycle that takes 0.2 s.
(a). Find the efficiency of the engine.
(b). How much heat QH is extracted from high temperature reservoir for
each cycle? And QL delivered to cold reservoir?
(c). What is the entropy change of the working substance for its energy
transfer to the high temperature reservoir? Transfer to the low
temperature reservoir.
7.13. If an inventor claims that he has invented an engine that has an efficiency
of 75% when operated between boiling point and freezing point of water.
Is it possible?