Contents

Page

- i -

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

Page

- ii -

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

QQ

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.

07 Thermodynamics

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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?