1. Introduction and Definition of Terms

2014-06-20 1

2014-06-20

Syllabus

재료열역학 (Thermodynamics of Materials)

담 당 교 수 : 유 상 임 (연구실: 131동 407호)

전화: 880-5720, E-mail: siyoo@snu.ac.kr, 홈페이지: http://emdl.snu.ac.kr/

강의시간 및 강의실: 월, 수 14:00 – 15:15, 33동 327호

강의조교 : 유재형 (연구실:131동 414호) 전화: 880-7443, E-mail: jh31330@snu.ac.kr

 주 교 재

▷ Introduction to the Thermodynamics of Materials, Talyors & Francis(2008), 5^th ed., David R. Gaskell

 참 고 서 적

▷ An introduction to thermal physics, Addison Wesley Longman(2000), 1^st ed., D. V. Schroeder

 성 적 평 가

▷ Quiz #1 10%, Midterm Exam 25%, Quiz #2 10%, Final Exam 35%, Home Problems 20%

* 4회 이상 무단 결석 시 F 학점 처리

Syllabus

- Chap. 1 Introduction and Definition of Terms (Lect. #1)

- Chap. 2 The First Law of Thermodynamics (Lect. #2~ Lect. #4) - Chap. 3 The Second Law of Thermodynamics (Lect. #4~ Lect. #6) - Quiz #1 (2014.03.24)

- Chap. 4 The Statistical Interpretation of Entropy (Lect. #7~ Lect. #9) - Chap. 5 Auxiliary Function (Lect. #10~ Lect. #11)

- Chap. 6 Heat Capacity, Enthalpy, Entropy, and the Third Law of Thermodynamics (Lect. #12~ Lect. #14)

- Midterm Exam (2014.04.16)

- Chap. 7 Phase Equilibrium in a One-Component System (Lect. #15) - Chap. 8 The Behavior of Gases (Lect. #16)

- Chap. 9 The Behavior of Solutions (Lect. #17~ Lect. #18)

- Chap. 10 Gibbs Free Energy Composition and Phase Diagrams of Binary Systems (Lect. #19~ Lect. #20)

- Quiz #2 (2014.05.14)

- Chap. 11 Reactions Involving Gases (Lect. #21~ Lect. #22)

- Chap. 12 Reactions Involving Pure Condensed Phases and a Gaseous Phases (Lect. #23~ Lect. #24)

- Chap. 13 Reaction Equilibria in Systems Containing Components in Condensed Solution (Lect. #25~ Lect. #26)

- Final Exam (2014.06.16)

 Lecture Plan

1. Introduction and Definition of Terms

Lecture Contents

1. Introduction and Definition of Terms 1.1 Introduction

1.2 The Concept of State 1.3 Simple Equilibrium

1.4 The Equation of State of an Ideal Gas 1.5 The Units of Energy and Work

1.6 Extensive and Intensive Properties

1.7 Phase Diagrams and Thermodynamics Components

Thermodynamics

- Thermodynamics is concerned with the behavior of matter and the transformation between different forms of energy on a macroscopic scale, where matter is anything that occupies space. The matter which is the subject of a thermodynamic analysis is called a system.

- In materials science and engineering, the systems to which thermodynamic principles are applied are usually chemical reaction systems.

- The aim of applied thermodynamics is thus the establishment of the relationships which exist between the equilibrium state of existence of a given system and the influences which are brought to bear on the system from environment or surroundings, generally determined as P(pressure) exerted on the system and T(temperature) of the system.

1.1. Introduction

1.2. The Concept of State

( , ) V = V P T

2 1

V V V

∆ = −

2 1

1 2

(

) (

) V V V

V V V V

∆ = −

= − + −

Figure 1.1

The equilibrium states of existence of a fixed quantity of gas in P-V-T space.

Consider the volume V of a fixed quantity of a pure gas as a property, the value of which is dependent on the values of P and T.

The mathematical relationship of V to P and T for a system is called an equation of state for that system.

Consider a process which moves the gas from state 1 to state 2. (Figure 1.1)

Consider the path 1 → a → 2, the change in volume is

(1.1)

1.2. The Concept of State

2

1

1

(

)

T P

V V V dT

T

 ∂ 

− = ∫   ∂  

Where 1 → a occurs at constant pressure P₁ and a → 2 occurs at the constant temperature T₂;

and

2

1

2

(

)

P T

V V V dP

P

 ∂ 

− = ∫   ∂  

Thus

2 2

1 1

1 2

T P

T P

P T

V V

V dT dP

T P

∂ ∂

   

∆ = ∫   ∂   + ∫   ∂   ^(1.2)

1.2. The Concept of State

Similarly for the path 1 → b → 2,

2

1

1

(

)

P T

V V V dP

P

 ∂ 

− = ∫   ∂  

2

1 2

(

)

T P

V V V dT

T

 ∂ 

− = ∫   ∂  

and

And, hence, again

2 2

1 1

1 2

P T

P T

T P

V V

V dP dT

T P

∂ ∂

   

∆ = ∫   ∂   + ∫   ∂   ^(1.3)

1.2. The Concept of State

Eqs. (1.2) and (1.3) are identical and are the physical representations of what is obtained when the complete differential of Eq. (1.1) i.e.,

T P

V V

dV dP dT

P T

∂ ∂

   

=   ∂   +   ∂  

is integrated between the limits P₂, T₂ and P₁, T₁.

The change in volume caused by moving the state of the gas from state 1 to state 2 depends only on the volume at state 1 and the volume at state 2 and is independent of the path taken by the gas between the states 1 and 2. This is because the volume of the gas is a state function.

(1.4)

cf) isothermal, diathermic, adiabatic, isobaric, isochoric

1.3 Simple Equilibrium

Figure 1.2

A quantity of gas contained in a cylinder by a piston.

A particularly simple system in Figure 1.2.

This system is at rest, i.e., is at equilibrium when 1. The pressure exerted by the gas on the

piston equals the pressure exerted by the piston on the gas.

2. The temperature of the gas is the same as the temperature of the surroundings.

1.4. The Equation of State of an Ideal Gas

The pressure-volume relationship of a gas at constant temperature was determined experimentally in 1660 by Robert Boyle, who found that, at constant T.

P 1

∝ V

which is known as Boyle’s law.

Similarly, the volume-temperature relationship of a gas at constant pressure was first determined experimentally by Jacques-Alexandre- Cesar Charles in 1787. This relationship, which is known as Charles’

law.

V ∝ T

1.4. The Equation of State of an Ideal Gas

Figure 1.3

(a) The variations, with pressure, of the volume of 1mole of ideal gas at 300 and 1000 K.

1.4. The Equation of State of an Ideal Gas

In 1802 Joshep-Luis Gay-Lussac observed that the thermal coefficient of what were called “permanent gases” was a constant. The coefficient of thermal expansion, α, is defined as the fractional increase, with temperature at constant pressure, of the volume of a gas at 0℃; that is

0

1

P

V V T α = ^   ^∂ ∂ ^  

where V₀ is the volume of the gas at 0℃.

It is convenient to invent a hypothetical gas which obeys Boyle’s and Charles’ laws exactly at all temperatures and pressures. This hypothetical gas is called the ideal gas, and it has a value of α of 1/273.15.

The existence of a finite coefficient of thermal expansion sets a limit on the thermal contraction of the ideal gas; that is, α as equals 1/273.15 then the fractional decrease in the volume of the gas, per degree decrease in temperature, is 1/273.15 of the volume at 0℃. Thus, at -273.15 ℃ the volume of the gas is zero, and hence the limit of temperature decrease, -273.15 ℃, is the absolute zero of temperature. This defines as absolute scale of temperature, called the ideal gas temperature scale.

1.4. The Equation of State of an Ideal Gas

T(degrees absolute) = T(degrees celsius) + 273.15

1.4. The Equation of State of an Ideal Gas

Combination of Boyle’s law

0

( ,

) ( , ) PV T P = PV T P

and Charles’ law

0 0 0

0

( , ) ( , ) V P T V P T

T = T

where

P₀ = standard pressure (1 atm)

T₀ = standard temperature (237.15 degrees absolute) V(T,P) = volume at temperature T and pressure P

gives

0 0 0

constant PV PV

T = T = (1.5)

From Avogadro’s hypothesis the volume per gram-mole of all ideal gases at 0℃ and 1 atm pressure (termed standard temperature and pressure - STP) is 22.414 liters. Thus the constant in Eq. (1.5) has the value

1.4. The Equation of State of an Ideal Gas

0 0 0

1 atm 22.414 liters 273.15 degree mole PV

T

= ×

⋅

= 0.082057 liter∙ atm/degree ∙ mole

This constant is termed R, the gas constant, and being applicable to all gases, it is a universal constant. Eq. (1.5) can thus be written as

PV = RT

which is thus the equation of state for 1 mole of ideal gas. Eq. (1.6) is called the ideal gas law.

(1.6)

1.5. The Units of Energy and Work

Work is done when a force moves through a distance, and work and energy have the dimensions force × distance.

Pressure is force per unit area, and hence work and energy can have the dimensions pressure × area × distance, or pressure × volume.

The unit of energy in S.I. is the joule, which is the work done when a force of 1 newton moves a distance of 1 meter.

1 atm= 101,325 newtons/meter² Multiplying both sides by liters (10^-3 m³)gives

1 liter ∙ atm = 101.325 newtons/meters = 101.325 joules

and thus

R = 0.082057 liter ∙ atm/degree ∙ mole = 8.3144 joules/degree ∙ mole

1.6. Extensive and Intensive Properties

Properties (or state variables) are either extensive or intensive.

Extensive properties – depend on the size of the system. (volume)

Intensive properties – independent of the size of the system. (temperature, pressure)

The values of extensive properties expressed per unit volume or unit mass of the system, have the characteristics of intensive variables; e.g., the volume per unit mass (specific volume) and the volume per mole (the molar volume).

For a system of n moles of an ideal gas, the equation of state is

where V ’ is the volume of the system.

where V, the molar volume of the gas, equals V ’/n.

'

PV = nRT

PV = RT

1.7. Phase Diagrams and Thermodynamic Components

Figure 1.4

Schematic representation of part of the phase diagram for H₂O.

Of the several ways to graphically represent the equilibrium states of existence of a system, the phase diagram is the most popular and convenient.

The complexity of a phase diagram is determined primarily by the number of components which occurs in the system, where components are chemical species of fixed composition.

e.g., one-component (unary) systems, two-component (binary) system,

three-component (ternary) system,

four-component (quaternary) system, etc.

Phase : a finite volume in the physical system within which the properties are uniformly constant.

Homogeneous, Heterogeneous

1.7. Phase Diagrams and Thermodynamics Components

Figure 1.5

The phase diagram for the system Al₂O₃-Cr₂O₃.

2. The 1st Law of Thermodynamics

Lecture Contents

2. The 1st Law of Thermodynamics 2.1 Introduction

2.2 The Relationship between Heat q and Work w

2.3 Internal Energy U and the First Law of Thermodynamics 2.4 Constant-Volume Processes

2.5 Constant-Pressure Processes and the Enthalpy H 2.6 Heat Capacity C_v & C_p

2.7 Reversible Adiabatically Processes

2.8 Reversible Isothermal Pressure or Volume Changes of an Ideal Gas

2.1 Introduction

Kinetic energy is conserved in a frictionless system of interacting rigid elastic bodies. A collision between two of these bodies results in a transfer of kinetic energy from one to other, the work done by the one equals the work done on the other. If the kinetic system is in the influence of a gravitational field, then the sum of the kinetic and potential energies of the bodies is constant. Kinetic energy may be converted to potential energy and vice versa, but the sum of the two remains constant.

If, however, friction occurs in the system, then the total dynamic energy of the system decreases and heat is produced with interaction among the bodies.

It is thus reasonable to expect that a relationship exists between the dynamic energy dissipated and the heat produced as a result of the effect of friction.

The development of thermodynamics from its early beginning to its present state was achieved as the result of the invention of convenient thermodynamic functions of state. Among many thermodynamic functions of states, U and H are introduced in this chapter.

2.2 The Relationship between Heat and Work

The relation between heat and work was first suggested in 1789 by Count Rumford, who, during the boring of cannon at the Munich Arsenal, noticed that the heat produced during the boring was roughly proportional to the work performed during the boring.

“the caloric theory of heat”: the temperature of a substance was considered to be determined by the quantity of caloric gas which it contained.

James Joule conducted experiments in which work was performed in a certain quantity of adiabatically contained water and measured the resultant increase in the temperature of the water, He observed that a direct proportionality existed between the work done and the resultant increase in temperature and the same proportionality existed no matter what means were employed in the work production.

Joule’s experiment resulted in the statement that “the change of a body inside an adiabatic enclosure from a given initial state to a given final state involves the same amount of work by whatever means the process is carried out.”

2.3 Internal Energy and the First Law of Thermodynamics

When a body of mass m is lifted in a gravitational field from height h₁ to height h₂, the work w done on the body is given by

w = force × distance = mg × (h₂ – h₁) = mgh₂ – mgh₁

= potential energy at position h₂ minus potential energy at position h₁.

f ma m dv

= = dt

The work done on the body is obtained by integrating

dw fdl

dv dl

m dl m dv mv dv

dt dt

=

= = =

2

1

1

w mv dv mv mv

∴ = ∫ = −

According to Newton’s Law,

(The work done on the body is dependent only on its final and initial positions and is independent of the path taken by the body between the two states.)

(independent of the path)

In the case of work being done on an adiabatically contained body of constant potential and kinetic energy, the pertinent function which describes the state of the body, or the change in the state of body, is the internal energy U. Thus the work done on, or by, an adiabatically contained body equals the change in the internal energy of the body.

2.3 Internal Energy and the First Law of Thermodynamics

(

) w = − U − U

The heat q equals to the change of internal energy:

(

) q = U − U

If work w is done on the body, then U_B > U_A and if the body itself performs work, then U_B < U_A.

When heat flows into the body (an endothermic process), q is a positive and U_B > U_A, whereas if heat flows out of the body (an exothermic process), U_B < U_A and q is a negative.

2.3 Internal Energy and the First Law of Thermodynamics

The change in the internal energy of a body which simultaneously performs work and absorbs heat is:

A → B

B A

U U U q w

∆ = − = −

This is a statement of the First Law of Thermodynamics.

For an infinitesimal change of state, Eq. (2.1) can be written as a differential.

(2.1)

dU = δ q − δ w (2.2)

While U is a state function, the heat and work effects depend on the path taken between the two states, implying that δq and δw cannot be evaluated without a knowledge of the path.

d : a differential element of a state function or state property

δ : a differential element of some quantity which is not a state function

2.3 Internal Energy and the First Law of Thermodynamics

The First Law of Thermodynamics

:

( , )

T V

U

U U V T

U U

dU dV dT

V T

=

∂ ∂

   

=   ∂   +   ∂   state function

Figure 2.1

Three process paths taken by a fixed quality of gas in moving from the state 1 to the state 2.

In the case of a cyclic process 1 → 2 → 1 in Fig. 2.1, the change in U is zero; i.e.,

2 1

2 1 1 2

1 2

( ) ( ) 0

U dU dU U U U U

∆ = ∫ + ∫ = − + − =

The vanishing of a cyclic integral is a property of a state function.

If V and T are chosen as the independent variables,

2.4 Constant-Volume Processes

If the volume of a system is maintained constant during a process (isochoric or isometric processes), then the system does no work (∫ 𝑃𝑃𝑃𝑃𝑃𝑃 = 0), and from the First Law,

dU = δ q + δ w = δ q

U q

∆ =

Where the subscript v indicate constant volume. Integration Eq. (2.3) gives

(2.3)

For such a process, the increase or decrease in the internal energy of the system equals, respectively, the heat absorbed or rejected by the system.

2.5 Constant-Pressure Processes and the Enthalpy H

( )

2 2

2 1

1 1

w = ∫ PdV = P ∫ dV = P V − V

( )

2 1 p 2 1

U − U = q − P V − V

( U

+ PV

) ( − U

+ PV

) = q

Where the subscript p indicates constant pressure and, as the expression (U+PV) contains only state functions, the expression itself is a state function. This is termed the enthalpy H

H = + U PV

Hence, for a constant-pressure process,

2 1 p

H − H = ∆ = H q

If the pressure is maintained constant during a process (isobaric processes), then the work done by the system is given as

and the First Law gives

The enthalpy change during a constant-pressure process is equal to the heat admitted to or withdrawn from the system during the process.

(2.5)

(2.4)

2.6 Heat Capacity

C q

T C q

dT δ

= ∆

=

v v v

V V

p p p

P P

q dU

C dU C dT dU q

dT dT

q dH

C dH C dT dH q

dT dT

δ δ

δ δ

   

=     =     = =

   

=     =     = =



 or

or

p p p

v v v

C nc c

C nc c

=

=

: the molar heat capacity at constant pressure : the molar heat capacity at constant volume

The heat capacity, C, of a system is the ratio of the heat added to or withdrawn from the system to the resultant change in the temperature of the system.

(2.6)

(2.7)

Thus a heat capacity at constant volume, C_v, and a heat capacity at constant pressure, C_p, are defined as

If the process is carried out at constant pressure, then, in addition to raising the temperature by the required amount, the heat added is required to provide the work necessary to expand the system at the constant pressure. The work of expansion against the constant pressure per degree of temperature increase is calculated as

or

P

PdV V

dT P T

 ∂ 

 ∂ 

 

p v

P

c c P V

T

 ∂ 

∴ − =   ∂  

v

V

c U

T

 ∂ 

=   ∂  

p

P P P

H U V

c P

T T T

∂ ∂ ∂

     

=   ∂   =   ∂   +   ∂  

p v

P P V

U V U

c c P

T T T

∂ ∂ ∂

     

∴ − =   ∂   +   ∂   −   ∂  

The difference between c_p and c_v can be calculated as follows:

Hence it might be expected that

2.6 Heat Capacity

T V

P T P V

U U

dU dV dT

V T

U U V U

T V T T

∂ ∂

   

=   ∂   +   ∂  

∂ ∂ ∂ ∂

       

∴   ∂   =   ∂     ∂   +   ∂  

p v

T P V

U V U

c c

V T T

∂ ∂ ∂

     

∴ − =   ∂     ∂   +   ∂  

V U

P T T

∂ ∂

   

+   ∂   −   ∂  

P T

V U

T P V

 

∂ ∂

   

=   ∂     +   ∂    

The difference between two expressions for c_p – c_v is

V U

T V

∂ ∂

   

 ∂   ∂ 

   

but

Hence,

(2.8)

2.6 Heat Capacity

And therefore

Joule’s Experiment

Figure

A schematic diagram of the apparatus used by Joule in an attempt to measure the change in internal energy when a gas expands isothermally. The heat absorbed by the gas is proportional to the change in temperature of the bath.

w = 0 (expansion into the vacuum) q = 0 (The bath temp. wasn’t change)

0 0

T

U U V

∴∆ =

 ∂ 

∴   ∂   =

High pressure

gas

Vacuum Thermometer

2.6 Heat Capacity

Joule experiment – to evaluate the term

0

∆ = U

p v

P

c c P V

T

 ∂ 

∴ − =   ∂  

T

U V

 ∂ 

 ∂ 

 

As the system was adiabatically contained and no work was performed, then from the First Law,

0

T V

U U

dU dV dT

V T

∂ ∂

   

=   ∂   +   ∂   =

Hence,

Thus as dT = 0 and dV is not 0, then the term (𝜕𝜕𝑈𝑈/𝜕𝜕𝑃𝑃)_𝑇𝑇 must be zero.

However, for real gases

0

T

U V

 ∂  ≠

 ∂ 

 

2.6. Heat Capacity

Nevertheless, if 0

T

U V

 ∂  =

 ∂ 

 

which is the case for an ideal gas, then, from Eq. (2.8),

p v

P

c c P V

T

 ∂ 

− =   ∂  

and as, for one mole of ideal gas, PV = RT, then

p v

c c R P R

− = × = P

In real gases, is very small compared with external pressure contributions. ^T

U V

 ∂ 

 ∂ 

 

In liquids and solids, is very large in comparison with the work done against the internal pressure. ^T

U V

 ∂ 

 ∂ 

 

2.6 Heat Capacity

2.7 Reversible Adiabatic Processes

In an adiabatic process, q=0

For a system comprising one mole of an ideal gas

dU δ w

∴ = −

dU c dT

w PdV δ

=

=

v

c dT PdV RTdV

∴ = − = − V

Integrating between states 1 and 2 gives

2 1

1 2

2 1

1 2

2 1

1 2

ln ln

v

v

v

c R

R c

T V

c R

T V

T V

T V

T V

T V

   

  =  

   

   

  =  

   

   

   = 

   

For a reversible adiabatic process,

For an ideal gas,

,

1

p v

P v v

V R c R

c c P P R

T P c c

 ∂ 

− =   ∂   = × = ∴ − =

Let ^p

, then 1

v v

c R

c = γ c = − γ

1

2 1

1 2

T V

T V

γ−

   

   = 

   

1

2 2 2 1

1 1 1 2

2 1

1 2

T P V V

T PV V

P V

P V

γ

γ

 

= =  

 

 

=  

 

2 2 1 1

P V

PV

PV

∴ = = = constant

2.7. Reversible Adiabatic Processes

hence

From the ideal gas law,

(2.9)

2.8. Reversible Isothermal Pressure or Volume Changes of an Ideal Gas

dU = δ q − δ w

w q PdV RTdV δ δ V

∴ = = =

Integrating between state 1 and 2 gives

2 1

1 2

ln V ln P

w q RT RT

V P

   

= =   =  

   

From the First Law

and as dT = 0 (isothermal process), then dU = 0.

per mole of ideal gas

(2.10)

For an ideal gas, an isothermal process is one of constant internal energy during which the work done by the system equals the heat absorbed by the system.

2.8. Reversible Isothermal Pressure or Volume Changes of an Ideal Gas

Figure 2.2

Comparison of the process path taken by a reversible isothermal expansion of an ideal gas with the process path taken by a reversible adiabatic expansion of an ideal gas between an initial pressure of 20 atm and a final pressure of 4 atm.

For a given decrease in pressure, the work done by the reversible isothermal process (which is equal to the area under the curve) exceeds that done by the reversible adiabatic process.

For the reversible isothermal path, Eq. (2.10) gives the work done by the gas as

w = 13.38 kJ

For the reversible adiabatic path, the work done by the gas is

w = 5924 J

2.9 Summary

1. Internal energy, U

- U establishes the relationship between the heat q and the work w on or by a system.

- U is a state function. The difference between the values of U is not dependent on the path but on the states.

- The 1^st Law of Thermodynamics: ∆U = q – w or dU = δq - δw

2. Integral of δq and δw for different process paths

- Const. V (isochoric, ): w = ∫ δ𝑤𝑤 = ∫ 𝑃𝑃𝑃𝑃𝑃𝑃 = 0, ∆U = q^v, c_v = (δq/dT)V = (∂U/ ∂T)V : measurable ∆U = ∫ c₁^{2 v}𝑃𝑃𝑇𝑇: ∆U can be determined. (U cannot be determined.)

- Const. P (isobaric): w = ∫ δ𝑤𝑤 = P∫ 𝑃𝑃𝑃𝑃 = 𝑃𝑃∆𝑃𝑃, c^p = (δq/dT)_P = (∂H/∂T)_P : measurable since H = U + PV, a state function, ∆H = ∆U + P∆V = (q_p - P∆V) + P∆V = q_p

∆H = ∫ c₁² p𝑃𝑃𝑇𝑇 ∶ ∆H can be determined. (H cannot be determined.), cp - c_v = R for an ideal gas - Const. T (isothermal): For an ideal gas. w = q = RT ln(V₂/V₁) = RT ln(P₁/P₂)

- Adiabatic: q = 0, For an ideal gas, PV^γ = const. where γ = cp/c_v for a reversible adiabatic

3. The 2nd Law of Thermodynamics Lecture Contents

3. The 2nd Law of Thermodynamics 3.1 Introduction

3.2 Spontaneous or Natural Processes

3.3 Entropy and the Quantification of Irreversibility 3.4 Reversible Processes

3.5 An Illustration of Irreversible and Reversible Processes 3.6 Entropy and Reversible Heat

3.7 The Reversible Isothermal Compression of an Ideal Gas 3.8 The Reversible Adiabatic Expansion of an Ideal Gas

3.9 Summary Statements

3.1. Introduction

When a system undergoes a change of state, the consequent change in the internal energy of the system, which is dependent only on the initial and final states, is equal to the algebraic sum of the heat and work effects.

What magnitudes may the heat and work effects have, and what criteria govern their magnitudes?

Two obvious cases: w = 0 or q = 0, in which cases, respectively, q = ∆U and w =- _∆U.

But, if q ≠ 0 and w ≠ 0, is there a definite amount work which the system can do during its change of state?

The answers to these questions require an examination of the nature of processes.

Two classes: reversible or irreversible

B A

U U U q w

∆ = − = −

3.1. Introduction

S (entropy) : a state function

S will be introduced from two different starting points

(i) as a result of a need for quantification of the degree of irreversibility of a process (Sec. 3.2-3.8)

(ii) as a result of an examination of the properties of reversibly operated heat engines, there naturally develops a quantity which has all the properties of a state function. (Sec. 3.10-3.14)

3.2. Spontaneous or Natural Processes

Left to itself, a system will do one of two things:

1) It will remain in the state in which it happens to be or

2) it will move, of its own accord, to some other state.

That is,

1) If the system is initially in equilibrium with its surroundings, then it will remain in this, its equilibrium, state.

2) If the system is not the equilibrium state, the system will spontaneously move toward its equilibrium state.

A process which involves the spontaneous movement of a system from a non equilibrium state to an equilibrium state is called a natural or spontaneous process.

3.2. Spontaneous or Natural Processes

As natural or spontaneous process cannot be reversed without the application of an external agency, such a process is said to be irreversible.

ex) The mixing of gases, the flow of heat down a temperature gradient

In both examples, the simplicity of the system, along with common experience, allows the equilibrium states to be predicted without any knowledge of the criteria for equilibrium.

In complicate systems, the criteria governing equilibrium state cannot be predicted from common experience, and the criteria governing equilibrium must be established before calculation of the equilibrium state can be made.

In the initial nonequilibrium state of an isolated system, some of the energy of the system is available for doing useful work, and when the equilibrium state is reached, none of the energy of the system is available for doing useful work. Thus, as a result of the spontaneous process, the system has become degraded.

3.3. Entropy and the Quantification of Irreversibility

Two distinct types of spontaneous processes.

(1) The conversion of work to heat. (i.e., the degradation of mechanical energy to thermal energy)

(2) The flow of heat down a temperature gradient.

If it is considered that an irreversible process is one in which the energy of the system undergoing the process is degraded, then there is a possibility that the extent of degradation can differ from one process to another.

Thus, a quantitative measure of the extent of degradation, or degree of irreversibility can exist.

3.3. Entropy and the Quantification of Irreversibility

Figure 3.1

A weight-pulley-heat reservoir arrangement in which the work done by the falling weight is degraded to heat, which appears in the heat reservoir.

The system is at equilibrium when an upward force acting on the weight exactly balances the downward force, W, of the weight.

1) If the upward force is removed

→ the equilibrium is upset

→ weight falls

→ performs the work

→ heat enters the heat reservoir.

2) If the upward force is replaced, the equilibrium is retained.

Consider the weight-heat reservoir system in Fig. 3.1

3.3. Entropy and the Quantification of Irreversibility

The following three processes can be considered.

(proposed by Lewis & Randall)

1. The heat reservoir in the system is at the temperature T₂. The weight is allowed to fall, performing work, w, and the heat produced, q, enters the heat reservoir.

2. The heat reservoir at the temperature T₂ is placed in thermal contact with a heat reservoir at a lower temperature T₁, and the same heat q is allowed to flow from the reservoir at T₂ to the reservoir at T₁.

3. The heat reservoir in the system is at the temperature T₁. The weight is allowed to fall, performing work, w, and the heat produced, q enters the reservoir.

Each of these process is spontaneous and hence irreversible, and degradation occurs in each of them.

3.3. Entropy and the Quantification of Irreversibility

Since process (3) = process (1) + process (2), the degradation occurring in process (3) must be greater than the degradation occurring in each of the processes (1) and (2). Thus, process (3) is more irreversible than either process (1) or process (2).

Examination of the three processes indicates that both the amount of heat produced, q, and the temperatures between which this heat flows are important in defining a quantitative scale of irreversibility.

In the case of comparison between process (1) and process (3),

the quantity q/T₂ is smaller than the quantity q/T_1, which is in agreement with the conclusion that process (1) is less irreversible than process (3)

Thus, the quantity q/T is taken as being a measure of the degree of irreversibility of the process.

The value q/T is called the increase in entropy, S, occurring as a result of the process.

3.3. Entropy and the Quantification of Irreversibility

When the weight-heat reservoir system undergoes a spontaneous process which causes the absorption of heat q at the constant temperature T, the entropy produced by the system, ∆S, is given by

The increase in entropy is a measure of the degree of irreversibility of the process.

S q

∆ = T (3.1)

3.4. Reversible Processes

As the degree of irreversibility of a process is variable, it should be possible for the process to be conducted in such a manner that the degree of irreversibility is minimized. If a process is reversible, then the concept of spontaneity is no longer applicable. At all times during the process, the system is at equilibrium.

A reversible process is one during which the system is never away from equilibrium, and a reversible process which takes the system from the state A to the state B is one in which the process path passes through a continuum of equilibrium states. – not spontaneous process.

Such a path is, of course, imaginary, but it is possible to conduct an actual process in such a manner that it is virtually reversible.

Such an actual process is one which proceeds under the influence of an infinitesimally small driving force such that, during the process, the system is never more than an infinitesimal distance from equilibrium.

3.5. An Illustration of Irreversible and Reversible Processes

Figure 3.2

A thermo-stalled piston and cylinder containing water and water vapor.

The water vapor in the cylinder exerts a certain pressure P_H2O(T), which is the saturated vapor pressure of water at the temperature T.

The system is exactly at equilibrium when the external pressure acting on the piston, P_ext, equals P_H2O(T), and when the temperature of the water + water vapor equals the temperature T of heat reservoir.

Consider a system of water and water vapor at T in Fig. 3.2.

Assume that a piston is frictionless and the cylinder is in thermal contact with a heat

reservoir.

3.5. An Illustration of Irreversible and Reversible Processes

(1) If Pext is suddenly decreased by a finite amount ∆P, then the piston is accelerated rapidly out of the cylinder. →

(2) The expansion of the water vapor decreases its pressure to a value below its saturation value, and hence spontaneous evaporation occurs to reestablish equilibrium. →

(3) The spontaneous evaporation, being endothermic, decreases the temperature in the cylinder to a value lower than T.→

(4) The temperature gradient between the cylinder and the heat reservoir causes the spontaneous flow of heat. →

(5) If, when 1 mole of water vapor has evaporated, the external pressure is instantaneously increased to the original value, P_ext, then evaporation of the water vapor ceases, the flow of heat ceases, and complete equilibrium is reestablished.

The work done by the system during this process is (P_ext - ∆P)V.

Evaporation process

3.5. An Illustration of Irreversible and Reversible Processes

(1) If P_ext is suddenly increased by a finite amount ∆P, then the piston is accelerated rapidly into the cylinder. →

(2) The compression of the water vapor increases its pressure to a value greater than the saturation value, and hence spontaneous condensation occurs. →

(3) The spontaneous condensation, being exothermic, increases the temperature in the cylinder to a value higher than T.→

(4) The temperature gradient between the cylinder and the heat reservoir causes the spontaneous flow of heat. →

(5) If, when 1 mole of water vapor has been condensed, the external pressure is instantaneously decreased to the original value, equilibrium is reestablished.

The work done on the system during this process is (P_ext + ∆P)V.

The permanent change in the external agency caused by the cyclic process is

∆PV.

Condensation process

3.5. An Illustration of Irreversible and Reversible Processes

Consider, again, the evaporation process.

If the magnitude of P_ext is decreased by an infinitesimal amount δP (P_ext - δP)

→ The piston moves slowly out of the cylinder.

→ The slow expansion of the water vapor decreases its pressure.

→ When the pressure has fallen by an infinitesimal amount below the saturation value, evaporation of the water begins.

→ An infinitesimal temperature gradient between the heat reservoir and the cylinder is set up.

The smaller the value of δP, then the slower the process, the smaller the degree of undersaturation of the water vapor, and the smaller the temperature gradient.

The more slowly the process is carried out, the greater the opportunity afforded to the evaporation and heat flow process to “keep up” with equilibrium.

3.5. An Illustration of Irreversible and Reversible Processes

During the evaporation process, if the external pressure is instantaneously increased to its original value P_ext after 1 mole evaporation of water, the work done by the system equals (Pext - δP)V.

In the same manner, during the condensation process, if the external pressure is increased by δP, the work done on the system equals (P_ext + δP)V to condense 1 mole of water vapor..

The permanent change in the external agency caused by the cyclic process

= work done on the system – work done by the system = 2δPV.

Thus, the smaller the value of δP, then no permanent change occurs in the external agency, and hence the cyclic process has been conducted reversibly.

If a complete reversibility is approached, the process becomes infinitely slow.

3.6 Entropy and Reversible Heat

Consider only the evaporation process.

The maximum work done by the system during the evaporation of 1 mole is

max ext

w = P V

when the process was conducted reversibly.

Any irreversible process performs less work,

(

) w = P − ∆ P V

The maximum amount of heat, q_rev, is transferred from the reservoir to the cylinder when the process is reversible.

max

q

= ∆ + U w

If the process is carried out irreversibly, then less heat is transferred from the reservoir to the cylinder.

q = ∆ + U w

( q

− q ) = ( w

− w )

The heat produced by degradation:

3.6 Entropy and Reversible Heat

Thus, if the evaporation process is conducted reversibly, then qrev leaves the reservoir and enters the cylinder at the temperature T.

The change in the entropy of the heat reservoir is

heat reservoir rev

S q

∆ = − T

The change in the entropy of the water and water vapor in the cylinder is

water + vapor rev

S q

∆ = T

The change in the entropy of the combined water-vapor-heat reservoir system is

0

S S S

∆

= ∆

+ ∆

=

The zero change in the entropy ∵ reversible process (no degradation)

3.6 Entropy and Reversible Heat

If the evaporation process is conducted irreversibly, then heat q (< qrev) is transferred.

The change in the entropy of the heat reservoir is

heat reservoir

S q

∆ = − T

The total heat appearing in the cylinder equals the heat q transferred from the heat reservoir plus the heat which is produced by degradation of work due to the irreversible nature of the process. Thus degraded work, (w_max–w), equals (q_rev – q).

q q q S q

T T T

∆

= +

− =

The change in the entropy of the combined water-vapor-heat reservoir system is

S S S

q q q

q

T T T

∆ = ∆ + ∆

= − + = −

total reservoir water + vapor

rev rev

3.6 Entropy and Reversible Heat

As q_rev > q, this entropy change is positive, and thus entropy has been produced as a result of an irreversible process.

The entropy produced, (q_rev – q)/T is termed ∆Sirreversible(∆S_irr) and is the measure of the degradation which has occurred as a result of the process.

Thus for the evaporation process, irrespective of the degree of irreversibility,

q q

S S T

∆

= T + ∆

=

(3.2)

3.6 Entropy and Reversible Heat

Consider only the condensation process.

The work done on the system has a minimum value when the process is conducted reversibly, and, correspondingly, the heat transferred has a minimum value q_rev.

Thus, for a reversible condensation,

heat reservoir rev

S q

∆ = T

S q

∆

= − T

tot

0 : i.e., entropy is not created.

∆ S =

For an irreversible condensation,

3.6 Entropy and Reversible Heat

water + vapor

rev rev

heat leaving cylinder

heat produced in cylinder by degradation

S T

T

q q q

q

T T T

∆ =

+

= − + − = −

heat reservoir

S q

∆ = T

q q q q

q q

S T T T T

− −

 

∆ = − +     + =

rev rev

total

and

3.6 Entropy and Reversible Heat

As q > q_rev, entropy has been created as a result of an irreversible process.

The entropy created is ∆S_irr , and, thus, again, the change in entropy of the water and water vapor is given by

q S q

S T

∆

= − T + ∆

= −

From Eq. (3.2) and Eq. (3.3)

rev

water + vapor

q reversible evaporation

S T

∆ =

rev

water + vapor

q reversible condensation

S T

∆ = −

: state function

∴ S

irr rev

B A

q q

S S S S

T T

∆ = − = + ∆ =

∆Scan be measured only for reversible process, in which the measured heat flow is q_rev and ∆S_irr = 0.

3.7 The Reversible Isothermal Compression of an Ideal Gas

Consider the reversible isothermal compression of 1 mole of an ideal gas from the state (V_A,T) to the state (V_B,T).

As the compression is conducted isothermally, ∆U = 0 and thus the work done on the gas = the heat withdrawn from the gas, i.e.,

max rev

w = q

where

B B

ln

A A

V V

B

V V

A

V

w PdV RTdV RT

V V

= ∫ = ∫ =

max

As V_B < V_A, w_max is a negative quantity, in accordance with the fact that work is done on the gas.

3.7 The Reversible Isothermal Compression of an Ideal Gas

ln

A

q w V

S R

T T V

∆

=

=

=

The transfer of heat from the gas to the reservoir causes a change in the entropy of the gas

which is also negative quantity.

Consequently, as there is no change in the total entropy of the system during the reversible compression, the change in the entropy of the reservoir is given by

ln

B

S S R V

∆

= −∆

= V

3.8 The Reversible Adiabatic Expansion of an Ideal Gas

Consider the reversible adiabatic expansion of 1 mole of an ideal gas from the state (P_A,T_A) to the state (P_B,T_B).

The adiabatic process (q = 0) dictates that the process path across the P-V-T surface follows the line PV^γ = constant.

As process is reversible, no degradation occurs, and, as the process is adiabatic, no heat flow occurs.

A reversible adiabatic process is an isentropic process. 0

∆S_gas =

3.9 Summary Statements

Three points have emerged from the discussion so far:

1. The entropy of a system increases when the system undergoes an irreversible process.

2. Entropy is not created when a system undergoes a reversible process;

entropy is simply transferred from one part of the system to another part.

3. Entropy is a state function.