thermodynamics lecture notes 2014 1st part

CH1108 ThermodynamicsInstructor: Vincent ChanDept: Chemical and Biomolecular EngineeringE-mail: [email protected]: N1.2-B2-26 (Appointments thru email most welcomed)Telephone: 6790 6739Textbook: Engel and Reid, Thermodynamics, Statistical

Thermodynamics, Kinetics, Pearson (Main book for my part); Smith, Van Ness and Abbott, Introduction to Chemical Engineering Thermodynamics, McGraw Hill

Tutorial questions (Sample questions), on E-Adventure. Grading: Continuous Assessment and One Final Examination Quiz (for my part): 30 minutes, 19th, Feb, 2014 (Wednesday) Exam/Quiz Format: Closed Book Tutorials: Every week starting from 2nd teaching week

Fundamental of Thermodynamics: Learning Objectives

Basic concepts in thermodynamics: system, surroundings, intensive and extensive variables, adiabatic and diathermal walls, equilibrium, temperature, and thermometry.

Discuss the usefulness of equations of state, which relate the state variables of pressure, volume, and temperature.

Boyles Law The pressure of a sample is inversely proportional to its

volume, and the volume of a sample is inversely proportional to pressure.

Here are some plots depicting Boyles Law. Each plotted line corresponds to a different temperature, and are known as isotherms, as they depict the other state variables at a constant temperature:

Boyles Law Logistics Boyles law strictly only applies to ideal gases at

very low pressures, when there are very few molecular collisions and very few interactions between the molecules.

How do we rationalize Boyles law? Say we have an ideal gas in a container of

volume V, and then we reduce the volume of the container by half, to V/2. What happens to the pressure?

The pressure doubles, since now twice as many molecules are striking the sides of the container!

Charles (Gay-Lussacs) Law Charles (1746-1823), a French physicist,

constructed the first hydrogen balloons, making an ascent to over 3000 meters (1.9 mi) in 1783.

His name is chiefly remembered, however, for his discovery of Charles's law, which states that the volume of a fixed quantity of gas at constant pressure is proportional to its temperature.

Hence all gases, at the same pressure, expand equally for the same rise in temperature.

He communicated his early results to Joseph-Louis Gay-Lussac, who published his own experimental results in 1802, six months after Dalton had also deduced the law.

Charles Law

Or, on the thermodynamic temperature scale devised by Kelvin: P = Constant X T or V = Constant X T

The volume of a gas should extrapolate to zero near -273 oC. Plots of volume and pressure as a function of temperature, at constant pressure and volume, respectively, are shown below:

( )( ). 273V Con T C= +

Ideal Gas Law Most common gases like O2, N2 can be treated as ideal gas.

The P-V-T relationship is called the ideal gas law:

PV nRT=where P is pressure, V is volume, R is ideal gas constant (0.08206 L atm deg-1 mol-1 or 8.314 J K-1 mol-1) and n is the number of mole of gas molecule. The following is a typical V-Pplot at different temperatures:

M

o

l

a

r

V

o

l

u

m

e

(

m

l

/

M

o

l

e

)

Pressure (atm)

T increases

Ideal Gas Law At constant temperature, the following holds: PV

= constant. For example, the increase of P from 1 to 1000

atm will lead to the reduction of 1000 times in V(compared to 3% change in water).

The ideal gas law is a powerful tool for determining the work done in 1st Law. Ideal gas law works fine for most gases near room temperature and low pressure.

When we have a mixture of two or more gases, what contribution each of the member gases make to the overall pressure of the system?

1.4 Equations of State and the Ideal Gas Law

Pressure units related to Pascal as shown

1.4 Equations of State and the Ideal Gas Law

Values of the constant R with different combinations of units as shown.

Daltons Law for Ideal Gases The total pressure exerted by a homogeneous

mixture gases is equal to the sum of the partial pressures of the individual gases.

The partial pressure of a gas is the pressure it would exert if all the other gases in the mixture were absent.

...total A B ii

P P P P= + = If the partial pressure of gas A is PA, and the partial pressure of gas PB, etc. then the total pressures for gases in the same vessel is:

Daltons Law Logistics Pi is dependent on the # of mole (ni) of a particular

gas within the gas mixture is defined:

ii

totaltotal i

i

n RTPV

n RTP PV

=

= =For each component of a gaseous mixture i, the mole fraction, Xi, is the amount of i expressed as a fraction of the total number of molecules:

.....

ii

ii

i A Bi

nXn

n n n n

=

= = + +

Daltons Law: Partial Pressure If no i molecules are present: Xi = 0 If only i molecules are present: Xi = 1, therefore

....... 1i A Bi

X X X= + + =The partial pressure of gas i in a mixture is formally defined as:

i i totalP X P=It follows that for ideal gases:

( )...... .......

.....A B A total B total

total A B

total

P P X P X PP X XP

+ + = + += + +=

1.5 A Brief Introduction to Real Gases

Two assumptions for ideal gas: a) Atoms or molecules of an ideal gas do not

interact with one another,

b) Atoms or molecules can be treated as point masses.

Real Gases Real gases do not obey Ideal Gas Law exactly. Deviation from the law is significant at high pressure

and low temperatures, e.g., condensation. It is because molecules interact with one another,

repulsive forces assist expansion and attractive force assist compression.

Repulsion is critical when average separation between molecules is small, e.g., at high pressure when many molecules occupy small volume.

Attractive forces is critical when average separation is over several molecular diameter but not in contact

At low pressure, molecules occupy large volume, intermolecular forces are unimportant.

Gas behaves as Ideal Gas

1.5 A Brief Introduction to Real Gases

A real gas equation of state, is defined with van der Waals equation of state.

It takes both the finite size of molecules and the attractive potential into account.

22

Van

nbVnRTP =

Equation of State Other equation of states has been developed for

correcting the discrepancies between ideal gas and real gas, a classical equation is know as van der Waals:

( )22n aP V nb nRTV + =

Where a and b is the van der Waals constant (represents the attractive force between molecules) and intrinsic volume of the gas molecule, respectively.

If a and b are zero, the van der Waals equation of state became the ideal gas law.

Compression Factor Compression factor, Z, of a gas is the ratio of its

measured molar volume V/n, to the molar volume of ideal or perfect gas, Vm0, at the same pressure and temperature:

0m

m

VZV

= Because the molar volume of an ideal gas is equal to

0

0

m

m m

m

m

RTVPV VZ RTV

PPV ZRT

=

= =

=

First Law: Learning Objectives

Introduce the first law of thermodynamics. Discuss the concepts of the heat capacity,

the difference between state and path functions, and reversible versus irreversible processes.

Introduce enthalpy, H, as a form of energy. Show how , , q, and w can be

calculated for processes involving ideal gases.

U H

Introduce internal energy, U or E.

First Law: Outline(Chapter 2 of Textbook)

The Internal Energy and the First Law of ThermodynamicsWorkHeatHeat CapacityState Functions and Path FunctionsEquilibrium, Change, and Reversibility

2.4)

2.2)

2.3)

2.1)

2.5)

2.6)

2.7) Comparing Work for Reversible and Irreversible Processes

Outline

Determining and Introducing Enthalpy, a New State FunctionCalculating q, w, , and for Processes Involving Ideal Gases

HU

U

2.9)

2.8)

2.10) The Reversible Adiabatic Expansion and Compression of an Ideal Gas

First Law of Thermodynamics Thermodynamics deals with interchange among

different forms of energy. ENERGY IS CONSERVED! The energy of the universe is constant. Energy can be converted from one form to

another, but cannot be destroyed. The change in internal energy equals the heat

exchanged by the system plus the work done on or by the system.

Thermodynamics is the law of nature, collected from a wide range of experiment and theoretical studies.

Definition of System and Surrounding

1.2 Basic Definitions Needed to Describe Thermodynamics Systems

A system consists of all the materials involved in the process under study.

Open system can exchange matter with the surroundings, if not, it is a closed system.

Isolated systems exchange neither matter nor energy with the surroundings.

The interface between the system and its surroundings is called the boundary.

Definition of a System In our subject, the universe is divided into

two parts: system and surrounding. The system is part of the world in which we

have interests, e.g., reaction vessel, engine. The surrounding comprises the region

outside the system and are where we make our measurement.

The type of system depends on the characteristics of boundary between system and surrounding.

Definition of a System Opened or Closed or IsolatedOpen system, e.g., boiling water inside beaker in atmosphere.

H2O

Water Evaporates

Open System

Air Dissolves

Heat

In an open system, there are exchanges of mass, energy and matters.

Closed System In a closed system, there are no mass and no matter

exchanges. There is exchange of energy.

H2O

Closed System

Heat

In an isolated system, there is neither mass/matter exchange nor energy exchange.

If matters can be transferred through the boundary between system and surrounding, the system is classified as open. If matters cannot pass through the boundary, the system is classified as closed. Both open and closed system can exchange energy with the surrounding. A closed system can expand and thereby raise a weight in the surroundings. An isolated system is a closed system that has neither mechanical nor thermal contact with surrounding

Variables in First Law of Thermodynamics Change of energy is the key of

thermodynamics Energy can be basically classified into three

categories including:

Work (w)

U (internal energy of the system)

Heat (q)

2.1 The Internal Energy and the First Law of Thermodynamics

Internal energy is total energy for the system of interest.

First law of thermodynamics states that energy can be neither created nor destroyed, if both system and the surroundings are taken into account.

The internal energy, U, of an isolated system is constant. When changes occurs in U:

gssurroundinsystem

gssurroundinsystemTotal

UUUUU

==+= 0

For any decrease of Usystem, Usurroundings must increase by exactly the same amount.

This derive a second and useful formulation of the first law:

2.1 The Internal Energy and the First Law of Thermodynamics

wqU +=q = heatw = work

2.2 Work

Work is defined as any quantity of energy that flows across the boundary between the system and surroundings that can be used to change the height of a mass in the surroundings.

Pressure Pressure is the force per unit area, where A is the cross-sectional area of the piston whose orientation is perpendicular to the direction of the force.

exex

fPA

=What is the work done when the piston is moved by a infinite displacement of dx by the external force?

exex ex

ff dx Adx P AdxA

= =

Pressure-WorkThe change in the volume of the gas in the system can be determined as:

Therefore the total work done on the system can be expressed as:

2 2

1 1

x V

ex exx Vw f dx P dV= =

The negative sign of R.H.S. is kept for keeping the definition of work consistent with our original definition, e.g., w is positive when the surrounding is doing work on the system.

dV = A dx

Work (Gas Expansion) The sign of work is negative if system is doing work on the

surroundings, e.g., expansion of a gas inside a piston-cylinder system.

Force

Work (-)

h1h2

h1 < h2

Force

Work done by the system to the surrounding: w = F x (h1-h2) becomes negative.The sign convention must be kept for all thermodynamics parameters!

Work: Gas ExpansionDuring gas expansion, the work done is:

2 2

1 1

V V

gas exV Vw P dV P dV= =

In the case, dV is positive and w will be negative as work is done by the system to the surrounding.

It must be noted that Pex and Pgas are positive as the sign of the w is only addressed at the end of the derivation.

The analysis as mentioned above is important for the understanding of heat engine operations.

Work (Gas Compression) From classical physics, Work (w) is defined as: external force

X displacement The sign of work is positive if surroundings are doing work on

the systems, e.g., compression of gas inside a rigid cylinder by an external force.

Force

Work (+)

h1h2

h1 > h2Work done by surroundings to the system: w = F x (h1-h2) becomes positive.

Work done by the change of volumeFor gas containing in a piston-cylinder system as follows:

SystemV

fex

Pex

P

The system do work on the surrounding when the gas or liquid expand (Pex < P). Otherwise, the surrounding can do work on the system when the gas is compressed (V reduced, Pex > P).

One wall of the system is a massless, frictionless, rigid, perfectly fitting piston of area A. If external pressure is Pex, the magnitude of force acting on the outer face of piston is F = Pex A. When the system expands by a distance dz against external pressure Pex, the work done is

exdw P AdzAdz dV

= =

Suppose the external pressure is constant throughout the expansion. The piston is pressed on by the atmosphere, which exerts the same pressure during the whole process. An example is the expansion of gas formed in a chemical reaction. The work done is the area under the horizontal line of Pvs. V plot.

( )2 1p exw P V V=

James Watt first used this approach (indicator diagram)

Consider an isothermal, reversible expansion of an ideal gas. The work done is negative during gas expansion as system has done work on the surrounding. The rectangular area represents irreversible expansion against constant external pressure fixed at the same final level in reversible expansion. More work is obtained when the expansion is reversible because matching the external with internal pressure at each stage of the expansion ensures that none of the systemspushing power is wasted.

If two gases are in separate containers sharing a movable wall, the gas has higher pressure will tend to compress (reduce the volume of) the gas that has lower pressure. The pressure of the high pressure has will fall as it expands and that of the low pressure gas will rise as it is compressed. When the two pressures are equal, the wall has no further tendency to move. An mechanical equilibrium between the two gases is reached. Thus the pressure of the gas is an indicator of whether the system is in mechanical equilibrium.

Work (Gravitational field)Work is done from the system to the surrounding when an object is falling from higher height to lower height:

( )2 1w mg h h= w becomes negative. m is the mass of the object, gis the gravity (9.807ms-2)

When an object is raised up from lower level, w is positive as work is done by the surrounding to the system.

Work (Electric field)Electric field will produce a force on the charges which are moved. Then the current will flow. The work done on the system is shown as:

w E I t= where E is the voltage or electromotive force (V), I is the current (A), t is the time. 1 W s = 1 joule.

Total Work

Table 2.1 shows the expressions for work for four different cases.

Example Problem 2.1a. Calculate the work involved in expanding 20.0 L of an

ideal gas to a final volume of 85.0 L against a constant external pressure of 2.50 bar.

Example Problem 2.1 (Solution)

( )( ) kJ

LmLL

barPabar

VVPdVPw ifexternalexternal

3.16100.200.851050.2 335

==

==

a.

HeatHeat transfer occurs when the temperature on two adjacent bodies (originally with different temperatures) equilibrates upon the formation of contact.

When the energy is exchanged between the two bodies, the hot body will cool down (lose energy) and the cold body will warm up (gain energy).

The heat transfer is caused by the energy passing through the system-surrounding boundary.

The change of state can be interpreted as rising from flow of energy from one object to another. Temperature T is a property that indicates the direction of energy flow thru a thermally conducting, rigid wall. If energy flows from A to B when they are in contact, then A has a higher temperature than B. A boundary is diathermal if a change of state is observed when the two objects at different temperatures are brought into contact. A boundary is adiabatic if no change occurs even though the two objects have different temperatures

The heat capacity is a material-dependent property defined by the relation

Most common are constant volume or constant pressure, CV and CP, respectively.

2.4 Heat Capacity

dTdq

TTqC

ifT== lim0

where C is the SI unit of J K-1

Heat: FormulaIn a closed system, a small change in heat dq is accompanied by a change of temperature dT. The thermal response of any materials, e.g., gas or liquid, towards the change of temperature is characterized by:

dq CdT

=where C is the heat capacity.

C is a mass-dependent parameter and is different for different materials.

Total HeatThe total heat gained in parallel with the increase of temperature from T1 to T2 can be determined by

2

1

T

Tq CdT=

For a system with a given mass, C is sometime assumed constant against the change of temperature.

Heat is positive if it flows from the surrounding into the system and is negative when heat flows out from the system to the surrounding. Then the total heat capacity will be ( )2 1q C T T=

At constant pressure, heat flow between the system and surroundings can be written as

The molar heat capacity, Cm, is used in calculations with the units of J K-1 mol-1.

2.4 Heat Capacity

( ) ( )dTTCdTTCq fsurrisurr

fsys

isys

T

T

gssurroundinP

T

T

systemPp ==

,

,

,

,

Total HeatIt is more convenient to express heat capacity in terms of molar unit as J K-1 mol-1 or cal K-1 mol-1. For example,

C nC=

For an ideal gas:

Not all heat flow into the system can be used to increase Uin a constant pressure process, because the system does work on the surroundings as it expands.

2.4 Heat Capacity

or ,, RCCnRCC mVmPVP ==

Example Problem 2.3

The volume of a system consisting of an ideal gas decreases at constant pressure. As a result, the temperature of a 1.50-kg water bath in the surroundings increases by 14.2C.

Calculate qP for the system.


( )kJKKJgkg

TCdTTCqfsurr

isurr

T

T

gssurroundinP

gssurroundinPp

1.892.1418.450.1 11

,

,

==

==

INTERNAL ENERGY U Internal Energy (U) - property of a substance/system.

Origin of U: Nuclear binding energy, Proton -electron attractions, Bonding energy, Intermolecular forces, Translation, rotation and vibration of a molecule.

Energy is the capacity to do work or release heat.

The following events can change U: vary temperature, phase transition, chemical reactions.

Change of Internal EnergyOn changing a substance, e.g. by undergoing a chemical reaction, the Energy (internal energy) change is determined:

final initialU U U = A change in energy of a system/substance is always accompanied by an opposite change in the surroundings.

That is exactly the first law of thermodynamics, energy is transferred between system and the surrounding but the total energy of the system and the surrounding is constant.

First Law of ThermodynamicsFor a closed system, only heat and work are the main energy exchanged with the surroundings. Therefore the change of internal energy:

U q w = +Where q = net heat transferred to the system (heat in)w = net work done on the system (work in)

In general, two scenarios arisen from the change of internal energy as follows: Endothermic and exothermic process

Endothermic Process

Heat absorbed from the surroundings ( > 0) U

Exothermic Process

UHeat released from the system ( < 0).

K.E. P.E. (No change of kinetic and potential energy)

E UE U

= + + =

State Function Pressure (P) 1 atm = 760 torr = 1.01 X 105

Pa (N m-2) Temperature (T) K = C + 272.15 Volume (V) 1 L Internal Energy (U) 1J For system consisting of pure liquids, the

fixing of P, V and T leads to the fixing of all physical properties of the liquid including:

Density, surface tension, refractive index, etc.

State Function State Function depend on the

thermodynamics state of the system and not on the path of arriving at that state.

U is a state function or variable of state (a property of the system)

Work and heat transfer are not state functions because they depend on the path how one state is changed from one to another.

U is independent of the path between the initial and final states and depends only on the initial and final states.

For any state function, U must satisfy the equation,

For cyclic path such that the initial and final states are identical,

State Functions and Path Functions

==f

iif UUdUU

== 0if UUdU

As both q and w are path functions, there are no exact differentials for work and heat.

if

f

iif wwwqqdqq and


Types of State Variable For a cyclic process, when the system returns to

the initial state after several steps in between, and change of state variable are both zero.

There two types of variable of state including intensive and extensive parameters.

Extensive variable of state is dependent on the mass of the system, e.g., total volume of gas, total internal energy (U), and is changed by the sub-division of the system.

Intensive variable of state includes all parameters expressed in term of unit/mass or unit/mole, etc.

Density (mass/volume) is an intensive parameter of state.

As the height of the mass in the surroundings is lower after the compression,

w is positive and U is increased.


2.6 Equilibrium, Change, and Reversibility

A quasi-static process maintain internal equilibrium where rate of change of the macroscopic variables is negligibly small from the initial to the final state.

When an infinitesimal opposing change in the variable that drives the process in reversal direction, the process is reversible.

If an infinitesimal change in the driving variable does not change the direction of the process, the process is irreversible.

2.7 Comparing Work for Reversible and Irreversible Processes

Indicator diagram shows the relationship between P and V for the process graphically.

Total work in yellow area. Expansion work in red

area. Compression work in

total red and yellow area


The indicator diagram for reversible process shows that the areas under the PV curves are the same in the forward and reverse directions.


The work associated with the expansion is

====1

2exp ln V

VnRTVdVnRTPdVdVPw externalansion


When the process is reversed and the compression work is calculated, the following result is obtained:

2

1lnVVnRTw ncompressio =

Example Problem 2.4In this example, 2.00 mol of an ideal gas undergoes isothermal expansion along three different paths: (1) reversible expansion from Pi = 25.0 bar and Vi =4.50 L to

Pf = 4.50 bar; (2) a single-step expansion against a constant external

pressure of 4.50 bar, and (3) a two-step expansion consisting initially of an expansion

against a constant external pressure of 11.0 bar until P=Pexternal, followed by an expansion against a constant external pressure of 4.50 bar until P=Pexternal.

Calculate the work for each of these processes. For which of the irreversible processes is the magnitude of the work greater?

The processes are depicted in the following indicator diagram:


We first calculate the constant temperature at which the process is carried out, the final volume, and the intermediate volume in the two-step expansion:


LP

nRTV

LP

nRTV

KnRVPT

ff

ii

2.100.11

67700.210314.8

0.2550.4

67700.210314.8

67700.210314.8

50.40.25

2

intint

2

2

===

===

===

The work of the reversible process is given by


J

VV

nRTwi

f

3

1

103.1950.4

0.25ln677314.800.2

ln

==

=

We next calculate the work of the single-step and two-step irreversible processes:

The magnitude of the work is greater for the two-step process than for the single-step process, but less than that for the reversible process.


( )( )( )

J

VPw

JVPw

externalsteptwo

externalgle

3

35

35

335sin

109.12 102.100.251050.4

1050.42.10100.11

1023.91050.400.251050.4

===

===

For Compression: leirreversibreversible ww


that can be extracted from a process between the same initial and final states is that obtained under reversible conditions.

The maximum work

Reversible Process A process during which the system is never more

than infinitesimally far from equilibrium is reversible An infinitesimal change in external conditions can

reverse the process at any point - a change can be reversed by infinitesimal modification of a variable.

Consider a sample of gas in thermal and mechanical equilibrium with the surroundings; i.e., with Tgas = Tsurroundings and Pgas = Pexternal.

If the external pressure is decreased infinitesimally at constant T, the gas will expand infinitesimally; if the external pressure is increased infinitesimally at constant T, the gas will be compressed by an infinitesimal amount.

Reversible Process: Details Strictly speaking, a reversible process cannot be

achieved, because a finite transformation in a series of infinitesimal steps would require infinite time.

All real processes are therefore irreversible. A reversible process is an idealization (very useful).

In order for the gas to expand, Pex must be smaller than P (the pressure of the gas). At the same time, the system must be infinitesimally close to equilibrium throughout the expansion.

This means that Pex must be equal to P dP at all times during the expansion.

Internal energy of a substance increases when its temperature is raised. The increase depends on the conditions under which heating takes place. Here we assume the sample is confined to a constant volume. The slope of the tangent to the curve at any temperature is called the heat capacity. The heat capacity at constant volume is demoted as CV as follows:

VV

UCT

=

A partial derivative is a slope calculated with all variables except one held constant. In this case, internal energy varies with both temperature and volume of the sample. But we are only interested in its variation with temperature, the volume is held constant. Heat capacity is an extensive property (dependent on mass) The molar heat capacity at constant volume is an intensive property. A large heat capacity implies that, for a fixed quantity of heat, T is small.

Enthalpy Based on the three variables of state as

mentioned above, new variable of state can be developed since parameter depending on the variables of state must also be a variable of state.

For instance, a new variable of state, enthalpy Hcan be developed from U, P and V:

H U PV= +H has the same unit of U (J or kJ or cal or kcal).

When a system goes from an initial state to a final state, the same value of variable of state will be obtained irrespective of the actual processes which drive the change from the initial to the final state of the system.

2.8 Determining U and Introducing Enthalpy, a New State Function

When process under constant volume conditions and that non-expansion work is not possible.

U+PV is a state function and is called enthalpy, H.

At constant pressure,

vqU =

PVUH +

pqH =

Enthalpy The change in internal energy is not equal to heat supplied

when the system is free to change its volume. Under this situation, some of the energy supplied as heat

to the system is returned to the surrounding as expansion work. Thus dU is less than dq.

The change of enthalpy between any pair of initial and final states is independent of the path between them.

Enthalpy has important implication for thermochemistry because the heat of enthalpy is equal to heat supplied at constant pressure: or PdH dq H q= =

The result implicates that, when a system is subjected to a constant pressure, and only expansion work can occur, the change of enthalpy is equal to the energy supplied as heat.

Change of internal energy for a temperature range over which CV is constant,

2.1 Ideal Gas under Constant Pressure or Volumehttp://media.pearsoncmg.com/bc/bc_engel_physchem_1/simulations/content/heat_capacity/main.html

( )ifVv TTCqU ==


U is a function of T only for an ideal gas where V is constant.

H is also a function of T only for an ideal gas, which gives the following:

( )ifpp TTCqH ==


Enthalpy of a substance increase as temperature is raised. The relation between the increase in enthalpy and increase in temperature depends on condition, e.g., constant pressure. The slope of the tangent to a plot of H vs. T at constant pressure is called the heat capacity at constant pressure:

PP

HCT

= The heat capacity at constant pressure is the analogue of the heat capacity at constant volume. The increase in enthalpy is equated with heat supply at constant pressure

Gas Expansion (Isobaric) (Non-Isothermal)

Molar Volume (ml/Mole)

P

r

e

s

s

u

r

e

(

a

t

m

)

T1

T2

V1 V2

P

Under constant pressure, the work done during the gas expansion (the area if the P-V diagram) is

( )2 1p exw P V V=

Work of Gas Expansion In general, the external pressure (Pex) is kept a level

slightly lower than the gas pressure inside the cylinder (P).

As demonstrated earlier, P is directly proportional to T based on the ideal gas law:

1 1

2 2

1 2

PV nRTPV nRTPV nRTP P P

===

= =

Work Done in Isobaric Expansion Assuming that gas is expanding inside a piston

cylinder from V1 to V2 at Pext:

( )2 21 1

2

1

V V

exV V

V

V

w P dV P dP dV

PdV

= = =

For gas expansion at constant pressure, the maximum work done during gas expansion is

( ) ( )2 1 2 1ex

p

P Pw P V V nR T T

= =

Change of Internal Energy and Enthapy The change of internal energy (U) during the

isobaric process is

( ) ( )( ) ( )

2 1 2 1

2 1

p p

p

p

U q w

nC T T nR T T

n C R T T

= += + =

The change of enthalpy for isobaric gas expansion is

( )( )( ) ( )

( )2 1 2 1

2 1

p

p

H U PV

n C R T T nR T T

nC T T

= + = + =

Heating and Cooling of Gases Heat transfer of ideal gases at

constant pressure or volume is:

( )( )

2 2

2 2

p p

V V

q nC T T

q nC T T

= =

Consider an isobaric expansion (constant external pressure) of a gas inside a piston-cylinder system

Isothermal Expansion (Two Types) Consider an isothermal expansion of gas. Expansion

is isothermal if system is in contact with constant thermal surroundings (i.e., a constant temperature bath)

1. Each stage of expansion, P = nRT/V2. Temperature T is constant, so treated outside of the

integral

Isothermal Expansion (Reversible Work)

Here, w is equal to the area under the P = nRT/Visotherm, and this represents work of a reversible expansion at constant temperature, where external pressure is continually matched against the internal pressure

Yellow square: irreversible expansion against constant external pressure

More work and maximum pushing power is obtained from the isothermal expansion

Pushing power is wasted when P > Pex - more work also cannot be obtained by increasing Pex, since compression will result!

Work Done in Isothermal Gas Expansion

[ ]

2

1

2

1

2

1

2

1

2

1

2

1

1

ln

ln

V

exV

exV

V

V

V

V

V

V

V

w P dV

P P

w PdV

nRT dVV

nRT dVV

nRT V

VnRTV

=

=

=

= =

=

Work for Ideal Gas Since volume of the gas is inversely proportional to

the pressure of an ideal gas, the work done is

2 1

1 2

1 1 2 2

1 2

ln ln

since

TV Pw nRT nRTV P

P V P VT T

= = =

At constant temperature, U which is solely a function of Tbecomes zero and therefore

2 1

1 2

0

ln ln

T T

T T

T T

q w Uq w

V Pq w nRT nRTV P

+ = = =

= = =

Work Done Logistics When V2 > V1, gas expands, and work

done to expand the gas causes the system to lose internal energy, w < 0

Constant supply of energy at constant Treplenishes internal energy

More work is done for a given change in volume at higher temperature

Pressure of confined gas needs higher opposing pressure for reversibility

Path in Change of State Functions When a gas is changed from P1, V1, T1 to P2, V2,

T2, we can use the following path for calculating the change of internal energy during the process:

( ) ( ) ( )1 1 1 2 2 2, , ,T P T P T P Basically, it starts an isothermal process and then followed an isobaric process. The final volume of the gas will be V2 and is directly proportional to T2 (Pressure is constant in Step 2):

2 2

2

Ideal Gas Law:

PV nRTP VnR

T

==

Change of Internal Energy

( )

( ) ( )( ) ( )

( )

2 1 2 1

2 1

2 1

( )0 ( )0 p p

p

p

p

U U isothermal U isobaricU isobaric

q w

nC T T nR T T

n C R T T

H nC T T

= + = + = + += + = =

Effect of Changing Path What happen if we alter the path of the change of P,

V, T? For example,( ) ( ) ( )1 1 1 2 2 2, , ,T V T V T V ( )

( )( )

2 1

2 1

( )0 ( )0

0 0

where is the heat capacity at constant volume.

V V

V

V

V

U U isothermal U isochoricU isochoric

q w

nC T T

nC T T

C

= + = + = + += + +=

Effect of Changing Path

( )( ) ( )

( ) ( )2 1 2 1

2 1

The change of enthalpy during the process:

V

V

H U PV

nC T T nR T T

n C R T T

= + = + = +

Based on the First Law of thermodynamics, the path of change should not lead to the change of U and H.

By comparing the two expressions of H of different paths, the following will be determined:

p VC C R= +

The change in internal energy of an ideal gas when temperature is changed from Ti to Tf and volume is changed from Vi to Vf can expressed as the sum of two steps. In the first step, only the volume changes and the temperature is held constant at its initial value. Because internal energy of an ideal gas is independent of volume the molecules occupy. The overall change of internal energy solely comes from the second step, the change in temperature in constant volume.

Adiabatic Process When an ideal gas expands adiabatically, a

decrease of temperature should be expected It is because work is done and internal

energy falls, and the temperature of the working gas also falls.

In molecular terms, the kinetic energy of the molecule falls as work is done, so their average speed decreases, and hence the temperature falls.

Adiabatic Process Provided that heat capacity is independent of temperature,

the change of internal energy is ( )V f i VU C T T C T = = Because expansion is adiabatic, q=0, because

ad V

U q wU w C T

= + = =

Thus the work done during an adiabatic expansion is proportional to the temperature difference between the initial and final stage. Thats what we expect on molecular grounds, because mean kinetic energy is proportional to T. So a change in kinetic energy arising from T alone is also expected to be proportional to T.

Adiabatic Process

and (adiabatic)

ln ln

f f

i i

V

V

T VV

V T V

f fV

i i

c cf f i i

dw PdVdU dq dw dU dwdU C dTC dT PdVC dT nRdV dT dVC nR

T V T VT V

C nRT V

V T V T

= = + ==

= = =

= =

where V VC n CcR R

= =

1/ 1

1

Let 1

1 1

The initial and final stage of ideal gas mustsatisfy ideal gas law in adiabatic expansion:

p V

V V V

i i i

f f f

cf fi

f i i

fi i

f f i

C C R RC C C

c

PV TP V T

V VTT V V

VPVP V V

+= = = +

=

=

= = =

i i f fPV P V =

ConstantPV =

If CV is constant over the temperature interval Tf -Ti, then

Because Cp-Cv= nR for an ideal gas, Equation (2.42) can be written in the form

where .

2.10 The Reversible Adiabatic Expansion and Compression of an Ideal Gas

( )

=

=

1

ly,equivalent or, ln1lni

f

i

f

i

f

i

f

VV

TT

VV

TT

mVmp CC ,,=

i

f

i

fV V

VnR

TT

C lnln =

Substituting in the previous equation, we obtain

ffii VPVP =

iiffif VPVPTT /=

2.10 The Reversible Adiabatic Expansion and Compression of an Ideal Gas

Use of Heat Capacity From both paths, we can conclude that

( )( )

2 1

2 1

V

p

U nC T T

H nC T T

= =

Till now, we assumed that heat capacities are constant against the change of temperature.

For real gases, the heat capacity is a function of temperature as follow:

2 ...C a bT cT= + + +

Change of Energy under Cp(T) Therefore the change of enthalpy and internal

energy: 21

2

1

T

VT

T

pT

U n C dT

H n C dT

= =

For isochoric (constant volume) process, work done is zero and the change of internal energy is

2

1

0V V

VT

VT

U q wU q

n C dT

= + = +=

Enthalpy Change under Cp(T) For isobaric process with the presence of only P-V

work, the enthalpy is equal to heat transfer:( )

2

1

p p

p

T

p pT

H U PVU P V

q w P Vq P V P V

H q n C dT

= + = + = + + = + = =

As a result, enthalpy is same as the heat content of the system at constant pressure.

A cloud mass moving across the ocean at an altitude of 2000 m encounters a coastal mountain range. As it rises to a height of 3500 m to pass over the mountains, it undergoes an adiabatic expansion. The pressure at 2000 and 3500 m is 0.802 and 0.602 atm, respectively. If the initial temperature of the cloud mass is 288 K, what is the cloud temperature as it passes over the mountains? Assume that CP,m for air is 28.86 J K1 mol1 and that air obeys the ideal gas law. If you are on the mountain, should you expect rain or snow?

Example Problem 2.7

Because the process is adiabatic, q =0, and . Therefore,

You can expect snow.


( ) ( )

( ) ( )

( )

KTT

atmatm

PP

RCC

RCC

PP

TT

PP

TT

PP

TT

VV

TT

if

f

i

mP

mP

mP

mP

f

i

i

f

f

i

i

f

f

i

i

f

i

f

i

f

2659207.0

0826.0602.0802.0ln

314.886.2886.28

314.886.2886.28

ln1

ln1ln

ln1ln1

ln1ln1ln

,

,

,

,

==

=

=

=

=

=

=

=

wU =

Thermochemistry: Objectives(Chapter 4 of Textbook)

Discussion of Hesss Law. Derive property that allows H and U to

be calculated without experiments.

4.1 Energy Stored in Chemical Bonds Is Released or Taken Up in Chemical Reactions

The change in enthalpy or internal energy due to temperature result in heat flow and/or in the form of expansion or non-expansion work.

The focus is on using measurements of heat flow to determine changes in U and H due to chemical reactions.

The enthalpy of reaction, Hreaction, is the heat withdrawn from the surroundings as the reactants are transformed into products at constant T and P.

The standard enthalpy of formation, Hf, is the enthalpy associated with the reaction of 1 mol of the species under standard state conditions.

The enthalpy change associated with reaction


The enthalpy change associated with a chemical reaction is

ifi

ireaction HvH ,=


4.3 Hesss Law Is Based on Enthalpy Being a State Function

Hesss law states that the enthalpy change for any sequence of reactions that sum to the same overall reaction is identical.

It is useful to have tabulated values of Hffor chemical compounds at one fixed combination of P and T.

Hreaction can then be calculated for all reactions among these compounds at the tabulated values of P and T.

Hesss Law of Heat of Reaction Using Hesss Law, the heat of reaction can be

determined multiple steps. For example, the combustion of methane to CO2 and water:

4 2 2 22 2CH O CO H O+ +

The example demonstrates that the path of the reaction is not critical.

Path of Process and Enthalpy Supposed the enthalpy vaporization of water at

100C and 1 atm is known. How to calculate the enthalpy vaporization of water at 35C and 1 atm?

H2O (l), T2 = 35C, P =1 atm H2O (g), T2 = 35C, P =1 atmH (35 C)

H2O (l), T1 = 100C, P =1 atm

P is Constant

H2O (g), T1 = 100C, P =1 atmH (100 C)

P is Constant

The enthalpy change for the vaporization of water at 35 C is:

( ) ( )[ ] ( ) ( )[ ]( ) ( ) ( ) ( )

35 100 35 100 35 100

100 65

Vap p Vap p

Vap p p

H C nC l C H C nC g C

H C n C g C l C

= + + = +

Enthalpy vs. Change of Temperature

Generally, the phase transition enthalpy at a certain temperature T2 can be determined as

( ) ( ) ( )( ) ( ) ( )

( ) ( )

2 1 , 2 1

2 1 2 1

vap vap vap P

vap vap p

p p

H T H T C T T

H T H T C T T

C g C l

= + = + ,vap PC=

Energy Change in Chemical Reaction When a chemical reaction occurs, the internal

energy and enthalpy of the system can be changed. Lets consider a general chemical reaction:

A B C Dn A n B n C n D+ + nx is the number of mole of molecule X.

The state variables including P, V and Tbefore and after the reaction must be specified in order to determine the energy change during the process.

Enthalpy of Reaction The change of enthalpy during the reaction as

mentioned above can be listed as follows:

( ) ( )products reactants

Molar Enthalpy (J/mole)

r

r C C D D B B A A

X

H H H

H n H n H n H n H

H

= = +

= For a reaction under constant volume (e.g., inside a bomb calorimeter), the measured heat of a reaction is

0r V V VV

U q w qq

= + = +=

Enthalpy Under Constant Pressure For a reaction under constant pressure, the measured heat of a reaction is( )r r

r p p p

r p

H U PVH q w P V q P V P VH q

= + = + + = + =

The term (PV) is the difference between PV of the products and that of reactants. (PV) of solid and liquids can be ignored.

When rU or rH is negative, the reaction is known as exothermic. When rU or rH is positive, the reaction is known as endothermic.

Hess Law: Path The final product and initial reactant is important as

follows:

Hesss Law can be used to determine the enthalpy change of many reactions. Certain standard conditions must be defined so that heats of reaction can be compared and combined.

Standard Enthalpy Change The heat absorbed for a process of the system

when the reactants in their standard states are converted to the products in their standard states, ie 1 bar and 298 K.

The enthalpy of a certain reaction A ------- B can be determined from the summations of the enthalpies along the path of the reaction as follow:

H

H3

A B

C DH2

E

H1H4

Hess Law: Details Based on the principle of Hesss law, the overall

enthalpy of the last reaction is

1 2 3 4H H H H H = + + + Standard enthalpy change is always referred to the reaction carried in standard states including 1atm and 298 K. However, most reaction occur at higher temperature.

How to convert the standard enthalpy to the realistic thermodynamics parameter? Take an example of a simple reaction A --- B at T2:

Hess Law: Reaction at Different TempH (T2)

H (T1)

A (T2) B (T2)

A (T1) B (T1)

HA HB

( ) ( )[ ] ( ) [ ]

( ) [ ]( ) [ ]

2 1

2 1 1 2 1

1 2 1

1 2 1

r A r B

A Bp p

B Ar p p

r r p

H T H H T H

C T T H T C T T

H T C C T T

H T C T T

= + + = + +

= + = +

Enthalpy: Alternative Analysis It is shown that the standard enthalpy can be

converted into enthalpy at different temperatures. We can arrive at the same conclusion at a

different angle from the definition of enthalpy of reaction as follows:

p, B p, A

r B B A A

r B AB A

B A

H n H n H

d H d H d Hn ndT dT dTn C n C

= =

= r

r p r pd H n C C

dT = =

Enthalpy Analysis

( )( )

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

2 2

1 1,

2 1 , 2 1

2 1 , 2 1

2 1 , 2 1

Therefore the same solution is arrived

or in molar enthalpy:

r

r

H T T

r r PH T T

r r r p

r r r p

r r r p

d H C dT

H T H T C T T

H T H T C T T

H T H T C T T

=

=

= +

= +

HT at elevated temperature is used to determine a reaction to be endothermic or exothermic.

The enthalpy for each reactant and product at temperature T relating to the value at 298.15 K is

where

4.4 The Temperature Dependenceof Reaction Enthalpies

( ) += T298.15K

p15.298 ''C dTTHH KT

( ) ipi

ip CvTC ,' =

The correlation from last page is known as Kirchhoffs law. It is normally a good approximation to assume that

is independent of temperature.,r pC

Measuring Internal Energy: Calorimetry How do we study changes in energy due to addition

or subtraction of heat in a system? We can use an adiabatic bomb calorimeter

Adiabatic: Implies that the device is isolated from the outside world, such that no thermal transfer of energy can take place between system and surroundings

Bomb: The sturdy vessel with constant volume inside which very vigorous reactions, combustions and explosions can take place at high pressure

Calorimeter: From the cgs unit of energy, the calorie, implying the measurement of energy.

Calorimeter: Details1. Reaction is conducted in the constant volume bomb2. Bomb is immersed in a stirred water bath, forming the calorimeter3. Calorimeter is immersed in a outer water bath, the temperature of which is continuously adjusted to the temperature of the bomb (inner water) bath, thereby ensuring an adiabatic system (q = 0)

4.5 The Experimental Determination of U and H for Chemical Reactions

Ureaction can be determined through experiment in a bomb calorimeter, which carried out in constant volume.


0volume)(constant 0

g)surroundin the tosystem thefrom flowheat (no 0

bathr inner wate theof re temperatusame at thekept is bath whichr outer wate theis

gsurroundin ther vessel,calorimete 3) bath,r inner wate 2) r,calorimete in the rectants 1) of consists system The

System) (Overall

===

+=

Uwq

wqU


T = temperature change of the inner water bath

reactants ofenergy internal of Changesystem overall theofenergy internal of Change

rcalorimete ofcapacity Heat

waterofweight Molecular

bathinner in water of mass sample ofweight Molecular

sample of mass where

0T

,

0

0

0

0,

2

2

2

2

==

==

==

=

=++=

mreaction

rcalorimete

H

H

s

s

rcalorimetewaterH

Hmreaction

s

s

UU

C

MmMm

CTCMm

UMmU


When Ureaction is obtained, we can get

n = number of moles of gas change in reaction

For constant pressure calorimetry involving the solution of a salt in water (or solid):

nRTUH reactionreaction +=

0,0

0, 2

2

2 =++= TCTCMm

HMmH rcalorimetemOH

H

Hmsolution

s

sreaction


For constant pressure calorimetry involving the solution of a salt in water:

0,0

0, 2

2

2 =++= TCTCMm

HMmH rcalorimetemOH

H

Hmsolution

s

sreaction

Second Law of Thermodynamics Neither the sign nor the magnitude of U and H

from the first law of thermodynamics tell us which way a reaction will go.

However, we can get this information from the Second Law of Thermodynamics.

The first law gives us no clue what processes will actually occur and which ones will not. For that matter, why does anything ever happen at all?

The universe is an isolated system after all. There is no change in internal energy. There is no heat transferred in or out and no work is done on or by the system as a whole. q = 0; w = 0; U = 0.

No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work (Figure). In reality, some heat is always discarded into cold sink and not converted into work. The Kelvin statement is the generalization of another everyday observation: a ball at rest on a surface has never observed to leap spontaneously upwards. (requires conversion of heat from the surface into work)

Process 1: First Law fulfilled? A perfectly elastic ball in a vacuum is dropped from

some height with an initial potential energy. When it strikes the ground, all energy is converted

into kinetic energy The ball bounces back to the original height, where

all of the kinetic energy is converted back to the original potential energy

Process 2: Why doesnt the egg bounce? An egg is dropped from the same height as the ball Initial and final states are not the same - seem to have lost

energy! First Law: we cannot lose energy - where did it go? Converted into random molecular motion and heat We wound up with a more disorganized form of matter The universe tends to more random disorganized states.

2nd Law: Entropy The universe tends towards more random,

disorganized states. This is a rather loose statement of the

second law of thermodynamics Our way of quantitating the disorder and

randomized motion in one state versus another state is a state function known as entropy

Increasing entropy means increasing disorder and randomized motion.

Entropy: Facts The fall into disorder can result in highly order

substances such as crystals, proteins, life!! Organized structures and patterns can

emerge as energy and matter disperse (i.e., the entropy increases)

The meaning of life: to increase entropy in the universe

Spontaneous Change: It means a process that results in a change from one state to another in an irreversible way. Anything that happens in the universe that results in an irreversible change in state is spontaneous.

Reversible Process Truly reversible processes do not happen in reality,

because in a truly reversible process all forces would be perfectly balanced and there would be no driving force for the system to move.

By moving things very slowly always keeping forces in near perfect balance, we can approximate reversible processes to whatever degree we like.

For example, during reversible expansion of a gas, we keep the pressures essentially the same on the inside and the outside.

If this was strictly true, the gas would not have any driving force to expand and nothing would ever happen.

However, we can make it as close to true as we like by making the imbalance as small as we want.

Irreversible (Spontaneous) Change All processes that really happen are irreversible;

forces driving process are substantially out of balance

The 1st Law uses the internal energy to identify permissible changes

2nd Law uses the entropy to identify the spontaneous change among these permissible changes

2nd Law of thermodynamics can be expressed in terms of entropy

The entropy of an isolates system in the course of spontaneous change:

Entropy is a state function. To prove this assertion, we need to show that integral of dS is independent of path. To do so, it is sufficient to prove that integral of

revdqT

around an arbitrary cycle is zero. That guarantees that the entropy is the same at the initial state and final state of the system regardless of the path taken between them.

5.2 Heat Engines and the Second Law of Thermodynamics

Heat engines produce work produced by release heat in a combustion process

Contacting the cylinder with hot or cold reservoirs changes in temperature which generate mechanical to a rotary motion, which is used to do work.

The Carnot Cycle It focuses on the efficiency of a heat engine and is a

milestone of 2nd Law development. In a Carnot engine (an ideal machine), there are four

reversible steps during each cycle of operation as follows:

Carnot Cycle: Steps Reversible isothermal expansion from A to B at

Th. qh is heat supplied from a hot source, and is positive.

Reversible adiabatic expansion from B to C, temperature falls from Th to Tc, where Tc is the temperature of the cold sink.

Reversible isothermal compression from C to D. qc is heat released into a cold sink, and is negative.

Reversible adiabatic compression from D to A, temperature rises from Tc to Th.


The PV diagram of reversible Carnot cycle consists of two adiabatic and two isothermal segments.

The arrows indicate the direction in which the cycle is traversed.

Analysis of Carnot Cycle In Step 1, isothermal gas expansion (as shown in

the 1st Law Section) leads to U = 0. The work done during the gas expansion at constant temperature is

1 lnB

A

VB

hVA

Vw PdV nRTV

= = Since temperature is constant in Sept 1:

1

1 1 1

1 1

00 (First Law)

ln

B A

Bh

A

U U UU q w

Vq w nRTV

= = = + =

= =

Analysis of Step 2 In Step 2, heat transfer during the gas

expansion is zero (adiabatic) as follows:

2

2 2

0

C B

qw U U U

== =

( )2 V V c hw C T C T T= =

Step 2: Incorporation of Ideal Gas Law2

2

ln ln ln

c C

h B

V

V

V

T V

V T V

c C BV

h B C

dw PdVdU C dT

nRTC dT PdV dVV

C nRdT dVT V

dT dVC nRT VT V VC nR nRT V V

= =

= =

=

= = =

Step 3: Analysis In Step 3, an isothermal compression occurs and is

opposite to Step 1:

3

3 3 3

3

3 3

00 (First Law)

ln

ln

D

C

D C

V DhV

C

Dc

C

U U UU q w

Vw PdV nRTV

Vq w nRTV

= = = + =

= = = =

Step 4: Analysis In Step 4, an adiabatic compression occurs

and is opposite to Step 2:

( )4

4 4

4

4

4

0

A D

V V h c

V

V

qw U U Uw C T C T Tdw PdVdU C dT

nRTC dT PdV dVV

== = = = = =

= = VC nRdT dV

T V=

Step 4: More Analysis

ln ln

ln ln

h A

c D

V

V

V

T V

V T V

c DV

h A

V c D

h A

CnR

c D

h A

CR

c D

h A

CR

h D

c A

dT dVC nRT V

T VC nRT V

C T VnR T V

T VT V

T VT V

T VT V

= = =

= = =

C CV VR R

c D h AT V T V =


The efficiency of the Carnot cycle can be determined by calculating q, w, and for each segment of the cycle assuming that the working substance is an ideal gas.

Overall Analysis: Carnot Cycle New T-V correlation has been developed for

adiabatic process. The total heat absorbed during the cycle:

31 2 4q q q q q= + + +

ln 0 ln 0B Dh cA C

V Vq nRT nRTV V

= + + +

Total Work during Carnot Cycle

( ) ( )( ) ( )

31 2 4

ln ln

ln ln

ln ln

B Dh V c h c V h c

A C

B Dh V c h c V c h

A C

B Dh c

A C

w w w w w

V Vw nRT C T T nRT C T TV V

V Vw nRT C T T nRT C T TV V

V VnRT nRTV V

= + + + = + + = +

= Therefore the same conclusion of First Law is arrived. For cyclic process

1 3

0U q ww q q q

= + = = = +


5.1 The Reversible Carnot Cyclehttp://media.pearsoncmg.com/bc/bc_engel_physchem_1/simulations/content/reversible_carnot_cycle/main.html

The heat flow in the two isothermal segments of Carnot cycle is:

( ) ( )cdabcycle qqw 31 that so ,0 >

Change during Carnot Cycle Both work done and heat transfer are not equal to

zero in Carnot cycle because they are not variable of state.

Carnot made an important discovery that the sum of qrev/T during the cycle is equal to zero.

The symbol rev indicates reversible process. qrev/T is determined from the integration of from

infinitesimal heat change along a reversible path is

1

3

1 (Path 1: Isothermal Expansion)

1 (Path 3: Isothermal Compression)

revrev

h h

revrev

c c

dq qdqT T T

dq qdqT T T

= =

= =

Entropy Change during Carnot Cycle For Path 2 and Path 4, the process is adiabatic and

therefore dqrev = 0 and

0revdqT

= For the entire Carnot cycle, the sum of qrev/T is

1 30 0

ln ln

ln 0

rev

h c

B D

A C

B D

A C

dq q qT T T

V VnR nRV V

V VnRV V

= + + + = +

= =

Entropy Change: Continued

Therefore a new state variable qrev/T is defined herein. Similar to U and H which are state variables, U and Hare equal to zero for cyclic work.

Historically, heat engine is driven by the isothermal expansion of gas inside a Carnot engine when heat is absorbed from a heat source or heat reservoir at Th.

The same calculation just completed for gases applies to all types of materials and systems.

ln ln

ln ln 0

rev B D

C A

c cV V

h h

dq V VnR nRT V V

T TC CT T

= + = =


The efficiency, , of the reversible Carnot engine is defined as the ratio of the work output to the heat withdrawn from the hot reservoir.

Efficiency is always less than one shows that not all heat withdrawn from the hot reservoir can be converted to work.

0 and ,0, because 11 >

Carnot Engine Efficiency The efficiency, , of a heat engine is:

1hot

w wq q

= = is the ratio of work performed during the cycle and the heat adsorbed during the cycle. The greater the work output from a given supply of heat, the greater the efficiency of the engine. The work done during the complete cycle is

1 3 1 3

h c

w w w q qq q

= = += +

Carnot Engine Efficiency The overall energy flow diagram for a Carnot

Engine is as follow:

Therefore the engine efficiency is1 3 3

1 1 1

1 1 ch

w q q q qq q q q

+= = = + = +

q3

q1

Engine Efficiency is always smaller than one because q3 is

negative (heat released to the sink). Based on conclusion drawn from the entropy (qrev/T) balance during the cycle

31

31

3

1

0

Thus is

1

rev

h c

h c

c

h

c

h

dq qqT T T

qqT Tq Tq T

TT

= + =

=

=

=


The efficiency of the reversible Carnot heat engine with an ideal gas as the working substance is

( ) 11

1


KelvinPlanck formulation second law of thermodynamics:

It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservoir and the performance of an equivalent amount of work by the system on the surroundings.

Engine Efficiency Because the conclusion is contrary to experience, the initial assumption that engines A and B can have different efficiencies must be false. It follows that the relation between the heat transfers and the temperatures must also independent of the working material, and therefore that the following is always true:

1 ch

TT

=

The mathematical statement of the second law is

When the cyclic integral of is zero, the exact differential shows a state function called the entropy, S:

5.3 Introducing Entropy

= 0Tdqreversible

TdqdS reversible

Origin of Entropy Because all reversible engines have the same

efficiency, a new state function known as entropy is defined as

(For isothermal Process)

rev

rev

dqS dST

qT

= =

=

The law used to identify spontaneous change, and can be quantified in terms of a state function known as entropy, S. Using entropy, S, to identify the spontaneous (irreversible) changes among the permissible changes.

Entropy Analysis The entropy of an isolated system increases in the

course of spontaneous change ( )isolated system 0S

Unlike internal energy (U) or enthalpy (H), entropy is not conserved. The sum of the entropy changes of the system and the surrounding is always positive

( ) ( )system surroundings 0S S + In case there is a negative change of entropy of the system, there must be a positive change for the entropy of the surrounding. Entropy is a measure of molecular disorder in a system, letting us assess whether one state is accessible from another via spontaneous change.

Entropy: Concepts Entropy change in the extent to which energy

is dispersed in a random disorderly manner depends on the amount of energy transferred as heat (creates random thermal motion).

Work is not accompanied by increase in random motion, but rather, implies uniform motion, and is therefore not accompanied by changes in entropy.

What is the entropy change during the isothermal expansion of gas?

Similar step has been illustrated during the operation of Carnot engine.

5.4 Calculating Changes in Entropy

S must be calculated along a reversible path. Because S is a state function, S is necessarily

path independent, provided that the transformation is between the same initial and final states in both processes.

When U is zero reversible isothermal expansion or compression of an ideal gas,

i

freversible

reversible

VV

nRqTT

dqS ln1 ===

Entropy in Isothermal Expansion 1

0

ln

ln

ln

f f

i i

f frev

revi i

rev rev

V V frev V V

i

frev

i

frev

i

dqS dqT T

Uq w

VdVw PdV nRT nRTV V

Vq nRT

V

VqS nRT V

= = = =

= = = =

= =

Molecules will randomly occupy the space that is available to them - tendency of system to explore all available states.

5.4 Calculating Changes in Entropy For ideal gas that undergoes a reversible

change in T at constant V or P,

For a constant pressure process,

i

fmv

mvreversible

TT

nCT

dTnCT

dqS ln ,, ==

i

fmp

mpreversible

TT

nCT

dTnCT

dqS ln ,, ==

Temperature Dependence of Entropy The entropy change of a system at temperature T2

can be calculated from knowledge of initial temperature and heat supplied to make T

2

1

2

1

State 2

2 1State 1

2

1

since

(At constant Pressure)

ln

Trev

T

rev p

Tp

pT

dq CdTS S ST T

dH dq dq C dT

C dT TS CT T

= = == =

= =

It is assumed that Cp is heat capacity at constant P.

5.4 Calculating Changes in Entropy Any reversible or irreversible process for an

ideal gas described by Vi,Ti Vf,Tf can be treated a consisting of two segments, constant volume and constant temperature.

For this two-step process, is given by

Similarly, for any reversible or irreversible process for an ideal gas

i

fmV

i

f

TT

nCVV

nRS lnln ,+=

i

fmP

i

f

TT

nCPP

nRS lnln ,+=

5.4 Calculating Changes in Entropy

When liquid is converted to a gas at a constant boiling temperature through heat input at constant pressure

Similarly, for the phase change solid liquid,

onvaporizati

onvaporizati

onvaporizati

reversiblereversibleonvaporizati T

HTq

TdqS

===

fusion

fusion

fusion

reversiblereversiblefusion T

HT

qT

dqS===

The entropy change when the substance melts at Tf and boils at Tb:

( ) ( ) ( ) ( )

( )0

0

f b

f

b

T Tp fus p

f T

Tvap p

b T

C s dT H C l dTS T S

T T T

H C g dTT T

= + + +

+ +

All properties required, except S(0)can be measured calorimetrically, and the integral can be evaluated graphically. The area under the curve of Cp/Tagainst T is the integral required. Because dT/T = d lnT, an alternative approach is to evaluate the area under a plot of Cp against ln T.

Dependence on Temperature2

1

2

1

ln (At constant Volume)

TV

VT

C dT TS CT T

= = In other words, we can calculate the entropy at T2 if we know the entropy at T1

2

1

22 1 1

1

ln

TV

VT

C dT TS S S CT T

= + = +

Example

5.7 The Change of Entropy in the Surroundings and System

As H and U are state functions, the amount of heat entering the surroundings is independent of the path.

It is essential to understand this reasoning in order to carry out calculations for S and Ssurroundings.

gssurroundintotal SSS +=

One mole of an ideal gas at 300 K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because it is very large, the temperature of the water bath thermal reservoir in the surroundings remains essentially constant at 300 K during the process. Calculate , , and .

Example Problem 5.7

S totalSgssurroundinS

Because this is an isothermal process, , and . From Section 2.7,


0=U

J

VV

nRTVdVnRTwq

f

i

V

V i

freversible

31029.20.250.10ln300314.800.1

ln

==

=== wqreversible =

The entropy change of the system is given by

The entropy change of the surroundings is given by

The total change in the entropy is given by


13

62.7300

1029.2 ==== JKTqTdqS reversiblereversible

13

62.7300

1029.2 ==== JKT

qT

qS systemreversiblegssurroundin

062.762.7 =+=+= gssurroundintotal SSS

5.7 The Change of Entropy in the Surroundings and

If the system and the part of the surroundings with which it interacts are viewed as an isolated composite system, the criterion for spontaneous change is

gssurroundintotal SSS +=

0>+= gssurroundintotal SSS

5.8 Absolute Entropies and the Third Law of Thermodynamics

Under constant pressure the molar entropy of the gas can be expressed as

( ) ( )

+

++++=T

T

gasmP

b

onvaporizatiT

T

liquidmP

f

fusionT solid

mPmm

b

b

f

f

TdTC

TH

TdTC

TH

TdTC

KSTS

''

''

''

0

,

,

0

,

The heat capacity of O2 has been measured at 1 atm pressure over the interval 12.97 K < T < 298.15 K. The data have been fit to the following polynomial series in T/K,

Example Problem 5.9

094.46)(

:39.5466.43

10685.608384.0308.270.31)(

:76.4366.23

10943.910191.56927.0666.5)(

:6.2397.12

1011.2)(

:97.120

11,

3

34

2

2

11,

3

34

2

23

11,

3

33

11,

=

The transition temperatures and the enthalpies for the transitions indicated in Figure 5.8 are as follows:

Solid III solid II 23.66 K 93.8 J mol-1Solid II solid I 43.76 K 743 J mol-1Solid I liquid 54.39 K 445 J mol-1Liquid gas 90.20 K 6815 J mol-1

Example Problem 5.9

3

37

2

24

11,

3

35

2

2

11,

10819.110545.104093.071.32)(

:15.29820.90

10407.601516.01467.1268.81)(

:20.9039.54

KT

KT

KT

KJmolTC

KTK

KT

KT

KT

KJmolTC

KTK

mP

mP

+=

a. Using these data, calculate Sm for O2 at 298.15 K.b. What are the three largest contributions to Sm?

Example Problem 5.9

a.

There is an additional small correction for nonideality of the gas at 1 bar. (Linstrom, P. J., and Mallard, W. G., Eds., NIST Chemistry Webbook: NIST Standard Reference Database Number 69, National Institute of Standards and Technology, Gaithersburg, MD, retrieved from http://webbook.nist.gov.)


( )

11

15.298

20.90

,20.90

39.54

,39.54

76.43

,,

76.43

66.23

,,

66.23

0

,,

9.20427.3559.7506.27181.813.1098.1661.19964.3182.8

20.906815

39.54445

76.43743

66.238.9315.298

=++++++++=

++++

++++=

KJmol

TdTC

TdTC

TdTC

TdTC

TdTC

KS

gasmP

liquidmP

IsolidmP

IIsolidmP

IIIsolidmP

m

b. The three largest contributions to Sm are S for the vaporization transition, S for the heating of the gas from the boiling temperature to 298.15 K, and S for heating of the liquid from the melting temperature to the boiling point.


5.9 Standard States in Entropy Calculations

For an ideal gas at constant T where P = 1 bar, the molar entropy Sm is

( ) ( )= PbarPRSPS mm ln

Entropy Change of a Chemical Reaction Considering a common irreversible reaction as

follows

A B C Dn A n B n C n D+ + The entropy change as a result of the reaction

r C C D D A A B BS n S n S n S n S = + S of a reaction is dependent on the temperature and pressure. There a reference state of 25 C and 1 atm is applied to entropy, it is known as standard molar entropy (S0).

5.10 Entropy Changes in Chemical Reactions

The entropy change SR in a chemical reaction is a major factor in determining the equilibrium concentration in a reaction mixture.

However, it is often necessary to calculate entropy at other temperatures as follow:

=i

iv iR SS

However, it is often necessary to calculate at other temperatures as follow:

This equation is only valid if no phase changes occur in the temperature interval between 298.15 K and T.

5.10 Entropy Changes in Chemical Reactions

+= T298.15

15.298 'TC

SS pT

Entropy: Temperature Dependence Chemical reactions are often carried at

temperature other than 25C. As a result, we need to calculate the entropy

change during a reaction from that at the standard state.

S0 (T2)

S0 (25 C)

A (T2) B (T2)

A (25 C) B (25 C)

Entropy in Reaction( ) ( )

( )( )( ) ( )( )

2 2

2

2

2 2

2

2

0 02 , 25 , 25

250, A , B25

0, A , B25 25

0, B , A25

0

25

25

25

25

25

25

A T C B C T

C T

p pT C

T T

p pC C

T

p pC

T

pC

S T S C S S

dT dTS C C CT T

dT dTS C C CT T

dTS C C CT

dTS C CT

= + + = + +

= +

= +

= +

In general, Cp is the difference of the heat capacity of the products and that of the reactants.

The standard entropies of CO, CO2, and O2 at 298.15 K are

The temperature dependence of constant pressure heat capacity for of CO, CO2, and O2 is given by

Calculate for the reaction CO(g) + 1/2 O2(g) CO2(g) at 475 K.

Example Problem 5.10

( )( )( )

2

252

112

3

38

2

252

112

3

38

2

252

11

103968.210187.181.30,

104426.2107834.610937.786.18,

104973.1101415.310452.108.31,

KT

KT

molJKgOC

KT

KT

KT

molJKgCOC

KT

KT

KT

molJKgCOC

P

P

P

+=++=

+=

( )( )( ) 11215.298

11215.298

1115.298

138.205,

74.213,

67.197,

==

=

molJKgOSmolJKgCOS

molJKgCOS

S


( ) ( )( ) ( )

3

38

2

242

3

38

2

25

211

10940.310112.110983.9763.2

104973.14426.2103968.2211415.37834.6

10187.121452.1937.781.302

108.3186.18

KT

KT

KT

KT

KT

KT

molJKC P

++=

++++

+++=

The value is negative at both temperatures because the number of moles is reduced in the reaction.

Example Problem 5.10 (Solution)( ) ( ) ( )

( )11

11475

15.298

3

38

2

242

15.29815.298

11

215.29815.298215.298

26.88757.150.860594.1605.7654.17866.1250.86

10940.310112.110983.963.2750.86'

'

50.86138.2052167.19774.213

,21,,

==+++=

++

+=+=

==

=

molJK

molJKKTd

KT

KT

KT

KT

dTTCSS

molJK

gOSgCOSgCOSS

KTP

T

Pressure Dependence of Entropy We can assume that pressure change will not influence the

entropy for solids and liquids. Assume that pressure is changed under constant

temperature for an ideal gas, a small change of internal energy (dU) is defined by first law of thermodynamics as follows:

The small change of entropy is equal to

0 (isothermal process)

dU dq dw

dq dU dwdST T

dUdw PdVdST T

= +

= ==

= =

Pressure Dependence of Entropy Considering ideal gas law: PV = nRT, the R.H.S. is

constant at constant temperature. Thus the differential of PV against temperature will be

( ) 00

PV nRTd PV nRdTPdV VdPPdV VdP

PdV VdPdST T

== =

+ ==

= =

Application to Ideal Gas

( ) ( )

2

1

2 22 1

1 1

From ideal gas law:

ln ln

P

P

V nRT P

dPdS nRP

dPS nRP

P VS P S P nR nRP V

=

=

= = = =

If P2 < P1, the entropy of the gas will be increased at constant T.

Entropy Change: Mixing of two Gases Two gases (A and B) originally stored separately

in two flasks connected through a stopcock. By opening the stopcock, A and B will be mixed

together. It must be noted that A or B is at the same pressure P and different volume (VA and VB)before mixing.

There is no change of temperature during mixing. The pressure of A and B after mixing is maintained

at P and the volume after mixing is VA + VB. Furthermore, reaction is negligible during the

mixing.

Entropy Change in Gas Mixing

( )

(Before mixing) (Before mixing)

(After mixing)

A A

B B

A B

A B A B

AA A

A B

BB B

A B

PV n RTPV n RT

n RT n RTPV V

RTP n n P PV

nP P X Pn nnP P X P

n n

==

= +

= + = +

= =+= =+

Entropy Change Based on the entropy change induced by pressure

change, the change of entropy due to the mixing of

( )

( )

( ) ( )

2

1

2

1

ln ln

ln

ln ln

ln

ln ln

AA A A

A A

BB B B

B B

Mixing A B

A A B B

P X PS n R n RP P

n R X

P X PS n R n RP P

n R XS S S

n R X n R X

= = =

= = = = + =

the two gases is

Entropy of Mixing The entropy of the system is increased

when the two identical gases are mixed. For two different gases, the swapping of

molecules between the two flasks upon mixing changes the number of microscopic states of the system.

However, no entropy change occurs when two identical gases are mixed.

Entropy of Melting When ice is melted into water at equilibrium

transition temperature T, the change of entropy at constant pressure is

pmelt

Melt

dH dq

qS

T

=

=p meltq H=

Chemical Equilibrium: Objectives

Spontaneity is discussed in the context of a reactive mixture of gases.

The Gibbs free energy is introduced.

Gibbs Free Energy In order to know whether a reaction will occur

or not at constant P and T, just the enthalpy or internal energy will not be sufficient.

A measure of the reaction spontaneity will be necessary for the system which can exchange heat with the surrounding at constant temperature

The system can expand or contract in volume at constant pressure.

Therefore new state variable is needed for accessing reaction spontaneity

Gibbs Free Energy A new variable of state, Gibbs free energy G is

defined as follows

G H TS= G will be shown as a critical parameter for determining whether a process will occur spontaneously or not. Otherwise, the system can be at equilibrium. We are interested in correlating between Gwith the system capacity to do work. When a system has a change of state resulting in G, this results from changes in H, S and T.

Correlating G and other Parameters Again, for infinitesimal changes of G:

( ) ( )( )

(From Definition of H)

(From First Law)

(From Definition of S)

rev rev

rev

d G d H TSdG dH TdS SdT

d U PV TdS SdT

dG dU PdV VdP TdS SdTdq dw PdV VdP TdS SdT

dG TdS dw PdV VdP TdS SdT

= =

= +

= + + = + + +

= + + +

Correlating G and other Parameters Work can be either produced from volume-pressure

type or other forms:

*

*

*

rev

rev rev

rev

rev

dG dw PdV VdP SdTdw PdV dw

dG PdV dw PdV VdP SdTdG dw VdP SdT

= + + = +

= + + + = +

At constant pressure and temperature, G is equal to the non-expansion work only as follows:

*

*rev

rev

dG dwG w

= =

Compare this result with that of Example Problem 6.1. What can you conclude about the relative amounts of expansion and nonexpansion work available in these reactions?

Example Problem 6.2 (Solution)( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )

( )

( ) ( ) ( ) ( ) ( )

1

3

218822188

1

324224

218822188tan

5296

2.2052251.3610.7098.213815.298105471

,225,,9,8,

8182.20523.1860.7028.21315.29810891

,2,,2,,

,225,9,8,

=

+=

+=

=+=

+=

+=

kJmol

gOSlHCSlOHSgCOSTlHCHG

kJmol

gOSlCHSgOHSgCOSTgCHHG

gOSgHCSOHSgCOSTlHCUA

combustionmethane

combustionmethane

combustioneoc

The usefulness of G is in predicting the maximum nonexpansion work available through a chemical transformation and determine the direction of spontaneous change.

The condition for spontaneity is G

The following total differentials for U, H, A, and G can be formed

As U, H, A, and G are state functions, two different equivalent expressions can be formulated:

6.2 The Differential Forms of U, H, A, and G

VdPSdTSdTTdSVdPTdSdGPdVSdTSdTTdSPdVTdSdAVdPTdSVdPPdVPdVTdSdH

PdVTdSdU

+=+===+=++=

=

dVVUdS

SUPdVTdSdU

SV

+

==

Analysis of Gibbs Free Energy If the process is spontaneous at constant

temperature and pressure, the non-expansion work done by the system to the surrounding must be negative: 0G If no useful non-expansion is done when the system is at equilibrium:

0G =

Gibbs Free Energy For a closed system doing no non-expansion

work, = 0, the change of Gibbs free energy is

*revdw

dG VdP SdT= Change in G is proportional to changes in pressure and temperature. Because we control P and T, G is a very important quantity in chemical engineering.

Gibbs Free Energy: Temperature Dependence Determine G at a certain temperature T from G

at 25 C based on the assumptions that H and Sare not so dependent on T

( ) ( ) ( )( ) ( )

thermodynamics lecture notes 2014 1st part

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