cm1502 chapter 8 - second law

8/21/2019 CM1502 Chapter 8 - Second Law

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CM 1502 1

The Second Law of Thermodynamics

1. Why is there a need for the Second Law ofThermodynamics?

2. Spontaneous and Non-spontaneous Processes

3. Entropy

4. Molecular Interpretation of Entropy

5. Entropy of the Universe, System and Surroundings

6. T dependence of Entropy

7. Standard Entropies

8. Conversion of Heat into Work

9. Carnot Engine


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Why there is a need for the Second Law?

Limitations of the First Law of Thermodynamics

The total energy-mass of the universe is constant.

Euniverse= Esystem+ Esurroundings= 0

+Esystem= -Esurroundings -Esystem= +Esurroundingsor

The first law does not provide any information on the

spontaneityof a given process.

This is provided by the second law of thermodynamics.


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The first law states that when absorbed heat is converted

into work, the work done is equivalent to the heat absorbed

and so U = 0.

U = q + w

U = q + w = 0 then q = -w

Conversion of absorbed heat into work: q w

Why there is a need for the Second Law?

Limitations of the First Law of Thermodynamics


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Heat engine: a device that

converts heat energy into

work, by expansions andcompressions.

However, the second law says that the heat absorbed at

any one temperature cannot be completely transformed into

work without leaving some change in the system or its

surroundings.

*

q2

q1w

Hot SOURCE

Cold SINK


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5

Spontaneous Non-spontaneous

Spontaneous processes are processes which can take placewithout external intervention of any kind.

Examples of spontaneous physical and chemical processes:water runs downhill, heat flows from hot to cold,

iron + water + O2 rust. Non-spontaneous processes are processes which require

external intervention to take place (under normal conditions). Examples of non-spontaneous physical and chemical

processes: water running uphill, boiling and freezing of water.

Spontaneous Processes


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Non-spontaneous

Spontaneous

Non-spontaneous Spontaneous

Gas molecules in closed container

Throw bricks


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Mixing of gases

stopcock

closed

stopcock

opened1 atm Evacuated

vacuum

0.5 atm 0.5 atm

Free expansion

Non-spontaneous

Spontaneous

Non-spontaneous

Spontaneous


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Heat Transfer

Spontaneous


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Chemical Reactions

potassium and water

K + H2O

iron and water

Fe + H2O

Spontaneous Spontaneous
http://images.google.com/imgres?imgurl=http://dl.clackamas.cc.or.us/ch105-09/images/sponrxFe.JPG&imgrefurl=http://dl.clackamas.cc.or.us/ch105-09/spontane.htm&h=240&w=320&sz=9&hl=en&sig2=Y4Nc1n4nup13-N3ZbcIT0A&start=1&tbnid=zNp9vNgUU3k5CM:&tbnh=89&tbnw=118&ei=M8_qRKSzHauCJZr4-YUB&prev=/images%3Fq%3D%2Bsite:dl.clackamas.cc.or.us%2Bspontaneous%2Breaction%26ndsp%3D20%26svnum%3D10%26hl%3Den%26lr%3D%26rls%3DRNWE,RNWE:2005-40,RNWE:en


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Electrochemical Processes

www.public.asu.edu/.../lecture10.html

Galvanic cell Electrolytic cell

Non-spontaneousSpontaneous


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Entropy and the Second Law

The second law can be expressed as:

The entropy of the universe increases (Suniverse> 0)

when a spontaneous process occurs and is not

changed in an equilibrium.

So when Suniverse or total 0,

process isspontaneous

final states (products) more disorderedthan initial states (reactants)


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CM 150212

What is Entropy? Entropy (symbol = S) is a measure of disorder, mess or

randomness.

The greater the disorder, the higher the entropy, the higher the value

of S.

A microscopic interpretation is required for understandingentropy. e.g. gases have higher entropy compared to

solids.

S is a state function

[Ssystem= Sfinal states (products) Sinitial states (reactants)].

Feeling of entropy: Look at entropy of some systems.

Underline/Circle on the next seven slides.


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CM 1502 13http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter08/Text_Images/FG08_13.JPG

Physical States and Phase Changes

More (Ssystem> 0) or less (Ssystem< 0) entropy?


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CM 1502 14Figure 20.6

Pure solid

Pure liquid

Solution

MIX

Dissolution of a Solid



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Ethanol Water Solution of

ethanoland water

Dissolution of a Liquid

Figure 20.7



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O2gas

Dissolution of a Gas

Figure 20.8

O2gas dissolved in water



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Temperature Changes Increase

Heating leads to greater degree of thermal motion of the

molecules thus greater entropy.

Molecular motion includes translation, vibration and rotation.

stretch

bend



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Chemical Reaction

wps.prenhall.com/.../602/616516/Chapter_17.html




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CM 1502 19Figure 20.9

NO

NO2N2O4

Atomic Size or Molecular Complexity



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Predicting SsysValues

4. Temperature changes

1. Physical states and phase changes

2. Dissolution of a solid or liquid

5. Atomic size or molecular complexity

3. Dissolution of a gas

S increases as the temperature rises.

S increases as a more ordered phase changes to a less

ordered phase.

S of a dissolved solid or liquid is usually greater than the S

of the pure solid or solute. However, the extent depends

upon the nature of the solute and solvent.

A gas becomes more ordered when it dissolves in a liquid or

solid.

In similar substances, increase in mass relate directly to

entropy.

In allotropic substances (same element, different molecular

forms), increase in complexity (e.g. bond flexibility) relate

directly to entropy.


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Choose the member with the higher entropy in each of the

following pairs, and justify your choice [assume constant

temperature, except in part (e)]:

(a) 1mol of SO2(g) or 1mol of SO3(g)

(b) 1mol of CO2(s) or 1mol of CO2(g)

(c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3)

(d) 1mol of KBr(s) or 1mol of KBr(aq)

(e) Seawater in midwinter at 20

C or in midsummer at 230

C(f) 1 mol of CF4(g) or 1mol of CCl4(g)

(a) 1mol of SO3(g) - larger mass

and more complex molecule(b) 1mol of CO2(g) - gas > solid

(c) 3mol of O2(g) - larger number

of mols

(d) 1mol of KBr(aq) - solution > solid

(e) 23 C - higher temperature

(f) CCl4- larger mass

*


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Molecular Interpretation of Entropy

Microstates (W) position of atom.

Example: Expansion of a gas increase in number of microstates.

Increasing microstates happensFAST!

How does this relate to entropy?

Increase the number of ways of

arranging the particles Increase W

Increase disorder

Increase S

Figure 20.31 mol

2nNo. of

compartments

No. of

atoms


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Molecular Interpretation of Entropy

Entropy is a measure of how many different microscopicstates are consistent with a particular macroscopicstate.

Macroscopic states which consists of many different

microscopic states have large entropies.

Expressed mathematically:

Boltzmann equation: S = (R/NA) ln W = k ln W

where S is entropy, k is the Boltzmann constant = R/NA.

(R is the gas constant, NAis Avogadros number) =

1.38 x 10-23JK-1and W is the number of microscopic

states.


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1877 Ludwig Boltzmann

Boltzmann equation: S = k ln W

A system with relatively fewequivalent ways to arrange its

components, has smaller W, has relatively less disorder and low

entropy.

A system with manyequivalent ways to arrange its components,

has larger W, has relatively more disorder and high entropy.

S = k ln W

S = Sfinal Sinitial

S= k ln Wfinal k ln Winitial

S= k ln (Wfinal/Winitial)

*
http://en.wikipedia.org/wiki/File:Boltzmann2.jpg


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Is the melting of ice or freezing of ice a spontaneous process?

Is this process spontaneous in the freezer? Yes No

Depends on Stotal/universe, consider S of system andsurroundings.


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exothermic

- ve Ssys:system

becomes moreordered

exothermic

+ ve Ssys:

system becomesmore

disordered

endothermic

Exothermic and endothermic

reactions can be spontaneous.

Reactions that

have Ssys=

+ve or -ve can

be spontaneous.

Suniverse*

Stotal/universe= Ssystem+ Ssurroundings 0

Reactions that

have Ssurr

=

+ve or -ve can

be spontaneous.


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How to calculate Suniverse?

Stotal/universe

= Ssystem

+ Ssurroundings

1. Calculate Ssystem

2. Calculate Ssurroundings

Stotal/universe > 0: spontaneous.

Stotal/universe < 0: non-spontaneous.

Stotal/universe = 0: equilibrium process, which isspontaneous.

For a spontaneousprocess,

Suniverse= Ssys+ Ssurr 0

The Second Law requires thesum of Ssys+ Ssurrto be 0

for spontaneous processes.


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Ways to calculate Ssystem

Most straight forward way to calculate Ssystem

:

Ssystem = n Sproducts n Sreactants

Another way to calculate the difference in S between two states is to

find a reversible path between them and integrate the energy (heat)supplied at each stage of the path divided by the T at which heating

occurs:

At constant P, qp= H, thus Ssystem= Hsys/T.

T

qdq

T

1S

revf

irevsystem ==


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Another way to calculate Ssystem

Ssystem= qrev/T only applies if process is reversible, but what aboutirreversible processes?

From First Law, conversion of q to w is possible: q = -w

wrev= -nRT In(Vf/Vi) => qrev = nRT In(Vf/Vi) => from Ssystem= qrev/TSsys= nR In(Vf/Vi)

Because S, n and V are state functions, they do not depend on the

path. So, although the preceding calculation is derived for a reversible

process, S has the same value for any reversible or irreversibleisothermal (Ti= Tf) path that goes between the same initial and final V.

So, Ssysfor irreversible and reversible processescan be calculatedusing:

Ssys= nR In(Vf/Vi)

Expansion: Ssys> 0

Compression: Ssys< 0


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Ssurroundings

From First Law,Hsurr= -Hsys

From second law, for a process at constant P,

Ssys= Hsys/T

Applying to surroundings,

Ssurr= Hsurr/T =>

Ssurr = -Hsys/T

for all processes with the assumption that P of the surroundings isconstant.

At constant P, Stotal = 0 equilibrium.

For irreversible isolated systems, Stotal > 0


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At 298K, the formation of ammonia has a negative Ssys;

Calculate Suniverseand state whether the reaction occurs spontaneously at thistemperature.

N2(g) + 3H2(g) 2NH3(g) Ssys= -197J/K

Given Hf values: N2(g) = 0, H2(g) = 0 and NH3(g) = -45.9 kJmol

-1.

Hsys= [(2mol)(Hf NH3)] - [(1mol)(H

f N2) + (3mol)(H

f H2)]

Hsys= -91.8 kJ

Ssurr= -Hsys/T = -(-91.8x10

3 J/298 K) = 308 J/K

Suniverse= Ssys+ S

surr = (-197 J/K) + 308 J/K = 111 J/K

Suniverse> 0 so the reaction is spontaneous.

*


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T Dependence of S If process occurs at a particular T, for example, in the case of a phase

change:For example, fusion/melting, qp= Hfus(the enthalpy of fusion) and if

the process has been carried out at Tfus

=> Ssys= Hfus/Tfus. (1 of the 4 ways to express Ssys)

If process occurs over a range of Ts, then

Ssys= qrev/T 1

and qp= H = Cp x n x T 2

(refer to first law: coffee cup calorimeter)

=> substitute 2 into 1:

Ssys= Cp n T/T = Cpn ln(Tf/Ti)

This equation assumes that the T interval is small enough that the T

dependence of Cpcan be neglected.*

Integrate w.r.t. T


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V and T change

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Calculate the entropy change when argon at 25 C and 1 atm in acontainer of volume 0.5 L is allowed to expand to 1 L and is

simultaneously heated to 100 C. Cp,m= 20.79 J mol-1K-1

Ssys= nR In(Vf/Vi) = (PiVi/Ti) In(Vf/Vi)

Ssys= Cvn ln(Tf/Ti)

Step 1: Increase V (expand)

Step 2: Increase T (heat)

Ssys= Cvn ln(Tf/Ti) = (Cp R) (PiVi/RTi) In(Tf/Ti)

Overall,

Ssys= (PiVi/Ti) In(Vf/Vi) +

(Cp R) (PiVi/RTi) In(Tf/Ti)

= (0.17) In(2) +

(12.476) (0.02) In(1.252)

= 0.17 J K-1


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Entropy Change of Chemical Reactions

Most straight forward way to calculate Ssystem

Ssystem = n Sproducts n Sreactants

If standard conditions (1 atm, 298 K) apply, this is indicated with the

superscript o,So = n Soproductsn S

oreactants

So for example, the entropy change of the following reaction is

CaCO3(s) CaO(s) + CO2(g)

So

= [So

(CaO) + So

(CO2)] So

(CaCO3)= (39.75 + 213.8) 92.88 = 160.7 JK-1

How to determine So for substances?


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Standard Entropies (S)

How to determine So for substances?

This is different as for H and G which are derived from Hfand Gf,respectively. [G will be discussed in next lecture.]

There is NO entropy of formation.

From the Boltzmann equation: S = k ln W, a zero entropy can becalculated if W = 1. In a perfect crystal, each atom occupies oneof the

crystal lattice sites and has the lowest possible energy.

When is W = 1?

W = 1 at absolute zero temperature (0 K).

So at absolute zero of temperature (0 K), in a perfect crystal there is

only one distinguishable microscopic state and the entropy is zero.

This is used as the standard point. (Third law of thermodynamics)


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Calculation of Standard Entropies (S)

For example, calculate S of a

substance at 393 K. Need to

extrapolate from 0 K.

S393 K

=

CpIn(T2/T1) +

CpIn(T3/T2) +

CpIn(T4/T3)

T1 = 0 KT2 T3

T4 = 393 K

Hfus/Tfus+

Hvap/Tvap+

*

Equation assumes that the Tinterval is small enough that

the T dependence of Cpcan

be neglected.

T1has to be ~0 K, e.g. 0.0001 K


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37

Conversion of Heat into Work

In a heat engine, two reservoirs, one hot and the other cold are in turnsbrought into contact with the cylinder.

Hot reservoir (source) expansion, input of heat which is converted

into work.

Cold reservoir (sink) compression, outflow of heat which allows work.

Linear motion that is mechanically converted to rotary motion.

CM 1502


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Efficiency ()

Efficiency is the fraction of the heat that is transformed into work.

Efficiency () = w/qH

For the most efficient heat engine, qH= w, and so = 1.

Higher T

Source

qH

Lower T

Sink

qC

Heat engine

w


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The Second Law

Kelvin statement:

It is impossible for a system to

convert heat completely into work.

Clausius statement:

Heat cannot flow spontaneously

from a cooler to a hotter object.

*


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Carnot Engine

the most efficient heat engine cycle

Nicolas Lonard Sadi Carnot

Heat inWork done Heat out

qC


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Carnot Engine the most efficient heat engine cycle

universe-review.ca/option2.htm

4 reversible steps:

1. Rev. isothermal expansion from ato bat TH. S = qH/TH, q enter from hot source.

2. Rev. adiabatic expansion from bto c. S = 0 as q = 0, THdecreases to TC.

3. Rev. isothermal compression from cto dat TC. S = -qC/TC, q release to cold sink.

4. Rev. adiabatic compression from dto a. S = 0 as q = 0, TCincreases to TH.

TH

TC

E i h C C l


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Entropy in the Carnot Cycle

0T

q

T

qS

C

c

H

H

sys =+=

qHTH

TC

Rev. isothermal

expansion

S = qH/TH

Rev. isothermal

compression

S = -qC/TC

Rev. adiabatic

expansionq = 0

qC

Ssysonly involvesthe isothermal steps

C

c

H

H

sys

T

q

T

qS ==

For a reversible cyclicprocess,Rev. adiabatic

compression

q = 0

In a generalized cycle dS = 0


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H

C

H

C

H

CH

q

q1q

q1q

qqq

wabsorbedheat

performedwork =+=

+===

Efficiency () for a Carnot Cycle

H

C

H

C

C

H

C

H

T

T

q

q

T

T

q

q==

H

C

T

T-1 =

C

H

C

H

C

C

H

H

T

T

q

q

T

q

T

q==Ssys=


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100% Efficiency from a Carnot Cycle

The efficiency of a heat engine depends on the Ts of the

hot source and the cold sink.

The lower the Tcand the higher the Th, the greater the

efficiency.

To achieve 100% efficiency the Tc/Thterm has to be 0.

Theoretically, to do this:

Have the cold sink at absolute zero of T (0 K). Have the hot source at infinite T.


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Carnot Cycle

Any reversible cycle can be

approximated as a

collection of a large number

of Carnot cycles, with

alternating isothermal and

adiabatic segments.

So, for a cyclic process, S = 0

In fact, for any cyclic process, the

change in any state function is

zero.

Isotherms

Adiabatic

processes

cm1502 chapter 8 - second law

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