Chapter 5: Entropyscienide2.uwaterloo.ca/~nooijen/website_new_20_10... · Winter 2013 Chem 254:...

Winter 2013 Chem 254: Introductory Thermodynamics

46

Chapter 5: Entropy ........................................................................................................................ 46

Calculating ΔS for processes in ideal gas .................................................................................. 48

ΔS as a function of changes in P,T,V in pure substances (not mixtures or reactions) .............. 51

The Second law of thermodynamics......................................................................................... 54

Absolute Entropy (Pure Substances) ........................................................................................ 56

Microscopic (molecular) view of Entropy ................................................................................. 56

Heat Engine ............................................................................................................................... 58

Carnot Cycle .............................................................................................................................. 59

Midterm Review............................................................................................................................ 61

Chapter 5: Entropy

Entropy: the second and third law thermodynamics

First law: Energy can be converted from one form to another but the total energy for an

isolated system is constant (heat-work-chemical energies)

0U (is constant if all processes are included)

Can we obtain a law that describes what can or cannot happen?

Mixture of 2N ,

2H (raise T , catalyst) forms 3NH in equilibrium, once it reaches

equilibrium, it does not go back to reactants, again we observe an “arrow of time”.

For isolated system

- Entropy ( S ) 0S for any possible, irreversible process


Chapter 5: Entropy 47

0S for a reversible process

- Other quantities (Ch 6)

G Gibbs free energy, used when ,T P constant

A Helmholtz free energy, used when ,T V constant

Entropy is an abstract quantity for macroscopic systems, thermodynamics

History of discovery

- discovered by engineer Carnot

- improve efficiency of steam engines (1830s)

- it is still taught like this (see 5.2 in book)

Physical insight into entropy microscopic theory

- Quantum Mechanics

- Statistical Mechanics (molecular point of view, discussed at Ch. 5 end)

Consider a reversible process for an ideal gas rev revdu q w ,rev revq w inexact differentials (depends on path)

rev

V extC dT q P dV

rev

V

nRTq C dT dV

V rev

ext gas

nRTP P

V

Differential for revq

revq : NOT a exact differential

V

T V

nRTC

V T V

0nR

V

0nR

V therefore revq is not an exact differential

rev

VCq nRdT dV

T T V

is this an exact differential?

V

VT

C nR

V T T V

0 0 therefore is an exact differential

revq

dST

entropy, an exact diff, independent of path



revf

f ii

qS S S

T

entropy is a state function

1

T : is called an integrating factor changes inexact diff to exact diff

Eg. From math

dF ydx xdy

yx

y xy x

1 1 dF is an inexact diff

2 2

1dF ydG dx dy

xx x

2

1

yx

y

y x xx

2 2

1 1

x x dG is an exact diff

2

1

x: integrating factor

Calculating ΔS for processes in ideal gas

revq

dST

(exact differential)

S is a state function

, ,S T P V of single component systems

final initialS S S

Like ,H U , S independent of process or path

How to calculate S

revf

i

qS

T

What does this mean?

Choose a reversible path: at all stages, the system retains equilibrium (stays on

the ideal gas curve)



Even for irreversible process calculate S for a reversible path with same initial and

final states

Ideal gas PV nRT

a)1 1 2 2TV T V

( ) ( ) ( )C A BS S S

A. Constant T (reversible isothermal) B. Constant V

0U q w 0w U q

( )f

exti

nRTq w P dV dV

V

Vq dU C dT

nRT

q dVV

VCdS dT

T

q nR

ds dVT V

ln

f fVV

ii

TCS dT C

T T

lnf f

ii

VnRS dV nR

V V

,V mnC nRdS dT dV

T V

1 1 2 2TV T V , ln lnf f

V m

i i

T VS nC nR

T V

b) 1 1 2 2PV PV

( ) ( ) ( )S F S E S D



D. Constant P E. Constant V (same as before)

P Pq H C dT ln

f fVV

ii

TCS dT C

T T

PCdS dT

T Constant Volume ,

nRT

P constant

lnf

P

i

TS C

T

f f

i i

T P

T P

f i

f i

T T

V V since

nRTP

V constant ln

f

V

i

PS C

P

lnf

P

i

VS C

V

, ,P m V mC CdS n dV n dP

V P

, ,ln lnf f

P m V m

i i

V PS nC nC

V P

c) 1 1 2 2PT PT

f i

i f

P V

P V constant T

,P mC RdS n dT n dP

T P

, ln lnf f

P m

i i

T PS nC nR

T P



d) adiabatic reversible process

0revq 0revq

T

0S

1 1 2 2TV T V adiabatic reversible process 0S

, ln lnf f

v m

i i

T VnC nR

T V

,

ln lnf fv m

i i

T VC

R T V

(see section 2.11 in book)

ΔS as a function of changes in P,T,V in pure substances (not mixtures or reactions)

Dependence of S on ,T P

P T

S SdS dT dP

T P

exact differential

P

P

C VdT dP

T T

Why is this true?

1) At constant P rev

PCqdS dT dT

T T

2) T P

S V

P T

Maxwell relation, see later

Liquids and solids:

P

VV

T

f f

i i

T PP

T P

CS dT V dP

T

f

i

TP

T

CdT V P

T

Gases:

P

V

T

a little effort

nRT

VP

P

V nR

T P

(ideal gas)



,P mC RdS n dT n dP

T P

, ln lnf f

P m

i i

T PS nC nR

T P

Dependence of S on ,T V

V T

S SdS dT dV

T V

V

V

C PdT dV

T T

1) At constant V , rev

VCqdS dT

T T

2) Maxwell Relation T V

S P

V T

V T P

P P V

T V T

Liquids and solid:

f f

i i

T VV

T V

CS dT dV

T

f

i

TV

T

CdT V

T

Gases:

Calculate V

P

T

f

i

V

VV

PdV

T

Ideal gas V

nRT P nRP

V T V

f f

i i

V V

V VV

P dVdV nR

T V

,f f

i i

T Vv m

T V

C nRS n dT dV

T V

, ln lnf f

v m

i i

T VS nC nR

T V

(as we saw before)

S for phase change (single substance)

s l s g l g



Reversible process: example l g

Boiling point of water, 100oC at 1 atm

373

orevvap

vap

Hq

T T

Vaporization 373 0o

vapH

o

vap T

moles

vap

HS n

T

s g s l

o

sub T

moles

sub

HS n

T

o

fus T

moles

fus

HS n

T

Example: Raise T of water at 1 atm from 25oC to 125 oC

298 K 373 K heating liquid

At 373 K liquid gas

373 K 398 K heating gas

373 398

, ,

298 373

liquid gas

vapP m P m

vap

HC CS n dT n n dT

T T T

Other example

liquidS freezing + cooling

systembath

bath

bath bath

qqS

T T

system systemq H (constant P )

OR system systemq U (constant V )

surrounding system

surrounding

surrounding surrounding

q qS

T T

(constant surroundingT )

0system surroundingS S for any possible process. (see later)



The Second law of thermodynamics

Postulate by Rudolph Clausius (~1865)

- A Postulate is something you cannot derive but take to be true, because large

number of experiments support it.

0dq

T for any possible physical process

(If 0dq

T for any reproducible process we would have to rewrite

thermodynamics and all of its rules.)

Anything derived from postulate should be true also

Consequences of Postulate:

a) 2

1 2 1

dqS

T for any process

2 2

1 1

revdq dq

T T

2 1

1 20

revdq dq dq

T T T

2 2

1 10

revdq dq

T T

b) For isolated system 0S for a possible physical process

2

1

dqS

T For isolated system 0q

So 0S

c) We can always create a system and environment such that total

system is isolated

0total system surroundingS S S

If environment is kept at constant T

sysenv

env

env env

qqS

T T

We can calculate S for ANY process (allowed and unallowed)

If we calculate totalS for isolated system then:

0S irreversible, possible spontaneous (ice forming at -5oC)



0S reversible, possible (ice equilibrium at 0 oC)

0S impossible (ice forming at 5 oC)

Example:

0q

T T show it?

0total

q qdS

T T

q qT T

T T

Heat flows from hot to cold (another formulation of the second law)

ln 0f

total

i

VS nR

V (at constant T )

0U w q

final initalV V

0

dq

T

2 1

1 20

irrev revdq dq

T T



Absolute Entropy (Pure Substances)

Molar entropy, all 3 phases, all H and ,P mC can be measured

, , ,

0( ) (0 K)

fus vap f

fus vap

S l gT T Tfus vapP m P m P m

m mT T

tus vap

H HC C CS T S dT dT dT

T T T T T

( ) (0)m m mS S T S

Nernst and Richards (~1910)

(0 K) 0mS for perfect crystalline solids.

Today, we can calculate absolute entropy from first principles, quantum and statistical

mechanics for ideal gas. Confirms (0 K) 0mS

Non crystalline substances

Perfect arrangement, but in practice we can get random arrangements

In this case (0 K)mS 5.76 J/mol ( ln 2R )

Microscopic (molecular) view of Entropy

Diffusion of gas:

Laws of microscope physics are time reversible? Why do processes occur in particular

direction?



Difficult (impossible..) to get the initial conditions to reverse such reactions

Calculate for diffusion of gas

Pr 1 1 1 1

......1 1 1Pr 1

2 2 2 2

A

full

N

half

Pr 1

ln ln ln 2 ln 2Pr 2

aN

full

B B B A

half

K K K N R

Thus Pr

ln ln 2Pr

full

B

half

K R (for 1 mol)

S for 1 2V V ln ln 2

f

i

VnR R

V

Similar idea for the crystal, only 2 orientations

Pr 1

Pr 1

2

a

ordered

N

disordered

ln 2 ln 2AN

B B AS K K N

Fully ordered state has zero entropy, ln 1 0AN

BS K

Quantum and statistical mechanic picture of entropy:

Quantum: jE quantum energy levels

Statistical: Probability to find molecule at particular state (Boltzmann

distribution)

/

/( )

j B

j B

E K T

j E K T

j

eP T

e

( ) lnB j j

j

S T K P P

This is our best understanding of entropy



Heat Engine

History: Carnot investigated efficiency of heat (steam) engines ~1830

Carnot: run each cycle in reversible fashion to get max efficiency, recall that grain of

sand compression is more efficient than irreversible process.



Carnot Cycle

All reversible processes:

A: 0U 0hot hotw q

B: 0q B B v cold hotw U C T T

C: 0U 0cold coldw q

D: 0q D D v hot coldw U C T T

total A Bw w w C Dw w

hot coldq q

0 0 0 0hot cold

total

hot cold

q qS

T T

hot cold

hot cold

q q

T T cold

cold hot

hot

Tq q

T

cold

total hot hot

hot

Tw q q

T

1 cold

total hot

hot

Tw q

T


60

Efficiency = useful quantity

1costly quantity

total cold

hot hot

w T

q T

(best efficiency you can ever get)

Abstraction of Heat Engine

Carnot: heat cannot be converted to work perfectly

1 cold

hot

T

T

Without wasted heat 1 , then 0coldT K

Refrigerators (or heat pumps)

Can’t convert all work into heat 0S

coldq

w

How do you do it? Run Carnot in reverse

Expand at cold T (you gain very little)

Compress at hot T (takes a lot of work)

cold cold

hot cold

T q

T T w

There is a lot of waste heat, we can use that to heat our homes (Heat pump). We extract

heat from a cold environment (lake) and release the heat in our living room.


Midterm Review 61

Midterm Review

4 questions, 50 points each, multiple parts

1) Question to derive something

2) Calculate r H ,

rU for reaction at particular T , using heats of reaction. Constant

pC value

3) Ideal gas cycle. Calculate , ,P V T , , , , ,q w U H S

4) Calculate S for spontaneous process

systemS : reversible

system

environment

bath

HS

T

More details on each

1) Derivatives:

- Manipulate derivatives

1

z

z

x

yy

x

y xz

x x y

y z z

- Use an exact differential

, , , ,dF A P V T dT B P V T dV

T V

A B

V T


Midterm Review 62

Why? Exact differential CAN BE written as

V T

B FdF dT dV

T V

- Mixed derivatives are equal. Any variation is possible

,dT dV

,dT dP

,dV dP (in principle)

Read paragraph 6.2 for application

2) Calculating r H ,

rU

H U PV

U H PV

U H RT n

gasn n (only gases count)

Calculating r H :

298298

To

r i f i p

i i

H T H i C i dT

i in for products

in for reactants

- use general formula Calculate

r H

r H is given, calculate missing piece

- Or draw diagram for complicated reactions and process (I will ask for it if needed)

- Or balance reactions to put proper data together

[examples in book]

3) Ideal gas cycle


Midterm Review 63

Adiabatic: 0q

Isothermal: constant T

Reversible: follows ideal gas curve

Calculate , ,P V T

Important : reversible adiabatic

,ln ln

f fV m

i i

T VC

R T V

Or ,ln ln

fP m i

i f

TC P

R T P

,V mU nC T

,P mH nC T

extw P dV

q U w (sometimes known to be 0)

,ln

f

i

T fV m

Ti

VCS n dT nR

T V

,ln

f

i

T fP m

Ti

PCn dT nR

T P

0S for adiabatic process

4) Calculate S for spontaneous process

systemS : use reversible path to calculate S always

environmentS : Often keep at constant T

system

environment

bath

HS

T

- Calculate systemH for the actual process

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Chapter 5: Entropyscienide2.uwaterloo.ca/~nooijen/website_new_20_10... · Winter 2013 Chem 254:...

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