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Chap 3. The Second and Third Laws

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(1)

Homework

Chap 3. The Second and Third Laws

Problems: 3A.5, 3A.6, 3A.10, 3B5, 3C.3,

3D.1, 3D.6, 3D.7

(2)

Chap 3. The Second and Third Laws

• What is a driving force of chemical and physical change?

Spontaneous process, nonspontaneous process

• Entropy , S: (Greek, to give a direction) the index of energy dispersal

determines the direction of spontaneous change

• Gibbs free energy , G: in terms of the properties of a system G ≡ H − TS

Maximum nonexpansion work available

(3)

The universe has a natural direction of change

• Heat flows from hotter bodies to colder bodies.

• Gases mix rather than separates.

A metal rod initially at a uniform T can undergo a spontaneous transformation in which one end becomes hot and the other end becomes cold without being in conflict with the first law

• An ideal gas that is uniformly distributed in a rigid adiabatic container could undergo a spontaneous transformation such that all of the gas moves to one-half of the container, leaving a vacuum in the other half.

T hot T cold

Supplement:

• Experience tells us that there is a natural direction of change: spontaneous process

• The first law does not tells us the direction

• Entropy (S) can tell us the direction:

It is more important to learn how to calculate the changes in S than it is to understand the S.

(need the 2 nd law)

(4)

Heat engines and the 2 nd law of thermodynamics

Supplement:

• Work can be converted to heat with 100% efficiency

• What is the maximum theoretical efficiency of the conversion of heat to work?

There is a natural asymmetry in the efficiency of

converting work to heat and converting heat to work.

The efficiency of heat engine

→ Requires the 2 nd law.

(5)

(Recall) Isothermal reversible expansion of a perfect gas

 

ln ln

ln

f f

i i

f

i

V V

V V

V

f i

V

f i

w pdV nRT dV

V

nRT dV nRT V V

V nRT V

V

   

    

 

   

 

 

Reversible path: the maximum work !!!

(6)

Heat engines and the 2 nd law of thermodynamics

Supplement:

• Reversible path produces the maximum work

• A reversible engine has the maximum efficiency of the real engine

The efficiency of heat engine

(7)

Heat engines and the 2 nd law of thermodynamics

Supplement:

• The second law of thermodynamics

More heat is withdrawn from the hot reservoir than is

deposited in the cold reservoir

(8)

The 2 nd law of thermodynamics

Supplement:

Clausius: It is impossible for a system to

undergo a cyclic process whose sole effects are the flow of heat into the system from a cold

reservoir and the flow of an equivalent amount of heat out of the system into a hot reservoir

Kelvin-Planck: It is impossible 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

(9)

w q 2

q 1

q 2

• Clausius: It is impossible for any system to operate in a cycle that takes heat from a cold reservoir and transfers it to a hot reservoir without at the same time converting some work into heat.

T 1 (hot)

T 2 (cold)

q 2 > 0 w > 0 q 1 < 0

‒q 1 = w + q 2

OK!

T 1 (hot)

T 2 (cold)

q 1

q 2 > 0 q 1 < 0

‒q 1 = q 2

IMPOSSIBLE!

Different statement of the second law

Supplement:

Engine

(10)

Different statement of the second law

• Kelvin: It is impossible for any system to operate in a cycle that takes heat from a hot reservoir and converts it to work in the surroundings without at the same time transferring some heat to a colder reservoir.

q

w

q > 0 w < 0

‒w = q

IMPOSSIBLE!

T 1 (hot)

T 2 (cold)

q 1

q 2 w

q 1 > 0 w < 0 q 2 < 0

q 1 = ‒w ‒ q 2

OK!

Supplement:

(11)

• First law: showed the equivalence of work and heat

DU = q + w, for a cyclic process → q = ‒w

Suggests engine can run in a cycle and convert heat into useful work

0 dU

Supplement

• Second law:

– Puts restrictions on useful conversion of q to w

– Follows from observation of a directionality to natural or spontaneous processes

– Provides a set of principles for

• Determining the direction of spontaneous change

• Determining equilibrium state of system

• Heat reservoir: A very large system of uniform T , which does not

change regardless of the amount of heat added or withdrawn. Also called

heat bath. Real systems can come close to this idealization.

(12)

엔트로피는 절대 감소하지 않는다.

증기 기관 (steam engine)

http://en.wikipedia.org/wiki/Steam_engine

카르노 엔진

(13)

• Entropy, S (에너지 분산의 척도, state function)

• The 2 nd law: DS tot > 0 in a spontaneous process

dS total = dS sys + dS surr

• Statistical view:

• Different statement of 2nd law T

dSdq rev D S dq T rev

3장. 수업목표 1

surr surr

surr

T

Sq D

q w

IMPOSSIBLE!

T

1

(hot)

T

2

(cold) q

1

q

2

w

OK!

w

q

2

q

1

T

1

(hot)

T

2

(cold) OK!

l n

Sk W

(14)

3A.1 The Second Law

What determines the direction of spontaneous change?

T hot T cold

1. total energy? minimum energy? (energy is conserved) 2. distribution of energy (toward a more disordered form)

Puzzle??

• formation of a crystal

• synthesis of a protein

During a spontaneous change in an isolated system the total energy is

dispersed into random thermal motion of the particles in the system.

(15)

The 2 nd law of thermodynamics

Clausius: Heat does not flow from a cool body to a hotter body.

Kelvin-Planck: No process is possible in which the sole result is the absorption of heat from a reservoir and its

complete conversion into work.

(16)

dS total = dS sys + dS surr

The S of an isolated system increases in the course of a spontaneous change: DS tot > 0

The 2 nd law of thermodynamics

Entropy, S (a measure of energy dispersed in a process)

the signpost of spontaneous change

When there are no more possible spontaneous processes, S is at its max

(17)

3A.2 The definition of entropy

• Ex 3A.1 (p115): DS of a perfect gas when it expands isothermally from V

i

to V

f

0 ln

1 ln

f

rev rev

i

rev rev f

rev

i

U q w nRT V

V

dq q V

S dq nR

T T T V

 

D       

 

 

D      

 

 

(a) The thermodynamic definition of entropy

To calculate DS, find a reversible path between them

rev

dq rev

dS T

S dq D  T

 

(18)

 

For the surroundings: dq

surr

dU

surr

or dH

surr

dV

surr

 0 or dp

surr

 0

For an adiabatic change:

Since dq surr  0, D S surr  0

rev

r ve

dS dq

T

S dq D T

  

• ST 3A.2 (p116): DS

surr

when 1.00 mol N

2

O

4

(g) is formed from 2.00 mol NO

2

(g) under standard conditions at 298 K.

   

      

2 2 4

2

1

4 2

2

2 9.16 2 33.18 57.2

57.2 29 8 192

o

surr r

surr

surr

o o o

r f

su

f

rr

q H k

NO g N O g

H H N O g H NO

S J JK

g J

T

k

T K

D

D

D  D  D   

   

 

,

surr rev su

surr

surr urr

r s

dS r

T

dq T

dq

surr surr

surr surr

q q

S T T

 

 D 

(19)

3A.2(b) The statistical definition of entropy

Boltzmann's formula: Sk l n W

Statistical entropy

If the molecules in the system have access to a greater number of energy levels, then there are more microstates for a given total energy, W is greater and S is greater.

23 1

where is the

Boltzmann constant,

number of microstates 1.381 10

k JK

W

 

 

(20)

• For any reversible cycle

3장. 수업목표 2

D f

i

rev

T S dq

D D

T

T

p b

T vap T

p f

T p fus

b b

f f

T

dT g

C T

H T

dT l

C T

H T

dT s

S C T

S ( ) ( ) ( )

) 0 ( )

( 0

 

 

  D

i f

V nR V

S ln T

dSdq

rev

0 dq

T

1

c

1

c

h h

q T

q T

    

• Clausius inequality

• DS for isothermal expansion

DS tot =DS sys +DS surr > 0

• DS for heating at const p

• The third law:

S of all perfect crystalline substances is zero at T = 0

(21)

work performed

heat absorbed from hot sourc 1

e

c

h h

h h

w c q

q

q

q q

q

 

    

The efficiency of a heat engine:

1 c

h h

w q

q q

   

 

 

5 0.25 20

1 1 15 0.25

20

h

c h

w q

q q

  

  

     

q

h

> 0 w < 0 q

c

< 0

q

h

= ‒ w ‒ q

c

(22)

Refrigerator, Heat pump

T

h

(hot)

T

c

(cold) q

h

q

c

w

Heat engine

w q c

q h T h (hot)

T c (cold)

If cycle were run reverse, then q

h

< 0, q

c

> 0, w > 0 a refrigerator!

,

1 1

1 1

h c

h c h

c

c

h c

c c c

h h c

c c

c q w q

w q

q q q

c q q q

T T T

c T

w

c

T T T T

   

     

  

c c

h c

q T

c wT T

 

Coefficient of performance, c

(23)

3A.3 The entropy as a state function

To prove it, show dq rev 0 dST

  We will prove it by a Carnot cycle

Strategy

1. For a Carnot cycle involving a perfect gas 2. For whatever the working substance

3. For any cycle

,

, ,

,

1

, ,

For adiabat:

1

V m

R C

f i

i f

p m V m

V m V m

f

i f

V m

T V

i

T V

T V

C R C R

C

T V

C C

 

     

 

 

  

  

 

 

(24)

A→B: isothermal expansion at T h DU 1 = q h +w h B→C: adiabatic expansion (q = 0) DU 2 = w 2 C→D: isothermal compression at T c DU 3 = q c +w c D→A: adiabatic compression (q = 0) DU 4 = w 4

3A.3(a) The Carnot cycle

A typical heat engine

(all paths are assumed to be reversible)

rev h c

h c

q q q

dSTTT

 

, 0

h h

c c

q T

q T

If    dS

(25)

Carnot cycle for an ideal gas

ln

B

h

B

h A h

A

q w nR V

V T V

  pd  

  

 

 

ln

D

c

D

c C c

C

q w nR V

V T V

  pd  

  

 

 

1

1

 

 

 

 

 

 

D A C

B h

c

V V V

V T

T

re c

0

h c

v h

dS dq

T

q q T T

 

  

• A→B: DU = q

h

+ w

h

= 0

1

 

 

 

A D c

h

V V T

T

 

 

l ln

c

n

D

h h B A

c C

q T

T V

q

V V

V

1

 

 

 

C B h

c

V V T

T

 

 

 

 

 

 

 

 

 

 

A D

D B C

A

D A C

B

V V V

V

V V V

V

V V V

V

h

c

T

T

h c

0

h c

q q TT

• B→C: q=0; w

2

= C

V

(T

c

−T

h

)

• C→D:

• D→A:

state func n

: ti o

S

(26)

Efficiency of Carnot cycle

1 c

h h

w q

q q

   

, 1 1

Carnot ideal ga

h s

c h

q c

q

T

     T

For an ideal-gas Carnot cycle

h c

0

h c

q q TT

c h

c h

q q

T

   T

If 1 , the n

h c

0

h c

c h

q q T

T

T T

    

(27)

1

for any substance

involved in a Carnot cycle

c c c

h h h

T q T

T q T

     

All reversible Carnot engines have the same efficiency regardless of their construction

rev 0 dq

  T

, l g

If

Carnot idea as

, 1 1

c

c

h h

T T q

      q  

contradicts the 2

nd

law (Kelvin)

If η A > η B

,

1

c

Carnot ideal gas

h

T

   T

h c

0

h c

q q T T

  

(28)

For a reversible Carnot cycle:

For any reversible cycle:

rev

0 dq

T

rev

0 dq

T

Note that Entropy is a state function, but to calculate DS requires a reversible path

rev rev

0

all perimeter

q q

TT

 

Entropy: a state function!!!

Any reversible cycle can be approximated as a collection of Carnot cycles:

it becomes the sum of the integrals around each of the Carnot cycles.

(29)

3A.3(b) The thermodynamic temperature

1 c

h

T

   T

If we know the T h , T can be determined from the efficiency of the heat engine.

T = (1 ‒ η) T h

Kelvin scale: T 3 of water = 273.16K is defined as T h

(30)

For an isolated system dq = 0,

3A.3(c) The Clausius inequality

1 0 1  

 

  

 

c h

c h

total

T dq T

T dq T

dS dq

 0 dS

• ex, spontaneous cooling

dS dq

T

0

0

rev rev

rev rev

rev rev

rev rev

dw dw dw dw

dU dq dw dq dw dq dq dw dw

dq dq

dq dq dq

T T dS T

     

   

   

  

S의 정의가 제 2법칙과 일치함을 보여줌

(31)

dS dq

T

 0

dS

(32)

3A.4(a) D S for expansion

DU = 0

DS for isothermal expansion of a perfect gas

ln

f

rev rev

i

q w nRT V

V

 

    

 

ln f

v

i

re rev

S d

V

q q

T T

nR V D

 

 

 

  

(33)

DS for isothermal expansion of a perfect gas (DU = 0)

ln f

i

S nR V

V

 

D  

 

,

, surr irrev irrev

0

surr irrev

surr

q q

S T T

D    

,

, surr rev rev

ln

f

surr rev

surr i

q q V

S nR

T T V

 

D       

 

(ii) free expansion (irreversible) w = 0, q irrev = 0

(i) reversible:

dq surr = ‒ dq

T surr = T (isothermal)

rev rev

q   w

경로에 무관한 상태함수

DS total 는 경로에 무관한가?

DS

tot

=DS

sys

+DS

surr

, ,

0

total rev

S

surr rev

S D S

D   D 

,

0

total surr irrev

S D S S S

 D   D  D 

(34)

3A.4(b) D S for phase transitions

trs trs trs

trs p

T S H

H q

 D D

D

Note that D

trs

S is a molar quantity

At T trs , two phases in the system are in equilibrium

(35)

Trouton’s rule:

D

vap

S

o

~ 85 J/K mol

V

gas

are similar

(36)

• At const p, dq rev = C p dT

( ) ( )

f

( ) ( )

f

i i

T T

rev rev

f i T f i T

dq dq

S S T S T S T S T

T T

D       

3A.4(c) D S for heating

( ) ( )

f f

ln

i i

f V

T T

rev V

f i

T T i

dq C dT

S T S T T

T C

T

T  

 

 

    

( ) ( )

f f

ln

i i

f

i

T T f

p

i p T

T p

rev

f i T T

C dT

T

T dq C dT

T C

T T

S TS T      

 

 

 

• At const V, dq rev = C V dT

(37)

Isothermal expansion:

Heating at const

ln

t ln

an :

f f

i i

T T

V T V

i

T f

f V

i

C S nR V

V

dT dT

V T

T

S C T

T C

D   

 

D   

 

 

 

 

 

3A.4(d) D S for composite process

1 2 ,

3 2

1 2

, ln ln 3 ln

2

ln 3 ln ln

2

f f f

i i

V m

i i i i

f f f f

i i i i

i i i i i i

V T T

n p V S nR S nC n R

RT V T T

V T V T

p V p V

S S S

T V T T V T

     

 D    D      

     

 

         

D  D  D                         

• Ex 3A.2 (p124): Ar (25 ºC, 1.00 bar, 0.500 L) → (100 ºC, 1.000 L) isothermal expansion and heating at const V

(25 ºC, 1.00 bar, 0.500 L) → (25 ºC, 1.000 L) → (100 ºC, 1.000 L)

(38)

Reversible phase change at constant T and p

e.g. H 2 O (l, 0°C, 1 bar) → H 2 O (s, 0°C, 1 bar)

reversible

fus rev

p

H

dq

,

  D

fus fus

S H

D D T

(39)

Irreversible phase change at constant T and p

e.g. H 2 O (l, −10°C, 1 bar) → H 2 O (s, −10°C, 1 bar)

This is spontaneous and irreversible. ∴ We need to find a reversible path between the two states to calculate ΔS.

H 2 O (l, −10°C, 1 bar) H 2 O (s, −10°C, 1 bar)

H 2 O (l, 0°C, 1 bar) H reversible 2 O (s, 0°C, 1 bar)

irreversible

p

( )

rev

C

d ql d T d q

rev

C

p

( ) s T d

fus rev

p

H

dq

,

  D

1

( )

1

( )

fu

fus

s fus

fus

heating f

T reeze cooling

T p

p

T T

C l d

S S S S C s dT

T T

H T

D  D  D  D  T  D 

 

   

1 1

( ) ( ) ( ) ( ) ln

Tfus fus

fus fus

fus T p p fus p p

H H

T T

dT T

S C l C s T C l C sT

D  D   

   

D

 

(40)

• From the heat required to raise the temperature of a sample from T = 0

• Using Boltzmann’s statistical definition.

3B The measurement of entropy

( ) (0

0

) ) )

)

f

(

b

( (

f b

p fus p vap

T T T

T T

p

f b

dT dT dT

S T S

T T

C s H C l H

T C g

T T

  D 

 D  

  

3B.1 The calorimetric measurement of entropy

Heating + phase transition

(41)

S m of N 2 at 298.15 K

0

( ) ( ) ( )

( ) (0)

f b

f b

T T T

p fus p vap p

T T

f b

C s dT H C l dT H C g dT

S T S

T T T T T

D D

        

 

3

C

p

TaT

Debye extrapolation:

절대영도 근처에서

비금속 고체의 열용

량은 온도의 3승으로

잘 나타내진다.

(42)

0

( ) ( ) ( )

( ) (0)

f b

f b

T p fus T p vap T p

T T

f b

C s dT H C l dT H C g dT

S T S

T T T T T

D D

        

 

       

   

       

3

3

2 3

0 0

1 1

,

At low temperature,

0 0 0 1

3 0 1

3

0 1 0 0.14

3

p

T T

p

m m p m m

C T aT

S T S aT dT S a T dT S aT

T

S C T

S T S C T S JK mol

     

 

   

 

• Ex 3B.1 (p127): What is S m (4.2K) when C p,m (4.2K) = 0.43 JK -1 mol -1 ?

(43)

3B.2 The Third Law

S of all perfect crystalline substances is zero at T = 0

• At T = 0, no thermal motion, no spatial distribution (perfect crystal), S = 0

• At T = 0, only one way of arranging molecule, W = 1 and S = 0

3B.2(a) Nernst heat theorem 3B.2(b) Third-law entropies

• For ions, let S o (H + , aq) = 0, D f H o (H + , aq) = 0

solution of cations cannot be prepared in the absence of anions

If all the substances involved are perfectly crystalline, D S → 0 as T→0

S o (T): S based on S(0) = 0

(44)

Standard Third- Law entropies

S of all perfect crystalline substances is zero at T = 0

 D

J

o j m J o

r SS ,

Standard reaction entropy

참조

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