Homework
Chap 3. The Second and Third Laws
Problems: 3A.5, 3A.6, 3A.10, 3B5, 3C.3,
3D.1, 3D.6, 3D.7
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
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
ⅹ 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)
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.
(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 !!!
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
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More heat is withdrawn from the hot reservoir than is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
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
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
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:
• 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.
엔트로피는 절대 감소하지 않는다.
증기 기관 (steam engine)
http://en.wikipedia.org/wiki/Steam_engine
카르노 엔진
• 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
dS dq rev D S dq T rev
3장. 수업목표 1
surr surr
surr
T
S q D
q w
IMPOSSIBLE!
T
1(hot)
T
2(cold) q
1q
2w
OK!
w
q
2q
1T
1(hot)
T
2(cold) OK!
l n
S k W
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)
ⅹ
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.
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.
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
3A.2 The definition of entropy
• Ex 3A.1 (p115): DS of a perfect gas when it expands isothermally from V
ito V
f0 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
For the surroundings: dq
surr dU
surror dH
surrdV
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
surrwhen 1.00 mol N
2O
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
3A.2(b) The statistical definition of entropy
Boltzmann's formula: S k 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
• 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
dS dq
rev
0 dq
T
1
c1
ch 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
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
cRefrigerator, Heat pump
T
h(hot)
T
c(cold) q
hq
cw
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 w T T
Coefficient of performance, c
3A.3 The entropy as a state function
To prove it, show dq rev 0 dS T
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
iT V
T V
C R C R
C
T V
C C
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
dS T T T
, 0
h h
c c
q T
q T
If dS
Carnot cycle for an ideal gas
ln
Bh
B
h A h
A
q w nR V
V T V
pd
ln
Dc
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
Dh 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
hc
T
T
h c
0
h c
q q T T
• B→C: q=0; w
2= C
V(T
c−T
h)
• C→D:
• D→A:
state func n
: ti o
S
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 c0
h c
q q T T
c h
c h
q q
T
T
If 1 , the n
h c0
h c
c h
q q T
T
T T
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
cc
h h
T T q
q
contradicts the 2
ndlaw (Kelvin)
If η A > η B
,
1
cCarnot ideal gas
h
T
T
h c
0
h c
q q T T
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
T T
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.
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
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
dS dq
T
0
dS
3A.4(a) D S for expansion
DU = 0
DS for isothermal expansion of a perfect gas
ln
frev rev
i
q w nRT V
V
ln f
v
i
re rev
S d
V
q q
T T
nR V D
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
fsurr 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 revS D S
D D
,
0
total surr irrev
S D S S S
D D D
3A.4(b) D S for phase transitions
trs trs trs
trs p
T S H
H q
D D
D
Note that D
trsS is a molar quantity
At T trs , two phases in the system are in equilibrium
Trouton’s rule:
D
vapS
o~ 85 J/K mol
V
gasare similar
• At const p, dq rev = C p dT
( ) ( )
f( ) ( )
fi 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 fln
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 fln
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 T S T
• At const V, dq rev = C V dT
Isothermal expansion:
Heating at const
ln
t ln
an :
f fi 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)
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
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 q l 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 s T
D D
D
• 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
S m of N 2 at 298.15 K
0
( ) ( ) ( )
( ) (0)
f bf 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
3C
pT aT
Debye extrapolation:
절대영도 근처에서
비금속 고체의 열용
량은 온도의 3승으로
잘 나타내진다.
0
( ) ( ) ( )
( ) (0)
f bf 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