Thermodynamics of Materials
6th Lecture 2008. 3. 19 (Wed.)
SdT TdS
VdP PdV
dE
dG
= + + − −TS PV
E TS
H
G = − = + −
P
&
T const.
at S
Td
dG = −
i≤ 0
P
&
T const.
at S Td
i−
=
e e i
Td S Td S Td S at const. T & P
= − −
SdT S
Td S
Td
e − i −−
VdP PdV
PdV
dQ
− + +=
P
&
T const.
at S
Td
dG = −
i≤ 0
V
&
T const.
at S
Td
dF = −
i≤ 0
V
&
S const.
at S
Td
dE = −
i≤ 0
P
&
S const.
at S
Td
dH = −
i≤ 0
Why the term “free” energy?
dS
dS dS
dS
dS dS
dS
i
e i
e
surr sys
total
엔트로피 남은
→상쇄되고
=
− +
=
+
=
Energy
Free
Energy
Available
TdS
dG i
→
→
→
−
= 상쇄되고남은에너지
(
e i)
e
T S S
S
T Δ − Δ + Δ
=
S T H
G = Δ − Δ Δ
≤ 0 Δ
−
= T S
i. S
H 와 Δ 만 구하면 된다
Δ
∴
( )
P
&
T const.
at S T H
T S S T H TS
H G
Δ
− Δ
=
Δ
− Δ
− Δ
= Δ
− Δ
= Δ
∫
=
Δ H C
pdT Δ S = ∫ C T
pdT
. ility
irreversib , C
p있다 수 결정할 크기를
구동력의 그
방향과 반응의
주어진
알면 온도의존성만
∴ 의
Cp에 관한 더 자세한 것은 나중에 다룸
ΔG = ΔH – TΔS and dG = - SdT + VdP 이 두 식은 어떻게 다른가?
이 두 식에서 G는 어떻게 다른가?
= − = − +
dG dH d(TS ) SdT VdP
dG = - SdT + VdP
이식을 이용해서 앞의 문제를 풀어라.
Homework
At a pressure of 1 atm the equilibrium melting temperature of lead is 600 K, and, at this temperature, the latent heat of melting of lead is 4810 J/mole. Calculate Gibbs free energy change when 1 mole of supercooled liquid lead spontaneously freezes at 590 K and 1 atm pressure.
The constant-pressure molar heat capacity of liquid lead, as a function of temperature, at 1 atm pressure is given by
Cp(l)= 32.4 – 3.1ⅹ10-3T J/K
and the corresponding expression for solid lead is Cp(s)= 23.6 + 9.75ⅹ10-3T J/K
p,Pb( l ) Pb( l )
l
S C dT
T
. . dT . lnT . T C
T
− −
=
⎛ ⎞
= ⎜ − × ⎟ = − × +
⎝ ⎠
∫
∫ 32 4 3 1 10 3 32 4 3 1 103
dG
= −SdT
+VdP
= −SdT
( )
590 3
600 32 4 3 1 10
a K
b a b K l
G → SdT . lnT . − T C dT
Δ =
∫
− =∫
− + × −( )
3 2 590600
3 1 10
32 4 2
102306 8674 104358 3125 2051 445 10 2069 89 18 445 2051 445 10
b a l
l
l
. T
G . T lnT T C T
. . . C
. . . C
−
→
⎡ × ⎤
Δ = −⎢ − + − ⎥
⎣ ⎦
= − + = +
= − = +
혹은
( )
590 3
600 23 6 9 75 10
d K
c d c K s
G → SdT . lnT . − T C dT
Δ =
∫
− = −∫
+ × +( )
( )
3 2 590
600
9 75 10
23 6 2
76609 8137 78175 5239 1565 7102 10 1507 6977 58 0125 1565 7102 10
c d s
s
s
. T
G . T lnT T C T
. . . C
. . . C
−
→
⎡ × ⎤
Δ = −⎢ − + + ⎥
⎣ ⎦
= − − = +
= + = +
혹은
[ ]
1565 7102 10 2051 445 10 485 7 10
a d s l
l s
G . C . C
. C C
Δ → = + − −
= − −
Pb( l ) l K
Pb( s ) s
K
S ( K ) . lnT . T C
S ( K ) . lnT . T C
−
−
= − × + ⎦⎤
= + × + ⎤⎦
3
600 3
600
600 32 4 3 1 10 600 23 6 9 75 10
Pb( l ) l K
Pb( s ) s K
m
Pb( l ) Pb( s )
m
S ( K ) . lnT . T C
S ( K ) . lnT . T C
S ( K ) S ( K ) H .
T
−
−
= − × + ⎦⎤
= + × + ⎤⎦
− = Δ = =
3
600 3
600
600 32 4 3 1 10
600 23 6 9 75 10
600 600 4810 8 017
600
[ ]
485 7 10
485 7 10 56 6 80 3
a d l s
G . C C
. . .
Δ → = − −
= − × = −
l s 56 6 C −C = .
590 0 137 80 83
total
cf ) G T S . . J Δ = − Δ
= − ×
= −
The First Law of Thermodynamics
Conservation of Energy
W Q
E = Δ − Δ
Δ
The Second Law of Thermodynamics
?
2 법칙을 1 법칙 만큼 확실하게 이해하면 된다 .
다음 서술 중에서 엔트로피 효과와 관련된 것은 어느 것인가?
㉠ 서로 다른 두 종류의 기체는 혼합된다.
㉡ 열 에너지는 온도가 높은 곳에서 낮은 곳으로 전달된다.
㉢ 물과 기름을 한 용기에 넣고 흔들어 혼합하여도 둘은 섞이지 않고 분리된다.
① ㉠ ② ㉡ ③ ㉠㉢ ④ ㉠㉡ ⑤ ㉠㉡㉢
Ludwig Boltzmann (1844-1906)
http://www.corrosion-doctors.org/Biographies/BoltzmannBio.htm
Boltzmann’s atom
David Lindley, The Free Press 2001.
'Boltzmann was seconded by Felix Klein.
The battle between Boltzmann and Ostwald resembled the battle of the bull
with the supple fighter.
However, this time the bull was victorious ... . The arguments of Boltzmann carried the day.
We, the young mathematicians of that time, were all on the side of Boltzmann ...'
http://www.timelinescience.org/resource/students/matter/boltzmn.htm
Argument between Boltzmann and Ostwald in1895
http://www.timelinescience.org /resource/students/matter /boltzmn.htm