thermodynamics of materials
TRANSCRIPT
Thermodynamics of Materials
9th Lecture2008. 3. 31 (Mon.)
A rigid container is divided into two compartments of equal volume by a partition. One compartmentcontains 1 mole of ideal gas A at 1 atm, and the other contains 1 mole of ideal gas B at 1 atm.Calculate the increase in entropy which occurs when the partition between the two compartments is removed.
1 mol(1 atm)
A
1 mol(1 atm)
B
Calculate the increase in entropy
1)by Boltzman’s way 2)by volume change of each gas3)by pressure change of each gas
1 mol(1 atm)
A
1 mol(1 atm)
B
12
1221
PlnkPlnkSSS
−=−=Δ →
121
21
2
1
=
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛=
P
PNN
볼츠만 엔트로피로 구하는 법
42222
22221
2
lnRlnRlnRlnR
lnklnklnkPPlnk NNNN
==+=
+==⎟⎟⎠
⎞⎜⎜⎝
⎛=
1 mol(1 atm)
A
1 mol(1 atm)
B
21
2 lnRVVlnRdV
VR
VR
TPRTPVdV
TP
TqSA
===
=→===Δ
∫
∫ ∫δ
4lnRSSS BA =Δ+Δ=Δ∴
Isothermal mixing→ No change in E
PdVwqwqdE ==−== δδδδ0
21
2 lnRVVlnRdV
VRdV
TP
TqSB =====Δ ∫ ∫ ∫δ
Volume 변화로 생각해서 구하는 법
1 mol(1 atm)
A
1 mol(1 atm)
B
( )T constant at VPVPPPlnR
VVlnRdV
VRdV
TP
TqS
22112
1
1
2
==
====Δ ∫∫ ∫
∵
δ
4lnRSSS BA =Δ+Δ=Δ∴
Isothermal mixing→ No change in E
250
1
250
1
2
1
2
1
lnR.
lnRPPlnRS
lnR.
lnRPPlnRS
B
A
===Δ
===Δ
Pressure 변화로 생각해서 구하는 법
A,1 A,2
B,1 B,2
P 1, P 0.5P 1, P 0.5
= =
= =
If the first compartment has contained2 moles of ideal gas A, what would have been the increase in entropy when the partition was removed?
2 mol(2 atm)
A
1 mol(1 atm)
B
2 mol(2 atm)
A
1 mol(1 atm)
B
Calculate the change in entropy in three ways(Boltzman, volume change, pressure change).
( ) ( )NNlnkS 22 221 =Δ →
2 mol(2 atm)
A
1 mol(1 atm)
B
( ) 823 lnRlnR ==
821 lnRS =Δ∴ →
1
21221
2
2
1 121
21
PPlnkTSSS
PPNN
=−=Δ
=⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛=
→
볼츠만 엔트로피로 구하는 법
222221
1
1
2 lnRVVlnR
VVlnRSA ===Δ
21
2 lnRVVlnRSB ==Δ
823 lnRlnRSSS BA ==Δ+Δ=Δ∴
1
2
VVlnnRTdV
VnRTPdVwq ==== δδ
2 mol(2 atm)
A
1 mol(1 atm)
B
2 mol(2 atm)
A
1 mol(1 atm)
B Isothermal mixing→ No change in E
PdVwqwqdE ==−== δδδδ0
Volume 변화로 생각해서 구하는 법
2212222
2
1
1
2 lnRlnRPPlnR
VVlnRSA ====Δ
250
1
2
1 lnR.
lnRPPlnRSB ===Δ
23 lnRSSS BA =Δ+Δ=Δ∴
1
2
VVlnnRTdV
VnRTPdVwq ==== δδ
2 mol(2 atm)
A
1 mol(1 atm)
B
2 mol(2 atm)
A
1 mol(1 atm)
B Isothermal mixing→ No change in E
Pressure 변화로 생각해서 구하는 법
2
1
1
221 P
PVVVPVP T 21 =→=때일정할가
A,1 A,2
B,1 B,2
P 2, P 1P 1, P 0.5
= =
= =
Calculate the corresponding increase in entropy in each of the above two situations if both compartments had contained ideal gas A.
2 mol(2 atm)
A
2 mol(2 atm)
A
1 mol(1 atm)
A
3 molA
Calculate the change in entropy in three ways(Boltzman, Volume Change, Pressure Change).
2 mol(2 atm)
A
2 mol(2 atm)
A
1 mol(1 atm)
A
3 molA
NNNN
NN
ClnR
Clnk
PPlnkS
2323
2
1
221
822===Δ → !N!N
!NC NN 23
23 =
( )N Nln C N ln N N( N ln N N N lnN N )
= −
− − + −3 2 3 3 3
2 2 2
4272233 lnRlnNlnN =−= 27
3221 lnRS =Δ∴ →
121
21
2
2
231 =⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛= PCP
NN
NN
볼츠만 엔트로피로 구하는 법
3422
1
2 lnRVVlnRSA ==Δ
2732lnRSSS AA =Δ+Δ=Δ∴ ′
A, A,
A , A ,
2V , V ( )3
2V , V ( )3′ ′
= = →
= = →
1 2
1 2
1 2 32 3
2 21 13 3
몰이 2개의부피 1몰은 부피
몰은 부피에해당
32
1
2 lnRVVlnRSA ==Δ ′
2 mol(2 atm)
A
1 mol(1 atm)
A
3 molA
Volume 변화로 생각해서 구하는 법
3422
2
1 lnRPPlnRSA ==Δ
2732lnRSSS AA =Δ+Δ=Δ∴ ′
32
231
34
232
2
1
12
11
2
1
12
11
=⇒==
=⇒==
′′ PPRT
VP,RT
VP
PPRT
VP,RT
VP
,A,A
,A,A
2 mol(2 atm)
A
1 mol(1 atm)
A
3 molA
Pressure 변화로 생각해서 구하는 법
32
2
1 lnRPPlnRSA ==Δ ′
Statistical Thermodynamics(D.R. Gaskell Chap. 4 and R.T. DeHoff, Chap. 6)
Atomic description of thermodynamics in contrast withphenomenological description
microstate vs macrostate
Fundamental assumption or principle in ST
→ All microstates are equally probable.
→The probability of occurrence of any given macrostate (nj), which is macroscopically observable phenomenon, is proportional to the number of possible microstates.
A postulate of the quantum theory is that, if a particle is confined to move within a given fixed volume, then its energy is quantized, i.e., the particle may only have certain discrete allowed values of energy, which are separated by “forbidden energy bands.”
http://www.2ndlaw.com/entropy.html
http://www.2ndlaw.com/entropy.html
Quantized Energy → Energy Levels
The most probable distribution of molecules on various accessibleenergy levels
The maximum number of microstates
Equilibrium
one mole = 6 ×1023 molecules
Microstate
One arrangement in which the total energy of the system is distributed among energylevels and in space.
Irreversibility → change from less number of microstatesto more number of microstates.
W = number of microstates
S = k ln W
1
2
121221
WWlnk
WlnkWlnkSSS
=
−=−=Δ →
⎟⎟⎠
⎞⎜⎜⎝
⎛=Δ →
1
221 smicrostate of .no
smicrostate of .nolnkS
How many microstates are there in one mole of ice or water at 273 K?
How many microstates are there in one mole of ice or water at 273 K?
273 41 /Δ =KS for ice J K
273 63 /Δ =KS for water J K
Hint)
KK WlnkWlnk 273
273
1==
ice of mole one for s microstate of number→
K
KK W
WlnkS0
2732730 =Δ →
K/J ice for S K 412730 =Δ →
KK Wln.WlnkK/J 27323
273 104141 −×==
24273 1092 ×=→ .Wln K
2424 10311092273 10 ×× ==→ ..
K eW
water of mole one for s microstate of number→
K/J water for S K 632730 =Δ →
KK Wln.WlnkK/J 27323
273 104163 −×==
.KW ×→ =
242 0 10273 10
ice .Kcf . W ×=
241 3 10273 10
The solution’s entropy is greater than that of the puresolvent because solution has a higher density of states.
Solvent molecules less tend to “escape” from theirgreater entropy state in the solution to a vaporphase or to a solid phase than they “escape” from the pure solvent.
This lessened tendency to leave the solution results in a higher boiling point and lower freezing point for the solution compared to the pure solvent.
Colligative Effects (비점상승, 융점강하)
http://www.2ndlaw.com/entropy.html
Atomistic View
- What is pressure?- What is temperature?- What is internal energy?- What is heat or heat capacity?- What is entropy?
Kinetic Theory
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Maxwell-Boltzmann Velocity Distribution of Gas
Newton's Laws and Collisions
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
←
→
What is Gas Pressure?
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
= ⋅ = =2N 3 kN RTP kT T3V 2 V V
=PV RT
⎡ ⎤=⎢ ⎥
⎣ ⎦
21 3mv kT2 2
What is Temperature? → translational K.E.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
What is Internal Energy?
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
What is Heat Capacity?
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
→ capacity to store energy (heat)
CV =7/2 R
CV > 7/2 R
http://www.2ndlaw.com/entropy.html
단원자 기체 : translation다원자 기체 : translation, rotation, vibration
액체 : translation, rotation, vibration고체 : constrained vibration
vibration frequency ≅ 1012/sec
→ 물질이 에너지를 저장하는 방법
Vibration : large energy difference between levels
→ difficult to be excited.→ high E levels are not accessible.
Most liquid water molecules and gas phase water are in the lowest vibrational state.
Rotation : small energy difference between levels
→ easy to be excited
Water molecules rotate faster and faster with increasing temperature.
Translation : very small energy difference between levels
→ very easily excited
그림에서 첫번째 rotation level 근처
At RT, ~ 1000 miles an hour
http://www.2ndlaw.com/entropy.html
Quantized Energy → Energy Levels
Energy Difference between the Ground State and the First excited State
(CO at 300K in a box 10 cm on a side )
• Average thermal energy, kT = 0.59 kcal/mole• Translational energy, ΔEt = 1.7×10-20 kcal/mole• Rotational energy, ΔEr = 0.01104 kcal/mole• Vibrational energy, ΔEv = 6.21 kcal/mole• Electronic energy, ΔEe = 186.0 kcal/mole
Wavelengths to measure changes in the various modes of energy
(CO at 300K in a box 10 cm on a side )
• Translational energy, λt = 0.2 light years, none• Rotational energy, λr = 0.26 cm, far IR or Microwave• Vibration energy, λv = 4,610 μ, near IR• Electronic energy, λe = 0.154 μ, UV
Mode Wavelength Portion of spectrum