Post on 16-Dec-2015
Physics IPhysics I
Review & More ApplicationsReview & More Applications
Prof. WAN, Xin
xinwan@zju.edu.cnhttp://zimp.zju.edu.cn/~xinwan/
我的竺院寄语我的竺院寄语
今日的大学课堂内外正在发生颠覆性的变革。知识的获得变得平庸,上课时学生就可以通过无线网络搜索直接满足跳跃性思维的需求,课后更可以自由地去探索去加强和补充课堂的教学内容。课上和课下的界限即将消除,学习和研究的差别正在缩小,教师和学生的位置也开始模糊。重要的不是学过了什么,而是学到了什么,是 从只会做功课的孩子成长为会思考的人。
A Macroscopic ReviewA Macroscopic Review
For any process
– For reversible process
– For irreversible process
This limits the maximum work we can extract from a certain process.
pdVdUTdS
pdVdUTdS (1st law)
pdVdUdQTdS
Application 1: Available WorkApplication 1: Available Work
In a thermally isolated system at a constant T
|W| = F is the minimum amount of work to increase the free energy of a system by F, at a constant T.
WSTWQSTUTSUF )(
QST The 2nd law
More Available WorkMore Available Work
Since PV is free at a constant P
|Wother| = G is the minimum amount of other work (chemical, electrical, etc.) needed to increase the Gibbs free energy of a system by G, at a constant T and a constant P.
otherWVPWVPFPVTSUG )(
WF previously
ElectrolysisElectrolysis
)(O2
1)(H)(OH 222 ggl
kJ42
3OHO
2
1H 222
RTPVPVPVVP
0
fH (kJ) fG (kJ) S (J/K) CP (J/K)
H2O (l) -285.83 -237.13 69.91 75.29
H2 (g) 0 0 130.68 28.82
O2 (g) 0 0 205.14 29.38
ElectrolysisElectrolysis
The amount of heat (at room temperature and atmosphere) you would get out if you burned a mole of hydrogen (inverse reaction)
kJ282 PVHU f
)(O2
1)(H)(OH 222 ggl
kJ286)83.285(02
10 Hf
enthalpy
ElectrolysisElectrolysis
)(O2
1)(H)(OH 222 ggl
The maximum amount of heat that can enter the system
The minimum “other” work required to make the reaction go
kJ237 Gf
kJ49J702052
1131298
ST
ElectrolysisElectrolysis
)(O2
1)(H)(OH 222 ggl
U = 282 kJ
PV = 4 kJ(pushing atmosphere away)
TS = 49 kJ(heat)
G = 237 kJ(electrical work) System
At room temperature & atmospheric pressure
Fuel Cell (Reverse Process)Fuel Cell (Reverse Process)
2eOH2OH2H 22
U = -282 kJ
PV = -4 kJ
TS = -49 kJ(heat)
G = -237 kJ(electrical work) System
At room temperature & atmospheric pressure
OH22eOHO2
122
At – electrode:
At + electrode:
Fuel Cell (Reverse Process)Fuel Cell (Reverse Process)
2eOH2OH2H 22
Maximum electrical work produced: 237 kJ
Efficiency (ideal)
OH22eOHO2
122
At – electrode:
At + electrode:
otherWG kJ237
%83kJ286
kJ237e
benefit (G)
cost (H)
Fuel Cell (Reverse Process)Fuel Cell (Reverse Process)
2eOH2OH2H 22
Two electrons per mole of H2O
Voltage (ideal)
OH22eOHO2
122
At – electrode:
At + electrode:
Volt26.1Coul106.1
J1097.119
19
V
J1097.11002.62
kJ237electronperwork 19
23
practically, 0.6-0.9 Volt
Geometrical InterpretationGeometrical Interpretation
Surface U = U(S, V)
PdVTdSdU
TS
U
V
dVV
UdS
S
UdU
SV
VS S
P
V
T
VS
U
2
PV
U
S
(1st law)
Mixed second derivative
App. 2: Thermodynamic IdentitiesApp. 2: Thermodynamic Identities
Consider an arbitrary gas with equation of state p = p(T,V).
dVPV
STdT
T
STPdVTdSdU
TV
VVV T
ST
T
UC
dVV
SdT
T
SdS
TV
V
Nk
T
P
V
S B
T
gasideal
PV
ST
V
U
TT
Introducing Free EnergyIntroducing Free Energy
Introduce free energy F = U - TS
PdVSdTSdTTdSdUTSUddF )(
T
P
T
P
V
S gasideal
VT
0
gasideal
VTT
PT
PTP
V
ST
V
U
Maxwell relation
Van der Waals GasVan der Waals Gas
Equation of state
constNbVNkTCVTS V lnln,
TNkNbVV
aNP B
2
2
2
2
V
aNP
T
PT
V
U
VT
constV
aNTCVTU V
2
,
attractive
Van der Waals IsothermsVan der Waals Isotherms
dV
dPlarge
TC dP
dVT ,At
dV
dPsmall
Density fluctuation very large!
Application 3: Phase BoundariesApplication 3: Phase Boundaries
carbon dioxide
Supercritical fluid: It can effuse through solids like a gas, and dissolve materials like a liquid.
Superfluid Helium Can Climb WallsSuperfluid Helium Can Climb Walls
He-II (superfluid) will creep along surfaces in order to reach an equal level.
Clausius-Clapeyron RelationClausius-Clapeyron Relation
Along the phase boundary, the Gibbs free energies in the two phases must equal to each other.
dT
P
T
dP
gl dGdG
VT
L
dT
dP
lg
lg
VV
SS
dT
dP
dPVdTSdPVdTS ggll
Latent heat: L = T(Sg – Sl)
Volume difference: V = Vg – Vl
or
Clausius-Clapeyron RelationClausius-Clapeyron Relation
Along the liquid-gas phase boundary
Along the solid-liquid boundary dT
P
T
dP
0
VT
L
dT
dP
0
VT
L
dT
dP
0
VT
L
dT
dPnormally
Why?
for ice ls VV
ls VV
A Microscopic ReviewA Microscopic Review
Boltzmann’s formula
Suppose we are interested in one particular molecule in an isolated gas.
– The total number of the microstates (with the known molecule state r & v) is related to the possible states of the rest of the molecules.
WkS B ln
BRR
BR
BRksSsS
ksS
ksS
R
R ee
e
sW
sW
s
s /)()(/)(
/)(
1
2
1
2 12
1
2
)(
)(
)(Prob
)(Prob
A Microscopic ReviewA Microscopic Review
Thermodynamic identity
Total energy is conserved.
RRR PdVdUT
dS 1
TksEsEksSsS BBRR ees
s /)()(/)()(
1
2 1212
)(Prob
)(Prob
0
constsEsU R )()(
)()(1
)()(1
)()( 121212 sEsET
sUsUT
sSsS RRRR
A Microscopic ReviewA Microscopic Review
Thermodynamic identity
Total energy is conserved.
RRR PdVdUT
dS 1
TksE
TksE
B
B
e
e
s
s/)(
/)(
1
2
1
2
)(Prob
)(Prob
0
constsEsU R )()(
)()(1
)()(1
)()( 121212 sEsET
sUsUT
sSsS RRRR
Boltzmann factor
A Microscopic ReviewA Microscopic Review
Partition function
Normalized distribution
s
TkE BseZ /
TkB
1sBs ETkE
s eZ
eZ
P 11 /
ZZ
ZeE
ZP
PEE
s
Es
ss
sss
sln11
App. 4: Maxwell Speed DistributionApp. 4: Maxwell Speed Distribution
For a given speed, there are many possible velocity vectors.
v
v
vv
speedtoingcorrespond
velocitiesofnumber
velocityhaving
moleculeaofyprobabilit)(Prob
Tkmv Be 2/2 24 v
kTmvevkT
mNvN 2/2
2/32
24)(
App. 5: Vibration of Diatomic MoleculesApp. 5: Vibration of Diatomic Molecules
The allowed energies are E(n) = (n + 1/2)
e
eeeeZ
1
2/2/52/32/
Tke
ZE B
TkB
1
1
2
1ln
TkTkTk
TkTkTk
BBB
BBB
eee
eeeE
2/52/32/
2/52/32/ 2/52/32/
1/kBT
One More MysteryOne More Mystery
h
hh T
QS
'
0 dSSgas
c
cc T
QS
'
after a cycle
Q'h > 0
Q'c < 0
0'
0'
c
c
h
hcgash T
Q
T
QSSSS
The total entropy of an isolated system that undergoes a change can never decrease.
Force toward EquilibriumForce toward Equilibrium
With fixed T, V, and N, an increase in the total entropy of the universe is the same as a decrease in the (Helmholtz) free energy of the system.
At constant temperature and volume, F tends to decrease (no particles enter or leave the system).
– The total entropy (system + environment) increases.
T
dFTdSdU
TT
dUdSdS
R
Rtotal
1
T
-dU
App. 6: Why Different Phases?App. 6: Why Different Phases?
At low T, the system tends to lower the energy, forming ordered state.
At high T, the system tends to increase the entropy, forming disordered state.
TSUF
energy entropy
tends to decrease
Phase Transition: Order vs DisorderPhase Transition: Order vs Disorder
T decreases from top panel to bottom panel