Thermodynamics for Environmentology - University...
Transcript of Thermodynamics for Environmentology - University...
Thermodynamics!for Environmentology !
Susumu Fukatsu!
Thermodynamics and kinetics of natural systems�
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The University of Tokyo, Komaba Graduate School of Arts and Sciences Applied Quantum Physics Group!
Part I "(Fundamentals of thermodynamics*)#
"1) Thermodynamic system " States, Macroscopic variables" Diagram, Thermal equilibrium"2) Laws of Thermodynamics" Energy, Heat, Work, Entropy, "
"(Adiabatic, isothermal) processes"3) "Thermodynamic cycle," Heat engine, Carnot cycle, " Efficiency"4) Heat engines and Free energy" Joule-Thomson effect"5) Statistical mechanics"6) Phase transition and Kinetics"
Part II "(Natural systems)#
"7) Natural system as a heat engine "8) Laps rate, Climate change" Greenhouse effect"9) Air-water dynamics" Global circulation "10) Renewable energy"
Heat pipe, Thermosyphon"11) Carbon cycle,"
"Ozone depletion"12) Radiative forcing" Climate forcings" Natural variability ""
Table of Contents
*Honing the math skills that are needed?�
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Calendar 9/27, 10/4, 10/11, 10/18, 10/25, 11/1, 11/8"
Grading: Term-end exam. + Homework"
For future readings:"
Environmental "Physics"Claire Smith�
Elements of "Environmental "Engineering"Kalliat T. Valsaraj�
Environmental "Physics"Egbert Boeker�
Thermodynamics of"Natural Systems"Greg Anderson"
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Calendar 9/27, 10/4, 10/11, 10/18, 10/25, 11/1, 11/8"
One of the most important subjects"in science, technology and engineering, including physics, chemistry, biology, and many relevant fields of study. "
Thermodynamics?
We live in a world that is literally thermodynamic in many respects."
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Ice or water
Solid state"or liquid state."
Melting ice (=water) never returns to ice spontaneously!on a warm day. "
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Reversible?"
Solidification (freeze)"or liquefaction (thaw)."
Water or water vapor
http://rachelmeyerowitz.com/day-14-foggy-finger-prints/�
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Liquid state"or gas state."
Reversible?"
Vaporization (gasification) "versus "liquefaction"(condensation)"
More or less "
Raw egg or fried egg
Once fried, it never gets back to raw egg by chance.
Irreversible
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Thermodynamics & Statistical Physics!
http://www.almaden.ibm.com/vis/stm/images/stm15.jpg&imgrefurl=http://www.almaden.ibm.com/vis/"
Quantum Superposition "IBM Almaden “Stadium Corral” (permission granted)
Strange enough,"they have failed to "come to terms with"“quantum mechanics”"which is equally or even more important in modern physics."�
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1-1. Thermodynamic System Macroscopic�
Thermal equilibrium!Uniform T, No flow of mass or energy, No spontaneous change�
Microscopic"details neglected�
Water bath"@ Const. temperature� Gas in a "
He-tight container"""NA = 6.02 ×10
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1-1. Thermodynamic System
Thermal equilibrium is established eventually.�
Water bath "
T0
T0
TH
T∞
t0
Coin"
Coin"
Water bath"
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(Finite size, Thermally insulated)"
1-1. Thermodynamic System
Outside: Environment""Surroundings""Reservoir""(source, sink)""Constraints�
System""
{Open,Closed,"Isolated, …}"
"�
Configuration� Systems of interest�
Open�
Closed �
Isolated�
Mass flow"Heat�Q"Work W�
No mass flow"�
No mass flow"Self-contained."No heat in/out."No work in/out.�
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(Thermal) Equilibrium !
Non-equilibrium"Cf. Steady state"
Equilibrium " " Steady state"�
Uniquely specified with only a few parameters (P, V, T, N, U, S…).""State quantities are macroscopic while not materials-specific. ""No apparent change in macroscopic parameters over time."
1-2. Thermodynamic State
State quantities�
Q1. Give an example of non-equilibrium steady state. !
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Two systems sharing the same set of State Quantities"
1-2. Thermodynamic State
P, V, T, N…" P, V, T, N…"
No difference in the language of thermodynamics"
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Typical diagrams/plots frequently used in thermodynamics "
1-2. Thermodynamic State
P"
We must have points (states) on the diagrams."
V"
T"
V"
Isothermal expansion� Free Expansion"(Isothermal )�A"
B"
A" B"
Equilibrium"States�
Equilibrium "States�
Nonequilibrium"States"
Nonequilibrium"States"
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Transition from State A to B along path C"
1-2. Thermodynamic State
P"
V"
Isothermal expansion�A"
B"C"
Quasi-static process"" Slowly changing (moving) " in order to stay " in equilibrium at all times ��
Reversible�Quasi-static�
We must secure a reversible path to apply the appropriate math. "
Slowly�
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?�
?�
Dependent only on the current state (not on the path)"
1-3. State quantities (functions of states)
P"
V"
Pressure "P!Temperature "T!Volume "V!Number of moles "n!Heat capacity "C!Internal energy "U!Entropy "S!Enthalpy "H!Gibbs free energy "G!Free energy "F!(Helmholtz) ""
Work W�
Heat Q�
Heat and Work are not."(functions of path)�
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Extensive properties: Additive ( amount of material)"
1-4. Extensive and intensive properties not an exhaustive list
Mass " " " "m!Volume " " "V!Number of moles " "n!Energy" " " "U!Entropy " " "S!Enthalpy " " "H!Gibbs free energy " "G!
Intensive properties: Not additive "Temperature " " "T!Pressure " " "P!Density " " " "!Chemical potential " "!
∝
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ρµ
1-5. State variables and state postulate
P"
V"
State postulate(s):"The state of a simple compressible system is uniquely specified by two independent intensive quantities.�
State Variables� A set of state quantities that suffice to uniquely specify an equilibrium state"
P = P T ,V, n( )Equation of state�
e.g.,�
G = G T , p( )
A
B"
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Reversible!
P"
V"
Isothermal"expansion�
A
B"
(Quasi-static"=Equilibrium)"
Relaxation!(Nonequilibrium) !
1-6. Thermodynamic processes
The system can be brought back to the initial "state without dissipation (entropy production). "
Relaxation is an irreversible process�
Expl.1� Free expansion of gas �
T ,VL, n VR
Vacuum�Gas�
T ,VL, nT , ′V =VL +VR( ), n
Slowly�
T ,VL, nT , VL +VR( ), n
Cf. Adiabatic free expansion�
1-6. Thermodynamic processes
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Expl.2� Thaw (melting)� Expl.3� Friction (Mechanical) �Oscillation damping"
Expl.4� Diffusion (mixing)� Expl.5�
Q2. Give your own !example of an irreversible!
system and explain why it is so.!!
October 4, 2017�
�
mk > 0( ) x
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1-6. Thermodynamic processes
Reversibility"
Literally “Reversible”"(getting back to where it started) "but not necessarily “Quasi-Static” "is called “Cyclic” �
Thermodynamic "Reversibility" Reversible� Quasi-static�
A
Cycle�Cycle�
×
Quasi-static� PSFF ≥ PS
F ≤ PS ∴ F = PSExpansion�
Compression�
Balanced�
e.g., Joule-Thomson�
1-7. For the researcher in you
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Zeroth law:� “Thermodynamic syllogism” ""First law: "“Conservation of energy” (nontrivial)""Second law: "“Entropy never decreases” (conditional)""
Third law: "“Entropy must vanish at absolute zero”""
2. Laws of thermodynamics
2. Laws of thermodynamics
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Zeroth law:�“Thermodynamic syllogism” ""Two systems, A and B, "are in contact so that they are in thermal equilibrium."
Two systems, B and C, are likewise "in thermal equilibrium."
Then A and C must be in thermal equilibrium. "
"B"
"A�
"C�
"A�
"C�
To define the (absolute) temperature �
System A @T " System B @T "System C @ T "
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2. Laws of thermodynamics
Temperature"How can we define the absolute temperature?"
Measure V to find T"
β ≡ 1V∂V∂T
⎛⎝
⎞⎠ P
(Isobaric) volume expansion coefficient�
For ideal gas (or in the dilute-limit), �
β = 1V∂∂T
nRTP
⎛⎝
⎞⎠ P
= 1T
T =V ΔTΔV
⎛⎝
⎞⎠ P
Hence�
Reference point(s) necessary" Cf . F⎡⎣ ⎤⎦ =
95 C⎡⎣ ⎤⎦ + 32
0 C° = 273.15 [K] "
ΔTΔV
T
V
0 K"
1/273.15"Charles’s law�
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2. Laws of thermodynamics
The nonexistence of “Perpetual Motion Machines” "
If " " " , no doubt�′d Q = 0( ) ′d W = 0.
The 1st kind: Spontaneous motion without energy uptake.�
First law: “Conservation of energy” "
Energy balance" ΔU =Q +W
dU = ′d Q + ′d WAlternatively,�
dU = 0
?�
Beware: There are many fraud machines up there seemingly violating(?) the laws.�
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2. Laws of thermodynamics
Heat and Work are equivalent."
QW = −PΔV
1 calorie = 4.2 J�
A variant of "Joule’s experiment�
Thermal� Mechanical�
T1,V1 T0,V1T0,V0
Mechanical equivalent of Heat�
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2. Laws of thermodynamics
Heat entering the system:"
T1,V1T0,V1
Q
Q
The internal energy increases by"U
ΔU =U T1,V1( )−U T0,V1( ) =Q.
Work on the system:" W = −PΔV
W
T1,V1T0,V0
The internal energy increases by"
ΔU =U T1,V1( )−U T0,V0( ) =W .U
Heat and/or Work increase the internal energy "ΔU
(<0)�
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2. Laws of thermodynamics
Exact differential and Inexact differential�
dU = ′d Q + ′d W
Work d’W�
Heat d’Q�A
B"
States� Path (History)�
Exact "differential�
Inexact differential�
Exact differential is NOT path-dependent�
A
B
∫ dU =U B( )−U A( )
dU
C+ ′C∫ = 0 conservative�
C
′C
Net U vanishes.�
with prime�
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2. Laws of thermodynamics
Path-independent integral�
dU = 2xy dx + x2 dy
A
BExample�
A
B
∫ dU =U B( )−U A( )
Exact differential�
C2"
C1"
C3"
x�
y�
x1 x2
y1
y2C1"
C2"
C3"
dUx1
x2∫ = 2xy1 dxx1
x2∫ + x22 dy
y1
y2∫= x2y1⎡⎣ ⎤⎦x1
x2 + x22y⎡⎣ ⎤⎦y1
y2 = x22y2 − x1
2y1
dUx1
x2∫ = x12 dy
y1
y2∫ + 2xy2 dxx1
x2∫= x1
2y⎡⎣ ⎤⎦y1y2 + x2y2⎡⎣ ⎤⎦x1
x2 = x22y2 − x1
2y1
: y = f x( )
ddx x2 f x( )( )dx
x1
x2∫ = x2 f x( )⎡⎣ ⎤⎦x1x2
= x22y2 − x1
2y1 dU
C∫ = 0
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2. Laws of thermodynamics
Path-dependent integral�
′d U = xy dx + x2 dy
A
BExample�
′d UA
B
∫ ≠U B( )−U A( )
Inexact differential�
C2"
C1"x�
y�
x1 x2
y1
y2C1"
C2"
′d Ux1
x2∫ = xy1 dxx1
x2∫ + x22 dy
y1
y2∫= x2
2 y1⎡⎣⎢
⎤⎦⎥x1
x2+ x2
2y⎡⎣ ⎤⎦y1y2
= x22
2 y1 −x12
2 y11 + x22y2 − x2
2y1
′d U
C=C1+C2∫ ≠ 0
′d Ux1
x2∫ = x12 dy
y1
y2∫ + xy2 dxx1
x2∫= x1
2y2 − x12y1 +
x22
2 y2 −x12
2 y2
denoted by the prime�
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2. Laws of thermodynamics
Second law of thermodynamics!Clausius statement:!“Heat NEVER flows spontaneously " from a Low-T system to a High-T " system without the help of work”!
TH
TL
Q
Thomson (Kelvin) statement:!“One CANNOT convert heat "100%-efficiently "into work without influence "on the surroundings”!
Q W
100%-efficient "××
“Entropy never decreases” �
“Perpetual Motion Machines” of the 2md kind are forbidden. "
“One-way Evolution”�
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2. Laws of thermodynamics
A New State Quantity"
Entropy !
dU = ′d Q + ′d W
dS = ′d QT
The entropy " is defined only for quasi-static processes. "S
One of the most significant but elusive properties in TD."
TdS = dU + PdV.
′d W = −PΔVand�
The first law thus reads�
S = S U,V( ).
Is T doing a magic?"
dS = 1T dU + PT dV
This shows that �
∂S∂U
⎛⎝
⎞⎠V
= 1T ,∂S∂V
⎛⎝
⎞⎠U
= PT .so that�
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2. Laws of thermodynamics
F x, y( ) = xy
(Total) differential, partial derivative
dF x, y, z( ) = ∂F∂x y,z
dx + ∂F∂y x,z
dy + ∂F∂z x,y
dz
∴ dF x, y( ) = y dx + x dy
Partial derivatives
fixed" fixed"y, z x, z
ΔF x, y( ) = x + Δx( ) y + Δy( )− xy= xy + xΔy + yΔx + ΔxΔy − xy
= xΔy + yΔx +O(2)
fixed"x, y
⇔ ∂F∂x y
= y, ∂F∂y x
= x
xΔy
xy yΔx
ΔxΔy
Δx
Δy
x
y is fixed"y
x is fixed" 2nd order"(to be neglected)"
E.g."
Relation between "infinitesimals
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Entropy never decreases in an isolated system !Statistical mechanics provides a plausible explanation for this."
S UA,UB( ) = S UA( ) + S UB( )The energy is conserved:!
The first derivative! dS = 0
∂S
∂UA,B
⎛⎝⎜
⎞⎠⎟V
= 1TA,B
⎡
⎣⎢
⎤
⎦⎥
U =UA +UB
The entropy adds up:!
dS = ∂S∂UA
⎛⎝⎜
⎞⎠⎟VdUA +
∂S∂UB
⎛⎝⎜
⎞⎠⎟VdUB
= 1TA
− 1TB
⎛⎝⎜
⎞⎠⎟ dUA = 0
dUA + dUB = 0.with�
∴ TA = TB
S UB( )S UA( )
UA UB
Constant-volume vessels!Thermal equilibrium�
Entropy reaches its maximum value in thermal equilibrium�
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2. Laws of thermodynamics
S UA,UB( ) = SA U( ) + SB U0 −U( ).
An alternative route leading to the same conclusion!
The first derivative must vanish! dS = 0.
∂SA,B∂U
⎛⎝⎜
⎞⎠⎟V
= 1TA,B
⎡
⎣⎢
⎤
⎦⎥
U0 =UA +UB
The entropy is additive:!
dS = ∂SA∂U
⎛⎝⎜
⎞⎠⎟VdU + ∂SB
∂ U0 −U( )⎛⎝⎜
⎞⎠⎟Vd U0 −U( )
= 1TAdU + 1
TB−dU( ) = 0
∴ TA = TB
The total energy is fixed:!
dU −dU
Thermal equilibrium is reached (“Thermalized”).�
Entropy never decreases in an isolated system !
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2. Laws of thermodynamics
Entropy increases, why? Even without ceiling?"
dS = 1TLdQ + 1
TH−dQ( )
= 1TL
− 1TH
⎛⎝⎜
⎞⎠⎟ dQ > 0
TH > TL
The 2nd law of thermodynamics!
dQ−dQ
Clausius statement:! “Heat flows from ‘High’ to ‘Low’”!
The entropy change associated with the transfer of heat:!
simply because !TH TL
dQ
dQ
S reaches its maximum eventually.!Isolated system� Thermal equilibrium�
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2. Laws of thermodynamics
dA = ∂A∂X
⎛⎝
⎞⎠ YdX + ∂A
∂Y⎛⎝
⎞⎠ XdY .
∂∂Y
∂A∂X
⎛⎝
⎞⎠ Y
= ∂∂X
∂A∂Y
⎛⎝
⎞⎠ X
when and only when�
For a state quantity� A = A X,Y( ),
More about Exact differential�
the ED exists �
so that�
Clairaut-Schwarz�
dU = TdS − PdV dF = −PdV − SdT dG =VdP − SdT
∂T∂V
⎛⎝
⎞⎠ S
= − ∂P∂S
⎛⎝
⎞⎠V
∂P∂T
⎛⎝
⎞⎠V
= ∂S∂V
⎛⎝
⎞⎠ T
∂V∂T
⎛⎝
⎞⎠ P
= − ∂S∂P
⎛⎝
⎞⎠ T
Maxwell’s relations�
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2. Laws of thermodynamics
Specific heat of an ideal gas�
TdS = PdV + dU,
CP = ′d QdT
⎛⎝
⎞⎠ P
⎛⎝⎜
⎞⎠⎟= T ∂S
∂T⎛⎝
⎞⎠ P
= P ∂V∂T
⎛⎝
⎞⎠ P
+ ∂U∂T
⎛⎝
⎞⎠ P.
Since�
Specific heat at "constant pressure�
Starting from�
Specific heat at"constant volume�
∂U∂T
⎛⎝
⎞⎠ P
= ∂U∂T
⎛⎝
⎞⎠V
and �P ∂V∂T
⎛⎝
⎞⎠ P
= P ∂∂T
nRTP
⎛⎝
⎞⎠ P
= nR,
CP = nR + ∂U∂T
⎛⎝
⎞⎠V
= nR +CV Mayer’s relation�
(Defined along the QS path)�
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2. Laws of thermodynamics
P
V
Quasi-static isothermal processes!
The internal energy does not change.�
dT = 0 ∴ TdS = PdV
Upon integration�
′d W = nRTV dV.For an ideal gas, �
W = nRTV dV
V1
V2∫ = nRT lnV2V1
V2 >V1V2 <V1
Expansion " Work done by the system
"�Compression" Work done on the system
"�
WV2V1
Isotherm�
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2. Laws of thermodynamics
Quasi-static adiabatic processes!
Heat does not enter or leave the system.�
dU = CVdT =αnRdT
T( )dS = 0 ∴ dU = −PdV
On the other hand,�
dU = − nRTV dV.For an ideal gas, �
∴ V1αT = const.dV
V +α dTT = 0 PV
1+αα = const.⎛
⎝⎜⎞⎠⎟
3252
Monoatomic�
Diatomic�
Table of α values
Cf . PV = nRT
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2. Laws of thermodynamics
, so for an ideal gas*.�2) Show that� ∂U∂V
⎛⎝
⎞⎠ T
= 0
∂U∂T
⎛⎝
⎞⎠ P
= ∂U∂T
⎛⎝
⎞⎠V
3) Show that�
1) Prove that the Clausius and Thomson statements are equivalent. �
for an ideal gas*.�
PV = nRT*The EOS for an ideal gas:�
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U =U T( )