Thermodynamics for Environmentology - University …park.itc.u-tokyo.ac.jp/fkatz/_userdata/EnvTD_2017_1_41.pdfThermodynamics! for Environmentology ! Susumu Fukatsu! Thermodynamics

Thermodynamics!for Environmentology !

Susumu Fukatsu!

Thermodynamics and kinetics of natural systems�

1

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?�

2

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"

3

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."

4

Ice or water

Solid state"or liquid state."

Melting ice (=water) never returns to ice spontaneously!on a warm day. "

5

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

7

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."�

8

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

23

9

1-1. Thermodynamic System

Thermal equilibrium is established eventually.�

Water bath "

T0

T0

TH

T∞

t0

Coin"

Coin"

Water bath"

10

(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. !

12

Two systems sharing the same set of State Quantities"


P, V, T, N…" P, V, T, N…"

No difference in the language of thermodynamics"

13

Typical diagrams/plots frequently used in thermodynamics "


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"


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�

15

?�

?�

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�


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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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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


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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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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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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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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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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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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�

28


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

29


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�

30


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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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�

32


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

33

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�

34


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 !

35


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�

36


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�

37


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)�

38


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�

39


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

40


, 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:�

41

U =U T( )

Documents

Thermodynamics for Environmentology - University …park.itc.u-tokyo.ac.jp/fkatz/_userdata/EnvTD_2017_1_41.pdfThermodynamics! for Environmentology ! Susumu Fukatsu! Thermodynamics