Physics I Review & More Applications Prof. WAN, Xin [email protected] xinwan

Physics IPhysics I

Review & More ApplicationsReview & More Applications

Prof. WAN, Xin

[email protected]://zimp.zju.edu.cn/~xinwan/

Moments of JoyMoments of Joy

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


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


)(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


)(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:


2eOH2OH2H 22

Maximum electrical work produced: 237 kJ

Efficiency (ideal)

OH22eOHO2

122

At – electrode:

At + electrode:

otherWG kJ237

%83kJ286

kJ237e

benefit (G)

cost (H)


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


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


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


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

Specific Heat of Diatomic HSpecific Heat of Diatomic H22

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

The EndThe End

Thank you!

Documents

Physics I Review & More Applications Prof. WAN, Xin [email protected] xinwan