Relaxation in Glass: Review of Thermodynamics · 2nd Law of Thermodynamics – irreversible processes Exercise for Homework: Consider the two processes: Reversible (equilibrium) heating

Relaxation in Glass: Review of Thermodynamics

Lecture 10: Thermodynamic Functions

2nd Law of Thermodynamics – irreversible processes

Exercise for Homework: Consider the two processes: Reversible (equilibrium) heating of ice from -25 C to 0 C,

melted at 0 C and then reversibly warmed to 25 C Irreversible (non-equilibrium) melting of ice from -25C to Irreversible (non-equilibrium) melting of ice from -25C to

water at 25 C Exercise: What is SH2O, Ssurr and Suniv for each

process? Take Cp(ice) = 2.1J/g-oC, Cp(water) = 4.2J/g-oC,

H = 333 6 J/gHmelting = 333.6 J/g

[email protected] MITT: Relaxation in Glass - Lecture 10 Thermodynamic Functions 2

Exercise for Homework: Reversible process: Tsystem = Tsurroundings

dTT

CpK

qdTT

CpSK

watermeltK

iceOH

15.29815.273

15.2732

dTCpqdTCpSK

watermeltK

ice 15.29815.273

TKTKK 15.27315.248

15.273

dTT

CpK

qdTT

CpSK

watermelt

K

icegssurroundin

15.27315.24815.273

02

gssurroundinOHuniverse SSS


Exercise for Homework:

Irreversible process: Tsystem = Tsurroundings

dTT

CpK

qdTT

CpSK

K

watermeltK

K

iceOH

15.298

15.273

15.273

15.24815.2732

dTCpqdTCpqSKK

ice15.29815.273

00 SSS

dTCpqdTCpT

SK

watermeltK

icegssurroundin

icegssurroundin

15.27315.248

002

gssurroundinOHuniverse SSS


2nd Law of Thermodynamics – irreversible processes

For irreversible processes, the entropy change in the system is still the same so long as the system changes between the two same initial and final states the entropybetween the two same initial and final states, the entropy is a state function

However, the entropy change in the universe is now non-zero, we do more than exchange disorder between the system and surroundings, we create more disorder in the universe in the irreversible processesuniverse in the irreversible processes

TqdS

0 gssurroundinsystemirruniv SSS

T


2nd Law of Thermodynamics – Direction towards equilibrium

So for an isolated system, the entropy can be used to tell the direction of spontaneous change, the direction that increases the entropy of the systemincreases the entropy of the system

The challenge of course is to do science on isolated systems, we need a function that operates under conditions that are more common

Constant T and P?C t t T d V? Constant T and V?

Let’s see if we can “invent” a function that will give us the same ability to predict the position of equilibrium, but forsame ability to predict the position of equilibrium, but for a system that is contained under more commonly used conditions



Consider processes at constant T and Volume:

PdVqdU ATSU

PdVdUTdSqTdS

PdVdUq

0, VTdAThe Helmholtz Function decreases forspontaneous processes until it reaches a

00)(

dTconsiderTdSPdVdU

PdVdUTdS spontaneous processes until it reaches aminimum where after at equilibrium isis unchanged for processes taking place

t dT dP 0

0)()(

PdVTSUdTdSSdTTdSTSd T

at dT=dP =0

0)(0)(

0

TSUdTSUd

dVconsider

Hermann Ludwig Ferdinand von Helmholtz (August 31


0)( TSUdhttp://en.wikipedia.org/wiki/Hermann_von_Helmholtz

Hermann Ludwig Ferdinand von Helmholtz (August 31, 1821–September 8, 1894) was a German physician and physicist


So the Helmholtz Function, A(T,V), starts out high and decreases towards equilibrium…


2nd Law of Thermodynamics – Direction towards equilibriumConsider processes at constant Temperature and Pressure Consider processes at constant Temperature and Pressure

PdVqdU 0dPconsider

qTdSPdVdUqPdVqdU

0)()(

0

TdSPVUdVdPPdVPVd

dPconsider

0)(

TdSPdVdUPdVdUTdS

qTdS

0

)(

TdSdHPVUHrecall

)(0

TdSSdTTdSTSddTconsider

T

0)( TSHd

0

dGGTSH


0, PTdG


So the Gibbs function decreases for spontaneous processes and reaches a minimum at equilibrium for processes taking place at constant T andprocesses taking place at constant T andP, dT = dP = 0

0,

PTdGGTSH

Josiah Willard Gibbs (February 11, 1839 – April 28, 1903) was an American theoretical physicist, chemist, and mathematician. He devised much of the theoretical foundation for chemical thermodynamics as well as physical chemistry

htt // iki di / iki/J i h Will d Gibb


http://en.wikipedia.org/wiki/Josiah_Willard_Gibbs


So the Gibbs function decreases for spontaneous processes and reaches a minimum at equilibrium for processes taking place at constant T andprocesses taking place at constant T andP, dT = dP = 0

0

PTdGGTSH

0, PTdG


2nd Law of Thermodynamics – Thermodynamic Functions

Our thermodynamic functions so far…

PVUHwqU

TSHGTSUAPVUH

Their derivative forms…

TSHG

PdVTdSwqdU

PdVSdTTdSSdTPdVTdSdAVdPTdSVdPPdVPdVTdSdH

PdVTdSwqdU

VdPSdTTdSSdTVdPTdSdG


2nd Law of Thermodynamics – Thermodynamic Functions

Their “natural” variables:

dVVUdS

SUdUVSUU

),(

AA

dPPHdS

SHdHPSHH

VS

SP

SV

),(

dPPGdT

TGdGPTGG

dVVAdT

TAdAVTAA

TV

),(

),(

Exercise: What are the 8 gradients of these functions? For example…

PT TP

?

VSU


Thermodynamic Functions… 1st Derivatives

The first derivative functions

UU

HH

PVUT

SU

SV

PASA

VPHT

SH

SP

Some are more useful than others

VPGS

TG

PV

ST TV

others…

PT TP



Consider the Gibb’s Free-Energy Function…

For a gas phase…

gg

gg

VGSG

For a solid phase

VPGS

TG

TP

TP

VP

ST

For a solid phase…

ss

ss

VGSG

For a liquid phase…

TP PT

l

T

ll

P

l

VPGS

TG


TP


Consider the Gibb’s Free-Energy Function

P

P dTTGdG

22

)()()( 1212

TS

P

P

dTTSTGTGGGdG

SdTdG

11

)()()( 1212TS

P dTTSTGTGGGdG



Ss(T) < Sl(T) < Sg(T) gls GGG g

P

gl

P

ls

P

s

ST

GSTGS

TG

dTT

TCpPSPTSTmT s

oso )(),0(),(

__

0

__,

TCpTCpTCp

dTT

TCpSdTT

TCpPSPTS

T gTb lTm s

TbT

Tm

lomelt

Tm solo

)()()(

)()(),0(),(

___

_

0

__,

dTT

TCpSdTT

TCpSdTT

TCpPSPTST

Tb

govap

Tb

Tm

omelt

Tmovo )()()(),0(),(

__

0

__,



Temperature Dependence of the Entropy



Temperature dependence of the Gibb’s Free-Energy

sG

l

s

P

G

STG

g

l

P

G

STG

g

P

ST

G



Vs < Vl < Vgg

T

gl

T

ls

T

s

VP

GVPGV

PG


Thermodynamic Functions… 2nd Derivatives

The thermodynamic functions we use are state functions, and state functions (variables) have the property of being exactproperty of being exact

That is, the formation of their second derivative is independent of the order of differentiation:

yxFyxF ),(),(

yxFyxF

xyyx yxxy

),(),( 22

xyyx



When this is applied to thermodynamic functions, we call them the Maxwell relationships and we get the following:

VSSVVS SP

VT

VU

SSU

V

Exercise: There are four, write out the remaining three:

PdVTdSdUVdPTdSdHPdVTdSdU

VdPSdTdGPdVSdTdA



Maxwell Relations:

PdVTdSdU PT

VdPTdSdH

PdVTdSdU

VT

VS SV

PdVSdTdA

VdPTdSdH

PS SP

PS

VdPSdTdG

PdVSdTdA

VT TV

VS

PT TV

PS



Some are more useful than others…

VS

VPT

V

VTV

PS

22 VSP

T

V

dPTVdS

11

)()( 1212V VS

dPTVPSPSSSdS


Homework Exercise for next time: Derive an expression for the following quantity in

terms of easily measured quantities and apply it to li id B O li htl b it lti i t dliquid B2O3 slightly above its melting point and crystalline B2O3 slightly below its melting point

S

PVS


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Relaxation in Glass: Review of Thermodynamics · 2nd Law of Thermodynamics – irreversible processes Exercise for Homework: Consider the two processes: Reversible (equilibrium) heating