Thermodynamic Potentials

Why are thermodynamic potentials useful

Consider U=U(T,V)

Complete knowledge of equilibrium properties of a simple thermodynamicSystem requires in addition

P=P(T,V) equation of state

U=U(T,V) and P=P(T,V) complete knowledge of equilibrium properties

However

U(T,V) is not a thermodynamic potential

We are going to show: U=U(S,V) complete knowledge of equilibrium properties

U(S,V): thermodynamic potential

The thermodynamic potential U=U(S,V)

Consider first law in differential notation WdQddU

inexact differentials

Wd

Qdexpressed by exact differentials

PdVWd

TdSQd

2nd law

PdVTdSdU

Legendre Transformations

PdVTdSdU dU: differential of the function U=U(S,V)natural coordinates

Note: exact refers here to the coordinate differentials dS and dV. T dS and PdV are inexact as we showed previously.

PdVTdSdU

Legendre transformation Special type of coordinate transformation

Example:

coordinates

Partial derivatives of U(S,V) (vector field components)

Legendre transformation: One (or more) of the natural coordinates becomes a vector field component

while the associated coefficient becomes new coordinate.

PdVTdSdU

Back to our examplebecomes a coordinate

becomes a coefficient in front of dP

Click for graphic example

VdPTdSPVddU

VdPPVdTdSdU

easy check: PdVVdPPdVVdPVdPPVd

Productrule

VdPTdSPVUd

=:H (enthalpy)

H=H(S,P) is a thermodynamic potential

VdPTdSdH

Enthalpy

Geometrical interpretation of theLegendre transformation

- 1-dimensional example

)xx)(x(Y)x(Y)x(T 000

)x(Yx)x(Y)0(T 000

)Y(xd)xY(ddxYdY

)Y(xd)xYY(d

f:Note: natural variable of f is Y’ Y and X have to be expressed as Y=Y(Y’) and x=x(Y’)

YeY x

mapping between the graph of the function and the family of tangents of the graph

Legendre transformation:

Legendre transformation

from (S,V) to

(T,V): PdVTdSdU PdVSdT)TS(d

PdVSdT)TSU(d

F: Helmholtz free energy

(T,P): PdVSdTdF VdP)PV(dSdT

VdPSdT)PVF(d

G: Gibbs free energy

TSHPVTSUPVFG

equilibrium thermodynamics and potentials

complete knowledge of equilibrium properties

Consider Helmholtz free energy F=F(T,V)

PdVSdTdF

Differential reads:

VT

FS

and TV

FP

Entropy Equation of state

Response functions from 2nd derivatives

VV T

STC

V

2

2

T

FT

TT V

PVB

2

2

T

FV

V

andPT

TV T

V

V

PB

VT

P

VT

F2

etc.

thermodynamics potential

Maxwell relations

PdVSdTdF differential of the function F=F(T,V)

dF is an exact differential

V

22

T T

P

VT

F

TV

F

V

S

VT T

P

V

S

In general: relations which follow from the exactness of the differentials of thermodynamic potentials are called Maxwell relations

Properties of an ideal gas derived from the Helmholtz free energy

Helmholtz free energy F=U-TS

Reminder:

)V,T(SV

VlnnR

T

Tlnnc)V,T(S rr

rrV

0v nuTnc)V,T(U

r0rr

V TsunV

VlnnRT

T

Tln1Tnc)V,T(F

Equation of state:

TV

F)V,T(P

V

nRT

01 ln lnV rr r T

T Vnc T nRT n u Ts

V T V

T

r

V

V

VlnnRT

Equation of state derived from F

Note: U derived from F

S(T,V)VT

FS

obtained from )V,T(SV

VlnnR

T

Tlnnc)V,T(S rr

rrV

U(T,V) obtained from U=F+TS VT

FTF

0v nuTnc)V,T(U

Heat capacity at constant volume

VV

V ncT

UC

Isothermal bulk modulus

TT V

PVB

PV

nRT

V

nRTV

2

etc.

System

Heat Reservoir R

Systems in Contact with Reservoirs

Entropy statement of 2nd law: entropy always increased in an adiabatically isolated system

What can we say about evolution of systems which are not adiabatically isolated

T=const.

adiabatic wall

changes from initial state with

oooo PVTSUG

to final state with

ffff PVTSUG

VPSTUGGG 0f

remain constant

Consider system at constant temperature and pressure

VPSTUG From Entropy change of :T

GVPUS

Entropy change

Aim: Find the total entropy change Rtot SSS and apply 2nd law

of the reservoir:RS

L

RR T

QdS

LRQd

T

1

T

QR

Heat QR that, e.g., leaves the reservoir flows into the system Q = -QR

Rtot SSS T

Q

T

GVPU R

With 1st law:VPQWQU

T

Q

T

GQ R

totST

G

Heat reservoir: T=const.

Entropy statement of 2nd law: 0Stot

0T

G

for an adiabatically isolated system

0G

Gibbs free energy never increases in a process at fixed pressurein a system in contact with a heat reservoir.

Gibbs free energy will decrease if it can, since in doing so it causes the total entropy to increase.

(T=const, P=const.)

System with V=const. in contact with a heat reservoir Special case, very important for problems in solid state physics

STUF T

FUS

T

FQ Rtot SSS

T

QRT

F 0F (T=const, V=const.)

Q = -QR

T

FQ

Summary: Thermodynamic potentials for PVT systems

T=const,P=constT=const,V=constIsobaric process

1st law:Properties

Maxwell

relations

Vector field components

dG=-SdT+VdPdF=-SdT-PdVdH=TdS+VdPdU=TdS-PdVdifferential

Gibbs free energy

G(T,P)

G=U –TS+PV

Helmholtz free energy

F(T,V)

F=U -TS

Enthalpy

H(S,P)

H=U+PV

Internal energy

U(S,V)Potential

VS

UT

SV

UP,

PS

HT

SP

HV,

VT

FS

TV

FP,

PT

GS

TP

GV,

VS S

P

V

T

VT T

P

V

S

ST T

V

P

S

WQU QH 0F 0G

PS S

V

P

T

Open Systems and Chemical Potentials

Open system Particle exchange with the surrounding allowed

Heat Reservoir RT=const.

Thermodynamic potentials depend on variable particle number N

Example: U=U(S,V,N)

Particle reservoir

U( S, V, N) = 2 U(S,V,N)2 2 2

In general: )N,V,S(U)N,V,S(U

)N(

)N(

U)V(

)V(

U)S(

)S(

U

V,SN,SN,V

)N,V,S(U

S V N

holds and in particular for =1

)N,V,S(UNN

UV

V

US

S

U

V,SN,SN,V

(homogeneous function of first order)

)N,V,S(UNN

UV

V

US

S

U

V,SN,SN,V

keep N constant as in closed systems

TS

U

N,V

,S N

UP

V

:N

U

V,SChemical potential

NPVTS)N,V,S(U

dNN

UdV

V

UdS

S

UdU

V,SN,SN,V

dNPdVTdSdU

Intuitive meaning of the chemical potential μ

First law: WdQddU with TdSQd

WdTdSdU

mechanical work PdV work μdN required to change # of particles by dN

+

How do the other potentials change when particle exchange is allowed

Helmholtz free energy F=U-TS

SdTTdSdU)TS(ddUdF

dNPdVTdSdU

dNPdVSdTdF

Gibbs free energy G=U -TS+PV

VdPPdVdF )PV(ddFdG

dNPdVSdTdF

dNVdPSdTdG

P,TV,TV,S N

G

N

F

N

U

Properties of μ

With and NPVTSU PVTSUG

N

G both extensive )P,T( intensive (independent of N)

F

Equilibrium Conditions

Adiabatically isolatingrigid wall

System1:T1,P1, 1

System2:T2,P2, 2

Qd

From

differentials of entropy changes

dNPdVTdSdU

11

11

1

1

1

11 dN

TdV

T

P

T

dUdS

22

22

2

2

2

22 dN

TdV

T

P

T

dUdS

Total entropy change 1 2S S S 0

2nd law

In equilibrium 1 2dS dS dS 0

With conservation of

-total internal energy 1 2U U const. 1 2dU dU

-total volume 1 2V V const. 1 2dV dV

-total # of particles 1 2N N const. 1 2dN dN

1 2 1 21 1 1

1 2 1 2 1 2

1 1 P PdS dU dV dN 0

T T T T T T

1 2 1 21 1 1

1 2 1 2 1 2

1 1 P PdS dU dV dN 0

T T T T T T

small changes dU1, dV1, dN1

0 0 0

Equilibrium conditions

T1 = T2 , P1 = P2 ,1 = 2

Remark: )P,T(T1 = T2 , P1 = P2and 1 = 2

1 = 2 no new information for system in a single phase

but

Important information if system separated into several phases (see next chapter)