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 differentialsWd

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

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:

VTFS

and TV

FP

Entropy Equation of state

Response functions from 2nd derivatives

VV T

STC

V

2

2

TFT

TT V

PVB

2

2T

FVV

andPT

TV TV

VPB

VT

P

VTF2

etc.

thermodynamics potential

Maxwell relations

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

dF is an exact differential

V

22

T TP

VTF

TVF

VS

VT TP

VS

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

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

LRQdT

1TQR

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

Rtot SSS TQ

TGVPU R

With 1st law:VPQWQU

TQ

TGQ R

totSTG

Heat reservoir: T=const.

Entropy statement of 2nd law: 0Stot

0TG

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 TFU

S

TFQ

Rtot SSS TQR

TF

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

Q = -QR

TFQ

Summary: Thermodynamic potentials for PVT systems

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

1st law:Properties

Maxwellrelations

Vector field components

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

Gibbs free energyG(T,P)

G=U –TS+PV

Helmholtz free energyF(T,V)

F=U -TS

EnthalpyH(S,P)

H=U+PV

Internal energyU(S,V)Potential

VSUT

SV

UP,

PS

HT

SP

HV,

VT

FS

TV

FP,

PT

GS

TP

GV,

VS SP

VT

VT TP

VS

ST TV

PS

WQU QH 0F 0G

PS SV

PT

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(UNNUV

VUS

SU

V,SN,SN,V

(homogeneous function of first order)

)N,V,S(UNNUV

VUS

SU

V,SN,SN,V

keep N constant as in closed systems

TSU

N,V

,S N

U PV

:NU

V,SChemical potential

NPVTS)N,V,S(U

dNNUdV

VUdS

SUdU

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 NG

NF

NU

Properties of μ

With and NPVTSU PVTSUG

NG

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

TP

TdUdS

22

22

2

2

2

22 dN

TdV

TP

TdUdS

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 0T T T T T T

1 2 1 21 1 1

1 2 1 2 1 2

1 1 P PdS dU dV dN 0T 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