Thermodynamic Potentials Why are thermodynamic potentials useful Consider U=U(T,V) Complete...
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Transcript of Thermodynamic Potentials Why are thermodynamic potentials useful Consider U=U(T,V) Complete...
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)