Transcript Thermodynamic Potentials Why are thermodynamic potentials useful Consider U=U(T,V) Complete knowledge of equilibrium properties of a simple thermodynamic System requires in addition P=P(T,V) equation of state U=U(T,V)

Thermodynamic Potentials
Why are thermodynamic potentials useful
Consider U=U(T,V)
Complete knowledge of equilibrium properties of a simple thermodynamic
System requires in addition
P=P(T,V)
equation of state
U=U(T,V) and P=P(T,V)
complete knowledge of equilibrium properties
U(T,V) is not a thermodynamic potential
However
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
dU  dQ  dW
dQ  TdS
dQ
expressed by
inexact differentials
2nd law
exact differentials
dW  P dV
dW
dU  TdS  PdV
Note: exact refers here to the
coordinate differentials dS and dV.
T dS and PdV are inexact
as we showed previously.
Legendre Transformations
dU  TdS  PdV
dU: differential of the function U=U(S,V)
natural coordinates
Legendre transformation
Special type of coordinate transformation
Partial derivatives of U(S,V) (vector field components)
Example:
dU  TdS  PdV
coordinates
Legendre transformation: One (or more) of the natural coordinates becomes
a vector field component
while the
associated coefficient becomes new coordinate.
Click for
graphic example
Back to our example
becomes a coordinate
dU  TdS  PdV
becomes a coefficient in front of dP
dU  TdS  dP V  VdP
easy check:
Product
rule
 dP V  VdP  VdP  P dV VdP  P dV
dU  dP V  TdS  VdP
dU  P V  TdS  VdP
=:H (enthalpy)
Enthalpy
H=H(S,P) is a thermodynamic potential
dH  TdS  VdP
dY  Y dx  d(Yx)  xd (Y)
Geometrical interpretation of the
Legendre transformation
d(Y  Yx)  xd (Y)
- 1-dimensional example
: f
Note: natural variable of f is Y’
8
Y and X have to be expressed as Y=Y(Y’) and x=x(Y’)
2
T( x )  Y( x 0 )  Y( x 0 )(x  x 0 )
6
0
Y  e x  Y
-2
Y-Y'x
Y
4
2
-4
0
-6
-2
T(0)  Y(x 0 )  x 0 Y(x 0 )
-8
0.0
0.5
1.0
X
Legendre transformation:
1.5
2.0
0
1
2
3
4
5
6
7
Y'
mapping between the graph of the function and
the family of tangents of the graph
8
Legendre transformation
(T,V):
dU  TdS  PdV  d(TS)  SdT  P dV
d( U  TS)  SdT  P dV
from (S,V)
: F
to
(T,P):
Helmholtz free energy
dF  SdT  PdV  SdT  d(P V)  VdP
d(F  P V)  SdT  VdP
: G
Gibbs free energy
G  F  PV  U  TS  PV  H  TS
equilibrium thermodynamics and potentials
thermodynamics potential
complete knowledge of equilibrium properties
Consider Helmholtz free energy F=F(T,V)
Differential reads:
 F 
S  

 T  V
dF  SdT  PdV
Entropy
and
 F 
P  


V

T
Equation of state
Response functions from 2nd derivatives
  2F 
 S 
C V  T   T 2 
 T 
 T  V

V
2
 P 


F
B T   V
 V 

 V  T
V 2

T
2
 P   V    P     F
and  V BT     
 T V
 V  T  T  P  T  V
etc.
Maxwell relations
dF  SdT  PdV
differential of the function F=F(T,V)
dF is an exact differential
 2F
 2F
 S 
 P 


  
 
 V  T VT TV
 T  V
 S 
 P 

 

 V  T  T  V
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: U(T, V)  ncv T  nu0
S(T, V)  nc V ln

T
F(T, V)  ncV T1  ln
Tr

T
V
 nR ln
 S(Tr , Vr )
Tr
Vr

V
  nRT ln
 nu 0  Ts r 
Vr

Equation of state:
 F    
P(T, V)  

V
 V  T



T 
V
nc
T
1

ln

nRT
ln

n
u

Ts
 0 r 
 V 

T
V

r 
r

T
V


nRT
ln

Vr


V





nRT


V

T
F 

 T  V

S(T,V) obtained from S  
S(T, V)  nc V ln
 F 

F

T


U(T,V) obtained from U=F+TS

T

V
T
V
 nR ln
 S(Tr , Vr )
Tr
Vr
U(T, V)  ncv T  nu0
Heat capacity at constant volume
Note: U derived from F
 U 
CV  
  ncV
 T  V
Isothermal bulk modulus
Equation of state derived from F
 nRT  nRT
 P 
P
B T   V
  V  2  
V
 V  T
 V 
etc.
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
Consider system at constant temperature and pressure
adiabatic wall
 changes from initial state with
G o  U o  TS o  P Vo
System

T=const.
Heat Reservoir R
to final state with
G f  U f  TS f  P Vf
G  G f  G 0  U  TS  PV
remain constant
From G  U  TS  PV
Entropy change of :
S 
U  PV  G
T
Aim: Find the total entropy change Stot  S  SR and apply 2nd law
Entropy change SR of the reservoir:
S R 
dQ R
1
QR

d
Q

R
 T

T
T
L
L
Heat reservoir: T=const.
Stot  S  SR 
U  PV  G Q R

T
T

Q  G Q R

T
T
With 1st law:
U  Q  W  Q  PV
Heat QR that, e.g., leaves the reservoir flows into the system 
S tot 
G
T
Q = -QR
Entropy statement of 2nd law: Stot  0
G
0
T
for an adiabatically isolated system
G  0 (T=const, P=const.)
Gibbs free energy never increases in a process at fixed pressure
in 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.
System with V=const. in contact with a heat reservoir
Special case, very important for problems in solid state physics
F  U  TS
S 
U  F
Q  F

T
T
Q = -QR
Stot  S  SR
Q  F Q R


T
T

F
T
F  0 (T=const, V=const.)
Intuitive interpretation of Helmholtz free energy
What is “free” about the free energy
Consider a system at constant T
S  SR  0
S f  S 0 SR  0
System

T  S f  S 0   T S R  0
T=const.
T  S f  S 0   QR  0
Heat Reservoir R
Q = -QR
T S f  S 0   Q
with 1st law
U f U 0  Q W
T  S f  S 0   U f  U 0 W
TS f  U f  TS 0 U 0   W
W  F
Wmax  F
The maximum amount of work one can obtain from a system at
constant temperature is given by
the decrease in the Helmholtz free energy
If we keep V=const for a PVT system
F  0
F f  F0
W  0  F
as we have seen before
Summary: Thermodynamic potentials for PVT systems
Potential
Internal energy
U(S,V)
Enthalpy
H(S,P)
H=U+PV
Helmholtz free
energy
F(T,V)
F=U -TS
Gibbs free
energy
G(T,P)
G=U –TS+PV
differential
dU=TdS-PdV
dH=TdS+VdP
dF=-SdT-PdV
dG=-SdT+VdP
Vector field
components
 U 
 F 
T
 , P   U  T   H  , V   H  S   F  , P  

 S  V
 V  T
 V S
 T  V
 P S
 S 
P
Maxwell
relations
Properties
 T 
 P 

   
 V S
 S  V
1st law:
U  Q  W
 T   V 
  

 P S  S P
Isobaric
process
H  Q
 S 
 P 

  
 V  T  T  V
 G 
 G 
S  

, V  
 P  T
 T  P
 S 
 V 
   

 P  T  T S
T=const,V=const T=const,P=const
F  0
G  0
Open Systems and Chemical Potentials
Open system
Particle exchange with the surrounding allowed
Particle reservoir
T=const.
Heat Reservoir R
Thermodynamic potentials depend on variable particle number N
Example: U=U(S,V,N)
U(2 S,2 V,2 N) = 2 U(S,V,N)
In general:


U( S,  V,  N)   U( S, V, N)
(homogeneous function of first order)
 U 
(S)  U 
(V)  U 
(N)




 
 
 U(S, V, N)

(

S
)



(

V
)



(

N
)



 V,N

 S,N

 S,V
S
V
holds   and in particular for =1
 U 
 U 
 U 
S


 V
 N  U(S, V, N)
 S  V, N
 V S, N
 N S,V
N
 U 
 U 
 U 
S


 V
 N  U(S, V, N)
 S  V, N
 V S, N
 N S,V
keep N constant as in closed systems
 U 
T


 S  V, N
 U 

  P
 V S , N
 U 
: 


 N S,V
U(S, V, N)  TS  P V  N
dU  TdS  P dV dN
 U 
 U 
 U 
dU  
dS  

 dV  
 dN

S

V

N

 V, N

S, N

S,V
Chemical potential
Intuitive meaning of the chemical potential μ
First law:
dU  dQ  dW
with
dQ  TdS
dU  TdS  dW
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
dU  TdS  P dV dN
dF  dU  d(TS)  dU  TdS  SdT
dF  SdT  P dV dN
F
Gibbs free energy G=U -TS+PV
dG  dF  d(P V)  dF  PdV  VdP
dF  SdT  P dV dN
dG  SdT  VdP  dN
Properties of μ
 U 
 F 
 G 






 N S,V  N  T,V  N  T,P
With
G  U  TS  PV

G
N
and
both extensive
U  TS  P V  N
  (T, P) intensive (independent of N)
Equilibrium Conditions
Adiabatically isolating
rigid wall
System1:
T1,P1, 1
From
System2:
T2,P2, 2
dQ
dU  TdS  P dV dN
dS1 
differentials of entropy change
dS2 
dU1 P1

 dV1  1 dN1
T1 T1
T1
dU 2 P2

 dV2  2 dN 2
T2
T2
T2
Total entropy change
S  S1  S2  0
2nd law
In equilibrium
dS  dS1  dS2  0
With conservation of
-total internal energy U1  U2  const.
dU1  dU2
-total volume
V1  V2  const.
dV1  dV2
-total # of particles
N1  N2  const.
dN1  dN2
1 1
P P 
  
dS     dU1   1  2  dV1   1  2  dN1  0
 T1 T2 
 T1 T2 
 T1 T2 
1 1
P P 
  
S     dU1   1  2  dV1   1  2  dN1  0
 T1 T2 
 T1 T2 
 T1 T2 
0
0
 small changes dU1, dV1, dN1
0
Equilibrium conditions
T1 = T2
Remark: T1 = T2 , P1 = P2
and
, P1 = P2
  (T, P)
,1 = 2
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)