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Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���1
3 Equilibrium Between Two Systems
Chapter 2 Kittel&Kroemern+ bits and pieces e.g. from chapter 9 “Microcanonical methods”
Equilibrium between 2 systems Temperature, pressure, chemical potential
Definition
Ideal gas Thermodynamics identities
The three laws S,U,H,F,G A note about differentials
Chemical Potential Why is the Chemical Potential a potential Why is it call “Chemical”
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���2
Let us consider a gas Constraints: Energy U
Volume V Note: Additive Number of particles of species i : Ni
Take 2 systems and put them in contact Put them in weak interaction Isolated=> fixed U,V,N !!
!Theorem: at equilibrium !!!Formal definition of temperature, pressure, chemical potential
=> thermodynamic identity
U1,V1, Ni1 U2 ,V2 , Ni2
Thermal equilibrium between 2 Systems
∂σ 1
∂U1
= ∂σ 2
∂U2
≡ 1τ
∂σ 1
∂V1= ∂σ 2
∂V2≡ pτ
∂σ 1
∂Ni1
= ∂σ 2
∂Ni2
≡ − µτ
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���3
Let us consider a gas Constraints: Energy U
Volume V Note: Additive Number of particles of species i : Ni
Take 2 systems and put them in contact Put them in weak interactions Isolated=> fixed U,V,N !!
The configuration is described by U1,V1,Ni1
Weak interactions => (quantum) states are not modified multiplicity function =product of multiplicity functions !
!More states => Entropy increases
!!
U = U1 +U2V = V1 +V2Ni = Ni1 + Ni2
U1,V1, Ni1 U2 ,V2 , Ni2
Thermal equilibrium between 2 Systems 2
g U1,V1, Ni1( ) = g1 U1,V1, Ni1( )g2 U −U1,V −V1, Ni − Ni1( )
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���4
Thermal equilibrium between 2 Systems3
What is the most likely configuration (in equilibrium)? The probability of a configuration is proportional to the number of
its (quantum) states => Maximum probability is obtained for !!!!!!!
!Similarly !!!
!Since the distribution is very peaked, for all practical purpose we
can say that this is the “equilibrium configuration”!
∂g∂U1
= 0, ∂g∂V1
= 0, ∂g∂Ni1
= 0
�
∂g∂U1
= ∂g1
∂U1
g2 + g1∂g2
∂U1
= ∂g1
∂U1
g2 − g1∂g2
∂U 2 U2 =U−U1
= 0 <=> ∂ logg1
∂U1
= ∂ logg2
∂U 2
but logg1 =σ1 ⇒ ∂σ1∂U1
=∂σ 2∂U2
∂σ1∂V1
=∂σ 2∂V2
∂σ1∂Ni1
=∂σ 2∂Ni2
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���5
Graphic Representation
Watch out Number of states of the
combined system is the product of the number of states of each system
Most likely configuration = the configuration with the largest number of states
g1 g2
U1
g1 U1( )g2 U −U1( )
g1 U1( )g2 U −U1( )
UMost likely U1
log g1 log g2
U1
UMost likely
U1
∂ logg1 U1( )∂U1
= −∂ logg2 U −U1( )
∂U1
=∂ logg2 U2( )
∂U2 U2 =U −U1
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet
Calculation for an ideal gas
���6
http://cosmology.berkeley.edu/Classes/S2009/Phys112/Equilibrium2sysEquilbrium2sys.html
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���7
CommentsMicrocanonical methods
Compute number of states => entropy of configuration (U, V, Ni )=>T, p, µi Examples: gas system Kittel: systems of spins
Maximum probability <=> “equilibrium configuration” Strictly speaking we should be speaking of the equilibrium probability
distribution of configurations the system fluctuates around the configuration of maximum probability
Approximate language but does not matter because of narrowness of distribution <= Central limit theorem
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���8
CommentsAt equilibrium, the entropy of an isolated system is
maximum (an instance of the H theorem!) In this case:The total number of states accessible to the combined system
includes the product of the number of states initially accessible to each of the systems. This total number of states can only increase through the exchange of energy, volume, particles
!An approximate argument: the entropy of the “equilibrium configuration” is
maximum by definition. Total entropy even bigger!
dσdt
≥ 0
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���9
Temperature,Pressure,Chemical PotentialTemperature
Definition !
We have already checked that corresponds to ordinary T for ideal gas
Pressure Definition !
Have to check that corresponds to ordinary p
Chemical potential of species i Definition !
τ = kBT U =32NkBT
�
1τ
=∂σ∂U
=> at equilibrium τ 1 = τ 2
pτ=∂σ∂V
=> at equilibrium p1 = p2
PV = Nτ = NkBT
µiτ
= −∂σ∂Ni
=> at equilibrium µi1 = µi2
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���10
Simulationhttp://cosmology.berkeley.edu/Classes/S2009/Phys112/Diffusion/GaussDiffusion.html
Wall partially transparent to particles Initial state Final state
-5
-3
-1
1
3
5
-5 -3 -1 1 3 5-5
-3
-1
1
3
5
-5 -3 -1 1 3 5
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���11
Ideal Gas:Chemical Potential
We need to use full expression of entropy Notes chapter 2 slide 21
!!! µ is a measure of the concentration!
µ = −τ ∂σ∂N
= −τ∂ N log 2πM
h22U3N
⎛⎝⎜
⎞⎠⎟3/2 VN
⎛
⎝⎜⎞
⎠⎟+ 52N
⎛
⎝⎜
⎞
⎠⎟
∂N= τ log n
2πMh2
2U3N
⎛⎝⎜
⎞⎠⎟3/2
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
= τ log nnQ
⎛
⎝⎜⎞
⎠⎟
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet
What happens? !
���12
Expansion of Ideal Gas into VacuumA prototype Conceptually Important!
cf. end of chap. 6 in Kittel & Kroemer
Initial
FinalIsolated
Vi
Vf
Sudden!
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet
Isolated => ΔU=0, ΔQ=0, ΔW=0 => Tf=Ti Increase of entropy ? !Note Process is not a succession of equilibrium configurations (“irreversible”): T, p are
not defined during transition !
���13
Expansion of Ideal Gas into VacuumA prototype Conceptually Important!
cf. end of chap. 6 in Kittel & Kroemer
Initial
FinalIsolated
Vi
Vf
σ = log gt = NlogV +32N logU...⇒Δσ = Nlog
VfVi
⎛ ⎝ ⎜
⎞ ⎠ ⎟ ΔS = NkB log
VfVi
⎛ ⎝ ⎜
⎞ ⎠ ⎟
Sudden!
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���14
Thermodynamics IdentitiesZeroth Law
Two systems in equilibrium with a third one are in equilibrium with each other
First Law Heat transfer: Definition
Not an exact differential
Heat is a form of energy Fundamental Thermodynamic Identity: Apply only at equilibrium (or reversible
“quasi-static” processes)
!Second Law
When an isolated system evolves from a non equilibrium configuration to equilibrium, its entropy will increase
!Third Law
Entropy is zero at zero temperature ( or log of number of states occupied) => method to compute entropy
δQ = τdσ = TdS
dU = τdσ − pdV + µidNii∑ = TdS − pdV + µidNi
i∑
�
<= dσ =1τdU +
pτdV −
µi
τdN
i
i
∑
σ 0 = log(g10g20 ) constraints removed⎯ →⎯⎯⎯⎯⎯ σ f = log g1 U1( )g2 U −U1( )⎡⎣ ⎤⎦ ≥U1
∑ log g1g2( )max≥ σ 0
σ = − pss∑ log ps( ) only one state populated ⇒ po = 1 ⇒σ = 0
S =dQT0
T
∫ , σ =dQτ0
τ
∫
Succession of equilibria
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���15
Thermodynamics Identities (Gas)U,V, N
!S, V, N (e.g., for constant volume situations)
!
S, P, N (e.g., for constant pressure situations) Enthalpy (KK chap. 8)
!T, V, N (e.g., for constant volume situations)
Helmholtz Free Energy (KK chap. 3)
!T, P, N (e.g., for constant pressure situations)
Gibbs Free Energy (KKchap. 9)
will be derived later
dS =1TdU +
pTdV −
µiTdNi
i∑
dU = TdS
dQ!− pdV
dW" # $ + µidNi = dQ + dW + µidNi
i∑
i∑
H =U + pV ⇒ dH = TdS +Vdp + µidNii∑
F =U − TS
or τσ! ⇒ dF = −SdT
σdτ! − pdV + µidNi
i∑
G = F + pV ≡ µi T,P( )Ni
i∑ ⇒ dG = −SdT
σdτ! +Vdp + µidNi
i∑
Configuration Variables
Natural variables F T ,V ,Ni( )
Natural variables H S, p,Ni( )
Natural variables G T , p,Ni( )
Natural variables U S,V ,Ni( )
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���16
Exact DifferentialsExact Differential !
! independent of path Stokes theorem
!!!!
AB
g x, y( ) dg =∂g∂x
dx +∂g∂y
dy ⇒∂2g∂x∂y
=∂2g∂y∂x
⇔ dg = g B( )AB∫ − g A( )
a(x, y)dx + b(x + dx, y)dyb(x, y)dy + a(x, y + dy)dx
Difference = ∂b∂x
− ∂a∂y
⎡ ⎣ ⎢
⎤ ⎦ ⎥ dxdy
dy
dx
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���17
Exact Differentials 2≠ Non exact differential: dependent of path
!!
e.g. heat transfer depends on pathdg = a x, y( )dx + b(x, y)dy with ∂a
∂y≠∂b∂x
�
dQ = TdS⇒ dQ = dU + pdV = adU + bdV
clearly ∂a∂V U
= 0 ≠ ∂b∂U V
= ∂p∂U V
e.g. for an ideal gas pV = Nτ = 23U ⇒ ∂p
∂U V
= 23V
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���18
Differential identitiesConsequences of thermodynamic identities
non intuitive relationships that we will use often
Example: free energy (K.K. Chap. 3 p 70-71) !!!!
∂F∂T V ,Ni
= −S⇔ ∂F∂τ V ,Ni
= −σ ∂F∂V τ ,Ni
= − p ∂F∂Ni τ ,V
= µi
�
U = TS +F = −T ∂F∂T
+F = −T 2∂ FT⎛ ⎝ ⎜
⎞ ⎠ ⎟
∂T= −τ 2
∂ Fτ⎛ ⎝ ⎜
⎞ ⎠ ⎟
∂τ
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���19
Differential identities 2Maxwell identities (K.K. Chap. 3 p. 71): Advanced!
Consider e.g., F(T,V, N), S(T,V, N), p(T,V, N)∂2F∂T∂V
=∂2F∂V∂T
⇒∂S∂V
=∂p∂T
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���20
Why is the Chemical Potential a Potential?
Potential Recall: let us consider a force field . It derives from a potential if ! !independent of path (Stokes’ theorem) = potential energy difference between point 1 and 2 !
!!
!F = −
!∇Φ⇔ Φ2 − Φ1 = −
!F ⋅d!r
1
2
∫
1
2 ⇔!F.d!r is an exact differential
F!r( ) Φ
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���21
Why is the Chemical Potential a Potential?
Raising the potential energy of a system Let us consider an isolated system at zero potential energy !!Let us then raise it at uniform potential energy per particle
The entropy is not changed by uniform potential (number of states not changed)
Uo µo =∂Uo∂N σ ,V
Uo →U = Uo + NΔΦ
µtotal! =
∂U∂N σ ,V
= µointernal! + ΔΦ
external!
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���22
Conservation of Chemical PotentialEquilibrium with several species i
If the two systems are in equilibrium, each kind separately has to be in equilibrium
=> !
!Conservation of the Chemical (K&K chapter 9)
In a reaction between species, the number of disappearing particles or molecules is related to the number of produced particles or molecules
!
µi 1( ) = µi 2( ) ∀i21µi 1( ) µi 2( )
ν1A1 +ν2A2 ↔π 3A3 +π 4A4⇔ νi Ai
i∑ ↔ 0 with ν3 = −π3 , ν4 = −π 4
Theorem:At equilibriumν iµi
i∑ = 0
or ν iµiinitial∑ = π iµi
Final∑
Conservation of chemical potential
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���23
Conservation of Chemical Potential!
Conserved quantities (K&K chapter 9) From ! The probability distribution at equilibrium will be sharply peaked around the
configuration of maximum total entropy : !!or with the constraints
ν1A1 +ν2A2 ↔π 3A3 +π 4A4
or ν iAii∑ ↔ 0 with ν3 = −π 3 , ν4 = −π 4
δσ =∂σ∂NA1
δNA1 +∂σ∂NA2
δNA2 +∂σ∂NA3
δNA3 +∂σ∂NA4
δNA4 = 0
µ1δNA1 + µ2δNA2 + µ3δNA3 + µ4δNA4 = 0
δNA1ν1
=δNA2ν2
=δNA3ν3
=δNA4ν4
δNA1ν1
=δNA2ν2
=δNA3ν3
=δNA4ν4
⇒ ν iµii∑ = 0
or ν iµiinitial∑ = π iµi
Final∑
Conservation of chemical potential
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���24
Law of mass actionLaw of Mass Action (K&K chapter 9) !
Consider the reaction
! or
!=> !!!or
νiAii∑ ↔ 0
ν iµi
i∑ = 0 with µi = τ log
ninQi
⎛
⎝⎜⎞
⎠⎟ninQi
⎛
⎝⎜⎞
⎠⎟i∏
νi
= 1
niνi
i∏ = K τ( ) with K τ( ) = nQi( )
i∏ νi
niνi
initial i∏
njπ j
final j∏
= K τ( ) with K τ( ) =nQi( )νi
initial i∏
nQj( )π j
final j∏
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���25
Is the pressure the force per unit area? A last task: Show that the pressure we compute is indeed the
average force per unit area cf. Kittel and Kroemer Chapter 14 p. 391
Physics is simple: force from particles on the wall!
θ
density in d 3p = n p( ) p2dpdΩ
vΔt
dA
Phys 112 (S2014) 3 Equilibrium between 2 systems B. Sadoulet���26
Is the pressure the force per unit area? A last task: Show that the pressure we compute is indeed the
average force per unit area cf. Kittel and Kroemer Chapter 14 p. 391
Describe the particles by their individual density in momentum space (ideal gas)
If the particles have specular reflection by the wall, the momentum transfer for a particle arriving at angle θ is !!!!Integration on angles gives
that we would like to compare with the energy density
θ
�
non relativistic⇒ pv = 2ε ⇒ pressure P = 23u (energy density)
u = 32NVτ ⇒ P = N
Vτ = same pressure as thermodynamic definition = τ∂σ
∂V U,Nultra relativistic ⇒ pv = ε ⇒ P =
13u
2 pcosθ
�
23× 2π pv n p( )p 2dp
0
∞∫
!
�
u = ε n p( )d 3 p0
∞∫ = 4π ε n p( )p 2dp0
∞∫
density in d 3p = n p( ) p2dpdΩ
vΔt
dA
P =ForcedA
=d ΔpΔtdA
=1
dAΔt2pcosθ
Momentum transfer!"# $# vΔtdAcosθ
Volume of cylinder! "# $# n p( ) p2dpdΩ
density in cylinder! "## $##∫
= dϕ0
2π
∫ d cosθ 2pvcos2θ n p( ) p2dp0
∞
∫0
1
∫