chapter ii: thermodynamics - universidad de guanajuato · chapter ii: thermodynamics ... 4...
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Chapter II: Thermodynamics Section 1: Thermodynamics as driving forces.....................................................................2
Basic notions and definitions ...........................................................................................2 Fundamental Thermodynamic equations.........................................................................4
Homogeneous functions ..........................................................................................7 Mechanical and chemical equilibrium.............................................................................8 Methods of inference in thermodynamics .....................................................................15
First law of thermodynamics .................................................................................17 Method 1 ....................................................................................................................17 Method 2 ....................................................................................................................17
Carnot’s principle ..................................................................................................18 Absolute Temperature scale ..................................................................................20 Other statements of second law of thermodynamics .............................................20
Section 2: Laboratory conditions and free energy............................................................21 Helmholtz free energy ...................................................................................................22 Fundamental equation for F ...........................................................................................27 Enthalpy and Gibbs free energy ....................................................................................27
Table 2.2.1 – fundamental thermodynamic equations ...........................................32 Method 3: heat capacity.............................................................................................33
Third law of thermodynamics................................................................................36 Method 4: Using thermodynamic cycles ...................................................................38
Section 3: Maxwell’s relations & Mixtures .......................................................................42 Method 5 – Maxwell’s relations ................................................................................44
Table 2.3.1 Maxwell relations ...............................................................................46 Expansion versus compression..................................................................................49
Thermal expansion coefficient ..............................................................................49 Isothermal compressibility.....................................................................................50
Multicomponent systems ...............................................................................................53 Method 6 - homogeneous functions - Gibbs-Duhem equation..................................56
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Section 1: Thermodynamics as driving forces
Basic notions and definitions Thermodynamics = set of tools
• Predict tendencies of atoms and molecules to react, bind, absorb, partition, dissolve, change phase, shape or physical boundaries
• Three basic tools:
1. First law of thermodynamics 2. Second law of thermodynamics: maximum entropy principle 3. Multivariate calculus
Thermodynamics define forces acting on material systems:
• Pressure – tendency to exchange volume • Temperature – tendency to exchange energy • Chemical potential – tendency to exchange matter
Thermodynamic system: collection of matter delineated from surroundings by real or imaginary boundaries Boundaries determine what goes in or out:
• Open system – exchange of energy , mass or volume (living systems) • Closed system – energy can cross but mass do not, volume can change • Isolated system – Energy, mass and volume are constants (idealization) • Semi-permeable membranes – restrict the flow of some particles (dyalisis,
biological membranes) • Adiabatic boundaries – no heat flow allowed (thermo bottles)
Phase – homogeneous part of system mechanically separable from the rest Single system – Homogeneous single phase – uncharged ⇒ unaffected by E
ror B
r, large
enough such that surface effects is negligible In most cases, gravity is ignored (ambiguous definition of entropy in case of gravity)
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Properties of system can be extensive or intensive • Extensive ⇒ sum of properties of part of system = property of whole system (ex.
entropy, internal energy, mass or volume) • Intensive ⇒ independent of size of system (ex. temperature or pressure)
Extensive properties:
• Spatial extent - volume, area and length • Number of particles atoms or molecules • Internal energy and entropy
Number of moles:
(2.1.1) ii
AV
Nn
N=
Where iN is the number of molecules and AVN is Avogadro number 236.022045 10AVN = × molecules
Internal energy:
(2.1.2) 1
t
i ii
U N ε=
= ∑
Where iN is the number of particles in the energy level i , and iε is the energy of the level Entropy :
(2.1.3) 1
N
jj
S S=
= ∑
Where jS is the entropy of component j of the whole system
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Fundamental Thermodynamic equations Entropy and internal energy
• Multivariate: ( ), ,S S U V N→ and ( ), ,U U S V N→ • Predict equilibria – while main definition of thermodynamics originate from
energy equation, microscopic driving forces based on entropy Equation of state – specify interrelations among variables of system
• From experiments or microscopic models
For small changes in entropy and energy:
(2.1.4) 1, , , , i j
M
jjV N U N j V U N
S S SdS dU dV dN
U V N≠
=
∂ ∂ ∂ = + + ∂ ∂ ∂ ∑
(2.1.5) 1, , , , i j
M
jjV N S N j V S N
U U UdU dS dV dN
S V N≠
=
∂ ∂ ∂ = + + ∂ ∂ ∂ ∑
Where each partial derivative corresponds to measurable physical properties: Temperature :
(2.1.6) ,V N
UT
S∂ = ∂
, conjugate to S
Pressure :
(2.1.7) ,S N
Up
V∂ = − ∂
, conjugate to V
Chemical potential:
(2.1.8) , , i j
jj S V N
UN
µ≠
∂= ∂
, conjugate to jN
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From these definitions we rewrite the fundamental equation for energy as:
(2.1.9) 1
M
j jj
dU TdS pdV dNµ=
= − +∑
Rearranging equation 2.1.9 we can also write the fundamental equation for entropy as:
(2.1.10) 1
1 Mj
jj
pdS dU dV dN
T T T
µ
=
= + − ∑
Comparison of 2.1.10 with 2.1.4 yields 3 new definitions:
(2.1.11) , , , ,
1
i j
j
V N U N J V U N
S p S ST U T V T N
µ
≠
∂ ∂ ∂ = = = − ∂ ∂ ∂
• These sets of equations are general: deduced from functional dependencies
( ), ,S U V N and ( ), ,U S V N ⇒ function of extensive variables • And fundamental: completely specify all the changes that can occur in simple
thermodynamics systems ⇒ basis of predictions about equilibria Example 2.1.1: the ideal gas law deduced from lattice model
Start with one of equation in 2.1.11: ,U N
p ST V
∂ = ∂
Insert ( )S V from lattice model ( N particles in M lattices)
( ) ( )!
ln , ln! !
S MW N M
k N M N= =
−
Relation between V and M : There are M lattice sites per volume M M V⇒ ∝ or M Vα= where α is a constant
dM MdV V
α⇒ = = N N N
S S dM S MV M dV M V
∂ ∂ ∂ ⇒ = = ∂ ∂ ∂
Therefore ,U N N
p S S MT V M V
∂ ∂ = = ∂ ∂
Using Stirling approximation:
( )( ) ( )ln ln ln ln
M
M NN
S MM M N N M N M N
k N M N−≅ = − − − −
−
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But since ( )ln ln lnM M N M M N M= + −
( )ln lnS N M N
N M Nk M M
− = − − −
Taking the derivative:
( )1 ln ln ln 1N
S M N Nk M M N k
M M N M∂ − = + − − − = − − ∂ −
Which yields:
,
ln 1U N
p S M Nk
T V V M∂ = = − − ∂
Using series approximation:
2
ln(1 )2x
x x− − − +; L
,
1ln 1 1
2U N
p S M N M N Nk k
T V V M V M M∂ = = − − − − + + ∂
; L
Assuming 1NM
= and keeping only the first term
p NkT V
= or pV NkT= which is the ideal gas law
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T , p and µ are measurable quantities • They are intensive
• And equivalent to forces: 1T
the tendency for heat flows, pT
the tendency for
volume changes and Tµ
the tendency for particle changes
Homogeneous functions They are intensive because ( ), ,U S V N is a first orde r homogeneous function
( )1 2 3, ,U x x x where ix are extensive then i ix x U Uλ λ→ ⇒ =
But since T , p and µ are partial derivative i
Ux
∂∂ i i
U Ux x
λλ
∂ ∂⇒ =
∂ ∂
Important property of homogenous functions: Since ( ) ( )i if x f xλ λ=
Then ( ) ( )( )
1
Mi
ii i
xf fdf x f f
xλλ
λ λλ λ λ λ=
∂∂ ∂ ∂= + = =
∂ ∂ ∂ ∂∑
But since ( )i
i
xx
λλ
∂=
∂
( )1
M
ii i
ff x
xλ=
∂⇒ =
∂∑
However, since λ is arbitrary, we can choose 1λ = such that
( )1
M
ii i
fx f
x=
∂=
∂∑
From this last equation and equation 2.1.9, 1
M
j jj
dU TdS pdV dNµ=
= − +∑ , we deduce that
(2.1.12) 1
M
i ii
U TS pV Nµ=
= − +∑
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Mechanical and chemical equilibrium What defines the state of equilibrium is the maximum entropy principle 0totaldS = Example 2.1.2: Temperature describes energy exchange tendency
Two systems (isolated from surroundings) are brought into thermal contact ⇒ energy exchanged is allowed but volume and number of particles are kept constant
, , A A AA U S T→ and , , B B BB U S T→
Entropy measured indirectly through temperature: 1A
A A
SU T
∂=
∂ and
1B
B B
SU T
∂=
∂
What determines the temperature at equilibrium = maximum entropy principle Entropy of whole system (extensive): A BS S S= + Only the internal energies can change (degrees of freedom) ⇒ ( ),A BS U U With constraint (conservation of energy): constantA BU U U= + = Applying the maximum entropy principle: 0dS =
(2.1.13) , ,
0A BA B A B
A BV N V N
S SdS dS dS dU dU
U U ∂ ∂
= + = + = ∂ ∂
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Since constant 0A B B AU U U dU dU dU= + = ⇒ = ⇒ = −
, ,
0A BA
A BV N V N
S SdS dU
U U
∂ ∂⇒ = − = ∂ ∂
Which yields the condition at equilibrium:
(2.1.14) , ,
A B
A BV N V N
S SU U
∂ ∂= ∂ ∂
Equivalent by definition to:
(2.1.15) 1 1
A BA B
T TT T
= ⇒ =
⇒ At equilibrium the temperature are equals The energy moves in a way to increase the entropy:
(2.1.16) 1 1
0AA B
dS dUT T
= − >
If 0AdU > , the energy of system A increases, then 1 1
0 B AA B
T TT T
− > ⇒ > the energy
flows from the hot to the cold system
Other wise, 0AdU < , the energy of A decreases, and 1 1
0 A BA B
T TT T
− < ⇒ > which
means again that the energy flows from the hot to cold system
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Relation between energy, entropy, heat and temperature Energy = capacity to do work Heat = flow of energy = flow of capacity to do work Entropy = quantify tendency of heat flow S is a sort of potential and 1S U T∂ ∂ = is like a force – the higher the temperature and the higher the tendency for energy to escape
• Heat flows from the high entropy to the low entropy system • The amount of entropy gained by a cold system is higher than the amount of
entropy lost by the hot system
Heat flows to maximize entropy not to equalize the energy
Two systems at equilibrium have the same temperature (slope of S as function of U ) but not the same energy
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Example 2.1.3: Pressure = force to change volume
Cylinder partition into subsystems A and B by movable piston
V is the degree of freedom, U and N are constants
( ),A A AS U U and ( ),B B BS U U with constraints constantA BV V V= + = B AdV dV⇒ = − y
constantA BU U U= + = B AdU dU⇒ = − Applying the maximum entropy principle:
(2.1.17) 0A B A BA B A B
A B A B
S S S SdS dU dU dV dV
U U V V ∂ ∂ ∂ ∂
= + + + = ∂ ∂ ∂ ∂
(2.1.18) 1 1
0A BA A B
A B A B
S SdS dU dV dV
T T V V ∂ ∂
= − + + = ∂ ∂
At equilibrium A BT T T= =
(2.1.19) 0A B A BA B A
A B A B
S S S SdS dV dV dV
V V V V
∂ ∂ ∂ ∂= + = − = ∂ ∂ ∂ ∂
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Applying the definition p ST V
∂=
∂
(2.1.20) 0A BA
p pdS dV
T T = − =
A Bp p⇒ =
The trend to increase the volume:
(2.1.21) 0A BA
p pdS dV
T T = − >
For 0AdV > , the volume A increases, because the pressure is higher in this part of the cylinder The last example may seem ambiguous – the volume changes because a higher force is applied by the particles in the part of the cylinder having higher pressure But in reality the volume changes because of the increase of multiplicity (entropy) Example 2.1.4: Equalizing pressure = maximizing multiplicity (entropy)
Let consider the lattice model again
A BN N N= + particles and lattices A BM M M= +
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The multiplicity of the whole system:
(2.1.22) ( ) ( )
! !! ! ! !
A B
A A A B B B
M MW
N M N N M N
= ⋅ − −
Using the Stirling approximation and the fact that A BM M M− =
(2.1.23) ( )
( )( )
AA
A A A BA B
M MMAA
M N M M NN NA A A B A B
M MMW
N M N N M M N
−
− − −
− = ⋅ − − −
By definition, the entropy is equal to:
(2.1.24)
( ) ( )
( ) ( ) ( ) ( )
ln ln ln
ln ln
ln ln
A A A A A A
A A A B A B
A A B B
SW M M M N M N
kM M M M M M N M M N
N N N N
= = − − − +
+ − − − − − − − −
− −
Which is maximum for *A AM M=
Since , AM N and BN are constants
(2.1.25) ( ) ( )
( )
* * *
*
ln0 1 ln 1 ln 1 ln
1 ln
A A A AA
A B
d WM M N M M
dM
M M N
= = + − − − − − − +
+ + − −
Or
(2.1.26) * *
* *ln 0A A B
A A A
M M M NM N M M
− −⋅ = − −
* *
* * 1A A B
A A A
M M M NM N M M
− −⇒ ⋅ =
− − or
* *
* *A B A A
A A
M M N M NM M M− − −
=−
That is the fraction of B sites which are emptied is equal to the fraction of A sites which are emptied Equivalently the fraction of B sites which are filled is equal to the fraction of A sites
which are filled ⇒ the densities * *B A
B A
N NM M
= are the same
But since p NkT M
= ⇒ A B
A B
p pT T
=
Molecular model gives information on densities following maximum entropy (multiplicity) principle
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Example 2.1.5: Chemical potential = tendency for particles exchange
Two compartments separated by barrier allowing exchange of particles
AN and BN = degrees of freedom
If temperature and pressure are constants no energy exchange is involved
( )A AS N⇒ and ( )B BS N with the constraint
constantA B A BN N N dN dN= + = ⇒ = −
(2.1.27) 0A BA B
A B
S SdS dN dN
N N ∂ ∂
= + = ∂ ∂
(2.1.28) 0B AA
B A
dS dNT Tµ µ
= − =
At equilibrium A BT T T= = , such that A Bµ µ= The condition for increasing entropy is:
(2.1.29) 0B AA
B A
dS dNT Tµ µ
= − >
For 0AdN > , the number of particles in partition A increases, then 0B A
B AT Tµ µ
− > and for
A BT T= B Aµ µ⇒ > , the particles move from the region where the chemical potential is higher to the region where it is lower
µ⇒ describe the tendency for the particles to escape the volume • Move from high chemical potential to low chemical potential • Equivalent = moves from region of high concentration to region of low
concentration
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Methods of inference in thermodynamics Problem of thermodynamics = S and U are not directly measurable Observables: Temperature, pressure, Work, heat capacities, concentration, electric potentials etc. Thermodynamics ⇒ making inference about un-measurable quantities from observables There are six methods:
1. Inference on U from work ( w ) 2. Inference on S from heat (q ) 3. Using thermodynamic cycles 4. Using Maxwell’s relations 5. Using properties of homogenous functions
Work can be measured but is rarely predictable
• Depends on pressure, volume, temperature, velocity of piston motion, frictional forces, viscosity etc.
Under limited circumstances predictions are possible – needs to distinguish between state variables and process variables
• State variables: , , , ,p V T N U and S - characterize state of system without considering how they get there ⇒ do not depend on pathways
• Process variables: thermal gradients, friction , diffusion, piston velocity or any time-dependent – characterize how a system pass from one state to another – depends on pathways
In particular, heat and work depend on pathways while entropy and internal energy do not
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One important process = quasi-static process – process that is performed sufficiently slowly that properties of system are independent of time or speed
• More generally – significantly slower than relaxation time of system – time it takes for a system to establish equilibrium
Can be represented by state diagram: provides relationships between w and state variables
Example – gas in cylinder with piston – expanding volume ⇒ work against external pressure (2.1.30) extw p dV∂ = − If 0dV = then 0w∂ = If constantextp = then
(2.1.31) [ ]B
A
V
ext B AVw p dV p V V= − = − −∫
Convention: • 0w∂ > work is done on the system (internal energy is increasing) • 0w∂ < work is done by the system (internal energy is decreasing)
If constantT =
(2.1.32) lnB B B
A A A
V V VB
ext intV V VA
VNkTw p dV p dV dV NkT
V V= − = − = − = −∫ ∫ ∫
Thermodynamic cycle : series of steps that begin in one state pass through other states and then return to initial state to begin again
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First law of thermodynamics (2.1.33) dU q w= ∂ + ∂ Convention:
• 0q∂ > heat flows in the system (internal energy increases) • 0q∂ < heat flows out of the system (internal energy decreases)
(2.1.34) in out in outU q q w w∆ = − + −
Method 1 The first law allows relating w to U For working device with adiabatic boundaries 0q⇒ ∂ = then dU w= ∂ ( )w w U⇒ =
Method 2 Combining first law with fundamental energy equation: (2.1.35) dU TdS pdV q w= − = ∂ + ∂ For quasi-static processes: (2.1.36) w pdV∂ = − And
(2.1.37) q TdS∂ = or q
dST∂
=
This second relation (2.37) is known as the Thermodynamic definition of entropy - it allows measuring entropy variation from heat flow
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Reversible processes The concept of reversibility = an idealization (no real systems are reversible) The concept is useful – allows to define maximum work possible in system Reversible processes = subset of quasi-static processes
• Very slows – equilibrium is kept at all time ⇒ no production of entropy 0S∆ = Reversible heat and work sources
• Reversible work source – can change volume + perform work quasi-statistically o Enclosed in adiabatic boundaries 0q TdS∂ = = and dU w= ∂
• Reversible heat source – can exchange heat quasi-statistically o Enclosed in rigid boundaries 0w pdV∂ = − = and dU q TdS= ∂ =
Both are different from reversible processes because not necessarily we have 0S∆ = Reversible process = changes in whole system in which collection of reversible heat and work sources have 0S∆ = Carnot’s principle For heat engines (1820s): maximum idealized engine where processes are irreversible Heat source: h hU q∆ = − Some of the heat is conversed to work and remaining exit from system Cold exhaust source: c cU q∆ = (2.1.38) h cq q w= + or h cw q q= − Efficiency of engine :
(2.1.39) 1h c c
h h h
q q qwq q q
η−
= = = −
Entropy change of whole system: (2.1.40) 0h c wS S S S∆ = ∆ + ∆ + ∆ ≥ But since reversible work is adiabatic 0wS∆ = (2.1.41) 0h cS S S∆ = ∆ + ∆ ≥
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For reversible heat sources: hh
h
qS
T−
∆ = and cc
c
qS
T∆ =
The maximum efficiency ⇒ reversible process 0S∆ = ⇒ all other processes cause changes in entropy
(2.1.42) 1c c c
h h h
q T Tq T T
η= ⇒ = −
Larger possible value of efficiency ⇒ smaller possible value of c c
h h
q Tq T
=
In general:
(2.1.43) 1 c
h
TT
η ≤ −
Heat is efficiently converted to work when the heat source is at highest temperature and the exhaust has lowest possible temperature IMPORTANT CONSEQUENCES
• η depends only on temperature of heat source and exhaust • Only way to have 1η = is for 0cT = , which is impossible ⇒ inefficiency of heat
engine is intrinsic = do not depends on friction or other dissipative processes For automobile engines – Carnot efficiency of order 50-60% - real efficiency (including dissipation effects) is much lower ~ 25% Pollution – this is intrinsic to inefficiency of transforming heat into work
• solution = increasing efficiency of systems
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Absolute Temperature scale Perfect efficiency requires extracting all motion and internal energy from gas molecules Defines absolute temperature scale = Kelvin scale The definition of efficiency in other scale is not straightforward: T aT b′ = − ⇒ for Celsius 1a = and 273.15b =
( )( )
( )( )
1 1 1c cc
h h h
T b a T bTT T b a T b
η′ ′+ +
⇒ = − = − = −′ ′+ +
As 0cT ′ → ( )
1h
b
T bη → −
′ −
Other statements of second law of thermodynamics Heat does not flow from cold to hot object because this implies a decrease in entropy However, this is less general because one can decrease entropy through action of external agent (Y ):
(2.1.44) 1 1 A B
A A BA B A B
S SdS dU dY dY
T T Y Y ∂ ∂
= − + + ∂ ∂
The first difference can be less than zero if the two second terms (additional works) compensate ⇒ principle of refrigerator and heat pumps (possibly also gravity) When heat and work can cross the boundaries (open system) entropy can decrease – principle of apparition and evolution of life
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Section 2: Laboratory conditions and free energy Process in test tube – temperature and pressure are controlled at boundaries, not heat or work
• ⇒ Two new thermodynamics quantities = Free energy and Enthalpy • ⇒ New extremum principles: system tend towards minimum free energy
Choice of independent variable depends on boundaries
• When T , p and µ included as part of specification of a system ⇒ the system is in contact with large reservoir (bath) that can exchange (fluctuations) exclusive quantities U , V or N
Consider a process in test tube immersed in heat bath ⇒ T , N and V are kept constants Heat bath: any surroundings of a sys tem that keep the temperature constant T is controlled not U ⇒ independent variables ( ), ,T V N Natural variable of function ( ), ,F T V N = Helmholtz free energy
Extremum of ( ), ,F T V N predicts equilibrium of system constrained by T
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Helmholtz free energy Process in test tube could be complex:
• Vary in rate – quasi-static to explosive • Involve chemical or phase changes • Absorb heat
Process in test tube will influence heat bath only through heat exchange Combined system (isolated from surroundings) = test tube + heat bath ⇒ reach equilibrium when ( ), , maximumcombinedS U V N = : (2.2.1) 0combined system bathdS dS dS= + ≥ But since system is isola ted: (2.2.2) 0bath system bath systemdU dU dU dU+ = ⇒ = − Using fundamental equation:
(2.2.3) 1 1
bath bath bath
pdS dU dV dN dU
T T T Tµ
= + − =
Replacing with 2.2.2:
(2.2.4) 1
bath systemdS dUT
= −
The condition of equilibrium can be rewritten as:
(2.2.5) 1
0system systemdS dUT
− ≥
Or (2.2.6) 0system systemdU TdS− ≤ Define a new function – Helmholtz Free Energy: (2.2.7) F U TS= − Such that (2.2.8) dF dU TdS SdT dU TdS= − − = − And at equilibrium F is a minimum: (2.2.9) 0dF ≤
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The definition F U TS= − shows that the Helmholtz free energy is determined by a balance between internal energy and entropy
• Position of balance is determined by T • To minimize F the system must tend towards high entropy and low energy
depending on temperature – if T is high entropy dominates while at low enough temperature it is the contrary
Example 2.2.1: Model of dimerization
Consider two particles 2N = in V lattice sites arranged in row at temperature T
Dimmer state
Dissociated state
Consider the dimmer state: Since there is interaction (bounded state):
dimerU ε= − where 0ε > ⇒ attractive energy On V sites ⇒ 1W V= − arrangements ⇒ ln ln( 1)dimerS k W k V= = − The free energy of dimmer site: (2.2.10) ( )ln 1dimer dimer dimerF U TS kT Vε= − = − − − For the monomer state:
0monoU = And multiplicity:
( ) ( ) ( ) ( ) ( )1!
1 1 1 12! 2 ! 2 2mono total dimer
V VV VW W W V V V
V− = − = − − = − − = − − −
The free energy for the monomer state:
(2.2.11) ( )ln 1 12mono
VF kT V = − − −
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For 0T → , dimer monoF F< the dimmer state is favored by internal energy For high T , dimer monoF F> the monomer state is favored by entropy
For 0T T= dimer monoF F= ( ) ( )0 0 0ln 1 ln 1 ln 12V
kT V kT kT Vε ⇒ − − − = − − − −
(2.2.12) 0
ln 12
TVk
ε=
−
The stronger the interaction energy the higher the temperature necessary to weaken the bonding Although simple, the above model represents essentials of dimmer dissociation and many processes involving bond breakage – like boiling, melting or unfolding polymers Equilibrium = balance between energetic tendency of particles to stick together (dominating at low temperature) and entropic tendency to disperse (dominating at high temperature) The principle of maximum entropy used alone ⇒ no dimmers are possible, which is not consistent with observation
25
Example 2.2.2: Toy model for polymer collapse
When a polymer molecule is put into certain solvents (poor solvents) it collapses into a compact structure ⇒ attraction similar to dimmer 2D model = 4 monomers in test tube in water bath Polymer chain (compact state) with energy ε− 2 states possible – compact and open
For compact state, 1cW = , stabilized by internal energy cU ε= −
ln1c c cF U TS kTε ε⇒ = − = − − = − For the open state, 4oW = and 0oU =
0 ln4o o oF U TS kT= − = − For 0T → the compact state is favored by internal energy and at high temperature the open state is favor by entropy
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The collapse temperature: 0 l n 4T
kε
= which increase with the internal energy
Consider the collapse process = change from open state to compact state
( ) ( ) ln4c oF F T F T kTε∆ = − = − + The variation in energy is U ε∆ = − and the variation in entropy is ln4S k∆ = − Because V is constant no work is involved in the process The first law implies that 0U q ε∆ = = − < heat is given to the bath ⇒ exothermic for all temperature Process that takes heat = endothermic
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Fundamental equation for F (2.2.13) ( )dF d U TS dU TdS SdT= − = − − Substituting the fundamental energy equation for dU :
(2.2.14) 1 1
M M
j j j jj j
dF TdS pdV dN TdS SdT SdT pdV dNµ µ= =
= − + − − = − − +
∑ ∑
By definition:
(2.2.15) 1, , , , i j
M
jjV N T N j V T N
F F FdF dT dV dN
T V N≠
=
∂ ∂ ∂ = + + ∂ ∂ ∂ ∑
We deduce therefore three new thermodynamic relations:
(2.2.16) , , , ,
; ;i j
jV N T N j V T N
F F FS p
T V Nµ
≠
∂ ∂ ∂ = − = − = ∂ ∂ ∂
Enthalpy and Gibbs free energy Depending on the boundaries we have 3 useful functions:
1. ( ), ,S U V N which is maximum when U is controlled at the boundary
2. ( ), ,U S V N which is maximum when S is controlled at the boundary
3. ( ), ,F T V N which is maximum when T is controlled at the boundary There are two more fundamental functions possible: the Enthalpy ( ), ,H S p N and Gibbs
free energy ( ), ,G T p N The enthalpy is seldom used as an extremum principle because it is not convenient to control the entropy – but it is important because it can be measured during calorimetry experiments and it opens an experimental route to Gibbs free energy – central for chemistry and biology The Gibbs free energy is one of most important fundamental function – constant T and constant p are easiest constraints to impose (provided by atmosphere and room temperature)
28
To find ( ), ,H S p N and ( ), ,G T p N we can use the Legendre transform method Legendre transform method A function ( )y f x= can be described as a list of pairs ( ) ( )1 1 2 2, , , ,x y x y … You can express the same function as a list of different pairs - slope ( )c x and
intercepts ( )b x : ( ) ( )1 1 2 2, , , ,c b c b …
For a small change dx , the change dy can be described by the slope at that point
(2.2.17) ( )dydy dx c x dx
dx = =
The full function: (2.2.18) ( ) ( ) ( ) ( ) ( ) ( )y x c x x b x b x y x c x x= + ⇒ = − To see how small changes in slope lead to small changes in intercept, takes differential of 2.2.18 and substitute 2.2.17: (2.2.19) ( )db x dy cdx xdc xdc= − − = − For multivariate function ( )1 2 3, ,y y x x x= (2.2.20) 1 1 2 2 3 3dy cdx c dx c dx= + + Where
(2.2.21) 2 3
11 ,x x
yc
x ∂
= ∂ ,
1 3
22 ,x x
yc
x ∂
= ∂ ,
1 2
33 ,x x
yc
x ∂
= ∂
29
The intercept 1b along 1x axis
(2.2.22) ( )1 1 2 3 1 1, ,b c x x y c x= − Taking differential and substituting 2.2.20 (2.2.23) ( )1 1 2 3 1 1 1 1 1 1 2 2 3 3, ,db c x x dy cdx xdc xdc c dx c dx= − − = − + + Where
(2.2.24) 2 3
11
1 ,x x
bx
c ∂
= − ∂ ,
1 3
12
1 ,c x
bc
x ∂
= ∂ ,
1 2
13
3 ,c x
bc
x ∂
= ∂
Any of the original independent variables ix can be exchanged with their conjugate variables ic Transformation can be performed on any combination of conjugate pairs – there are total 2 1n − possible transformations To find ( ), ,H S p N start with ( ), ,U S V N , then replace dV with dp - this is done (the only way) using the internal energy – we define the enthalpy H as: (2.2.25) H U pV= +
dH dU pdV Vdp⇒ = + + Replacing by the fundamental equation for energydU :
1 1
M M
j j j jj j
dH TdS pdV dN pdV Vdp TdS Vdp dNµ µ= =
⇒ = − + + + = + +
∑ ∑
(2.2.26) 1
M
j jj
dH TdS Vdp dNµ=
= + + ∑
Therefore H is minimum at equilibrium when S , p and N are the independent variables To find ( ), ,G T p N start with ( ), ,H S p N and replace dS with dT (2.2.27) G H TS= −
dG dH TdS SdT⇒ = − −
30
Replacing by the fundamental equation for enthalpydH :
1 1
M M
j j j jj j
dG TdS Vdp dN TdS SdT SdT Vdp dNµ µ= =
⇒ = + + − − = − + +∑ ∑
(2.2.28) 1
M
j jj
dG SdT Vdp dNµ=
= − + +∑
Logic similar as Helmholtz free energy: if process occurs in test tube held at constant temperature and pressure, then equilibrium is reached when G is at minimum
• Important for chemical reactions, phase changes, biological or other physical processes
• Exchange of energy through change in volume or heat transfer Since
(2.2.29) 1, , , , i j
M
jjp N jT N p T N
G G GdG dT dp dN
T p N≠
=
∂ ∂ ∂ = + + ∂ ∂ ∂ ∑
We have three new definitions:
(2.2.30) , , , ,
; ;i j
jp N jT N p T N
G G GS V
T p Nµ
≠
∂ ∂ ∂ = − = = ∂ ∂ ∂
For equilibrium phase changes at constant T and p (and constant N ) 0dG =
31
Example 2.2.3: Melting of ice and freezing water
Reversible phase change occurring at 0T T= When 0T T< , G is minimum 0dG = , this is solidG Adding heat, the system change phase without changing T At equilibrium liquidG Free energy of melting: liquid solidG G G∆ = − Since 0 0dT dp dN dG= = = ⇒ = The entropy and enthalpy of melting balance
0H T S∆ = ∆
Melting, boiling and other phase changes involve increase in entropy compensated by increase in enthalpy resulting from breakage or weakening of molecular interactions
32
The limits in constructing thermodynamics Functions Four categories:
1. Fundamental and useful: the table below lists the main fundamental thermodynamics functions and their natural variables.
Table 2.2.1 – fundamental thermodynamic equations Function Equilibrium Fundamental equation Definition ( ), ,U S V N Minimum
1
M
j jj
dU TdS pdV dNµ=
= − +∑
( ), ,S U V N Maximum
1
1 Mj
jj
pdS dU dV dN
T T T
µ
=
= + − ∑
( ), ,F T V N Minimum
1
M
j jj
dF SdT pdV dNµ=
= − − + ∑ F U TS= −
( ), ,H S p N Minimum
1
M
j jj
dH TdS Vdp dNµ=
= + + ∑ H U pV= +
( ), ,G T p N Minimum
1
M
j jj
dG SdT Vdp dNµ=
= − + +∑ G H TS= −
2. Useful but not fundamental: ( ), ,U T V N , ( ), ,S T V N , ( ), ,H T p N and
( ), ,S T p N not function of natural variables ⇒ do not have extremum; useful
because components of ( ), ,F T V N and ( ), ,G T p N 3. Complete but not useful: rearrangements of dependent and independent
variables from fundamental thermodynamic functions, but not useful; ex. ( ), ,T F V N not useful because you cannot constrain F at the boundary of system
4. Incomplete: functions formed of conjugate pairs, ( ), ,U p V N or ( , , )S U Nµ
⇒ variables are missing and they do not specify the state of a system; cannot be obtained by Legendre transforms of the fundamental equations
33
Method 3: heat capacity Consider ( ) ( ) ( ), , , , , ,F T V N U T V N TS T V N= − The variables are natural for F but not for U or S ⇒ minimizing U or maximizing S individually does not predict state of equilibrium But U and S can be measured from calorimetry experiments
• Bomb calorimeter: a measured electrical current heats a sample – thermometer measures corresponding change in temperature;
0 0dV dw pdV dU q TdS= ⇒ ∂ = − = ⇒ = ∂ =
Similarly ( ) ( ) ( ), , , , , ,G T p N H T p N TS T p N= − The enthalpy plays the same role as U in F (balance with entropy)
34
Heat capacity: ability of material to absorb heat Definition = operational recipe:
(2.2.31) VV V V
q U SC T
dT T T∂ ∂ ∂ = = = ∂ ∂
Varies for different temperatures:
In general, to find U∆ from change A BT T→ need to know ( )VC T
(2.2.32) ( )B
A
T
VTU C T dT∆ = ∫
35
At constant pressure: ( )dH dU pdV Vdp q w pdV= + + = ∂ + ∂ +
For quasi-static processes w pdV∂ = − (2.2.33) dH q= ∂ Heat capacity (constant pressure):
(2.2.34) pp p p
q H SC T
dT T T∂ ∂ ∂ = = = ∂ ∂
Where H and pC have the same relation as U and VC
(2.2.35) ( )B
A
T
pTH C T dT∆ = ∫
Within given phase pC is independent from temperature Table 2.2.2 Heat capacities at 25 CT = o (Tinoco, Sauer & Wang, 1978, in Physical Chemistry Principles and Applications in Biological Science, Prentice Hall)
36
Example 2.2.4: Temperature dependence of entropy
For quasi-static processes, if VC independent of T
(2.2.36) V
dTdS C
T=
(2.2.37) lnB
A
TB
V VTA
TdTS C C
T T∆ = =∫
If pC is known instead of VC
(2.2.38) lnB
A
TB
p pTA
TdTS C C
T T∆ = =∫
In general, since p VC C≠ the entropy difference is different Sometimes it is useful to know entropy change starting from 0KT = Third law of thermodynamics For a perfect crystalline substance ( )0K 0S =
Equation 2.2.32 ( )B
A
T
VTU C T dT∆ = ∫ we get ( )U T∆ and from equation 2.2.38
B
A
T
pT
dTS C
T∆ = ∫ we get ( )S T∆
From this we can deduce ( )F T∆ : ( ) ( ) ( ), , , , , ,F T V N U T V N T S T V N∆ = ∆ − ∆
• We can predict the free energy change from heat capacity measurements • We can use heat capacities to predict equilibrium temperatures of objects in
thermal contact
37
Example 2.2.5: Equilibrium temperature of objects in thermal contact
Suppose objects A and B have different heat capacities A
VC and BVC
Suppose A BT T< , when they are brought in thermal contact they reach sameT Since the two objects are isolated from surroundings: (2.2.39) 0A BU U U∆ = ∆ + ∆ = Where
(2.2.40) ( )A
T A AA V V AT
U C dT C T T∆ = = −∫
And
(2.2.41) ( )B
T B BB V V BT
U C dT C T T∆ = = −∫
When VC is high (low) the temperature change is small (high) From 2.2.39: (2.2.42) ( ) ( )A B
V A V BC T T C T T− = − −
A B
V A V BA B
V V
C T C TT
C C+
=+
In particular, if 2
A B A BV V
T TC C T
+= ⇒ =
Usually heat capacities are given per mass or volume ⇒ need to multiply by these quantities to get total heat capacities of objects
38
Method 4: Using thermodynamic cycles Around complete cycle state functions must sum to zero around complete cycle ⇒ If you cannot determine thermodynamic changes directly you can compute it constructing fictitious cycles Example 2.2.6: Obtaining enthalpy for conditions that are not measurable directly
Boiling water at 0 Co at 1 atmosphere (impossible) - usefulness = ( )0 CboilH∆ o can tell us
something about vapor pressure Recipe: get boilH∆ from heat capacities of water and steam - get enthalpy of vaporization for more standard conditions – then use them in cycle Standard state enthalpy: ( ) 1100 C 540 cal gboilH −∆ =o , where 1 cal 4.186 J =
And heat capacities: For steam 1 10.448 cal K gs
pC − −= and for liquid 1 11.00 cal K glpC − −=
(2.2.43) ( )( )
0 1 1 1
100
0 1 1 1
100
1.00 cal K g ( 100K) 100 cal g
0.448 cal K g ( 100K) 44.8 cal g
lliquid p
ssteam p
H C dT
H C dT
− − −
− − −
∆ = = − = −
∆ = = − = −
∫∫
In a cycle: ( ) ( )0 C 100 Cboil liquid boil steamH H H H∆ = − ∆ + ∆ + ∆o o
( ) ( ) 1 10 C 100 540 44.8 cal g 595.2 cal gboilH − −∆ = + − =o
In general, even though a physical process can be complex (pathway is unknown) we can estimate values using (imaginary) pathways in cycle
39
Example 2.2.7: Ideal gas in adiabatic expansion
Model: gas expanding quasi-statistically and adiabatically in cylinder with piston Adiabatic assumption: common in real piston engines because heat transfer much slower than volume change When constantN = , then ( ),U U T V=
(2.2.44) T V
U UdU dV dT
V T∂ ∂ = + ∂ ∂
For an ideal gas ( )U U T= 0T
UV
∂ ⇒ = ∂ , because at low density a change in density do
not affect weak molecular interactions
(2.2.45) VV
UdU dT C dT
T∂ = = ∂
From first law: dU w pdV= ∂ = − (2.2.46) VC dT pdV= −
For ideal gas NkT
pV
= and VC independent of T
(2.2.47) V
NkTC dT dV
V= −
Rearranging the relation and integrating:
(2.2.48) 2 2
1 1
1 1T V
V T VC dT Nk dV
T V= −∫ ∫
(2.2.49) 2 1
1 2
ln lnV
T VC Nk
T V=
Therefore for a quasi-static volume change:
(2.2.50) 2 1
1 2
VN k CT VT V
=
Ideal gas adiabatic expansion: the temperature increases when the volume decreases and vice versa
40
Example 2.2.8: Combustion engine – Otto cycle
41
Efficiency = measure intake and exhaust temperature and heat capacity of fuel vapor Idealizations: fuel vapor = ideal gas and work performed adiabatically in reversible quasi-static steps Efficiency = net work performed by the system for given heat influx
(2.2.51) ( ) ( ) ( )
( )1 4 3 2 3 4
2 1 2 1
1f c V V
b V
w w C T T C T T T Tq C T T T T
η− + − − − − −
= = = −− −
For steps c and f :
3 1 13 2
2 2 2
V VN k C Nk CT V V
T TT V V
= ⇒ =
and 4 1 1
4 11 2 2
V VNk C N k CT V V
T TT V V
= ⇒ =
Therefore ( ) 13 4 2 1
2
VN k CV
T T T TV
− = −
And
(2.2.52) 1
2
1 1V
V
N k C
Nk CVr
Vη −
= − = −
Depends on compression ratio 2
1
Vr
V= and heat capacity of fuel VC
Typical ratio ~10r In real engines: step a involve a change of mole number when fuel is burned – steps c and f not quasi-static and vapor is not ideal ⇒ real efficiency lower than ideal one Because U is a state function:
0b c de fU U U U∆ + ∆ + ∆ + ∆ =
( ) ( ) ( ) ( )2 1 3 2 4 3 1 4 0VC T T T T T T T T⇒ − + − + − + − =
On the other hand, work and heat are not state functions ⇒ the engine perform work during each cycle and total b c de fw w w w w= + + + = area inside the cycle
42
Section 3: Maxwell’s relations & Mixtures We develop two other methods of thermodynamics 5) Maxwell relations – useful to find entropic and energetic contributions to the stretching of matter, expansion of surfaces or films, compression of bulk material 6) Mathematics of homogeneous functions – leads to the Gibbs-Duhem relationships , describing temperature and pressure dependence of chemical equilibria Design of fundamental equations (recipe):
• Each extensive degree of freedom in the fundamental energy equation is paired with its conjugate force
o [ ] [ ] [ ], , , , ,p V T S Nµ
o Solids and elastic materials [ ] [ ]force, length ,f l=
o Liquids and interfaces [ ] [ ]surface tension, area ,aγ=
o Charged particles [ ] [ ]electric potential, charge ,qψ=
o Magnetic field [ ] [ ]magnetic field, magnetic moment ,B I=
o General [ ] [ ]generalized force, extensive variable ,P X= • Generalize fundamental energy equation introducing relevant extensive variables
o ( ), , ,U U S V N X=
• Yields dU as a function of work terms∑
(2.3.1) j j j jj j
dU TdS pdV dN fdl da dq BdI PdXµ γ ψ= − + + + + + +∑ ∑
Where the general forces:
(2.3.2) , , , i j
jj S V N X
UP
X≠
∂= ∂
• Apply Legendre transform method
43
Example 2.3.1: fundamental equation when surface tension important Experiments on soap films in dish pan, on cell membranes, or on interfaces ⇒ everywhere where surface properties are important
• Ex. if spherical drop of water distorted to non-spherical shape keeping V constant ⇒ increase area ⇒ increase free energy
Consider process in which the surface tension is fixed and area can varies
• Langmuir trough - Area a can be varied at constant surface tension γ by pushing with constant force sideways on floating bar in a dish pan to keep the soap film corralled on one side
Find the function whose extremum identifies the state of equilibrium for constant , , ,T p N γ
Since ( ), , ,U U S V N a=
(2.3.3) j jj
dU TdS pdV dN daµ γ= − + +∑
Convert to independent variables , , ,T p N γ adding following exact differentials (choice of ± , cancel terms in dU to get appropriate independent variables)
( )( )( )
d TS TdS SdT
d pV pdV Vdpd a da adγ γ γ
− = − −
+ = +− = − −
(2.3.4) ( ) ( )j j
j
d U TS pV a SdT Vdp dN ad d G aγ µ γ γ− + − = − + + − = −∑
At equilibrium G aγ− is a minimum
44
Example 2.3.2: Surface area is independent variable Langmuir trough experiment - holding bar at fixed position - determining force applied ⇒ extremum identifies equilibrium at , , ,T p N a constants Starting with j j
j
dU TdS pdV dN daµ γ= − + +∑
Add only two terms: ( )( )
d TS TdS SdT
d pV pdV Vdp
− = − −
+ = +
(2.3.5) ( ) ( )j j
j
d U TS pV SdT Vdp dN da d Gµ γ− + = − + + + =∑
The relevant function is just the Gibbs free energy (where aγ is included) ⇒ at equilibrium with , , ,T p N a constants, the Gibbs function is minimum Ex. water droplet at constant , ,T p N , 0dG daγ= = so the surface area must be minimum ⇒ droplet takes the form of a sphere
Method 5 – Maxwell’s relations Relationships between partial derivatives that follow Euler’s reciprocal relations For example, if one wants to obtain ( ),U S V (not measurable) keeping N constant
Euler’s relation: 2 2U U U U
V S V S S V S V∂ ∂ ∂ ∂ ∂ ∂ = = = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
But since V
UT
S∂ = ∂
and S
Up
V∂ = − ∂
we get
(2.3.6) S V
T pV S
∂ ∂ = − ∂ ∂
45
Recipe to obtain Maxwell’s relations
Suppose the quantity of interest is ,T N
Sp
∂ ∂
, which describes the multiplicity of
microscopic states changes as you squeeze a material (model for force between atoms)
• Identify independent variables: ( ), ,T p N
• Find natural function (table 2.2.1 page 32): ( ), ,G T p N • Express total differential of the natural function
(2.3.7) j j
j
dG SdT Vdp dNµ= − + +∑
• Based on Euler relations set cross derivatives equal: 2 2G G
p T T p ∂ ∂
= ∂ ∂ ∂ ∂
(2.3.8) 2G S G
ST p p T
∂ ∂ ∂− = ⇒ − =
∂ ∂ ∂ ∂
And
(2.3.9) 2G V G
Vp T T p
∂ ∂ ∂= ⇒ =
∂ ∂ ∂ ∂
From which we deduce that
(2.3.10) ,, p NT N
S Vp T
∂ ∂ = − ∂ ∂
You cannot measure ,T N
Sp
∂ ∂
but it is easy to get ,p N
VT
∂ ∂
⇒ measure V change with
T keeping p constant
The slope is ,p N
VT
∂ ∂
, which is equivalent to ,T N
Sp
∂− ∂
For an ideal gas,
0p N
VT
∂ > ∂ , which implies that for 0 0p S∂ > ⇒ ∂ < the entropy
decreases as the pressure increases
46
Table 2.3.1 Maxwell relations (H.B. Callen 1985, in Thermodynamics and an introduction to thermostatistics, 2nd ed. Wiley)
47
Example 2.3.3: Type of work different than pV - thermodynamics of rubber band Question – is the retraction of a rubber band driven by enthalpy or entropy? Model of microscopic behaviors of polymer materials Assume quasi-static stretching force that increases with length l ⇒ force of retraction f = negative of force of stretching
Internal energy function ( ), ,U U S V l= (2.3.11) dU TdS pdV fdl= − + For constants ,T p ⇒ Gibbs free energy and ( ) ( )dG d H TS d U pV TS= − = + − Substituting in equation for energy: (2.3.12) dG SdT Vdp fdl= − + + The force can be defined as:
(2.3.13) , , ,T p T p T p
G H Sf T
l l l∂ ∂ ∂ = = − ∂ ∂ ∂
Taking the cross derivative in equation 2.3.12:
(2.3.14) , ,T p p l
S fl T
∂ ∂ = − ∂ ∂
The entropic component can be obtained from simple experiment: hold rubber band at fixed stretched length l and constant pressure p and measure retractive force variation with T
48
Retractive force of rubbery polymer – amorphous polyethylene (Mark et al. 1993, in Physical property of polymer, 2nd ed. American Chemical Society)
The slope is ,
0p l
fT
∂ > ∂ 0S⇒ ∂ < for 0l∂ > , the entropy decreases upon stretching
To get the enthalpy component substitute 2.3.14 in 2.3.13:
(2.3.15) , ,T p p l
H ff T
l T∂ ∂ = − ∂ ∂
Since ,
0p l
fT
∂ > ∂ , the retractive force gets stronger as the temperature increases
This behavior distinguish rubber from metal (J. Cough 1806): entropy of metal increases as temperature increase – stretching metal loosen the bonds increasing volume per atom leading to translational disorder – for rubber, increasing temperature decreases multiplicity of polymer conformation
49
Expansion versus compression
Thermal expansion coefficient
(2.3.16) 1
p
VV T
α∂ = ∂
For an ideal gas
(2.3.17) 1 1
p
V p NkV T NkT p T
α∂ = = ⋅ = ∂
Thermal expansion for polyethylene (Mark et al. 1993, in Physical property of polymer, 2nd ed. American Chemical Society)
At low temperature, polyethylene = hard crystalline plastic material – on melting at around 130 Co the specific volume increases significantly and α is large The thermal expansion coefficient is usually positive, because increasing the temperature causes a loosening up of the intermolecular bonds in the material
50
Isothermal compressibility
(2.3.18) 1
T
VV p
κ ∂
= − ∂
For ideal gas:
(2.3.19) 2
1 1
T
V p NkTV p NkT p p
κ ∂
= − = ⋅ = ∂
Relative volume for hexadecane ( 60C, 90C, 120 C)•o o oo W – compressibility increases with temperature (Orwoll & Flory 1967, Am Chem Soc 89, 6814)
Because slope is negative 0κ > ⇒ increasing the pressure increases compressibility
• Usually we assume solids and liquids are incompressible but this is not realistic • But gas are much more compressible than solids or liquids
51
Volume = clues of forces within material (A. Hvidt, 1978, Acta Chemica Scand A, 32, 675)
a) From bottom-up: water, benzene, n-pentane and diethyl-ether; Organic liquids have larger volume expansion – bonding in water more rigid than in organic liquids
b) Symbols similar than in a; Water less compressible than organic liquids – also changes with temperature unusual for water
Example 2.3.4: Entropy change with pressure
(2.3.20) ,T N
SdS dp
p ∂
= ∂
Use Maxwell relation: ,, p NT N
S Vp T
∂ ∂ = − ∂ ∂ and combine it with
1
p
VV T
α∂ = ∂
(2.3.21) ,p N
VdS dp Vdp
Tα
∂ = − = − ∂
We can measure entropy variation from thermal expansion experiment
(2.3.22) ( ) ( )2
1
p
pS p V p dpα∆ = −∫
52
Example 2.3.5: Energy change with pressure
T
UV
∂ ∂
describe forces in materials
Start with ( ),U T V and ( ),S T V :
T V
U UdU dV dT
V T∂ ∂ = + ∂ ∂
and T V
S SdS dV dT
V T∂ ∂ = + ∂ ∂
Substitute in dU TdS pdV= −
T V T V
U U S SdV dT T dV dT pdV
V T V T ∂ ∂ ∂ ∂ + = + − ∂ ∂ ∂ ∂
When T is constant this simplify to:
T T
U ST p
V V∂ ∂ = − ∂ ∂
Using the Maxwell relation: T V
S pV T
∂ ∂ = ∂ ∂
T V
U pT p
V T∂ ∂ = − ∂ ∂
The measurable quantity is V
pT
∂ ∂
= thermal pressure coefficient
For an ideal gas: 0V T
p Nk UT V V
∂ ∂ = ⇒ = ∂ ∂ the energy does not depend on the volume
For liquids:
0T V
U pT p
V T∂ ∂ = − < ∂ ∂
at high densities and 0> at low densities
When you squeeze a material ( )0V∂ < that is very compact you are pushing against
repulsive force 0 0 for 0U
U VV
∂< ⇒ ∂ > ∂ <
∂
When you pull on a material ( )0V∂ > at low density you are pulling against attractive
force 0 0 for 0U
U VV
∂> ⇒ ∂ > ∂ >
∂
53
Multicomponent systems Systems having more than one chemical component (mixtures and alloys) Consider a system with n moles of one type
Molar volume : V
vn
= and molar free energy: G
gn
=
Consider a system formed of M different species: 1 2, , , Mn n n n= … Partial molar volume :
(2.3.23) , , i j
jj T p n
Vv
n≠
∂= ∂
Describes how much V increase when adding jdn moles of type j species
(2.3.24) , , i j
j j jj jj T p n
VdV dn v dn
n≠
∂= = ∂
∑ ∑
Related quantity = partial specific volume
(2.3.25) , , i j
jj j T p w
VdV dw
w≠
∂= ∂ ∑
Where jw is the weight of species j Simple cases ⇒ partial molar volume independent of composition of mixture ⇒ molar volume of pure components Not true in general: ex. mixture of alcohol in water a wV V V≠ + , for small concentration of alcohol, the alcohol molecules pull the molecules of water towards them
a wV V V⇒ < + If we fix the molar volume of pure water wv
(2.3.26) w wapp
a
V n vV
n−
=
At low concentration a wn n= , appV V<
54
Adding alcohol to water first decreases the total volume. Adding more alcohol increases the apparent volume (S. G. Bruun & A. Hvidt 1977, Ber Bunsenges Phys Chem 81, 930) Ions (ex. magnesium sulfate) have negative partial molar volumes in water (electrostriction)
• Strong attraction through electrostatic interactions • Adding salt to water makes V to shrink
Chemical potentials = partial molar free energies In terms of , ,U F H and G :
j jj
j jj
j jj
j jj
dU TdS pdV dN
dF SdT pdV dN
dH TdS Vdp dN
dG SdT Vdp dN
µ
µ
µ
µ
= − +
= − − +
= + +
= − + +
∑
∑
∑
∑
There are consequently several definitions of the chemical potential
(2.3.27) , , , , , , , ,i j i j i j i j
jj j j jV S N T p N T V N S p N
U G F Hu
N N N N≠ ≠ ≠ ≠
∂ ∂ ∂ ∂= = = = ∂ ∂ ∂ ∂
Partial molar quantity at constants temperature and pressure:
(2.3.28) , , i j
jj T p n
Gn
µ≠
∂= ∂
Divided in partial molar enthalpy and partial molar entropy:
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(2.3.29) , , , ,i j i j
j j jj jT p N T p N
H ST h Ts
N Nµ
≠ ≠
∂ ∂= − = − ∂ ∂
Using this expression in j j
j
dG SdT Vdp dNµ= − + +∑
(2.3.30) j j jj
dG SdT Vdp h Ts dN = − + + + ∑
The entropy appears two times: • SdT− describes how G change with T • j jh Ts+ describes how G change with jdN ⇒ changes in chemical composition
in absence of thermal changes When quantities are linked ⇒ change of one partial molar quantity affect all the others
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Method 6 - homogeneous functions - Gibbs-Duhem equation Describes linkage relationships between partial molar quantities Extensive properties = first order homogenous functions of mole numbers Consider jV n⇒ independent variables assuming temperature and pressure are constants If all mole numbers increase uniformly by λ : (2.3.31) ( ) ( )1 1 1 1, , , , , ,M MV n n n V n n nλ λ λ λ=… …
(2.3.32) 1
, , i j
M M
i j ji j jj T p n
VV n n v
n≠
= =
∂= = ∂ ∑ ∑
In general for any extensive property (2.3.33) ( ) ( )1 1 1 1, , , , , , , , , ,M MY T p n n n Y T p n n nλ λ λ λ=… … T and p are intensive and not subject to the scaling The partial molar value of Y
(2.3.34) ( )1 2
, ,
, , , , ,i j
j Mj T p n
Yy T p n n n
n≠
∂= ∂
…
This is the change in Y that occurs when a small amount of the species j is added to a system
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To derive the linkage between the partial molar volumes of two components of one system we combine two results: The first one is:
(2.3.35) 1 2
1 211 2i i i M
M
M j jjMn n n
V V VdV dn dn dn v dn
n n n≠ ≠ ≠
=
∂ ∂ ∂= + + + = ∂ ∂ ∂
∑L
The second, taking the derivative of 1
, , i j
M M
i j ji j jj T p n
VV n n v
n≠
= =
∂= = ∂ ∑ ∑ and comparing with
previous definition
(2.3.36) ( )1 1
M M
j j j j j jj j
dV n dv v dn v dn= =
= + =∑ ∑
We deduce therefore that:
(2.3.37) 1
0M
j jj
n dv=
=∑
The partial molar volume of different components iv and jv are linked together
Because U is a first order homogenous equation it obeys the relation ( )1
M
jj j
fx f
xλ=
∂=
∂∑
The independent variables are ,S V and N , the corresponding equation is:
(2.3.38) 1, , , , i j
M
jjV N S N j S V N
U U UU S V N
S V N≠
=
∂ ∂ ∂ = + + ∂ ∂ ∂ ∑
Substituting ,T p− and jµ for the partial derivatives
(2.3.39) 1
M
j jj
U TS pV Nµ=
= − +∑
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Taking the derivative of this last equation:
(2.3.40) 1 1
M M
j j j jj j
dU TdS SdT pdV Vdp dN N dµ µ= =
= + − − + +∑ ∑
Subtracting the fundamental equation 1
M
j jj
dU TdS pdV dNµ=
= − +∑
We get the Gibbs-Duhem equation:
(2.3.41) 1 1
0M M
j j j jj j
N d Vdp SdT N dµ µ= =
= − ⇒ =∑ ∑
For constant temperature and pressure
• This equation gives the relation between the chemical potentials of the different species
• Helps in understanding how molecular interaction are coupled in osmosis, ligand bindings, and phase transitions
• Helps in describing how the chemical potential s depend on temperature and pressure
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