1
The Second Law of Thermodynamics
Statements of the 2nd Law of Thermodynamics
Kelvin’s statement – No process is possible in which the sole result is the absorption
of heat from a reservoir and its complete conversion to work.
Picture of heat engine attempting to violate Kelvin’s statement
Hot Reservoir
qH
w
Cold Reservoir
qH + w ≠ 0
Clausius’ statement – It is not possible for heat to flow from a colder body to a
warmer body without any work having been done.
Picture of heat engine attempting to violate Clausius’ statement
Hot Reservoir
qH
qC
Cold Reservoir
qH + qC ≠ 0
Standard chemist’s statement – The entropy of an isolated system increases in the
course of a spontaneous change. ∆STotal > 0 where STotal is the total entropy of the
isolated system of interest.
Defining the isolated system is vital to using the second law of thermodynamics as a
predictor of change. We assume universe is isolated so the second law becomes
∆SUniverse > 0
2
nd
Picture of heat engine satisfying the 2 Law.
Hot Reservoir
w
qH
qC
Cold Reservoir
Some energy must be wasted in the form of work (if we desire cooling from our
energy input) or heat (if we desire work from our energy input).
Ideal heat engines have ∆STotal = 0
Other statements (for fun)
Lucretius – Greek philosopher (1st century B. C.)
"For we see that anything can be more speedily disintegrated than put together again.
Hence, what the long day of time, the bygone eternity, has already shaken and
loosened to fragments could never in the residue of time be reconstructed."
Mother Goose
"Humpty Dumpty sat on a wall
Humpty Dumpty had a great fall
All the King's horses, and all the King's men
Couldn't put Humpty together again"
Gilbert N. Lewis (Notable Nebraskan☺)
1.
"Every system which is left to itself will, on the average, change toward a
condition of maximum probability"
2.
"Every system left to itself changes, rapidly or slowly, in such a way as to
approach a definite final state of rest. No system, except through the influence
of external agencies, will change away from the state of equilibrium"
3.
"Gain in information is loss in entropy"
T. S. Eliot – (poet: The Hollow Men – 1925)
"... this is the way the world ends
not with a bang, but with a whimper"
J. A. V. Butler (British electrochemist)
"There is a tendency in nature for energy to pass from more available to less available
forms."
3
Max Planck (Father of quantum mechanics)
1. "It is impossible to diminish the entropy of a system of bodies without thereby
leaving behind changes in other bodies."
2.
"A transformation whose only final result is to transform into work heat
extracted from a source which is at the same temperature throughout is
impossible."
Enrico Fermi (Father of the nuclear reactor)
"The state of maximum entropy is the most stable state of an isolated system."
Arthur Eddington (Father of astrophysics)
"Entropy is time's arrow"
dSTotal
>0
dt
Adiabatic Processes
Relationship between Volume and Temperature
dU = ñq + ñw ⇒ dU = ñw ⇒ C v dT = − pex dV ⇒
⇒
∫ C dT = − ∫
v
⇒ C v ln
nRT
dV ⇒
V
γ=
Cp
Cv
T2  V1 
= 
T1  V2 
( Cp − Cv )
Cv
Let
γ−1
is the heat capacity ratio.
v
ex
dV = − ∫ p dV
Cv
nR
T
V
dT = − ∫
dV ⇒ C v ln 2 = − nR ln 2
T
V
T1
V1
T2
V
V
= nR ln 1 = ( C p − C v ) ln 1
T1
V2
V2
V 
T
⇒ ln 2 = ln  1 
T1
 V2 
⇒
∫
∫ C dT = − ∫ p
⇒ ln
T2 ( Cp − C v ) V1
=
ln
T1
Cv
V2
V 
T
= γ ⇒ ln 2 = ln  1 
Cv
T1
 V2 
Cp
γ−1
4
Relationship between Volume and Pressure
T2  V1 
= 
T1  V2 
γ−1
⇒
p 2 V2  V1 
= 
p1V1  V2 
γ−1
⇒
p 2  V1 
= 
p1  V2 
γ
⇒ p1V1γ = p 2 V2γ
Let compare the shape of an isotherm to the shape of an adiabat on a pV plot.
p
isotherm
p= k
V
adiabat
k
p= γ
V
V
Note that the slope of the adiabat is always greater since the heat capacitity ratio, γ is
always greater than one.
Adiabatic Internal Energy
dU = ñq + ñw ⇒ dU = ñw ⇒ ∆U = − ∫ p ex ⋅ dV
V 
Why not ∆U = − nRT ln  2  for a reversible ideal gas?
 V1 
- Not necessarily isothermal conditions
- Can’t pull T out of integral
Adiabatic Enthalpy
H = U + pV ⇒ dH = dU + d ( pV ) = ñq + ñw + p dV + v dp
⇒ dH = − p dV + p dV + v dp ⇒ dH = v dp
5
Carnot Cycle
We have supposed that entropy is state function. However we have not proven it.
We can use a thermodynamic cycle constructed by the French military engineer, Sadi
Carnot to demonstrate that entropy is indeed a state function.
We will use Carnot’s inspired choice of closed path to
dq
I. Demonstrate that ∫ rev = 0 for an ideal gas on Carnot’s path.
T
dq rev
= 0 for an arbitrary substance on Carnot’s path.
II. Demonstrate that ∫
T
dq
III. Demonstrate that ∫ rev = 0 for an arbitrary substance on an arbitrary path.
T
p
A isotherms
T=T h
B
adiabats
D
T=Tc
C
V
Proof I: Ideal gas with Carnot’s path
Step 1. Reversible isothermal expansion at temperature Th from point A to point B.
+ qh
∆S1 =
Th
Step 2. Reversible adiabatic expansion from point B to point C (or Th → Tc).
∆S2 = 0
Step 3. Reversible isothermal compression at temperature Tc from point C to point D.
− qc
∆S3 =
Tc
Step 4. Reversible adiabatic compression from point D to point A (or Tc → Th).
∆S4 = 0
6
For the cycle A → B → C → D → A.
∆STotal = ∆S1 + ∆S2 + ∆S3 + ∆S4 = ∆S1 + ∆S3
∆STotal = ∆S1 + ∆S3 =
− w rev,1
Th
+
− w rev,3
Tc
V
nRTh ln  B
 VA
=
Th
 VD 


 nRTC ln 
+
 VC 
Tc
V 
V V 
V 
= nR ln  B  + nR ln  D  = nR ln  B D 
 VA 
 VC 
 VA VC 
Recall a thermodynamic expression for adiabatic expansions: T2 V2 γ−1 = T1V1γ−1
T2 V2 γ−1 = T1V1γ−1 ⇒ Th VA γ−1 = Tc VD γ−1 and Th VB γ−1 = Tc VC γ−1
⇒
Th VD γ−1
=
Tc VA γ−1
⇒
VD VC
=
VA VB
VB VD = VA VC
and
Th VC γ−1
=
Tc VB γ−1
⇒
VD γ−1 VC γ−1
=
VA γ−1 VB γ−1
⇒ VB VD = VA VC
V V 
V V 
⇒ ∆STotal = nR ln  B D  = nR ln  A C  = nR ln (1) = 0
 VA VC 
 VA VC 
End of Proof I.
Efficiency of Heat Engines
Definition: efficiency - ε
ε=
w
qh
The efficiency of heat engine in terms of the heat bath temperatures can be found
with the following derivation.
w
q − q in
q
ε=
= out
= 1 − in
q out
q out
q out
w = q out − q in
V 
q out = − nRT ln  B 
 VA 
V 
V 
V 
V 
= − nRTh ln  B  − nRTc ln  D  = − nRTh ln  B  − nRTc ln  A 
 VA 
 VA 
 VB 
 VC 
V
= − nRTh ln  B
 VA

 VB
 + nRTc ln 

 VA

 VB 
 = − nR ( Th − Tc ) ln 


 VA 
7
V
− nR ( Th − Tc ) ln  B
w
 VA
ε=
=
q out
V 
− nRTh ln  B 
 VA 
ε = 1−


 = Th − Tc = 1 − Tc
Th
Th
Tc
Th
Example: An engine using superheated steam at 550 °C has heat transferring to a
thermal sink at 100 °C. What is the maximum efficiency of the engine?
ε = 1−
TC
373K
= 1−
= 1 − 0.453 = 0.547
TH
823K
Therefore the maximum efficiency of the engine is 54.7%. 45.3% of the energy must
be wasted due to the second law of thermodynamics.
Example: If an internal combustion engine cycles between 3200 K and 1400 K, what
is its maximum efficiency?
ε = 1−
TC
1400 K
= 1−
= 1 − 0.44 = 0.56
TH
3200 K
Thus the maximum efficiency is 56%. In reality the efficiency is more like 25%.
Example: a) If a heat engine has a temperature of 800 K during the hot part of its
cycle, at what temperature must the heat sink be to have 95%
efficiency?
ε = 1−
TC
TH
⇒ ε −1 = −
TC
TH
⇒ TC = TH (1 − ε ) = 800 K (1 − 0.95 ) = 40 K
a) At what temperature must the heat sink be to have 100% efficiency?
TC = TH (1 − ε ) = 800 K (1 − 1) = 0 K
Thus the only time a heat engine can be 100% efficient is if we can cool our engine to
absolute zero. Shortly we’ll see that we can’t cool our engine that far.
8
We will soon show that two engines operating reversibly between two temperatures,
Th and Tc must have the same efficiency.
Heat Bath, Th
qin
qout
qo ut
w
w
qin
Heat Bath, Tc
Recall Kelvin’s statement of the 2nd Law: No process exists that has the sole result of
converting heat to work.
Proof II Carnot’s cycle with an arbitrary substance
We prove that the entropy change of an arbitrary substance going through Carnot’s
cycle is zero by proving Carnot’s Theorem below.
Carnot’s Theorem: Efficiencies of reversible processes operating between Th and
Tc must be the same regardless of whether the working substance is an ideal or a real
gas.
Let us assume that we have two different engines, A and B, operating between the
same Th and Tc and that the two engines have different efficiencies. (We will work
toward showing an obvious contradiction that will demonstrate that this initial
assumption is false.)
Furthermore let us assume that the efficiency of engine A is higher than efficiency of
engine B.
Engine A
100 J
qout
Heat Bath, Th
w 90 J
Engine B
100 J q’out
qin 10 J
Hea t Bath, T c
w’ 80 J
q’in 20 J
9
Engine B is not able to perform as much work as Engine A. To make Engine B
perform as much work as Engine A, we can increase the size of Engine B (by using
more of the working substance).
Engine A
Heat Bath, Th
w 90 J
100 J qout
Engine B
112.5 J q’out
qin 10 J
w’ 90 J
q’in 22.5 J
Heat Bath, Tc
(Careful: Temperatures Th and Tc remain the same. Do not let second graphic
deceive you.)
At this point, let us run Engine B in reverse and couple it to Engine A, so that Engine
B releases heat to the heat bath at Th and absorbs heat at the heat bath at Tc.
Engine A
100 J qout
Heat Bath, Th
w 90 J
Engine B
112.5 J q’out
qin 10 J
w’ 90 J
q’in 22.5 J
Heat Bath, Tc
During each cycle, the more efficient Engine A absorbs qout at Th and emits -qin at Tc.
The less efficient engine B, (going backwards), is emitting heat - q′out at Th and
absorbing heat q′out at Tc.
The work performed by Engine A equals the work performed by Engine B.
q out − qin = q′out − q′in
However, the efficiency of Engine A is greater than Engine B.
w
w′
q out − q in
q′ − q′in
ε > ε′ ⇒
>
⇒
> out
q out
q′out
q out
q′out
Since the work (the numerator) is equal on both sides of the inequality,
q out − q in
q′ − q′in
1
1
> out
⇒
>
⇒ q′out > q out
q out
q′out
q out
q′out
10
But also using the same inequality,
q out − q in
q′ − q′in
q
q′
> out
⇒ 1 − in > 1 − in
q out
q′out
q out
q′out
⇒
q in
⇒
q out
<
q′in
q′out
q in q′out < q′in q out
If under any conditions, q′out > q out , then the equality above can be rewritten as
qin q′out < q′in q out
q in < q′in
⇒
The inequality, q′out > q out , implies that the heat emitted by engine B is greater than
the heat absorbed by engine A.
The inequality, qin < q′in , implies that the heat emitted by engine A is greater than
the heat absorbed by engine B.
Both the above statements imply that as the coupled engine cycle, heat is being
transferred from the low temperature bath to the high temperature bath. That is
thermal energy is spontaneously flowing from cold to hot. This heat flow violates
what we know about heat flow (Fourier’s Law of Heat Conduction).
Therefore the initial assumption that Engine A and Engine B can have different
efficiencies while cycling between the same temperatures must be false.
Thus the efficiency of a heat engine depends on temperatures it cycles between
and not on the nature of the working substance in the engines.
Continuing with proof II.
ε = 1−
qc
qh
= 1−
Tc
Th
⇒
qc
qh
=
Tc
Th
From the Carnot cycle
∫ dS =
End of proof II.
qh
Th
−
qc
Tc
=0
⇒
qh
Th
=
qc
Tc
11
Proof III Arbitrary cycle with an arbitrary substance
The proof for an arbitrary cycle is rather ingenious.
1. Tile a chosen arbitrary closed path with Carnot cycles.
p
V
2. Make Carnot cycles smaller to fit as many as possible within the boundary
enclosed by the path.
3. As the Carnot cycle get infinitesimally small, the entire boundary is contributing
to at least one Carnot cycle on its interior.
4. The entropies of all the Carnot cycle paths in the interior of the boundary cancel
each other.
Path b
Path a
∆Spath a = −∆Spath b
5. The total entropy of all of the Carnot cycles is zero; therefore, the entropy for the
arbitrary closed path is zero.
12
Entropy as a Statisitical Measure
Classical Thermodynamic View of Entropy
Clausius’ definition of entropy
dq rev
T
dq is an inexact differential ⇒ line integral of dq depends on path
dS =
1/T is an integrating factor so that dS is an exact differential ⇒ line integral of dS is
independent of path ⇒ S is a state function.
Statistical View of Entropy
System evolves so that it can occupy a greater number of energy states.
Boltzmann’s definition of entropy
S = k ln ω
where ω is the total number of possible arrangement of states for a given energy.
- Note if ω = 1, then S = 0.
k – Boltzmann’s constant = 1.3806503 × 10-23 J/K
k is a fundamental constant of nature and has a simple relationship with the Ideal Gas
Constant, R.
R = k ⋅ NA
where NA is Avogadro’s number.
Equivalence of Views
To rigorously demonstrate that the two views are equivalent takes a lot of
sophisticated mathematical machinery. We’ll avoid such sophistication in this class.
Let us at least demonstrate that the views are equivalent for the simple system of the
isothermal expansion for an ideal gas.
Thermodynamic View (19th century)
Consider the entropy change for an isothermal expansion of an ideal gas
f
f
dq
1
∆S = ∫ rev = ∫ dq rev
T
Ti
i
⇒ ∆S =
q rev
T
13
For an ideal gas, the molar internal energy is, U =
3
RT .
2
Therefore for an isothermal process,
∆T = 0 ⇒ ∆U = 0 ⇒ ∆U = q rev + w rev
f
f
f
i
i
i
w rev = − ∫ p ex dV = − ∫ p dV = − ∫
⇒ q rev = − w rev
f
V 
nRT
1
dV = −nRT ∫ dV = − nRT ln  f 
V
V
 Vi 
i
Thus the entropy for an isothermal expansion of an ideal gas can be written as
∆S =
− w rev
T
V 
nRT ln  f 
 Vi  = nR ln  Vf 
=−
 
T
 Vi 
Statistical View (20th century)
Consider N ideal gas particles in a volume V. The volume that one particle occupies
can be considered a peculiar type of density. volume per particle = v.
1. How many places can I put a single gas particle in the volume?
V
ω=
v
2. How many places can I put two gas particles in the volume?
 V  V   V 
ω =    =  
 v  v   v 
3. How many places can I put N gas particles?
2
 V  V   V   V 
ω =   ⋯   =  
 v  v   v   v 
N
Using Boltzmann’s definition, we can calculate the entropy change when we
isothermally change the volume.
N
N
V 
V 
V
∆S = Sf − Si = k ln ωf − k ln ωi = k ln  f  − k ln  i  = Nk ln  f
 v 
 v
 v
 V
= Nk ln  f
 v
 Vf 
  v 
    = Nk ln  
  Vi  
 Vi 

 Vi 
 − Nk ln  

 v
14
Calculation of Entropy Changes
Constant volume entropy changes
dq
C dT
dS = rev = v
T
T
Sf
Tf
C dT
⇒ ∫ dS = ∫ v
T
Si
Ti
Tf
⇒ S ( Tf ) = S ( Ti ) + ∫
Ti
C v dT
T
If Cv is assumed to be constant over the temperature range from Ti to Tf,
S ( Tf ) − S ( Ti ) =
f
T 
C v dT
dT
C
=
= C v ln T |TTif = C v ln  f 
v∫
∫T T
T
 Ti 
Ti
i
Tf
T
Note that the key step in the calculation is noting how we are going to calculate the
reversible heat.
Constant pressure entropy changes
C dT
dq
dS = rev = p
T
T
⇒
Sf
Tf
Si
Ti
∫ dS =
∫
Cp dT
T
Tf
⇒ S ( Tf ) = S ( Ti ) + ∫
Ti
If Cp is assumed to be constant over the temperature range from Ti to Tf,
S ( Tf ) − S ( Ti ) =
Tf
∫
Ti
C p dT
T
Tf
= Cp ∫
Ti
T 
dT
= Cp ln T |TTfi = C p ln  f 
T
 Ti 
Cp dT
T
15
Example: Calculate the entropy change in tetrahydrofuran from 273 K to 333 K
given the following expression for the constant pressure heat capacity:
Cp = 63.393J mol ⋅ K + 0.40257 J mol ⋅ K 2 T − 1.2686 × 10−3 J mol ⋅ K 3 T 2
+ 1.8275 ×10−6 J mol ⋅ K 4 T 3
∆S = S ( 333K ) − S ( 273K ) =
333K
∫
C p dT
T
273K
63.393J mol ⋅ K + 0.40257 J mol ⋅ K 2 T

dT
333K 
3
2
4 3
−3
−6
1.2686
10
J
mol
K
T
1.8275
10
J
mol
K
T
−
×
⋅
+
×
⋅

= ∫ 
T
273K
63.393J mol ⋅ K dT
+ ∫ 0.40257 J mol ⋅ K 2 dT
∫
T
273K
298K
333K
∆S =
340 K
333K
−
∫
333K
−3
∫
1.2686 ×10 J mol ⋅ K T dT +
3
273K
1.8275 ×10−6 J mol ⋅ K 4 T 2 dT
273K
2
333K
∆S = ( 63.393J mol ⋅ K ) ln T |333K
273K + ( 0.40257 J mol ⋅ K ) T |273K
−6
− (1.2686 ×10−3 J mol ⋅ K 3 ) T 2 |333K
J mol ⋅ K 4 ) T 3 |333K
273K + (1.8275 × 10
273K
∆S = ( 63.393J mol ⋅ K ) ln
333K
+ ( 0.40257 J mol ⋅ K 2 ) ( 333K − 273K )
273K
(
)
J mol ⋅ K ) ( ( 333K ) − ( 273K ) )
− (1.2686 ×10−3 J mol ⋅ K 3 ) ( 333K ) − ( 273K )
+ (1.8275 ×10−6
4
2
2
3
3
∆S = 12.594 J mol ⋅ K + 24.154 J mol ⋅ K − 23.063J mol ⋅ K + 10.100 J mol ⋅ K
= 23.784 J mol ⋅ K
Example: Calculate the entropy change of poly(dimethylsiloxane)-5 from 298 K to
340 K at constant atmospheric pressure.
Cp ( EtOH ) = 98.39 + 0.5368 ( T ( K ) − 273.25) J / mol ⋅ K
The heat capacity depends on temperature, so we must do the integral.
S ( 340 K ) − S ( 298 K ) =
340 K
∫
298K
340 K
=
∫
298K
C p dT
T
98.39 + 0.5368 ( T − 273.25 )  dT
T
298K
340 K
=
∫
[98.39 − 147.5] dT + 340 K 0.5368dT = −49.1ln T |340K +0.5368T |340 K
T
∫
298K
= −6.47 J mol ⋅ K + 22.55 J mol ⋅ K = 16.08 J mol ⋅ K
298K
298K
16
Chemical Processes
Standard entropies
Standard entropies are tabulated for substances at 1 bar and 298.15 K. For the
elements in their standard state, [Carbon (graphite), O2 (g), Hg (l), P4 (s)] ; the
standard entropies are not zero in contrast to the standard enthalpies of formation
for the elements.
Entropies can be given an absolute reference based on the third law of
thermodynamics. Standard enthalpies of the elements can’t be given an absolute
reference; therefore, the arbitrary reference of zero is free for us to choose.
Constant volume or constant pressure changes
For a chemical reaction, the standard entropy of the reaction can be calculated using a
stoichiometric sum of the standard entropies.
∆Srxn = ∑ ν iSi0
i
The temperature dependence of the entropy can be found using the stoichiometric
sum of the heat capacity, dividing by the temperature and integrating through the
temperature range.
∆Srxn = ∑ ν S + ∑
0
i i
i
i
T
∫
298
νi C0p,i
T
dT
17
Phase transitions
Phase transitions occur at constant temperature and pressure. (much more about this
later in the course) Thus we can relate the entropy change of a phase transition to its
enthalpy change.
∆S = ∫
dq rev
T
∆T = 0

→ ∑S =
q rev
T
∆p = 0

→ ∆S =
∆H
.
T
Plot of the heat capacity and entropy as temperature of a substance increases from
zero to the gaseous region.
C p/T
T
S
Tm
Tb
T
Items to note from plots
1. The heat capacities and the entropies are discontinuous at the phase transitions
temperatures.
2. The heat capacities for solids generally increase with increasing temperature.
3. Cp/T near absolute zero has a form of Cp/T = aT2.
4. The heat capacities for liquids and gases are roughly constant over large
temperature regions so that Cp/T steadily decreases with temperature.
5. The entropy of a substance always increases with increasing temperature.
18
Trouton’s rule
As an undergraduate Frederick Trouton, spending an afternoon playing with
thermodynamics tables, found that the molar entropy of vaporization for most
substances is very close to the value of 87 J/mol⋅K.
∆Sɶ vap ≈
87 J
mol ⋅ K
Stated another way, Trouton’s rule shows that for most substances, the difference
between the entropy of the liquid phase and the entropy of the gaseous phase is the
same. Important exceptions to Trouton’s rule include hydrogen bonded substances,
like H2O and substances with molecules that have low moments of inertia (low
rotational energies) like CH4.
The Third Law of Thermodynamics
Statement
If the entropy of every element in its most stable state at T = 0 K is taken to be zero,
then every substance has a positive entropy with may go to zero as T → 0 K, and
does so for all perfect crystalline substances including compounds.
Nernst Heat Theorem
Consider a process that takes place at some temperature T with some entropy change
∆S(T). Then ∆S(T) → 0 as T → 0.
A consequence of the third law is that absolute zero will never be achieved (thus the
name, absolute zero).
Another consequence is that a heat engine’s has 100% efficiency only if it can be
cooled to absolute zero.
ε = 1−
qc
T
= 1− c
qh
Th
⇒
 T
ε ( Tc → 0 ) = lim 1 − c
Tc → 0
 Th

 =1

19
Third law entropy calculations
For a substance in the gaseous phase, the third law (absolute) entropy can be
calculated using several steps.
S ( T ) = S( 0) +
Tmelt
∫
0
dq rev ∆H melting boil dq rev ∆H boiling
dq rev
+
+ ∫
+
+ ∫
T
Tmelt
T
Tboil
T
Tmelt
Tboil
T
T
For a constant volume process, the calculation is messy since the volume of a
substance almost always changes during a phase transition. However assuming
constant pressure the calculation is tractable.
S ( T ) = S( 0)
+
Tmelt
∫
C p ( solid ) dT
0
T
+
∆H melting
Tmelt
+
Tboil
∫
Tmelt
Cp ( liquid ) dT
T
+
∆H boiling
Tboil
T
+
∫
Cp ( gas ) dT
Tboil
T
The only other issue to consider is the technological problem that we are not able to
measure the heat capacities of substances with temperature close to absolute zero.
We have to appeal to theory rather than experiment to find such low temperature
entropies.
Einstein-Debye law
3
12π4  T 
3
Cv =
R
 = aT
5
 ΘD 
where ΘD is the Debye temperature of a material. The Debye temperature is a
characteristic of a material that depends on the different modes in which the atoms in
a solid lattice vibrate. The vibrational modes depend on the particulars of the
bonding (metallic or intermolecular) thus each substance has its own Debye
temperature.
Below is a short table of Debye temperatures.
Al
Cd
Cr
Cu
Au
α - Fe
426 K
186 K
610 K
344.5 K
165 K
464 K
Pb
α - Mn
Ni
Pt
Si
Ag
96 K
476 K
440 K
240 K
640 K
225 K
Sn (white)
Ti
W
Zn
Diamond
Ice
195 K
420 K
405 K
300 K
2200 K
192 K
The Einstein-Debye model deviates from experiment at high temperatures; however,
at low temperatures (< 20 K), the theory gives an excellent approximation.
20
Thus a general third law entropy calculation can be completed as follows.
S ( T ) = S ( 0) +
Tlowest
∫
0
+
∆H melting
Tmelt
aT 3dT melt C p ( solid ) dT
+ ∫
T
T
Tlowest
T
+
Tboil
∫
Tmelt
Cp ( liquid ) dT
T
+
∆H boiling
Tboil
T
+
∫
Tboil
C p ( gas ) dT
T
Note that because of Nernst’s heat theorem, S(0) for all substances is zero.
Free Energy and the 2nd Law
The 2nd Law under constant volume conditions
dU = dq v since dw = 0 for constant volume conditions (assuming no non-pV work)
From the Clausius inequality
dq
dq
dU
dS ≥
⇒ dS −
≥ 0 ⇒ dS −
≥0
T
T
T
At this stage note that if U is constant, then dS ≥ 0
And if S is constant, then dU ≤ 0
At this stage it’s convenient to construct a new energy function.
Helmholtz Free Energy – A ≡ U – TS
Under isothermal conditions, dA = dU − TdS
Return to the Clausius inequality
dU
≥ 0 ⇒ TdS − dU ≥ 0 ⇒ dU − TdS ≤ 0 ⇒ dA
T
≤≤≤≤
dS −
0
Therefore when T and V are constant, according to the 2nd law, a system loses
Helmholtz free energy spontaneously.
21
nd
The 2 Law under constant pressure conditions
For constant pressure conditions, dH = dq p (assuming no non-pV work)
Using the analogous argument to that used in the constant volume case.
dS ≥
dq
T
⇒ dS −
dq
dH
≥ 0 ⇒ dS −
≥0
T
T
Now construct another new energy function.
Gibbs Free Energy – G ≡ H – TS ≡ U + PV – TS
≤≤≤≤
Under isothermal conditions, dG = dH − TdS
dH
dS −
≥ 0 ⇒ TdS − dH ≥ 0 ⇒ dH − TdS ≤ 0 ⇒ dG
T
0
Therefore when T and P are constant, according to the 2nd law, a system loses Gibbs
free energy spontaneously.
Free Energy and Work
Reconsider the second law under constant volume conditions.
dq
⇒ TdS ≥ dq ⇒ TdS ≥ dU − dw
T
dw ≥ dU − TdS ⇒ dw dA
≥≥≥≥
dS ≥
The Helmholtz energy is “free” energy because it permits us to find the maximum
amount of work that can be done by a system at constant volume. This maximum
work is possible only under reversible conditions.
When the energy that is used to increase the entropy of the universe is subtract from
the total energy, the remaining energy is “free” to do work.
Now, reconsider the second law under constant pressure and reversible conditions.
dU = dq + dw total
⇒ dH = dU + d ( pV ) = dq + dw total + d ( pV )
dG = dH − TdS = dq + dw total + d ( pV ) − TdS
= dw total + d ( pV )
= dw total + Vdp + p dV
= dw non − pV + dw pV + Vdp + p dV
= dw non − pV − p dV + 0 + p dV
dG = dw non-pV
Thus the Gibbs free energy for a reversible process is the maximum non-pV work that
is available from a process.
22
Maxwell’s Relations
From internal energy
In the previous chapters, we thought of the natural thermodynamic variable used to
describe the internal energy as volume and temperature; that is, U ≡ U ( T, V ) , since
dU = dq + dw = Cv dT − p dV .
However with our thermodynamic definition of entropy, we can also write dU as
dU = dq + dw = T dS − p dV . Therefore, we can also consider the internal energy as a
function of entropy and temperature. U ≡ U ( S, V ) .
 ∂U 
 ∂U 
If U ≡ U ( S, V ) , then dU = 
 dS + 
 dV . Comparing to dU = T dS − p dV ,
 ∂S V
 ∂V S
yields the following relationships.
 ∂U 

 =T
 ∂S  V
 ∂U 

 = −p
 ∂V S
U is also state function; therefore, dU is an exact differential. If dU is exact, then the
following relationships are true:
∂  ∂U 
∂  ∂U 
 ∂T 
 ∂p 

 = 
 ⇒ 
 = − 
∂V  ∂S V ∂S  ∂V S
 ∂V S
 ∂S V
 ∂T 
 ∂p 

 = −   is the first of Maxwell’s relations, a set of relationships between
 ∂V S
 ∂S V
thermodynamic partial derivatives that are enormously helpful.
From enthalpy
Reconsider enthalpy,
dH = dU + d ( pV ) = dq + dw + p dV + V dp
= T dS − p dV + p dV + V dp
Thus H ≡ H ( S, p ) and
= T dS + V dp
 ∂H 

 =T
 ∂S p
 ∂H 

 =V
 ∂p S
Since H is a state function, dH is exact; therefore, the Maxwell relation is
 ∂T 
∂  ∂H 
∂  ∂H 
 ∂V 
 ⇒   = −

 = 

∂p  ∂S p ∂S  ∂p S
 ∂S  p
 ∂p S
23
From Helmholtz free energy
Reconsider Helmholtz free energy,
dA = dU − d ( TS) = dq + dw − T dS − SdT
Thus A ≡ A ( V, T ) and
= T dS − p dV − T dS − SdT
= −p dV − SdT
 ∂A 

 = −p
 ∂V T
 ∂A 

 = −S
 ∂T  V
Since A is a state function, dA is exact; therefore, the Maxwell relation is
∂  ∂A 
∂  ∂A 

 =


∂T  ∂V T ∂V  ∂T V
 ∂p 
 ∂S 
⇒ −  = −

 ∂T  V
 ∂V T
From Gibbs free energy
Reconsider Gibbs free energy,
dG = dA − p dV − SdT + p dV + V dp + d ( pV )
= −SdT + V dp
Thus G ≡ G ( p, T ) and
 ∂G 

 =V
 ∂p T
 ∂G 

 = −S
 ∂T  p
Since G is a state function, dG is exact; therefore, the Maxwell relation is
∂  ∂G 
∂  ∂G 

 = 

∂T  ∂p T ∂p  ∂T  p
 ∂S 
 ∂V 
⇒ 
 = − 
 ∂T p
 ∂p T
24
Pressure Dependence of Gibbs Energy
Reconsider Gibbs free energy, dG = −SdT + V dp
At constant temperature, dG = V dp ⇒ G f − G i = ∫ V dp
Assume an ideal gas, G f − G i = ∫ V dp = ∫
p 
nRT dp
dp
= nRT ∫
= nRT ln  f 
p
p
 pi 
p 
G f = G i + nRT ln  f 
 pi 
How about assume constant volume, (as in a geological process)
dG = V dp ⇒ G f − G i = V ∫ dp ⇒ ∆G = V ∆p
Fugacity
In using the thermodynamic variable pressure thus far, we been assuming that gases
nRT
are ideal. What about the case where
≠ p . For a real gas, the reading that a
V
pressure transducer takes is called the fugacity. Thus at this point we make the
distinction between the measurement that is actually made (the fugacity) and the
measurement that would have been made in the gas was ideal (the pressure). This
distinction is admittedly confusing.
The roots of the word fugacity mean escape. Thus fugacity is sometimes referred to
as an “escaping tendency” of gas. The words fugitive and fugue (melodies chasing
each other) have the same root.
Fugacity is related to the pressure via a fugacity coefficient, φ.
f =φp
For ideal gas, φ equal one.
Thermodynamic equilibrium constant
Once the necessity of using fugacities in our thermodynamic work become apparent,
we can now use them to better define the equilibrium constant.
(f
K=
(f
p0 ) ( f D p0 )
c
C
p0 ) ( f B p0 )
a
A
d
b
25
Calculating fugacity coefficients for a real gas
Fugacity coefficients can be calculated theoretically from a particular real gas
equation of state or experimentally from compressibility, Z, data.
To see how this is done, we need to be clever in defining the standard state of real
gas. We will make the standard state of real gas the state that makes the gas ideal.
The most consistent way to make a real gas ideal is to examine the behavior of the
gas as the pressure approaches zero.
f
lim = 1
p →0 p
Recall that pressure dependence of the chemical potential was derived from an
integral of the volume with respect to pressure.
f
ɶ = RT ln  pf 
dG = Vdp ⇒ µf − µi = ∫ Vdp
i
 pi 
Now consider the differences between the real gas and the ideal gas.
 ff 
 pf 
 f f pi 
 pi f f 
 − RT ln   = RT ln 
 = RT ln 

i 
 pi 
 fi pf 
 fi pf 
∫ ( Vɶ − Vɶ ) dp = RT ln  f
f
i
ideal
If the initial state in both cases is the standard state, then pi = fi.

f
∫ ( Vɶ − Vɶ ) dp = RT ln  p
f
ideal
i

f
f

 φpf 
 = RT ln 
 = RT ln φ

 pf 
ɶ = ZRT , then
Thus if we have compressibility data such that V
p
∫
f
i
fZ
 ZRT RT 
1
−

 dp = RT ∫i  −  dp = RT ln φ
p 
 p
 p p
∫
f
i
 Z −1 

 dp = ln φ
 p 
Let’s find the activity coefficients for van der Waals gas in terms of parameters “a”
and “b”.
26
ɶ.
First let modify the van der Waals equation to get an expression for V
p=
RT
a
1  RT
a
+ 2 = 
+ 
ɶ
ɶ
ɶ
ɶ
V−b V
V 1− b V


ɶ
 V

Now lets use a Maclaurin series to simplify the expression with covolume.
−1
−1
 b
= 1 − 
⇒ (1 − x )
ɶ
b  V
1−
ɶ
V
1
2
f ( x ) = f ( x 0 ) + f ′ ( x 0 )( x − x 0 ) + f ′′ ( x 0 )( x − x 0 ) + ⋯
2
1
−1
−2
−3
2
= (1 − x 0 ) − ( −1)(1 − x 0 ) ( x − x 0 ) + ( −2 )( −1)(1 − x 0 ) ( x − x 0 ) + ⋯
2
1
−1
−2
−3
2
= (1 − 0 ) − ( −1)(1 − 0 ) ( x − 0 ) + ( −2 )( −1)(1 − 0 ) ( x − 0 ) + ⋯
2
2
= 1+ x + x +⋯
1
Thus
1
1−
b
ɶ
V
2
≈ 1+
b b
+
+⋯
ɶ  V
ɶ 
V
Therefore we can rewrite the van der Waals equation as follows:
p=
1  RT
a
ɶ = RT  1 + a  = RT  1 + b  + a 
+  ⇒ V

ɶ
ɶ
ɶ  p   V
ɶ  RTV
ɶ 
V 1− b V
p  1 − b RTV




ɶ
ɶ
 V

 V

f
∫ Vɶ − Vɶ ideal dp = RT ln φ
i
∫
f
i
f
i
f
i
)
 RT  
b
a  RT 
−

 dp = RT ln φ
1 + ɶ  +

ɶ
 p   V  RTV  p 
∫
∫
(
 1 
b
a  1
−  dp = ln φ
  1 + ɶ  +
ɶ
 p   V  RTV  p 
f1 b
a 
∫i  p  Vɶ + RTVɶ   dp = ln φ
f 1 
 1 
a 
a 
  dp = ∫i 
b+
  dp = ln φ
 ɶ b+
RT  
RT  
 RT 
 pV 
 b
a 
ln φ = 
+
∆p
2 
 RT ( RT ) 
27
Notes on the fugacity coefficient
1. Z < 1 for attractive forces, thus φ < 1
2. Z > 1 for repulsive forces, thus φ > 1
3. Usually fugacity coefficient decreases with increasing pressure.
Gibbs-Helmholtz Equation
We can use the relationships to find other relationships. For example, let us find a
relationship that describes the temperature dependence of the Gibbs free energy.
G = H − TS ⇒
G H TS H
= −
= −S
T T T T
Now take the derivative of the expression w.r.t. T, holding p constant.
G 1  H
H
 ∂  G  ∂G  1 G 1  ∂G  G  1 
=
 
 − 2 = 
 −  =  −S −  =  −  = − 2
T  ∂T p T  T 
T T T
T
 ∂T p T  ∂T p T T
∆H
 ∂  ∆G
=− 2
Because G, H and T are all state functions,  
T
 ∂T  p T
Refrigeration
Thermodynamic efficiency
Let us consider a reversible refrigeration cycle ⇒ ∆S = 0
Recall from the discussion of the Carnot cycle the relationship between heat and
temperature.
− qc qh
∆S =
+
=0
Tc
Th
Let us now define a measure of efficiency in a refrigeration cycle known as the
coefficient of performance, c. (The fraction of cooling that can be done per unit
amount of work supplied.)
q
c= c
w
28
Return to the first law of thermodynamics.
qc − q h + w = 0 ⇒
c=
− qc
Tc
+
c=
qh
Th
qc
w
qc
qh − qc
qc
=0 ⇒
qc
q h − qc
Compare to efficiency of heat engine ε =
Thus c =
=
w = qh − qc
=
qh
=
Tc
Th
Tc
Th − Tc
Th − Tc
Tc
1
ε
Note that the performance of a refrigerator worsens as T → 0 K.
Also note that c > c rev .
29
Practical problems with refrigeration
Insulation is imperfect. Because thermal energy is constantly leaking into a system;
power (energy per time) must be supplied to keep the temperature constant. The
leaking is well described by Newton’s law of cooling,
Power =
dq c
= A ( Th − Tc ) ,
dt
where Th is the temperature of the surroundings and Tc is the temperature of the
refrigerator.
The coefficient A depends on the
a) heat conductivity of the insulator
b) the size of the refrigerator (power is an extensive quantity)
c) shape (surface area) of the refrigerator (A increases with higher surface area).
The power need to run the refrigerator can be stated as the amount of work supplied
per unit time.
dw
P=
dt
q
We can relate the power to the cooling via the coefficient of performance. w = c
c
 T − Tc 
( Th − Tc )
1 d qc 1
P=
= A ( Th − Tc ) =  h
 A ( Th − Tc ) = A
c dt
c
Tc
 Tc 
2
The squared factor means that electrical power usage (for air conditioners) increases
exponentially during a heat wave.
30
Ultralow temperatures (adiabatic demagnetization)
To produce ultra-low temperatures, a multi-step process must be used.
1. Cool systems with many Joule-Thompson expansions (chlorofluorocarbons, argon
and helium are some of the refrigerants used). Approximately 4K can be
achieved.
2. Liquid helium evaporation to 1K. (Old-fashioned cooling makes a comeback!)
3. Below 1K adiabatic demagnetization is used. (Refrigerant is a gadolinium salt
such as Ga2(SO4)3 • 8 H2O or Gd5Si2Ge2.)
a) Electron spin of gadolinium (8 unpaired e-!!) starts randomized.
b) System is placed in an isothermal bath
c) Adiabatic magnetization: Magnetic field is applied to align electron spin (and
reduce entropy and heat capacity).
d) Bath is removed, sample is isolated
e) Adiabatic demagnetization: magnetic field is removed. As magnetic field is
removed, temperature decreases.
f) Repeat
4. To get beyond the nK range nuclear adiabatic demagnetization must be used.
(rhodium metal is refrigerant.)
Current world record cold temperature: 100 pK.
Heat pumps
A heat pump is a refrigerator (air conditioner) working in reverse. Thus during
operation, the heat pump cools the outside (in winter!) and heats the inside.
c=
qc
q h − qc
=
Tc
Th − Tc
The coefficient of performance shows that heat pumps are more practical for Florida
and not at all for Alaska. (Your house may be chilly during a cold snap in January in
Nebraska.)
31
Linde refrigerator
The Laws of Thermodynamics
- the wiseacre version.
1st Law
2nd Law
3rd Law
0th Law
You can't win
You can't break even
You can't get out of the game
But you do know how to keep score
32
Summary
Adiabatic Expansions (or Compressions)
T2  V1 
= 
T1  V2 
γ−1
⇒
p1V1  V1 
= 
p 2 V2  V2 
γ−1
⇒
2nd Law of Thermodynamics (irreversible)
∆SUniverse > 0
2nd Law of Thermodynamics (reversible)
∆STotal = 0
p1  V1 
= 
p 2  V2 
dq rev
T
Thermodynamic definition of entropy
dS =
Statistical definition of entropy
S = k ln ω
Definition of ideal gas constant
R = k ⋅ NA
Entropy change of isothermal expansion of ideal gas.
V 
∆S = nR ln  f 
 Vi 
Entropy is a state function
dq
∫ Trev = 0
Efficiency of a heat engine
w
q
T
ε=
ε = 1− c
ε = 1 − in
qh
Th
q out
Carnot’s Theorem
ε = 1−
qc
qh
= 1−
Tc
Th
⇒
qc
qh
=
Tc
Th
⇒
Entropy change at constant volume
Tf
S ( Tf ) = S ( Ti ) + ∫
Ti
C v dT
T
Entropy change at constant pressure
Tf
S ( Tf ) = S ( Ti ) + ∫
Ti
Cp dT
T
qh
Th
=
qc
Tc
γ
33
Entropy change for constant heat capacity
T 
S ( Tf ) − S ( Ti ) = C ln  f 
 Ti 
Entropy change during phase transition
∆St rans =
∆H trans
Ttrans
Einstein-Debye heat capacity
C v = aT 3
Coefficient of performance (refrigeration)
c=
qc
w
=
Helmholtz Free Energy
A = U – TS
Gibbs Free Energy
G = H – TS
Natural Variables of Energies
Internal Energy
Enthalpy
Helmholtz Free Energy
Gibbs Free Energy
U ≡ U ( V,S, n )
H ≡ H ( p,S, n )
A ≡ A ( V, T, n )
G ≡ G ( p, T, n )
Tc
Th − Tc
34
Gibbsians
dU = T dS − p dV
dH = T dS + V dp
dA = −p dV − SdT
dG = −SdT + V dp
 ∂U 

 =T
 ∂S  V
 ∂H 

 =T
 ∂S p
 ∂U 

 = −p
 ∂V S
 ∂H 

 =V
 ∂p S
 ∂A 

 = −p
 ∂V T
 ∂G 

 =V
 ∂p T
 ∂A 

 = −S
 ∂T  V
 ∂G 

 = −S
 ∂T  p
Maxwell’s Relations
 ∂T 
 ∂V 
  = −

 ∂S p
 ∂p S
 ∂S 
 ∂V 

 = − 
 ∂T  p
 ∂p T
 ∂T 
 ∂p 

 = − 
 ∂V S
 ∂S V
 ∂p 
 ∂S 
  =

 ∂T V  ∂V T
Gibbs-Helmholtz Equation
H
 ∂  G
=− 2
 
T
 ∂T p T
∆H
 ∂  ∆G
=− 2
 
T
 ∂T  p T
Pressure Dependence of Gibbs Free Energy
p 
G f = G i + nRT ln  f 
 pi 
Fugacity Coefficient
f =φp
Thermodynamic Equilibrium Constant
(f
K=
(f
p0 ) ( f D p0 )
c
C
p0 ) ( f B p0 )
a
A
Calculation of Fugacity Coefficient
∫
f
i
d
 Z −1 

 dp = ln φ
 p 
b