Download Chapter 3. The Second Law

Chapter 3.
The Second Law
2011 Fall Semester
Physical Chemistry 1
(CHM2201)
Contents
The direction of spontaneous change
3.1 The dispersal of energy
3.2 The entropy
3.3 Entropy changes accompanying specific processes
3.4 The Third Law of thermodynamics
Concentrating on the system
3.5 Helmholtz and Gibbs energies
3.6 Standard molar Gibbs energies
Combining the First and Second Laws
3.7 The fundamental equation
3.8 Properties of internal energy
3.9 Properties of the Gibbs energy
The Second Law of thermodynamics
No process is possible in which the sole result is the
absorption of heat from a reservoir and its complete
conversion into work
The spontaneous change
•  A spontaneous process is one that can occur in a
system left to itself; no action from outside the system
is necessary to bring it about.
•  A non-spontaneous process is one that cannot take
place in a system left to itself.
•  If a process is spontaneous, the reverse process is nonspontaneous, and vice versa.
The spontaneous change
Water falling (higher to lower
potential energy) is a
spontaneous process.
H2 and O2 combine
spontaneously to form
water (exothermic)…
Conclusion: enthalpy alone is
not a sufficient criterion for
prediction of spontaneity.
liquid water vaporizes
spontaneously at room
temperature; an
endothermic process.
The spontaneous change
When the valve is
opened …
• 
• 
• 
the gases mix spontaneously.
There is no significant enthalpy change.
Intermolecular forces are negligible.
So … why do the gases mix?
3.1 The dispersal of energy
Key points
1.  During a spontaneous change in an isolated
system the total energy is dispersed into random
thermal motion of the particles in the system
We look for the direction of
change that leads to dispersal
of the total energy of the
isolated system
3.2 Entropy
Key points
1.  The entropy acts as a signpost of spontaneous
change
2.  The Clausius definition for the entropy
3.  The Boltzmann formula for the entropy
4.  Entropy is a state function (Carnot cycle)
5.  Entropy increases in a spontaneous change (The
Clausius inequality)
3.2 Entropy
(a) Thermodynamic definition of entropy
dqrev
dS =
T
qrev is the heat supplied reversibily
ΔS =
∫
f
ii
dqrev
T
To calculate the difference in entropy, we find a reversible
path between two states, and integrate the energy supplied
as heat at each stage of the path divided by T
The Second Law of thermodynamics :
The entropy of an isolated system increases in the course
of a spontaneous change : ΔStot > 0
3.2 Entropy
(a) Thermodynamic definition of entropy
dSsur
dqsur,rev dqsur
=
=
Tsur
Tsur
1.  The surroundings consist of a reservoir of constant V.
2.  The energy supplied as heat can be identified with the
change in internal energy, ΔUsur.
3.  U is a state function and independent of the path.
4.  So we can drop “rev”.
ΔSsur
qsur
=
Tsur
Regardless of how the change is brought about in the system,
reversibly or irreversibly, we can calculate the change of
entropy of the surroundings by using the above equation.
For an adiabatic change,
ΔSsur = 0
3.2 Entropy
(b) The statistical view of entropy
Boltmann formula for entropy
S = k lnW
• 
• 
• 
• 
• 
k = 1.38×10-23 J/K: the Boltzmann constant
W : the number of microstates : the ways in
which the molecules of a system can be
arranged while keeping the total energy
constant
When W = 1, S = 0
When molecules can access more
microstates for a given energy (e.g. as the
system volume increases), the entropy
increases
Molecules in a system at high T can occupy
a large number of the available energy
levels, so a small additional transfer of
energy as heat will lead to a relatively small
change in the number of accessible energy
3.2 Entropy
(c) The entropy as a state function
Entropy is a state function
dqrev
∫ Tsur = 0
1.  We show that the above equation is true
for a ‘Carnot cycle’ involving a perfect
gas
2.  We show that the result is true whatever
the working substance is
3.  We show that the result is true for any
cycle
3.2 Entropy
(c) The entropy as a state function
A Carnot Cycle
1.  A ⟶ B : Reversible isothermal expansion at
Th ; The entropy change is qh/Th
2.  B ⟶ C : Reversible adiabatic expansion : No
energy leaves the system as heat so the
change in entropy is zero
3.  C ⟶ D : Reversible isothermal compression
at Tc ; the change in entropy of the system is
qc/Tc
4.  D ⟶ A : Reversible adiabatic compression ;
No energy enters the system as heat so the
change in entropy is zero
qq qc
∫ dS = Th + Tc
3.2 Entropy
(c) The entropy as a state function
A Carnot Cycle
qq qc
∫ dS = Th + Tc
qh
Th
=−
qc
Tc
dqrev
∫ Tsur = 0
3.2 Entropy
(c) The entropy as a state function
We need to show that the following equation applies to any materials
dqrev
∫ Tsur = 0
Efficiency (η)
η=
work performed
heat absorbed from hot source
w
=
qh
qh − qc
qc
η=
= 1−
qh
qh
qh
Th
Tc
Because
= − , η = 1−
qc
Tc
Th
3.2 Entropy
(c) The entropy as a state function
•  Suppose two reversible engines are
coupled together and run between the
same two reservoirs
•  The working substances and details of
construction of the two engines are
entirely arbitrary
•  Suppose that engine A is more efficient
than B
•  Choose a setting of the controls that
causes engine B to acquire energy as
heat qc from the cold reservoir and to
release a certain quantity of energy as
heat into the hot reservoir
•  Because engine A is more efficient, not
all the work that A produces is needed
for this process
3.2 Entropy
(c) The entropy as a state function
•  The net effect of the processes is the
conversion of heat into work without
there being a need for a cold sink :
Contrary to the Kelvin statement of the
Second Law
•  Initial assumption that engines A and B
can have different efficiencies must be
false
Tc
η = 1− is true for any substance
Th
3.2 Entropy
(c) The entropy as a state function
•  Any reversible cycle can be
approximated as a collection of
Carnot cycles
•  The integral around an arbitrary path
is the sum of the integrals around
each of the Carnot cycle
•  This approximation becomes exact as
the individual cycles are allowed to
become infinitesimal
qrev
qrev
∑ T = ∑ T =0
all
perimeter
•  dS is an exact differential
•  S is a state function
3.2 Entropy
(d) Thermodynamic temperature
•  Thermodynamic temperature scale
T = (1− η )Th
•  The temperature of the engine can be inferred from the
measured efficiency
•  Kelvin scale is defined by using water at its triple point as the
notional hot source and defining that temperature as 273.16K
exactly
•  If the efficiency of an engine is 0.20, the temperature of the
cold sink is 0.80×273.16K = 220K
•  This result is independent of the working substance of the
engine
3.2 Entropy
(e) The Clausius inequality
•  More work is done when a change is reversible than when it
is irreversible
dwrev ≥ dw ⇒ −dwrev ≥ −dw
dU = dq + dw = dqrev + dwrev
dqrev − dq = dw − dwrev ≥ 0 ⇒ dqrev ≥ dq
dqrev dq
dS =
≥
; the Clausius inequality
T
T
dS ≥ 0 for an isolated system due to dq = 0
In an isolated system the entropy cannot decrease when a
spontaneous change occurs!!!
3.3 Entropy changes of specific processes
Key points
1.  The isothermal expansion of a perfect gas
2.  The entropy change at the transition
temperature
3.  Entropy change can be estimated in terms of the
heat capacity
3.3 Entropy changes of specific processes
(a) Expansion
Vf
ΔS = nR ln
Vi
•  The total change in entropy does depend on how the
expansion takes place
•  A reversible change ; dqsur = -dq
Vf
qsur
qrev
ΔSsur =
=−
= −nR ln
T
T
Vi
ΔStot = 0
•  Free expansion ; w = 0 and q = 0
Vf
ΔStot = nR ln > 0
Vi
3.3 Entropy changes of specific processes
(b) Phase transition
•  The transition of a substance accompanies a change in
entropy
•  For example, when a substance vaporizes, a compact
condensed phase changes into a widely dispersed gas and
the entropy increases
•  At the transition temperature, any transfer of energy as heat
between the system and the surrounding is reversible
because the two phases in the system are in equilibrium
•  Because q = ΔtrsH at constant pressure,
Δ trs H
Δ trs S =
Ttrs
3.3 Entropy changes of specific processes
(b) Phase transition
•  Trouton’s rule ; a wide range of liquids give approximately the
same standard entropy of vaporization about 85 J/K⋅mol
3.3 Entropy changes of specific processes
(c) Heating
S (Tf ) = S (Ti ) +
∫
T f dq
rev
T
Ti
logarithm dependence
At constant pressure,
S (Tf ) = S (Ti ) +
∫
T f C dT
p
T
Ti
When Cp is independent of T in the range
S (Tf ) = S (Ti ) + C p ∫
Tf
Ti
dT
T
= S (Ti ) + C p ln
Tf
Ti
3.3 Entropy changes of specific processes
(d) The measurement of entropy
Sm (Tf ) = Sm (Ti ) +
+
∫
Tf C
p,m (s,T )
T
0
dT +
Δ vap H
Tb
+
∫
Δ fus H
+
Tf
∫
Tb C
p,m (1,T )
T
Tf
dT
T C
p,m (g,T )
T
Tb
dT
Debye extrapolation ; Cp,m=aT3 near T = 0
3.4 The Third Law of Thermodynamics
Key points
1.  The Nernst heat theorem
2.  The Third Law allows us to define absolute
entropies of substances
•  At T = 0, all energy of thermal motion has been quenched
•  In a perfect crystal all the atoms or ions are in a regular uniform array
3.4 The Third Law of Thermodynamics
(a) The Nernst heat theorem
•  Nernst heat theorem : The entropy change accompanying any
physical or chemical transformation approaches zero as the
temperature approaches zero. Provided all the substances involved
are perfectly ordered,
ΔS → 0 as T → 0
•  If we arbitrarily ascribe the value zero to the entropies of elements
in their perfect crystalline form at T = 0, then all perfect crystalline
compounds also have zero entropy at T = 0
•  The Third Law of Thermodynamics : The entropy of all perfect
crystalline substances is zero at T = 0
•  S = klnW
•  Residual entropy : imperfectness
3.4 The Third Law of Thermodynamics
(b) Third-Law entropies
•  Entropies reported on the basis that S(0) = 0 are
“Third-Law entropies”
•  The standard (Third-Law) entropy SΘ(T) : the
entropy of the substance in its standard state at T
•  The standard reaction entropy
Δr SΘ =
∑
vSmΘ −
Products
∑
Reactants
Δ r S Θ = ∑ vJ SmΘ (J)
J
S Θ ( H + , aq ) = 0
vSmΘ