Thermodynamics:
the Second Law
자연과학대학 화학과
박영동 교수
The Direction of Nature and spontaneity
Thermodynamics: the Second Law
4.1 Entropy
4.1.1 The direction of spontaneous change
4.1.2 Entropy and the Second Law
4.1.3 The entropy change accompanying expansion
4.1.4 The entropy change accompanying heating
4.1.5 The entropy change accompanying a phase transition
4.1.6 Entropy changes in the surroundings
4.1.7 Absolute entropies and the Third Law of thermodynamics
4.1.8 The statistical entropy
4.1.9 Residual entropy
4.1.10 The standard reaction entropy
4.1.11 The spontaneity of chemical reactions
4.2 The Gibbs energy
4.2.12 Focusing on the system
4.2.13 Properties of the Gibbs energy
The direction of spontaneous
change
the chaotic dispersal of energy
the dispersal of matter
Can we convert heat to work
completely?
ch04f03
The Second Law
1.
Reversible vs Irreversible processes
Reversible:
A process for which a system can be restored to its
initial state, without leaving a net influence on the
system or its environment.
*
idealized, frictionless;
*
proceeds slowly enough for the system to remain in
thermodynamic equilibrium.
Irreversible:
Not reversible
*
natural;
*
proceeds freely, drives the system out of thermodynamic
equilibrium;
*
interacts with environment, can not be exactly reversed
Example: Gas-piston system under a constant temperature
(  dU  0 )
* Slow expansion and compression ( pex  p )
 q    w   pdv  (w
12
 w21 )  0
* Rapid expansion ( pex  pe  p ) and compression ( pex  pc  p )
2
1
  q     w   pe dV   pc dV
1
2
2
2
2
  pe dV   pc dV   ( pe  pc )dV  0
1
1
1
The Carnot cycle and entropy
The cycle operates with one mole ideal gas.
1) 1 →2: Isothermal expansion
T2  T1  Thot ,
V2  V1
U12  0
q12  w12  
2
1
 V2 
pdV  RThot ln  
 V1 
2) 2 → 3: Adiabatic expansion
Tcold  T3  T2  Thot ,
V3  V2
q23  0
w23  U 23  cV (Tcold  Thot )
3) 3 → 4: Isothermal compression
T4  T3  Tcold ,
V4  V3
U 34  0
V 
q34  w34   pdV  RTcold ln  4 
3
 V3 
4
4) 4 → 1: Adiabatic compression
Thot  T1  T4  Tcold ,
V1  V4
q41  0
w41  U 41  cV (Thot  Tcold )
from chap. 2
4. Work for adiabatic expansion
For an adiabatic process for an ideal gas,
CV dT   pdV  
V2 1
CV T2 1
dT   
dV
V1 V
nR T1 T
T 
ln  2 
 T1 
V2T2
CV
nR
CV
nR
CV 1
1
dT   dV
nR T
V
V 
CV  T2 

ln     ln  2 
nR  T1 
 V1 
nRT
dV
V

V 
T 
 ln  1    2 
 V2 
 T1 
 V1T1
CV
nR
CV
nR
V 
 1 
 V2 
V2T2 c  V1T1c where, c 
T cV  const "
( pV )c V  const '
p cV c 1  const '
pV
CV
nR
c  1  CV
 1  CV  nR  nR CV  nR C p

 1  




c
nR
c
nR
C
C
C



 V
V
V
c 1
c
 const '
pV   const
The net heat transfer and the net work over the Carnot cycle are:
 V4 
 V2 
   w    q  RThot ln    RTcold ln  
 V1 
 V3 
 Tf 
 
 Ti 
CV
nR
 Vi

V
 f



2→3,  Tcold 


T
 hot 
CV
nR
V  V 
1   2   4  ,
 V3   V1 
V 
 2 
 V3 

4→1,  Thot 


T
 cold 
CV
nR
V 
 4 
 V1 
V2 V3

V1 V4
Then,
 V2 
   w    q  R(Thot  Tcold ) ln    0
 V1 
So the system absorbs heat and performs net work in the Carnot cycle,
which behaves as a heat engine.
The Carnot cycle
-summary
q
w
∆U
∆H
∆S
1→2
RThln(V2/V1)
-RThln(V2/V1)
0
0
Rln(V2/V1)
2→3
0
cV(Tc-Th)
cV(Tc-Th)
cP(Tc-Th)
0
3→4
-RTcln(V2/V1)
RTcln(V2/V1)
0
0
-Rln(V2/V1)
4→1
0
cV(Th-Tc)
cV(Th-Tc)
cP(Th-Tc)
0
0
0
0
cycle
R(Th-Tc)ln(V2/V1) -R(Th-Tc)ln(V2/V1)
For the Carnot cycle, we can also have:

  V2
q12 q34


 R ln 
T Thot Tcold
  V1
q
 V4  

  ln     0

 V3  
This relationship also hold for the reversed Carnot cycle.
This is called the Carnot’s Theorem:
 dS  0 ,
 q 
dS   
 T  rev
(4.1)
The change of S is independent of path under a reversible process.
S is a state function, means
a system property. It is called entropy.
3. The Second Law and its Various Forms
To get the second law, we use the Clausius Inequality, i.e.,
for a cyclic process,
q
T
0
which indicates during the cycle,
1) heat must be rejected to the environment;
2) heat exchange is larger at high temperature than at low temperature
under reversible conditions;
3) the net heat absorbed is smaller under the irreversible condition
than under the reversible condition.
Now, consider two cycles as shown in the plot.
For the cycle which contains one
reversible process and one irreversible
process,

2
1
 q 
  
 T  rev
 q 
2  T irrev  0
1
(4.2)
For the cycle which have two reversible
processes,

2
1
 q 
  
 T  rev
 q 
2  T rev  0
1
(4.3)
The difference between (4.2) and (4.3) gives
 q 
2  T rev 
1
1

2
dS 
 q 
2  T irrev
1
Because states 1 and 2 are arbitrary, we have the second law,
dS 
q
(4.4)
T
Combine (4.1) and (4.4), we have
q  q 
 
T  T  rev
or
 w   w 


T
 T rev
It indicates that the heat absorbed by the system during a process
has an upper limit, which is the heat absorbed during a reversible
process.
The first law relates the state of a system to work done on it and heat
it absorbs.
The second law controls how the systems move to the thermodynamic
equilibriums, i.e., the direction of processes.
The Second Law
of Thermodynamics
The Second Law of Thermodynamics establishes that all
spontaneous or natural processes increase the entropy of the
universe
Stotal = Suniv = Ssys + Ssurr
In a process, if entropy increases in both the system and the
surroundings, the process is surely spontaneous
16
Several simplified forms of the second law:
1)
For an adiabatic process, (4.4) becomes
dS  0
(4.5)
If the adiabatic process is reversible, then
dS  0
It is also isentropic (S is constant).
2) For an isochoric process, (4.1) becomes
 dT 
 dT 
dSV  cV 
  cV 

 T  rev
 T V
(4.6)
Because only state variables are involved, it holds for either
reversible or irreversible processes.
(4.5) and (4.6) show that :
Irreversible work can only increase entropy;
heat transfer can either increase or decrease entropy.
5. Thermodynamic Equilibrium
*
Consider an adiabatic process,
the second law becomes
dS  0
For an irreversible condition,
dS  0
or
s0
S  S0
is the entropy at the initial state.
When s reaches the maximum,
the state is in thermodynamic
equilibrium because the entropy
can not increase anymore.
Calculate the changes in entropy as a result of the transfer of
100 kJ of energy as heat to a large mass of water (a) at 0°C (273
K) and (b) at 100°C (373 K).
(a) ΔS at 0°C (273 K)
qrev 100 103 J
S 

 366 J K 1
T
273 K
(b) ΔS at 100°C (373 K)
qrev 100 103 J
S 

 268 J K 1
T
373 K
Heat engines
S 
qH  qL

Thot Tcold
The engine will not operate spontaneously
if this change in entropy is negative, and
just becomes spontaneous as ΔStotal
becomes positive. This change of sign
occurs ΔStotal = 0, which is achieved when
Tcold
qL 
 qH
Thot
the efficiency, η, of the engine, the ratio of
the work produced to the heat absorbed, is

q
work produced qH  qL

 1 L
heat absorbed
qH
qH
 1
Tcold
Thot
Refrigerators, and heat pumps
COPheating 
Thot
qH
qH


win qH  qL Thot  Tcold
qL 
COPcooling 
Tcold
 qH
Thot
Tcold
qL
qL


win qH  qL Thot  Tcold
Refrigerator power
No thermal insulation is perfect, so there is always a flow of energy as
heat into the sample at a rate proportional to the temperature difference.
The rate at which heat leaks in can be written as A(Th – Tc ), where A is a
constant. Calculate the minimum power, P, required to maintain the
original temperature difference? Assume the refrigerator is operating at
100% of its theoretical efficiency. Express P in terms of A, Th, Tc.
The entropy change with
isothermal expansion
1
dq
1 T
2 1
2 p
2 nR
   dw   dV  
dV
1 T
1 T
1 V
V 
p 
 nR ln  2   nR ln  2 
 V1 
 p1 
S  
ch04f04
2
The entropy change with heating
In case heat capacity is constant,
1
dq
1 T
2C
 T2 
x

dT  Cx ln  
1 T
 T1 
S  
ch04f05
2
The entropy change with heating
In case heat capacity is temperature
dependent,
1
dq
1 T
2C
  x dT
1 T
S  
ch04f06
2
The entropy change with a phase
transition
At phase transition, the temperature
stays constant, T = Ttr.
1
S   dq
1 T
 H
1 2
  dq  tr
Ttr 1
Ttr
2
When a solid (a), melts, the molecules form
a more chaotic liquid, the disorderly array of
spheres (b). As a result, the entropy of the
sample increases.
Absolute entropies and the Third
Law of thermodynamics
S (0) = 0 for all perfectly ordered
crystalline materials.
ch04f08
T
Cp
0
T
S (T )  
dT
nonmetallic solids, Debye T
At temperatures T << TD,
Cv,m = aT 3 ,
and
Sm (T) = 13 Cv,m
ch04f09
3
-law:
Standard Molar Entropies
In general, the more atoms in its molecules, the greater is the
entropy of a substance
ch04f10
The statistical entropy
U = 4ε
The 19 arrangements(W ) of four molecules in a system
with three energy levels and a total energy of 4ε.
A simple 4-particle system
Energy
(in equal-spaced energy levels)
U = nε
U=0
W=1
U=ε
W=4
U = 2ε
W = 10
W = C(n+3, n) = (n+3)(n+2)(n+1)/6
U = 3ε
W = 20
U
0
1
2
3
4
5
6
7
8
9
10
W
1
4
10
20
35
56
84
120
165
220
286
S = kB ln(W )
A simple 4-particle system
W = C(n+3, n) = (n+3)(n+2)(n+1)/6
U = nε
S = kB ln(W )
Two–Level system
a very simplified version
Energy
In case when ε >> kT, almost no particle can reach upper state.
According to Boltzmann, the probability of a particle at level 1 at T is
p1 = exp(-ε/kT ).
U=0
U=ε
p1 = exp(-ε/kT )
A two-level system
Two–Level system
a very simplified version
U(T) = Nε exp(-ε/kT )
Cv,m = (∂U/ ∂T)
= Nε exp(-ε/kT ) (-ε/k) (-1/T 2)
= N (ε 2/kT 2) exp(-ε/kT ) ≈ 0
ε
kB
T, K
0.001
1
10
100
1000
1.66E-19"=100kJ/mol"
1.36E-23
J/K
U(T), J/mol Cv(T), J/mol
0
0
0
0
0
0
9.00E-49
1.10E-48
4.96E-01
6.06E-03
1. This is why we do not consider heat capacity contributions
from subatomic particles such as electrons, etc.
2. This is why heat capacity approaches 0 when T approaches 0
for every material.
W and Energy Level Spacing
At a given temperature, the number of arrangements corresponding
to the same total energy is greater when the energy levels are closely
spaced than when they are far apart.
Residual entropy
for CO molecules, there are 2N possible arrangements at T= 0.
Sm = NAk ln 2 = R ln 2 = 5.8 J K−1 mol−1
Ice at 0 K
W =(32)N, S = k ln(32)N = Nk ln(32), Sm = R ln(32) = 3.4 J K−1 mol−1.
The spontaneity of chemical
reactions
S surr
q

T
q
Suniv  S  Ssurr  0
at constant p, qp=ΔH.
Suniv  S  S surr
H
T
H
 S 
0
T
S surr  
H  T S  0
if we define G ≡ H – TS, Gibbs Free Energy,
ΔG = ΔH –TΔS at constant T.
So, at constant T and p,
ΔG = ΔH –TΔS ≤ 0.
Gibbs Energy, G
Spontaneity criterion
ΔSuniv > 0. at constant pressure and temperature, ΔGsystem < 0.
Calculate the change in molar entropy when one mole of argon gas
is compressed from 2.0 dm3 to 500 cm3 and simultaneously heated
from 300 K to 400 K. Take CV,m = (3/2)R
p
T = 400 K
T = 300 K
V
Calculate the change in entropy when 100 g of water at 80°C is
poured into 100 g of water at 10°C in an insulated vessel given that
Cp,m = 75.5 J K−1 mol−1.
The enthalpy of vaporization of chloroform (trichloro-methane),
CHCl3, is 29.4 kJ mol−1 at its normal boiling point of 334.88 K. (a)
Calculate the entropy of vaporization of chloroform at this
temperature. (b) What is the entropy change in the surroundings?
Suppose that the weight of a configuration of N molecules in a gas
of volume V is proportional to VN. Use Boltzmann’s formula to
deduce the change in entropy when the gas expands isothermally.
Without performing a calculation, estimate whether the standard
entropies of the following reactions are positive or negative: