1. Thermodynamics: the Second Law
자연과학대학 화학과
박영동 교수
The Direction of Nature and spontaneity

2. 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
The standard reaction entropy
The spontaneity of chemical reactions
4.2 The Gibbs energy
Focusing on the system
Properties of the Gibbs energy

3. The direction of spontaneous change
the chaotic dispersal of energy
the dispersal of matter
One of spontaneous process is the chaotic dispersal of matter. This tendency accounts for the spontaneous tendency of a gas to spread into and fill the container it occupies. It is extremely unlikely that all the particles will collect into one small region of the container. (In practice, the number of particles is of the order of 1023.) Another spontaneous process is the chaotic dispersal of energy (represented by the small arrows). In these diagrams, the orange spheres represent the system and the purple spheres represent the surroundings. The double-headed arrows represent the thermal motion of the atoms.

4. Can we convert heat to work completely?
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The Second Law denies the possibility of the process illustrated here, in which heat is changed completely into work, there being no other change. The process is not in conflict with the First Law, because the energy is conserved.

5. 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

6. Example: Gas-piston system under a constant temperature
* Slow expansion and compression
* Rapid expansion and compression

7. The Carnot cycle and entropy
The cycle operates with one mole ideal gas.
1) 1 →2: Isothermal expansion
2) 2 → 3: Adiabatic expansion

8. 3) 3 → 4: Isothermal compression
4) 4 → 1: Adiabatic compression

9. 4. Work for adiabatic expansion
from chap. 2
4. Work for adiabatic expansion
For an adiabatic process for an ideal gas,

10. The net heat transfer and the net work over the Carnot cycle are:
2→3,
4→1,
Then,
So the system absorbs heat and performs net work in the Carnot cycle,
which behaves as a heat engine.

11. The Carnot cycle -summaryq w ∆U ∆H ∆S
1→2
RThln(V2/V1)
-RThln(V2/V1)
Rln(V2/V1)
2→3
cV(Tc-Th)
cP(Tc-Th)
3→4
-RTcln(V2/V1)
RTcln(V2/V1)
-Rln(V2/V1)
4→1
cV(Th-Tc)
cP(Th-Tc)
cycle
R(Th-Tc)ln(V2/V1)
-R(Th-Tc)ln(V2/V1)

12. For the Carnot cycle, we can also have:
This relationship also hold for the reversed Carnot cycle.
This is called the Carnot’s Theorem:
(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.

13. 3. The Second Law and its Various Forms
To get the second law, we use the Clausius Inequality, i.e.,
for a cyclic process,
which indicates during the cycle,
heat must be rejected to the environment;
heat exchange is larger at high temperature than at low temperature
under reversible conditions;
the net heat absorbed is smaller under the irreversible condition
than under the reversible condition.

14. Now, consider two cycles as shown in the plot.
For the cycle which contains one
reversible process and one irreversible
process,
(4.2)
For the cycle which have two reversible
processes,
(4.3)

15. The difference between (4.2) and (4.3) gives
Because states 1 and 2 are arbitrary, we have the second law,
(4.4)
Combine (4.1) and (4.4), we have or
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.

16. The Second Law of Thermodynamics
The Second Law of Thermodynamics establishes that all spontaneous or natural processes increase the entropy of the universe
DStotal = DSuniv = DSsys + DSsurr
In a process, if entropy increases in both the system and the surroundings, the process is surely spontaneous

17. Several simplified forms of the second law:
1) For an adiabatic process, (4.4) becomes
(4.5)
If the adiabatic process is reversible, then
It is also isentropic (S is constant).
2) For an isochoric process, (4.1) becomes
(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.

18. 5. Thermodynamic Equilibrium
* Consider an adiabatic process,
the second law becomes
For an irreversible condition, or
is the entropy at the initial state.
When reaches the maximum,
the state is in thermodynamic
equilibrium because the entropy
can not increase anymore.

19. 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)
(b) ΔS at 100°C (373 K)

20. Heat engines
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
the efficiency, η, of the engine, the ratio of the work produced to the heat absorbed, is

21. Refrigerators, and heat pumps

22. 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.

23. The entropy change with isothermal expansion
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24. The entropy change with heating
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In case heat capacity is constant,

25. The entropy change with heating
In case heat capacity is temperature dependent,
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26. The entropy change with a phase transition
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.
At phase transition, the temperature stays constant, T = Ttr.

27. Absolute entropies and the Third Law of thermodynamics
S (0) = 0 for all perfectly ordered crystalline materials.
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28. nonmetallic solids, Debye T 3 -law: At temperatures T << TD,
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nonmetallic solids, Debye T 3 -law:
At temperatures T << TD,
Cv,m = aT 3 ,
and
Sm (T) = Cv,m  

29. Standard Molar Entropies
In general, the more atoms in its molecules, the greater is the entropy of a substance
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30. The statistical entropy
The 19 arrangements(W ) of four molecules in a system with three energy levels and a total energy of 4ε.
U = 4ε

31. A simple 4-particle system (in equal-spaced energy levels)
U = nε
U = 0
U = ε
U = 2ε
U = 3ε U W 1 4 2 10 3 20 35 5 56 6 84 7
120 8
165 9
220
286
W = 1
W = 20
W = 4
W = 10
W = C(n+3, n) = (n+3)(n+2)(n+1)/6
S = kB ln(W )

32. A simple 4-particle system
W = C(n+3, n) = (n+3)(n+2)(n+1)/6
U = nε
S = kB ln(W )
From S = kB ln W, U = nε, T = dU/dS Cv = dU/dT

33. Two–Level system a very simplified version
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 ).
Energy
U = 0
U = ε
p1 = exp(-ε/kT ) 
A two-level system

34. Two–Level system a very simplified versionε
1.66E-19
"=100 kJ/mol" kB
1.36E-23
　J/K
T, K
U(T), J/mol
Cv(T), J/mol
0.001 1 10
100
9.00E-49
1.10E-48
1000
4.96E-01
6.06E-03
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
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.

35. 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.

36. 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

37. Ice at 0 K
W = ( 3 2 ) N, S = k ln ( 3 2 ) N = Nk ln ( 3 2 ) , Sm = R ln ( 3 2 ) = 3.4 J K−1 mol−1.

38. The spontaneity of chemical reactions
at constant V, qV = ΔU.
∆𝑆 𝑠𝑢𝑟𝑟 =− ∆𝑈 𝑇 q
∆𝑆 𝑢𝑛𝑖𝑣 = ∆𝑆+ ∆𝑆 𝑠𝑢𝑟𝑟 =∆𝑆− ∆𝑈 𝑇 ≥0
∆𝑈 −𝑇∆𝑆 ≤0
if we define A ≡ U – TS, Helmholtz Free Energy,
ΔA = ΔU –TΔS at constant V.
So, at constant T and V,
ΔA = ΔU –TΔS ≤ 0.

39. The spontaneity of chemical reactions
at constant p, qp = ΔH. q
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.

40. Gibbs Energy, G Spontaneity criterion
ΔSuniv > 0. at constant pressure and temperature, ΔGsystem < 0.

41. Standard molar Gibbs Energies
Δr 𝐺 ⊖ = Δr 𝐻 ⊖  TΔr 𝑆 ⊖
The Born equation
∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r
𝜀 r = relative permittivity
For water, 𝜀 r = at 25 ℃
∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑟 i /pm ×6.86× kJ mol −1

42. Gibbs Energies of solvation
The Born equation
∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r
𝜀 r = 𝜀 𝜀 0 = relative permittivity
For water, 𝜀 r = at 25 ℃
∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑟 i /pm ×6.86× kJ mol −1

43. Gibbs Energies of solvation
The Born equation
𝑉(𝑟)= 𝑄 1 𝑄 2 4π𝜀𝑟
𝜀= permittivity of the medium
For water, 𝜀 r = at 25 ℃
∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑟 i /pm ×6.86× kJ mol −1
∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r

44. Gibbs Energies of solvation
The Born equation
𝑉(𝑟)= 𝑄 1 𝑄 2 4π𝜀𝑟
𝜀= permittivity of the medium 𝜀 𝜀
𝑉 𝑟 = 𝑄 1 𝜙 𝑟 𝜙(𝑟)= 𝑄 2 4π𝜀𝑟
𝑟 i
𝑤= 0 𝑧 i 𝑒 𝜙 𝑟 i 𝑑𝑄 = 1 4𝜋𝜀 𝑟 i 0 𝑧 i 𝑒 𝑄𝑑𝑄 = 𝑧 i 2 𝑒 2 𝑒 2 8π𝜀 𝑟 i
∆ solv 𝐺 ⊖ = 𝑧 i 2 𝑒 2 𝑁 A 8πε 𝑟 i − 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 𝜀 0
=− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r

45. Combining the First and Second Laws
𝑑𝑈=𝑇𝑑𝑆 −𝑝𝑑𝑉
𝑈(𝑆, 𝑉)
𝑑𝑈= 𝜕𝑈 𝜕𝑆 𝑉 𝑑𝑆 − 𝜕𝑈 𝜕𝑉 𝑆 𝑑𝑉
𝜕 2 𝑈 𝜕𝑉𝜕𝑆 = 𝜕 2 𝑈 𝜕𝑆𝜕𝑉
𝜕𝑈 𝜕𝑆 𝑉 =𝑇 𝜕𝑈 𝜕𝑉 𝑆 =−𝑝
𝜕𝑇 𝜕𝑉 𝑆 =− 𝜕𝑝 𝜕𝑆 𝑉

46. The Maxwell relations
𝑈 𝑆, 𝑉 ;𝑑𝑈=𝑇𝑑𝑆 −𝑝𝑑𝑉;𝑇= 𝜕𝑈 𝜕𝑆 𝑉 ;𝑝=− 𝜕𝑈 𝜕𝑉 𝑆 ; 𝜕𝑇 𝜕𝑉 𝑆 =− 𝜕𝑝 𝜕𝑆 𝑉
𝐻 𝑉, 𝑇 ;𝑑𝐻=𝑇𝑑𝑆+𝑉𝑑𝑝; 𝑇= 𝜕𝐻 𝜕𝑆 𝑝 ; 𝑉= 𝜕𝐻 𝜕𝑝 𝑆 ; 𝜕𝑇 𝜕𝑝 𝑆 = 𝜕𝑉 𝜕𝑆 𝑝
𝐺 𝑉, 𝑇 ;𝑑𝐺=𝑉𝑑𝑝 −𝑆𝑑𝑇; 𝑉= 𝜕𝐺 𝜕𝑝 𝑇 ; 𝑆=− 𝜕𝐺 𝜕𝑇 𝑝 ; 𝜕𝑉 𝜕𝑇 𝑝 = − 𝜕𝑆 𝜕𝑝 𝑇
𝐴 𝑉, 𝑇 ;𝑑𝐴=−𝑝𝑑𝑉 −𝑆𝑑𝑇;𝑝=− 𝜕𝐴 𝜕𝑉 𝑇 ;𝑆=− 𝜕𝐴 𝜕𝑇 𝑉 ; 𝜕𝑝 𝜕𝑇 𝑉 = 𝜕𝑆 𝜕𝑉 𝑇

47. Properties of the Gibbs energy
𝑑𝐺=𝑉𝑑𝑝 −𝑆𝑑𝑇
𝐺(𝑝, 𝑇)
𝜕𝐺 𝜕𝑇 𝑝 =−𝑆 𝜕𝐺 𝜕𝑝 𝑇 =𝑉
𝜕𝐺 𝜕𝑇 𝑝 = 𝐺−𝐻 𝑇
𝜕(𝐺/𝑇) 𝜕𝑇 𝑝 =− 𝐻 𝑇 2
𝜕(∆𝐺/𝑇) 𝜕𝑇 𝑝 =− ∆𝐻 𝑇 2

48. the Gibbs energy vs p 𝜕𝐺 𝜕𝑝 𝑇 =𝑉 𝜕 𝐺 𝑚 𝜕𝑝 𝑇 = 𝑉 𝑚 𝑑𝐺=𝑉𝑑𝑝 −𝑆𝑑𝑇 𝐺(𝑝, 𝑇)
𝜕𝐺 𝜕𝑝 𝑇 =𝑉
𝜕 𝐺 𝑚 𝜕𝑝 𝑇 = 𝑉 𝑚
𝑑𝐺=𝑉𝑑𝑝 −𝑆𝑑𝑇
𝐺(𝑝, 𝑇)
𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i + 𝑝 i 𝑝 f 𝑉 𝑚 𝑑𝑝
𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i + 𝑉 𝑚 ( 𝑝 f − 𝑝 i ) ⊖
𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i +𝑅𝑇 ln 𝑝 f 𝑝 i
𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i +𝑅𝑇 ln(𝑓 𝑝 f 𝑝 i )
𝐺 𝑚 𝑝 = 𝐺 𝑚Θ +𝑅𝑇 ln(𝑓𝑝⊖)

49. The Gibbs energy and the fugacity
𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i +𝑅𝑇 ln 𝑝 f 𝑝 i
𝐺 𝑚 𝑝 f = 𝐺 𝑚 𝑝 i +𝑅𝑇 ln(𝑓 𝑝 f 𝑝 i )
𝐺 𝑚 𝑝 = 𝐺 𝑚 Θ+𝑅𝑇 ln(𝑓𝑝Θ)
𝑓:fugacity
𝑓=𝜙𝑝
𝜙 :fugacity coefficient ⊖
ln 𝜙 = 0 𝑝 𝑍−1 𝑝 d𝑝

50. dSU,V  0 (b) dUS,V  0 dSH,p  0 (b) dHS,p  0 A = U – TS G = H – TS
Property
Equation
Comment
Criteria of spontaneity
dSU,V  0
(b) dUS,V  0
Constant volume (etc)*
dSH,p  0
(b) dHS,p  0
Constant pressure (etc)
Helmholtz energy
A = U – TS
Definition
Gibbs energy
G = H – TS

51. Property Equation Comment (a) dSU,V  0 (b) dUS,V  0
Criteria of spontaneity
(a) dSU,V  0 (b) dUS,V  0
Constant volume (etc)*
(a) dSH,p  0 (b) dHS,p  0
Constant pressure (etc)
Helmholtz energy
A = U – TS
Definition
Gibbs energy
G = H – TS
(a) dAT,V  0 (b) dGT,p  0
Constant temperature (etc)
Equilibrium
dAT,V = 0
Constant volume (etc)
Maximum work
dwmax = dA,
wmax = A
Constant temperature
dGT,p = 0
Maximum non-expansion work
dwadd,max = dG,
wadd,max = G
Constant temperature and pressure
Standard Gibbs energy of reaction
Δr 𝐺 ⊖ = Δr 𝐻 ⊖  TΔr 𝑆 ⊖
∆ r 𝐺 ⊖ = J 𝜈 J ∆ f 𝐺 m ⊖ (J)
Practical implementation
Ions in solution
Δf 𝐺 ⊖ (H+, aq) = 0
Convention
Born equation
∆ solv 𝐺 ⊖ =− 𝑧 i 2 𝑒 2 𝑁 A 8π 𝜀 0 𝑟 i 1− 1 𝜀 r
Solvent a continuum

52. Property Equation Comment
Fundamental equation
dU = TdS  pdV
No additional work
Fundamental equation of chemical thermodynamics
dG = Vdp  SdT
Variation of G
(G/p)T = V and (G/T)p = S
Composition constant
Gibbs–Helmholtz equation
((G/T)/T)p = H/T 2
Pressure dependence of G
Gm(pf) = Gm(pi) + VmΔp
Incompressible substance
G(pf) = G(pi) + nRT ln(pf/pi)
Perfect gas
𝐺 m = 𝐺 m ⊖ +𝑅𝑇 ln(𝑝/ 𝑝 ⊖ )
Fugacity
𝐺 m = 𝐺 m ⊖ +𝑅𝑇 ln(𝑓/ 𝑝 ⊖ )
Definition
Fugacity coefficient
f = p
ln 𝜙 = 0 𝑝 𝑍−1 𝑝 d𝑝
Determination

53. 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 V
T = 300 K
T = 400 K

54. 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.

55. The enthalpy of vaporization of chloroform (trichloro-methane), CHCl3, is 29.4 kJ mol−1 at its normal boiling point of K. (a) Calculate the entropy of vaporization of chloroform at this temperature. (b) What is the entropy change in the surroundings?

56. 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.

57. Without performing a calculation, estimate whether the standard entropies of the following reactions are positive or negative: