Physical Chemistry
21. Second Law of Thermodynamics

Ruifeng ZHOU

Faculty of Science &Institute for the Advancement of Higher Education
Hokkaido University

Table of Content

1. Second law of thermodynamics
2. Entropy2

3

1. Second law of thermodynamics• A heat engine is a system
operating a cyclic processthat converts heat to work.
• A Carnot cycle iscomposed of reversible
isothermal processes andadiabatic processes.
• A Carnot engine is a heatengine based on Carnot
cycle.
• By defalt, 𝑇𝑇1 > 𝑇𝑇2 .

𝑄𝑄1Heatengine

𝑄𝑄2

𝑇𝑇1

𝑊𝑊𝑇𝑇2

A
𝑄𝑄1
D𝑄𝑄2BC𝑇𝑇1𝑇𝑇2

4

1. Second law of thermodynamics• Suppose the system is anideal gas.
𝑉𝑉B
𝑄𝑄1 = 𝑛𝑛𝑛𝑛𝑇𝑇1 ln
𝑉𝑉A
𝑉𝑉D
𝑄𝑄2 = 𝑛𝑛𝑛𝑛𝑇𝑇2 ln
𝑉𝑉C
𝑄𝑄1 𝑄𝑄2𝑉𝑉B 𝑉𝑉D+= 𝑛𝑛𝑛𝑛 ln
𝑇𝑇1 𝑇𝑇2𝑉𝑉C 𝑉𝑉A• In an adiabatic process,
𝑇𝑇𝑉𝑉 𝛾𝛾−1 = const
𝑇𝑇1 𝑉𝑉B 𝛾𝛾−1 = 𝑇𝑇2 𝑉𝑉C 𝛾𝛾−1A
𝑄𝑄1
D

𝑄𝑄2B

𝑇𝑇1 > 𝑇𝑇2C

𝑇𝑇1𝑇𝑇2

5

1. Second law of thermodynamics• Similarly,

𝑇𝑇2
𝑉𝑉B=𝑇𝑇1𝑉𝑉C

𝛾𝛾−1𝑇𝑇2 𝑉𝑉D 𝛾𝛾−1 = 𝑇𝑇1 𝑉𝑉A 𝛾𝛾−1𝛾𝛾−1𝑇𝑇1𝑉𝑉D=𝑇𝑇2𝑉𝑉A• For a 2-temperature Carnot cycle, the Carnot
equation is
𝑄𝑄1 𝑄𝑄2+=0
𝑇𝑇1 𝑇𝑇2

61. Second law of thermodynamics• The efficiency of a heat engine
is−𝑊𝑊 𝑄𝑄1 + 𝑄𝑄2𝑄𝑄2𝜂𝜂 ≡==1+𝑄𝑄1𝑄𝑄1𝑄𝑄1• The Carnot theorem states
that
𝑇𝑇1 − 𝑇𝑇2
𝑇𝑇2=1−
𝜂𝜂Carnot =𝑇𝑇1
𝑇𝑇1
• An engine can run reversely
as a refrigerator or heat pump.

𝑄𝑄1Carnot
engine

𝑄𝑄2

𝑊𝑊𝑄𝑄1
𝑄𝑄2

𝑇𝑇1

𝑊𝑊𝑇𝑇2
𝑇𝑇1𝑇𝑇2

71. Second law of thermodynamics• The efficiency of a refrigerator isdefined as
𝑄𝑄2 Carnot 𝑇𝑇2𝜂𝜂r ≡𝑊𝑊
𝑇𝑇1 − 𝑇𝑇2
• The efficiency of a heat pump isdefined as−𝑄𝑄1 Carnot 𝑇𝑇1𝜂𝜂p ≡𝑊𝑊𝑇𝑇1 − 𝑇𝑇2
• Obliviously,1𝜂𝜂p,Carnot =𝜂𝜂Carnot𝑊𝑊𝑊𝑊

𝑄𝑄1

𝑇𝑇1Fridge

𝑄𝑄2
𝑄𝑄1

𝑇𝑇2
𝑇𝑇1Heatpump

𝑄𝑄2

𝑇𝑇2

8

1. Second law of thermodynamics• Clausius’ statement: “No process is possible whosesole result is the transfer of heat from a colder bodyto a hotter body.”• When 2 bodies are at different temperatures, heattransfers from the hotter body to the colder bodyspontaneously as an irreversible process.
HotCold𝑄𝑄𝑇𝑇Q𝑇𝑇

1. Second law of thermodynamics

9• Heat transfer between bodies at identicaltemperature is a reversible process.
• Carnot’s statement (theorem): “Of all the heatengines working between 2 given temperatures,
none is more efficient than a Carnot engine.”• Kelvin’s statement: “No process is possible whosesole result is the complete conversion of heat intowork.”• The 3 statements are equivalent.

10

1. Second law of thermodynamics• Carnot’s statement is equivalent to Clausius’statement.
• Proof: suppose there is a Carnot violator withefficiency 𝜂𝜂 > 𝜂𝜂Carnot .
• Its work is outputted to a Carnot pump.
𝑄𝑄1

Carnotviolator

𝑄𝑄2

𝑊𝑊𝑄𝑄3

𝑇𝑇1Carnotpump

𝑄𝑄4

𝑇𝑇2

1. Second law of thermodynamics

11• Use the absolute values of work and heat.
𝑊𝑊𝑊𝑊𝜂𝜂 =>= 𝜂𝜂Carnot
𝑄𝑄1
𝑄𝑄3
• So,
𝑄𝑄1 < 𝑄𝑄3
• This means that, there is a net heat flow from low
temperature reservoir to high temperature
reservoir. It violates Clausius’ statement.
• Corollary: “All reversible engines working between
2 temperatures have the same efficiency, 𝜂𝜂Carnot ”.121. Second law of thermodynamics• Then, suppose there is a Clausius violator. Itpumps heat (𝑄𝑄3 ) from low temperature reservoir tohigh temperature reservoir.
𝑇𝑇1
𝑄𝑄3

𝑄𝑄1Clausiusviolator

Carnotengine

𝑄𝑄3

𝑄𝑄2

𝑊𝑊𝑇𝑇2

• The overall efficiency of combined engine is
𝑄𝑄2 − 𝑄𝑄3
𝑄𝑄2𝜂𝜂 = 1 −>1−= 𝜂𝜂Carnot
𝑄𝑄1 − 𝑄𝑄3𝑄𝑄1
• It violates Carnot’s statement.

131. Second law of thermodynamics• Clausius’ and Kelvin’s statements are equivalent.
• Proof: suppose there is a Kelvin violator.
𝑄𝑄1

Perpetual motion machine Kelvin 𝑊𝑊violatorof the second kind

𝑄𝑄2

𝑇𝑇1Carnotpump

𝑄𝑄3

𝑇𝑇2

• The Kelvin violator outputs work to a Carnot pump.
• The overall process violates Clausius’ statement.
• Perpetual motion machine of the second kind isimpossible.

14

1. Second law of thermodynamics• Then, suppose there is a Clausius violator.
𝑄𝑄2

𝑇𝑇1

CarnotengineClausius
violator

𝑄𝑄2

𝑄𝑄1

𝑇𝑇2

𝑄𝑄2

𝑊𝑊• If the heat flow form the Carnot engine to the cooler
reservoir is the same as heat flow form the cooler
reservoir to the Clausius violator (𝑄𝑄2 ), the overallsystem converts heat 𝑄𝑄1 − 𝑄𝑄2 to work, which
violates Kelvin’s statement.

1. Second law of thermodynamics

15• According to Carnot statement, the efficiency of aheat engine cannot be greater than a Carnot
engine.
𝑄𝑄1 + 𝑄𝑄2 𝑇𝑇1 − 𝑇𝑇2≤= 𝜂𝜂Carnot𝜂𝜂 =𝑄𝑄1𝑇𝑇1• We get the Carnot inequality.
𝑄𝑄1 𝑄𝑄2+≤0
𝑇𝑇1 𝑇𝑇2• 𝑇𝑇1 and 𝑇𝑇2 are the temperatures of the reservoirs(surroundings) .

16

1. Second law of thermodynamics• For example, a Sterlingengine is based on Sterlingcycle, composed ofisothermal and isochoricprocesses.
𝑄𝑄1 𝑄𝑄3
𝑉𝑉B 𝑉𝑉D+= 𝑛𝑛𝑛𝑛 ln=0𝑇𝑇1 𝑇𝑇2𝑉𝑉C 𝑉𝑉A
𝑄𝑄4 𝑄𝑄2+𝑇𝑇1 𝑇𝑇2𝐶𝐶𝑉𝑉 𝑇𝑇1 − 𝑇𝑇2
𝐶𝐶𝑉𝑉 𝑇𝑇2 − 𝑇𝑇1=+<0𝑇𝑇1𝑇𝑇2A

𝑄𝑄4D

𝑇𝑇1 > 𝑇𝑇2𝑄𝑄1𝑄𝑄3𝑇𝑇1𝑇𝑇2

B
𝑄𝑄2C1. Second law of thermodynamics

17• Therefore, in Sterling engine,
𝑄𝑄1 + 𝑄𝑄4 𝑄𝑄2 + 𝑄𝑄3
System Surroundings+<0𝑇𝑇
𝑇𝑇2𝑇𝑇1𝑇𝑇2𝛿𝛿𝑄𝑄
𝑄𝑄2 + 𝑄𝑄3𝑇𝑇2< 1 − = 𝜂𝜂Carnot𝜂𝜂 = 1 +𝑄𝑄1 + 𝑄𝑄4𝑇𝑇1• Explanation: during the isochoricprocesses, the system temperature
is different from the surroundingtemperature (irreversible).
System Surroundings
𝛿𝛿𝑄𝑄 𝛿𝛿𝑄𝑄 𝛿𝛿𝑄𝑄𝑇𝑇𝑇𝑇1𝛿𝛿𝑄𝑄>
>
𝑇𝑇2𝑇𝑇
𝑇𝑇1

1. Second law of thermodynamics

18• A multitemperature Carnot cycle consists of a few
isothermal and adiabatic processes.
• It can be broken into several 2-temperature Carnotcycles.
𝑄𝑄1 𝑄𝑄4+=0𝑇𝑇1
𝑇𝑇1 𝑇𝑇0
𝑄𝑄2 𝑄𝑄5+=0
𝑇𝑇2 𝑇𝑇0
𝑄𝑄4𝑄𝑄3 𝑄𝑄6𝑇𝑇2+=0𝑄𝑄5𝑇𝑇3 𝑇𝑇0
𝑄𝑄6 𝑇𝑇3𝑇𝑇0

1. Second law of thermodynamics

19

𝑄𝑄1 𝑄𝑄2 𝑄𝑄3 𝑄𝑄4 + 𝑄𝑄5 + 𝑄𝑄6+++=0
𝑇𝑇1 𝑇𝑇2 𝑇𝑇3𝑇𝑇0• Note that,
𝑄𝑄4 + 𝑄𝑄5 + 𝑄𝑄6 = 0• So,
𝑄𝑄1 𝑄𝑄2 𝑄𝑄3++=0
𝑇𝑇1 𝑇𝑇2 𝑇𝑇3• This result can be generalized to any Carnot cycle.
𝑄𝑄𝑖𝑖
� =0𝑇𝑇𝑖𝑖
𝑖𝑖

20

1. Second law of thermodynamics• Clausius theorem states that, in anarbitrary cycle,
𝛿𝛿𝑄𝑄�
≤0𝑇𝑇surr• Suppose reversible paths 𝑅𝑅 and 𝑅𝑅𝑅
share the initial and final states.
𝛿𝛿𝑄𝑄
𝛿𝛿𝑄𝑄
𝛿𝛿𝑄𝑄
�
=�
−�=0𝑇𝑇
𝑅𝑅 𝑇𝑇
𝑅𝑅𝑅 𝑇𝑇𝛿𝛿𝑄𝑄𝛿𝛿𝑄𝑄�=�𝑅𝑅 𝑇𝑇𝑅𝑅′ 𝑇𝑇f

𝑅𝑅i𝑅𝑅′2. Entropy• The change of entropy is defined so that in areversible process,f𝛿𝛿𝑄𝑄
∆𝑆𝑆 ≡ �i 𝑇𝑇• ∆𝑆𝑆 is independent of path.
f𝐼𝐼• Entropy is a state function.
𝛿𝛿𝑄𝑄
𝑑𝑑𝑑𝑑 ≡i
𝑇𝑇𝑅𝑅• Suppose a reversible path, 𝑅𝑅, and an irreversible
path, 𝐼𝐼, share the initial and final states.

21

22

2. Entropy
You cannot reverse the calculation because
any quantity related to 𝐼𝐼 is irreversible.

𝛿𝛿𝑄𝑄
𝛿𝛿𝑄𝑄
𝛿𝛿𝑄𝑄
�
<0= �−�
𝑇𝑇surr𝐼𝐼 𝑇𝑇surr
𝑅𝑅 𝑇𝑇i𝐼𝐼𝑅𝑅
𝛿𝛿𝑄𝑄
𝛿𝛿𝑄𝑄
�
= � 𝑑𝑑𝑑𝑑 > �
𝑅𝑅 𝑇𝑇𝑅𝑅𝐼𝐼 𝑇𝑇surr• It is still true when the initial and final states areclose to each other.
• Therefore, in an irreversible process,𝛿𝛿𝑄𝑄
𝑑𝑑𝑑𝑑 >𝑇𝑇surrf

23

2. Entropy• In an irreversible adiabatic process,𝑑𝑑𝑑𝑑 > 0• It leads to the Plank’s statement of
the second law of thermodynamics:
the entropy of an isolated systemtends to a maximum.
• For example, heat flow between 2bodies at 𝑇𝑇1 and 𝑇𝑇2 (𝑇𝑇1 > 𝑇𝑇2 ). 𝑑𝑑𝑑𝑑 ofthe (isolated) overall system is𝛿𝛿𝑄𝑄 𝛿𝛿𝑄𝑄
𝑑𝑑𝑆𝑆1 + 𝑑𝑑𝑆𝑆2 = −+>0
𝑇𝑇1𝑇𝑇2

𝑑𝑑𝑑𝑑 > 0𝑇𝑇1𝑄𝑄𝑇𝑇22. Entropy24• If a system is always in internal thermodynamicequilibrium and is in thermodynamic equilibriumwith its surroundings,
1. The system and the surroundings have the same
uniform temperature (Clausius’ statement & 0th lawof thermodynamics).
2. The heat engine has the highest possible efficiency(Carnot’s statement).
3. …(Kelvin’s statement).
4. The entropy of isolated system is maximized, sothat 𝑑𝑑𝑑𝑑 = 0 (Planck’s statement).

2. Entropy

25• The first law of thermodynamics for reversible
process with only PV work can be revised as
𝑑𝑑𝑑𝑑 = 𝑇𝑇𝑇𝑇𝑇𝑇 − 𝑃𝑃𝑃𝑃𝑃𝑃• 𝑆𝑆 and 𝑉𝑉 are the natural variables of 𝑈𝑈.
• With the natural variables of some type of energy,
the thermodynamic properties of a system can be
determined by just one function. This function iscalled the characteristic function of that energy.
• For instance, the exact differential of 𝑈𝑈 𝑆𝑆, 𝑉𝑉 is𝜕𝜕𝑈𝑈𝜕𝜕𝜕𝜕
𝑑𝑑𝑑𝑑 =𝑑𝑑𝑑𝑑 +𝑑𝑑𝑑𝑑
𝜕𝜕𝑆𝑆 𝑉𝑉
𝜕𝜕𝑉𝑉 𝑆𝑆26

2. Entropy• Therefore,

𝜕𝜕𝜕𝜕𝑇𝑇 =𝜕𝜕𝜕𝜕 𝑉𝑉

𝜕𝜕𝜕𝜕
𝑃𝑃 = −𝜕𝜕𝜕𝜕 𝑆𝑆• 𝑑𝑑𝑑𝑑 can be expressed in such a way.
1𝑃𝑃𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 + 𝑑𝑑𝑑𝑑
𝑇𝑇𝑇𝑇
• So,1
𝜕𝜕𝑆𝑆=𝑇𝑇𝜕𝜕𝑈𝑈 𝑉𝑉𝑃𝑃𝜕𝜕𝑆𝑆=𝑇𝑇
𝜕𝜕𝑉𝑉 𝑈𝑈

2. Entropy

27• For example, ∆𝑆𝑆 in an isochoric process of ideal
gas isff
1
𝐶𝐶𝑉𝑉 𝑑𝑑𝑑𝑑
𝑇𝑇f= 𝐶𝐶𝑉𝑉 ln∆𝑆𝑆 = � 𝑑𝑑𝑑𝑑 = �𝑇𝑇i
𝑇𝑇i 𝑇𝑇i
 Entropy increases with temperature at constantvolume.
• ∆𝑆𝑆 in an isothermal process (𝑑𝑑𝑑𝑑 = 0) of ideal gas is,
fff
𝑃𝑃𝑃𝑃𝑃𝑃𝑑𝑑𝑑𝑑
𝑉𝑉f= � 𝑛𝑛𝑛𝑛= 𝑛𝑛𝑛𝑛 ln∆𝑆𝑆 = � 𝑑𝑑𝑑𝑑 = �𝑉𝑉i𝑇𝑇𝑉𝑉iii
 Entropy increases with volume at constanttemperature.

28

2. Entropy• According to the symmetry of mixed partialderivatives,𝜕𝜕 𝜕𝜕𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕=𝜕𝜕𝑉𝑉 𝜕𝜕𝜕𝜕 𝑉𝑉
𝜕𝜕𝑆𝑆 𝜕𝜕𝑉𝑉 𝑆𝑆• Thus,𝑆𝑆𝜕𝜕𝑇𝑇𝜕𝜕𝑃𝑃=−
𝜕𝜕𝑉𝑉 𝑆𝑆𝜕𝜕𝑆𝑆 𝑉𝑉𝑉𝑉• This is one of the equations in Maxwell’s relations.

2. Entropy

29• For reversible process, 𝑑𝑑𝐻𝐻 can be expressed as
𝑑𝑑𝐻𝐻 = 𝑇𝑇𝑇𝑇𝑇𝑇 + 𝑉𝑉𝑉𝑉𝑉𝑉• Therefore, 𝑆𝑆 and 𝑃𝑃 are the natural variables of 𝐻𝐻.
• 𝐻𝐻 𝑆𝑆, 𝑃𝑃 is a characteristic function.
𝜕𝜕𝐻𝐻
𝜕𝜕𝐻𝐻𝑇𝑇 =𝑉𝑉 =𝜕𝜕𝜕𝜕 𝑃𝑃
𝜕𝜕𝑃𝑃 𝑆𝑆• They leads to another one of Maxwell’s relations.
𝜕𝜕𝑇𝑇
𝜕𝜕𝑉𝑉=
𝜕𝜕𝑃𝑃 𝑆𝑆𝜕𝜕𝑆𝑆 𝑃𝑃2. Entropy• In isochoric process,
𝑑𝑑𝑈𝑈 = 𝑇𝑇𝑇𝑇𝑇𝑇 = 𝐶𝐶𝑉𝑉 𝑑𝑑𝑑𝑑
• Thus,𝜕𝜕𝜕𝜕
𝐶𝐶𝑉𝑉 = 𝑇𝑇𝜕𝜕𝜕𝜕 𝑉𝑉• In isobaric process,𝑑𝑑𝐻𝐻 = 𝑇𝑇𝑇𝑇𝑇𝑇 = 𝐶𝐶𝑃𝑃 𝑑𝑑𝑑𝑑
• Thus,𝜕𝜕𝜕𝜕𝐶𝐶𝑃𝑃 = 𝑇𝑇𝜕𝜕𝜕𝜕 𝑃𝑃30

31

2. Entropy• Heat of a reversible
process is the area
below the curve of
S-T plotff� 𝛿𝛿𝑄𝑄 = � 𝑇𝑇𝑑𝑑𝑑𝑑
ii• Heat of a reversible
cycle is the area ofthe cycle in S-T plot.
𝑄𝑄 = � 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇1𝑇𝑇2

A

B

Carnot cycle

D

𝑆𝑆2B

𝑆𝑆2𝑆𝑆