Physical Chemistry
22. Third Law of Thermodynamics

Ruifeng ZHOU

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

Table of Content

1. Third law of thermodynamics
2. Absolute entropy2

1. Third law of thermodynamics• The technique to cool matters to extreme lowtemperature is magnetic refrigeration.
• The first law of thermodynamics of magnetic
material can be expressed as (𝑑𝑑𝑉𝑉 = 0)
𝑑𝑑𝑑𝑑 = 𝑇𝑇𝑇𝑇𝑇𝑇 + 𝑉𝑉ℋ𝑑𝑑ℬ• 𝑉𝑉ℋ𝑑𝑑ℬ is the infinitesimal work.
• ℋ is magnetic field strength.
• ℬ is magnetic flux density.
𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕
𝑇𝑇 =𝑉𝑉ℋ =𝜕𝜕𝜕𝜕 ℬ𝜕𝜕𝜕 𝑆𝑆3

1. Third law of thermodynamics• Use the symmetry of mixed
second order derivatives.
𝜕𝜕𝜕𝜕𝜕𝜕ℋ𝜕𝜕𝜕𝜕𝜕𝑇𝑇= 𝑉𝑉= 𝑉𝑉
𝜕𝜕𝜕 𝑆𝑆𝜕𝜕𝑆𝑆 ℬ
𝜕𝜕𝑇𝑇 ℬ 𝜕𝜕𝜕𝜕 ℬ• Electromagnetics tells us,
ℬ
ℋ=−ℳ
𝜇𝜇0
• 𝜇𝜇0 is vacuum permeability.
• ℳ is magnetization (magneticmoment per unit volume).

41. Third law of thermodynamics• For paramagnetic anddiamagnetic materials,ℳ = 𝜒𝜒𝜒• 𝜒𝜒 is magnetic susceptibility.
• Paramagnetism: 𝜒𝜒 > 0.
• Diamagnetism: 𝜒𝜒 < 0.
• Curie’s law states that, forparamagnetic material,𝒞𝒞𝜒𝜒 =𝑇𝑇• 𝒞𝒞 > 0 is Curie constant.

5

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1. Third law of thermodynamicsℋ=• Meanwhile,• Thus,ℬ𝒞𝒞
𝜇𝜇0 1 +𝑇𝑇𝜕𝜕𝜕>0𝜕𝜕𝑇𝑇 ℬ𝜕𝜕𝑇𝑇>0𝜕𝜕𝜕𝜕 ℬ𝜕𝜕𝜕𝜕>0𝜕𝜕𝜕 𝑆𝑆

1. Third law of thermodynamics• In adiabatic
magnetization, 𝑇𝑇rises.
• In isomagnetic
enthalpic
transfer, 𝑄𝑄 < 0, 𝑇𝑇returns.
• In adiabatic
demagnetization,
𝑇𝑇 lowers.

Magnetic entropy decreases.
Heat entropy increases.

78

1. Third law of thermodynamics• Magnetic
refrigeration
resembles
mechanicalrefrigeration (e.g.,vapour cyclerefrigeration usedin home fridge
and airconditioner)

T+HT
Adiabatic
process+PT+ Tad

T+ TadIsomagnetic
process-QT

T- TadIsomagnetic
process+QT

Magnetic refrigeration

Isochoric
process

T
Adiabatic
process

H=0-Q-PT- Tad+Q

Isochoric
process

TVapor cycle refrigeration

1. Third law of thermodynamics

9• The third law of thermodynamics.
• Nernst-Simon statement: the entropy change
associated with any condensed system undergoinga reversible isothermal process approaches 0 as
the temperature at which it is performed
approaches 0K.
𝜕𝜕𝜕𝜕lim=0
𝑇𝑇→0 𝜕𝜕𝑋𝑋
𝑇𝑇• 𝑋𝑋 is an arbitrary state function, e.g., 𝑃𝑃 and ℬ.
 At absolute zero, the entropy of a system is aconstant.

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1. Third law of thermodynamics• It is impossible to cool to 𝑇𝑇 = 0 in a finite number ofsteps.
Isothermalmagnetization
Adiabaticdemagnetization• Planck’s statement: the entropy of a purecrystalline substance is 0 at the absolute zero oftemperature.
𝑆𝑆 0K = 0

1. Third law of thermodynamics

11• The heat capacity is 0 at the absolute zero.
𝜕𝜕𝜕𝜕lim 𝐶𝐶𝑃𝑃 = lim 𝑇𝑇= 𝑑𝑑𝑑𝑑 = 0𝑇𝑇→0𝑇𝑇→0𝜕𝜕𝜕𝜕 𝑃𝑃
𝜕𝜕𝜕𝜕lim 𝐶𝐶𝑉𝑉 = lim 𝑇𝑇= 𝑑𝑑𝑑𝑑 = 0𝑇𝑇→0𝑇𝑇→0𝜕𝜕𝜕𝜕 𝑉𝑉• These can be derived using statistical mechanics.
• Thermal expansion stops at absolute zero.
𝜕𝜕𝑉𝑉𝜕𝜕𝑆𝑆= − lim=0lim
𝑇𝑇→0 𝜕𝜕𝑇𝑇
𝑇𝑇→0 𝜕𝜕𝑃𝑃
𝑃𝑃𝑇𝑇

1. Third law of thermodynamics

12• According to previous chapters, 𝑃𝑃, 𝑉𝑉, 𝑇𝑇 and 𝑆𝑆 areimportant parameters of a system.
• Unlike 𝑃𝑃, 𝑉𝑉 and 𝑇𝑇, 𝑆𝑆 cannot be measured directly.
• The change of 𝑆𝑆 in an isobaric process can bemeasured if there is no first order phase transition.
𝑇𝑇𝐶𝐶𝑃𝑃𝑑𝑑𝑑𝑑𝑆𝑆 𝑇𝑇 = 𝑆𝑆 𝑇𝑇0 + �
𝑇𝑇0 𝑇𝑇• If the reference point is at absolute zero,𝑇𝑇𝑇𝑇
𝐶𝐶𝑃𝑃
𝐶𝐶𝑃𝑃
𝑑𝑑𝑑𝑑 = �𝑑𝑑𝑑𝑑
𝑆𝑆 𝑇𝑇 = 𝑆𝑆 0K + �0K 𝑇𝑇
0K 𝑇𝑇

2. Absolute entropy

13• In a 1st order phase transition, 𝑑𝑑𝑇𝑇 = 0.
𝛿𝛿𝑄𝑄=∞
𝐶𝐶𝑃𝑃 =𝑑𝑑𝑑𝑑 𝑃𝑃∆trs 𝐻𝐻 𝐿𝐿=∆trs 𝑆𝑆 =𝑇𝑇𝑇𝑇• ∆trs 𝐻𝐻 and 𝐿𝐿 are the enthalpy and latent heat ofphase transition.
• If there are 𝑛𝑛 1st order phase transitions at 𝑇𝑇1 , ⋯ , 𝑇𝑇𝑛𝑛 ,𝑛𝑛𝑇𝑇𝑖𝑖
𝑇𝑇
𝐶𝐶𝑃𝑃
𝐿𝐿𝑖𝑖
𝐶𝐶𝑃𝑃
𝑑𝑑𝑑𝑑 +𝑑𝑑𝑑𝑑
𝑆𝑆 𝑇𝑇 = � �+�𝑇𝑇𝑖𝑖𝑇𝑇𝑖𝑖−1 𝑇𝑇
𝑇𝑇𝑛𝑛 𝑇𝑇𝑇𝑇0 = 0K

𝑖𝑖=1

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2. Absolute entropy

Molar absolute entropy (JK-1mol-1)• The entropy
measured andcalculated in thisway is theabsolute entropy.
• The molar
absolute entropy
at standard
condition is thestandard molar
entropy (𝑆𝑆°).

200

35.61 77.36
63.15
10298.15

N2 at 1bar

39.25

150

Vaporization

72

100

11.46
11.2

Fusion50

Transition
00𝑇𝑇050

𝑇𝑇1 𝑇𝑇2 𝑇𝑇3100

150

200

Absolute temperature (K)

250

23.41
6.43
25.79
2.05300

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2. Absolute entropy• For a general reaction,� 𝜈𝜈𝑖𝑖 A𝑖𝑖 = 0𝑖𝑖• Note: the stoichiometric numbers of the reactants
are negative.
• The stoichiometric numbers of the products arepositive.
• The standard entropy of reaction is∆r 𝑆𝑆° = � 𝜈𝜈𝑖𝑖 𝑆𝑆𝑖𝑖 °𝑖𝑖

2. Absolute entropy

16• For example,
2H2 O l → 2H2 g + O2 g
• Thus,∆r 𝑆𝑆° = 2𝑆𝑆° H2 + 𝑆𝑆° O2 − 2𝑆𝑆° H2 O
= 2 × 130.68 + 205.15 − 2 × 69.95 J � K −1 � mol−1= 326.61J � K −1 � mol−1• Note: 𝑆𝑆° ≠ 0 for any substance (including elements)
at room temperature.
• ∆r 𝑆𝑆° of reactions that increase the of amount ofsubstance of gases are usually positive.

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3. Entropy
Reactants
𝑃𝑃°, 𝑇𝑇′∆𝑆𝑆1̅

Reactants
𝑃𝑃°, 𝑇𝑇∆r 𝑆𝑆° 𝑇𝑇′

Products𝑃𝑃°, 𝑇𝑇′∆r 𝑆𝑆𝑆 𝑇𝑇

Products
𝑃𝑃°, 𝑇𝑇∆r 𝑆𝑆° 𝑇𝑇= ∆r 𝑆𝑆𝑆 𝑇𝑇𝑇 + ∆𝑆𝑆1̅ + ∆𝑆𝑆2̅

∆𝑆𝑆2̅

𝐶𝐶𝑃𝑃̅ A
∆𝑆𝑆1̅ = �𝑑𝑑𝑑𝑑′′ 𝐶𝐶𝑃𝑃̅ A = − � 𝜈𝜈𝑖𝑖 𝐶𝐶𝑃𝑃̅ A𝑖𝑖
𝑇𝑇𝑇𝑇
𝑇𝑇𝑖𝑖
𝑇𝑇 ̅
𝐶𝐶𝑃𝑃 B∆𝑆𝑆2̅ = �𝑑𝑑𝑑𝑑′′ 𝐶𝐶𝑃𝑃̅ B = � 𝜈𝜈𝑗𝑗 𝐶𝐶𝑃𝑃̅ B𝑗𝑗
𝑇𝑇′ 𝑇𝑇𝑇𝑇𝑇𝑇′𝑗𝑗

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3. Entropy

𝐶𝐶𝑃𝑃̅ B − 𝐶𝐶𝑃𝑃̅ A∆r 𝑆𝑆° 𝑇𝑇 = ∆r 𝑆𝑆° 𝑇𝑇𝑇 + �𝑑𝑑𝑑𝑑′′ =𝑇𝑇𝑇𝑇𝑇𝑇′𝑇𝑇∆r 𝐶𝐶𝑃𝑃
𝑑𝑑𝑑𝑑′′= ∆r 𝑆𝑆° 𝑇𝑇𝑇 + �𝑇𝑇𝑇 𝑇𝑇𝑇𝑇
• ∆r 𝐶𝐶𝑃𝑃 is a constant in a small range of temperature.
𝑇𝑇1𝑇𝑇∆r 𝑆𝑆° 𝑇𝑇 − ∆r 𝑆𝑆° 𝑇𝑇𝑇 = ∆r 𝐶𝐶𝑃𝑃̅ �𝑑𝑑𝑑𝑑′′ = ∆r 𝐶𝐶𝑃𝑃̅ ln
𝑇𝑇𝑇
𝑇𝑇𝑇 𝑇𝑇′′• Generally speaking, ∆r 𝐶𝐶𝑃𝑃̅ ≠ 0.
• ∆r 𝑆𝑆° is temperature dependent.
𝑇𝑇193. Entropy• For example, calculate ∆r 𝑆𝑆𝑆 at 373K.
2H2 O l → 2H2 g + O2 g
• Given that, ∆r 𝑆𝑆° = 326.61J � mol−1 � K −1 at 298K.
𝐶𝐶𝑃𝑃̅ l H2 O = 75.35J � K −1 � mol−1g̅
𝐶𝐶𝑃𝑃 H2 = 28.84J � K −1 � mol−1g̅
𝐶𝐶𝑃𝑃 O2 = 29.4J � K −1 � mol−1• Solution:∆r 𝐶𝐶𝑃𝑃 = −63.62J � K −1 � mol−1373K∆r 𝑆𝑆° 373K = ∆r 𝑆𝑆° 298K + ∆r 𝐶𝐶𝑃𝑃 ln298K= 312.33J � K −1 � mol−1