Physical Chemistry
24. Phase Equilibria

Ruifeng ZHOU

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

Table of Content

1. Chemical potential
2. Phase transition

23

1. Chemical potential• Consider a one-component open
system whose amount of substance,𝑛𝑛, is not fixed.
• The system has 3 degrees offreedom.
• 𝐺𝐺 is the characteristic function of 𝑇𝑇, 𝑃𝑃and 𝑛𝑛.
𝑑𝑑𝑑𝑑𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕=
𝑑𝑑𝑑𝑑 +
𝑑𝑑𝑑𝑑 +𝑑𝑑𝑛𝑛
𝜕𝜕𝜕𝜕 𝑃𝑃,𝑛𝑛𝜕𝜕𝜕𝜕 𝑇𝑇,𝑛𝑛
𝜕𝜕𝑛𝑛 𝑃𝑃,𝑇𝑇𝑇𝑇, 𝑃𝑃, 𝑛𝑛

4

1. Chemical potential• When 𝑑𝑑𝑑𝑑 = 0, the system is closed. The partial
derivatives with 𝑛𝑛 held constant are 𝑆𝑆 and 𝑉𝑉.
𝜕𝜕𝜕𝜕𝑆𝑆 = −𝜕𝜕𝜕𝜕 𝑃𝑃,𝑛𝑛𝜕𝜕𝜕𝜕𝑉𝑉 =𝜕𝜕𝜕𝜕 𝑇𝑇,𝑛𝑛• The chemical potential is defined as
𝜕𝜕𝜕𝜕
𝜇𝜇 ≡
𝜕𝜕𝑛𝑛 𝑃𝑃,𝑇𝑇𝑑𝑑𝑑𝑑 = −𝑆𝑆𝑆𝑆𝑇𝑇 + 𝑉𝑉𝑑𝑑𝑑𝑑 + 𝜇𝜇𝑑𝑑𝑛𝑛

51. Chemical potential• Since• So

𝐻𝐻 = 𝐺𝐺 + 𝑆𝑆𝑆𝑆

𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 + 𝑆𝑆𝑆𝑆𝑆𝑆 + 𝑇𝑇𝑇𝑇𝑇𝑇 = 𝑇𝑇𝑇𝑇𝑇𝑇 + 𝑉𝑉𝑉𝑉𝑉𝑉 + 𝜇𝜇𝑑𝑑𝑑𝑑• Therefore, another expression of 𝜇𝜇 is
𝜕𝜕𝐻𝐻𝜇𝜇 =𝜕𝜕𝜕𝜕 𝑆𝑆,𝑃𝑃• Similarly,𝜕𝜕𝜕𝜕𝜇𝜇 =𝜕𝜕𝜕𝜕 𝑆𝑆,𝑉𝑉

𝜕𝜕𝜕𝜕𝜇𝜇 =𝜕𝜕𝜕𝜕 𝑇𝑇,𝑉𝑉1. Chemical potential• 𝐺𝐺 is an extensive quantity.
𝜆𝜆𝐺𝐺 𝑇𝑇, 𝑃𝑃, 𝑛𝑛 = 𝐺𝐺 𝑇𝑇, 𝑃𝑃, 𝜆𝜆𝑛𝑛
• Differentiate the equation wrt 𝜆𝜆.
𝜕𝜕𝜕𝜕𝐺𝐺 =𝑛𝑛 = 𝜇𝜇𝑛𝑛𝜕𝜕𝜕𝜕 𝑇𝑇,𝑃𝑃• So,𝜇𝜇 = 𝐺𝐺̅� 𝜇𝜇 ≠ 𝐻𝐻
� and 𝜇𝜇 ≠ 𝐴𝐴.̅• Note! 𝜇𝜇 ≠ 𝑈𝑈,

6

1. Chemical potential

7• The 𝜇𝜇 of the standard state is the standard
chemical potential, 𝜇𝜇°.
𝐺𝐺𝐺
𝜇𝜇𝜇 =𝑛𝑛• The standard state of an ideal gas is the gas whose
partial pressure is standard pressure, 𝑃𝑃°.
𝑃𝑃𝐺𝐺 𝐺𝐺𝐺 + 𝑛𝑛𝑛𝑛𝑇𝑇 ln 𝑃𝑃°
𝑃𝑃= 𝜇𝜇𝜇 + 𝑅𝑅𝑅𝑅 ln𝜇𝜇 = =𝑛𝑛𝑛𝑛𝑃𝑃°• For real gas,𝑓𝑓
𝑃𝑃
𝜇𝜇 = 𝜇𝜇𝜇 + 𝑅𝑅𝑅𝑅 ln = 𝜇𝜇𝜇 + 𝑅𝑅𝑅𝑅 ln + 𝑅𝑅𝑅𝑅 ln 𝛾𝛾𝑃𝑃°𝑃𝑃°

1. Chemical potential

8• Consider 2 phases of a substance in equilibrium.
• Each phase is an open system.
𝑑𝑑𝑈𝑈 = 𝑇𝑇𝑇𝑇𝑇𝑇 − 𝑃𝑃𝑃𝑃𝑃𝑃 + 𝜇𝜇𝑑𝑑𝑑𝑑
A• 𝑑𝑑𝑆𝑆 in the phases are
BA
A
A
A
A
𝑑𝑑𝑈𝑈 + 𝑃𝑃 𝑑𝑑𝑉𝑉 − 𝜇𝜇 𝑑𝑑𝑛𝑛
A𝑑𝑑𝑆𝑆 =𝑇𝑇 A
B + 𝑃𝑃 B 𝑑𝑑𝑉𝑉 B − 𝜇𝜇 B 𝑑𝑑𝑛𝑛B𝑑𝑑𝑈𝑈
𝑑𝑑𝑆𝑆 B =𝑇𝑇 B• The overall 𝑑𝑑𝑆𝑆 is𝑑𝑑𝑈𝑈 A 𝑑𝑑𝑈𝑈 B 𝑃𝑃A 𝑑𝑑𝑉𝑉 A 𝑃𝑃B 𝑑𝑑𝑉𝑉 B 𝜇𝜇A 𝑑𝑑𝑛𝑛A 𝜇𝜇B 𝑑𝑑𝑛𝑛B𝑑𝑑𝑑𝑑 = A + B ++−−A
B
A𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 B

1. Chemical potential

9• If the overall system is isolated,𝑑𝑑𝑈𝑈 A + 𝑑𝑑𝑈𝑈 B = 0𝑑𝑑𝑉𝑉 A + 𝑑𝑑𝑉𝑉 B = 0𝑑𝑑𝑛𝑛A + 𝑑𝑑𝑛𝑛B = 0𝑑𝑑𝑑𝑑
A
B
AB
11𝑃𝑃
𝑃𝑃𝜇𝜇𝜇𝜇A
A
A=−𝑑𝑑𝑈𝑈+−𝑑𝑑𝑉𝑉−−𝑑𝑑𝑛𝑛
𝑇𝑇 A 𝑇𝑇 B
𝑇𝑇 A 𝑇𝑇 B
𝑇𝑇 A 𝑇𝑇 B• If the overall system is in equilibrium, 𝑑𝑑𝑑𝑑 = 0.
• Note, 𝑑𝑑𝑈𝑈 A , 𝑑𝑑𝑉𝑉 A and 𝑑𝑑 𝑛𝑛A are independent.
𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
=0=0=0AAA𝜕𝜕𝑈𝑈 𝑉𝑉 A ,𝑛𝑛A
𝜕𝜕𝑉𝑉 𝑈𝑈 A ,𝑛𝑛A
𝜕𝜕𝑛𝑛 𝑉𝑉 A ,𝑈𝑈 A

1. Chemical potential• Thus,10

𝑇𝑇 A = 𝑇𝑇 B
• Thermal equilibrium: the phases in equilibrium haveidentical temperature.
𝑃𝑃A = 𝑃𝑃B
• Force equilibrium: the phases in equilibrium haveidentical pressure.
𝜇𝜇A = 𝜇𝜇B
• Phase equilibrium: the phases in equilibrium haveidentical chemical potential.

1. Chemical potential

11• Suppose the phases are in force equilibrium andphase equilibrium but not thermal equilibrium.
𝑃𝑃A = 𝑃𝑃B = 𝑃𝑃
𝜇𝜇A = 𝜇𝜇B = 𝜇𝜇𝑑𝑑𝑑𝑑
11𝑃𝑃𝑃𝑃𝜇𝜇𝜇𝜇A
AA−𝑑𝑑𝑈𝑈+−𝑑𝑑𝑉𝑉−−𝑑𝑑𝑛𝑛=𝑇𝑇 A 𝑇𝑇 B
𝑇𝑇 A 𝑇𝑇 B
𝑇𝑇 A 𝑇𝑇 B
11A + 𝑃𝑃𝑃𝑃𝑉𝑉 A − 𝜇𝜇𝑑𝑑𝑛𝑛A=−𝑑𝑑𝑈𝑈
𝑇𝑇 A 𝑇𝑇 B• Note that𝑑𝑑𝑑𝑑 = 𝛿𝛿𝑄𝑄 − 𝑃𝑃𝑃𝑃𝑉𝑉 + 𝜇𝜇𝑑𝑑𝑛𝑛

121. Chemical potential

11A
𝑑𝑑𝑑𝑑 =−𝛿𝛿𝑄𝑄
𝑇𝑇 A 𝑇𝑇 B• According to the Planck’s statement of the second
law of thermodynamics, 𝑑𝑑𝑑𝑑 > 0.
• If 𝑇𝑇 A > 𝑇𝑇 B , 𝛿𝛿𝑄𝑄 A < 0.
Hot𝑄𝑄Cold Heat flows from hot phase to cold phase.
• It is equivalent to the Clausius’s statement of thesecond law of thermodynamics.

1. Chemical potential

13• Suppose the system is in thermal equilibrium andphase equilibrium but not force equilibrium.
𝑇𝑇 A = 𝑇𝑇 B = 𝑇𝑇
𝜇𝜇A = 𝜇𝜇B = 𝜇𝜇AB
1 1𝑃𝑃𝑃𝑃𝜇𝜇 𝜇𝜇
A
A−𝑑𝑑𝑉𝑉 −−𝑑𝑑𝑈𝑈 +− 𝑑𝑑𝑛𝑛A
𝑑𝑑𝑑𝑑 =𝑇𝑇𝑇𝑇
𝑇𝑇 𝑇𝑇
𝑇𝑇 𝑇𝑇
𝑃𝑃A − 𝑃𝑃B
𝑑𝑑𝑉𝑉 A > 0=𝑇𝑇High 𝑃𝑃Low 𝑃𝑃• If 𝑃𝑃A > 𝑃𝑃B , 𝑑𝑑𝑉𝑉 A > 0 The phase with higher 𝑃𝑃 expands (Newton’s laws).

1. Chemical potential

14• Suppose the system is in thermal equilibrium andforce equilibrium but not phase equilibrium.
𝑇𝑇 A = 𝑇𝑇 B = 𝑇𝑇
𝑃𝑃A = 𝑃𝑃B = 𝑃𝑃A
B
1 1𝑃𝑃
𝑃𝑃𝜇𝜇𝜇𝜇−𝑑𝑑𝑛𝑛A−
𝑑𝑑𝑈𝑈 A +−𝑑𝑑𝑉𝑉 A −𝑑𝑑𝑑𝑑 =𝑇𝑇𝑇𝑇
𝑇𝑇 𝑇𝑇
𝑇𝑇 𝑇𝑇
𝜇𝜇B − 𝜇𝜇A A
𝑑𝑑𝑛𝑛 > 0=𝑇𝑇High 𝜇𝜇Low 𝜇𝜇• If 𝜇𝜇A > 𝜇𝜇B , 𝑑𝑑𝑛𝑛A < 0.
 Matter transits from high 𝜇𝜇 phase to lower 𝜇𝜇 phase.

15

2. Phase transition• A molecular point of view of phases.
𝐺𝐺 = 𝐻𝐻 − 𝑆𝑆𝑆𝑆
• The attraction between particles tends to packthem into smallest regular pattern (e.g., crystal) tominimize its enthalpy.
�Low 𝐻𝐻

�
Medium 𝐻𝐻

�High 𝐻𝐻

16

2. Phase transition• However, as the particles gain thermal energy(vibration, rotation and translation), they tend tomove more freely.
• 𝑆𝑆𝑆𝑆 increases with 𝑇𝑇 and overcomes 𝐻𝐻 at somepoint.
Low 𝑆𝑆High 𝑆𝑆

17

2. Phase transition• As one state function changes, theother state functions change with it.
• In most cases, the state functions 𝐵𝐵change continuously with eachother (like “reform”).
• However, in some conditions, statefunctions have discontinuous
relationships (like “revolution”),
where the phase transitions occur.
• The conditions of phase transitions
depend on the particles’ interaction.

Phasetransition𝐴𝐴2. Phase transition18• Ehrenfest classification: if the 𝑛𝑛th order derivative of𝜇𝜇 wrt a state function (e.g., 𝑇𝑇, 𝑃𝑃) is discontinuous,
the transition is a 𝑛𝑛th order phase transition.
• In isobaric process,𝜕𝜕𝜇𝜇= −𝑆𝑆̅𝜕𝜕𝜕𝜕 𝑃𝑃• In 1st order phase transition, the 1st order derivativeof 𝜇𝜇 wrt 𝑇𝑇, 𝑆𝑆, is discontinuous.
𝜕𝜕𝑆𝑆=∞
𝜕𝜕𝜕𝜕 𝑃𝑃19

2. Phase transition𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕= 𝐶𝐶𝑃𝑃 = 𝑇𝑇=∞𝜕𝜕𝜕𝜕 𝑃𝑃𝜕𝜕𝜕𝜕 𝑃𝑃𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕
𝜕𝜕𝑆𝑆==∞
𝜕𝜕𝜕𝜕 𝑃𝑃𝜕𝜕𝑆𝑆 𝑃𝑃 𝜕𝜕𝜕𝜕 𝑃𝑃• In 1st order phase transition, 𝜇𝜇 is continuous; 𝑆𝑆, 𝑉𝑉and 𝐻𝐻 are discontinuous; 𝐶𝐶𝑃𝑃 is infinity.
μV

TC

TH

TC

TCP

STC

TTC

T∞TCT

20

2. Phase transition

𝜕𝜕 2 𝜇𝜇
𝜕𝜕𝜕𝜕
𝐶𝐶𝑃𝑃
==2𝑇𝑇
𝜕𝜕𝑇𝑇 𝑃𝑃𝜕𝜕𝜕𝜕 𝑃𝑃• In 2nd order phase transition, 𝜇𝜇, 𝑆𝑆, 𝑉𝑉 and 𝐻𝐻 arecontinuous; 𝐶𝐶𝑃𝑃 is discontinuous.
• 2nd and higher order phase transitions are also
called continuous phase transitions.
μVTCTHTCTCP

STC

TTC

TTCT

21

2. Phase transition• Example of 2nd order ferro/para magnetic transition.
• 𝑇𝑇C is Curie temperature.

FerroPara22

2. Phase transition• This chapter focuses on 1st order phase transitions
between solid, liquid and gas phases, andcontinuous phase transition involving supercriticalfluid.
SupercriticalfluidGas

Solid

Melting (fusion)Freezing

Liquid23

2. Phase transition• 𝜇𝜇 − 𝑇𝑇 and 𝜇𝜇 − 𝑃𝑃 plots
are helpful for theunderstanding ofphase transitions.
𝜕𝜕𝜇𝜇= −𝑆𝑆̅𝜕𝜕𝜕𝜕 𝑃𝑃
𝜕𝜕𝜇𝜇= 𝑉𝑉�𝜕𝜕𝑃𝑃 𝑇𝑇𝜇𝜇SolidLiquidGas𝜇𝜇𝑇𝑇

Solid
Liquid

Gas

𝑃𝑃

24

2. Phase transition
CO210,000

P (bar)• A phase diagram
is used to show
the existence to 1,000
phases in certain
conditions.
100• 𝑇𝑇 and 𝑃𝑃 areusually used as
10horizontal andvertical axis for
one component
1phase diagram.
200

CO2

Solid

Supercriticalfluid
LiquidGas
Triple point: 3 phases
in equilibrium.
250

300

T (K)

350

400

25

2. Phase transition
Solid
Supercriticalfluid
LiquidP• The gas that canbe condensed ordeposited byincreasing pressure
is called vapour.
• 𝑃𝑃 at which thevapour and thecondensed phaseare in equilibrium isthe vapour
pressure (𝑃𝑃∗ ).

Critical point: no 1storder transition
above 𝑇𝑇𝑐𝑐 .

𝑃𝑃∗GasVapourT2. Phase transition• For example, 𝑃𝑃∗ of solid and liquid chlorine aregiven by simplified empirical Antoine equation.
𝑃𝑃s3777K
ln= 24.320 −
torr
𝑇𝑇𝑃𝑃l
2669K
ln
= 17.892 −torr𝑇𝑇• Find the 𝑇𝑇 and 𝑃𝑃 at the triple point.

• Solution: at triple point, 𝑃𝑃s = 𝑃𝑃l .
3777K
2669K
= 17.892 −24.320 −𝑇𝑇𝑇𝑇
𝑇𝑇 = 172K
𝑃𝑃 = 11torr

26

27

2. Phase transition
172K

10000

LiquidPressure (torr)1000

100Solid

10Vapour

10.1
14011torr

Triple point

160

180

200

Temperature (K)

220

240

2. Phase transition

28• At one point on the line of equilibrium, 𝜇𝜇 of 2
phases are identical.
𝜇𝜇A = 𝜇𝜇B
• At another point on the line of equilibrium nearby,
𝜇𝜇A + 𝑑𝑑𝜇𝜇A = 𝜇𝜇B + 𝑑𝑑𝜇𝜇B
𝑑𝑑𝑑𝑑• Therefore,
𝑑𝑑𝑑𝑑AB
𝑑𝑑𝜇𝜇 = 𝑑𝑑𝜇𝜇
𝑉𝑉� A 𝑑𝑑𝑑𝑑 − 𝑆𝑆̅ A 𝑑𝑑𝑑𝑑 = 𝑉𝑉� B 𝑑𝑑𝑑𝑑 − 𝑆𝑆̅ B 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 𝑆𝑆̅ B − 𝑆𝑆̅ A ∆trs 𝑆𝑆= B=A𝑑𝑑𝑑𝑑 𝑉𝑉� − 𝑉𝑉�∆trs 𝑉𝑉• This is the Clausius-Clapeyron relation.

2. Phase transition29• ∆trs 𝑉𝑉 and ∆trs 𝑆𝑆 are the 𝑉𝑉� and 𝑆𝑆̅ changes during thephase transition.
• During a 1st order phase transition at constant 𝑃𝑃,∆trs 𝐻𝐻 𝐿𝐿�
∆trs 𝑆𝑆 ==𝑇𝑇𝑇𝑇� during phase transition.
• ∆trs 𝐻𝐻 is the change of 𝐻𝐻• 𝐿𝐿� is the molar latent heat of phase transition.
• The Clausius-Clapeyron relation can be written as
𝑑𝑑𝑑𝑑 ∆trs 𝐻𝐻 1𝐿𝐿�==𝑑𝑑𝑑𝑑𝑇𝑇 ∆trs 𝑉𝑉 𝑇𝑇∆trs 𝑉𝑉

2. Phase transition

30• Melting point is the temperature when solid andliquid phases are in equilibrium at given pressure.
• For example, the densities of solid and liquid acetic
acid (𝑀𝑀 = 60.05g � mol−1 ) are 𝜌𝜌s = 1.27g � cm−3 and𝜌𝜌l = 1.05g � cm−3 , respectively.
• At 1bar, its melting point is 16.7℃. Δfus 𝐻𝐻 = 11.7kJ �mol−1 . Find its melting point at 100bar.
𝑀𝑀 𝑀𝑀Δfus 𝑉𝑉 = l − s = 9.91 × 10−6 m3 � mol−1𝜌𝜌 𝜌𝜌𝑃𝑃f𝑑𝑑𝑑𝑑Δfus 𝐻𝐻 𝑇𝑇f 𝑑𝑑𝑑𝑑
��
𝑑𝑑𝑑𝑑 =Δfus 𝑉𝑉 𝑇𝑇i 𝑇𝑇𝑃𝑃i 𝑑𝑑𝑑𝑑

312. Phase transition

𝑃𝑃f − 𝑃𝑃i = 9.9 × 106 Pa ≈ 1.18 × 109 Pa ln

𝑇𝑇f
289.85𝐾𝐾

𝑇𝑇f = 292.29K = 19.14℃• For most substances, volume increases during
endothermic phase transitions (𝐿𝐿 > 0).
𝑑𝑑𝑑𝑑>0𝑑𝑑𝑑𝑑• However, during the melting of water, the volume
decreases (19.6ml � mol−1 vs. 18.0ml � mol−1 ).
𝑑𝑑𝑑𝑑<0𝑑𝑑𝑑𝑑32

2. Phase transition
0K
1 TPa

100 K

200 K

Temperature

300 K

400 K

500 K

600 K

700 K
1 MbarPressure

1 GPa
Critical point
647 K, 22.064 MPa

Solid

1 kbarSupercriticalfluid

Liquid1 MPaFreezing point at 1 atm
273.15 K

1 PaGasVapor

1 kPa

1 bar

Boiling point at 1 atm
373.15 K
Triple point273.16 K, 611.657 Pa-200 °C -100 °C0 °C

100 °C

200 °C

1 mbar

300 °C

400 °C2. Phase transition• During vaporization, ∆vap 𝑉𝑉 is approximately 𝑉𝑉 ofthe vapour (𝑉𝑉 of condensed phase is negligible).
• Suppose the vapour is an ideal gas,
𝑅𝑅𝑅𝑅
∆vap 𝑉𝑉 ≈ ∗𝑃𝑃∗
𝑑𝑑𝑃𝑃∗ 𝑃𝑃 ∆vap 𝐻𝐻=𝑅𝑅𝑇𝑇 2𝑑𝑑𝑑𝑑• This is the Clausius-Clapeyron equation.
• If ∆vap 𝐻𝐻 is independent of 𝑇𝑇,
𝑃𝑃f∗ ∆vap 𝐻𝐻 1 1ln ∗ =−𝑃𝑃i𝑅𝑅
𝑇𝑇i 𝑇𝑇f33

2. Phase transition

34• Boiling point is the temperature when 𝑃𝑃∗ is equal tothe surrounding pressure.
• For example, at 4000m altitude above sea level,
𝑃𝑃 = 61.6kPa. Find the boiling point of water.
∆vap 𝐻𝐻 = 40.657kJ � mol−1
61.6kPa
40.657kJ � mol−111ln=−−1−1
101.325kPa 8.314J � mol � K
373.15K 𝑇𝑇
𝑇𝑇 = 356K ≈ 83℃• In a pressure cooker, 𝑃𝑃 = 200kPa. Find the boilingpoint of water.
𝑇𝑇 = 473K ≈ 120℃

2. Phase transition• If ∆vap 𝐻𝐻 depends on 𝑇𝑇,�l� g 𝑑𝑑 𝐻𝐻
�g�l𝑑𝑑∆vap 𝐻𝐻 𝑑𝑑 𝐻𝐻
𝜕𝜕𝐻𝐻
𝜕𝜕𝐻𝐻=
≈−−𝑑𝑑𝑇𝑇
𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝜕𝜕𝜕𝜕 𝑃𝑃𝜕𝜕𝜕𝜕 𝑃𝑃• Consider the following relations.
�𝜕𝜕𝐻𝐻= 𝐶𝐶𝑃𝑃̅𝜕𝜕𝜕𝜕 𝑃𝑃• So,𝑑𝑑∆vap 𝐻𝐻g̅≈ 𝐶𝐶𝑃𝑃 − 𝐶𝐶𝑃𝑃̅ l ≡ ∆vap 𝐶𝐶𝑃𝑃
𝑑𝑑𝑑𝑑

35

2. Phase transition• Suppose ∆vap 𝐶𝐶𝑃𝑃 is independent of 𝑇𝑇.
∆vap 𝐻𝐻 = 𝐻𝐻0 + ∆vap 𝐶𝐶𝑃𝑃 𝑇𝑇 − 𝑇𝑇036• Insert it back to the Clausius-Clapeyron equation.
𝑑𝑑𝑃𝑃∗ 𝐻𝐻0 + ∆vap 𝐶𝐶𝑃𝑃 𝑇𝑇 − 𝑇𝑇0 ∗=𝑃𝑃2𝑑𝑑𝑑𝑑
𝑅𝑅𝑇𝑇• Integrate from 𝑇𝑇0 to 𝑇𝑇,
𝑃𝑃∗ ∆vap 𝐶𝐶𝑃𝑃 𝑇𝑇0 − 𝐻𝐻0 ∆vap 𝐶𝐶𝑃𝑃 𝑇𝑇 ∆vap 𝐶𝐶𝑃𝑃 𝐻𝐻0+ln −+ln ∗ =𝑃𝑃0
𝑅𝑅𝑅𝑅
𝑅𝑅
𝑅𝑅
𝑅𝑅𝑇𝑇0𝑇𝑇0𝑃𝑃∗
𝐵𝐵𝑇𝑇= 𝐴𝐴 − + 𝐶𝐶 lnln
Pa
K𝑇𝑇

37

2. Phase transition• For example, thevapour pressure ofwater.
Pressure (kPa)100• At 100℃,
𝑃𝑃∗ = 101.325kPa
𝐿𝐿� vap = 40.657kJ � mol−1∆vap 𝐶𝐶𝑃𝑃= −38.75kJ � mol−1 � K −1

10Experimental
constant ∆vapH
constant ∆vapCP

1020

4060

Temperature (°C)

80100

2. Phase transition

38• The critical point is the end point of a phaseequilibrium curve defined by the criticaltemperature 𝑇𝑇c and the critical pressure 𝑃𝑃c .
• At higher temperatures than 𝑇𝑇c , the gas cannot beliquefied by increasing pressure alone. The phase
above 𝑇𝑇c and 𝑃𝑃c is called supercritical fluid.
• Gas - supercritical fluid phase transition is acontinuous phase transition.
• The phase transition and existence of critical point
can be explained qualitatively by van der Waals
equation.

39

2. Phase transition• Van der Waals equation modifies the ideal gas law
by introducing the interactions between gasmolecules.
• The space in which molecules move freely is
reduced due to their own volume.
𝑉𝑉 − 𝑛𝑛𝑛𝑛

𝑉𝑉𝑛𝑛𝑛𝑛𝑉𝑉 − 𝑛𝑛𝑛𝑛

40

2. Phase transition• Particles attract each other with Van der Waals
forces.
• The speed of the particles colliding the wall slows
down. The pressure is lowered.
• The strength of attraction depends on the type ofgas particles and the distance between them.
van der Waals forces

41

2. Phase transition• The attraction is proportional
to the concentration of gas
particles.
• The number of the gas
molecules which collide withthe wall is also proportional
to the concentration of gas
particles.
• The reduction of 𝑃𝑃 is 𝑎𝑎𝑛𝑛 2.
𝑉𝑉42

2. Phase transition• The state ismechanicallyunstable.

12285K
295K
304.19K 𝑇𝑇c305K
315K

CO210Pressure (MPa)• Van der Waals
equation is
𝑅𝑅𝑅𝑅𝑎𝑎− 2𝑃𝑃 =𝑉𝑉� − 𝑏𝑏 𝑉𝑉�
• At low 𝑇𝑇, there isan area in which
𝜕𝜕𝜕𝜕>0�𝜕𝜕𝑉𝑉 𝑇𝑇

Critical point
8𝑃𝑃c = 7.38MPa

64

Unstable
0.1

0.2

0.3

Volume (L)

0.4

0.5

43

2. Phase transition• Consider the 𝜇𝜇 change in an isothermal process.
�∆𝜇𝜇 = � 𝑉𝑉𝑑𝑑𝑑𝑑• ∆𝜇𝜇 is the area to the left of curve in P-V plot.
• Integrate it by part.
𝑅𝑅𝑅𝑅𝑎𝑎� − � 𝑃𝑃𝑃𝑃 𝑉𝑉� = ∆ 𝑉𝑉𝑃𝑃
� −�
− 2 𝑑𝑑 𝑉𝑉�
∆𝜇𝜇 = ∆ 𝑉𝑉𝑃𝑃
𝑉𝑉� − 𝑏𝑏 𝑉𝑉�
1 1
𝑉𝑉�f − 𝑏𝑏� − 𝑅𝑅𝑅𝑅 ln− 𝑎𝑎−= ∆ 𝑉𝑉𝑃𝑃
𝑉𝑉�i − 𝑏𝑏𝑉𝑉�f 𝑉𝑉�i

44

2. Phase transition• Path ABCDEF does not have the lowest 𝜇𝜇.
• Path ABEF has the lowest 𝜇𝜇. The volume changes
discontinuously from B (gas) to E (liquid), which isa first order phase transition.
FH2O, 560K200CEB

G

100

D00.0C4000

0.40.6

Molar volume (L/mol)

0.8

1.0

B,ED

3000

2000

1000

A0.2

F

5000

Chemical potential (J/mol)

Pressure (bar)

300

0A
50

100

150200Pressure (bar)250300452. Phase transition
𝐵𝐵𝐵𝐵

𝐵𝐵

�
� � − � 𝑃𝑃𝑃𝑃 𝑉𝑉� = 0𝜇𝜇𝐵𝐵 − 𝜇𝜇𝐸𝐸 = � 𝑉𝑉𝑑𝑑𝑑𝑑= 𝑉𝑉𝑃𝑃
Pressure (bar)

300

𝐸𝐸

𝐸𝐸

F

H2O, 560K

𝐸𝐸

200

supercoolingCEsuperheating

B

G

100

D
00.0

A
0.2

0.40.6

Molar volume (L/mol)

0.81.0• Maxwell construction: the area of BCG is equal toarea of GDE.

46

2. Phase transition• In region BC,

𝜕𝜕𝑃𝑃<0�𝜕𝜕𝑉𝑉 𝑇𝑇• The state is mechanically stable even if 𝜇𝜇 is notminimized. Gas remains as gas though it shouldhave condensed. It is called the supercooling gasstate. It can be achieved in experiment.
• Similarly, at region DE, the system is at thesuperheating liquid state.
• Reason: the absence of seeds around which a newstructure can form slows down phase transitions.

47

2. Phase transition• At 𝑇𝑇c , 1st order phase transition disappears.
𝜕𝜕𝑃𝑃𝑅𝑅𝑅𝑅
2𝑎𝑎=−+ 3=02��
𝜕𝜕𝑉𝑉 𝑇𝑇𝑉𝑉 − 𝑏𝑏𝑉𝑉�
𝜕𝜕 2 𝑃𝑃
2𝑅𝑅𝑅𝑅
6𝑎𝑎=− 4=023
𝜕𝜕𝑉𝑉� 𝑇𝑇𝑉𝑉� − 𝑏𝑏𝑉𝑉�

• Thus,𝑉𝑉�c = 3𝑏𝑏
8𝑎𝑎
𝑇𝑇c =27𝑏𝑏𝑅𝑅𝑎𝑎
𝑃𝑃c =27𝑏𝑏 2

2. Phase transition• Supercritical fluids have low viscosity and highdiffusivity like gas.
• At the critical point,𝜕𝜕𝑉𝑉�=∞𝜕𝜕𝜕𝜕 𝑇𝑇48• The density, as well as other properties related todensity, e.g., viscosity, diffusivity and solubility,change drastically due to a small change of 𝑃𝑃.
• The supercritical fluids have special industrialapplications.

492. Phase transition• For example: supercritical fluid extraction.

Low 𝑃𝑃,
Low solubility

Citrus peels

Essential oil (90% limonene)
High 𝑃𝑃,High solubility