Chapter 8
Phase Diagrams and the
Relative Stability of
Solids, liquids, and gases
Engel & Reid
1
Objectives
• Introduce phases of substances which is in solid, liquid and gaseous phases.
• Criteria to determine which of phases is favored at a given temperature and pressure.
• Discuss the conditions under which two or three phases of a pure substance can coexist in equilibrium. • Supercritical fluids
• Summarize P–T, P–V, and P–V–T phase diagrams.
2
Outline
8.1 What Determines the Relative Stability of the Solid, Liquid,
and Gas Phases? 決定三相的相對穩定性之要素
8.2 The Pressure-Temperature Phase Diagram P‐T 相圖
8.3 The Pressure-Volume and Pressure-Volume-Temperature
Phase Diagram P‐V 相圖 與 P‐V‐T 相圖
8.4 Providing a Theoretical Basis for the P-T Phase Diagram 提供
P-T 相圖的理論基礎
8.5 Using the Clapeyron Equation to Calculate Vapor Pressure as a
Function of T 利用 Clapeyron 方程計算蒸氣壓為溫度的函數
8.6 The Vapor Pressure of a Pure Substance Depends on the
Applied Pressure 純物質蒸氣壓為施加外壓的函數
8.7 Surface Tension 表面張力
8.8 Chemistry in Supercritical Fluids 超臨界流體的化學
8.9 Liquid Crystals and LCD Displays 液晶與液晶顯示
3
8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases?
•
•
•
Phase refers to a form of matter that is uniform with respect to chemical composition.
For pure substance, μ varies with changes in P and T can be determined by:
  
  
   Sm and   Vm
 P T
 T P
µ decrease as temperature increases and increase as pressure increase.
4
8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases?
•
As heat capacity is always positive for a solid, liquid, or gas, the entropy of the three phases follows this order:
S
gas
m
•
S
liquid
m
S
solid
m
Chemical potential in the solid, liquid and gaseous states can be seen from µ vs T plot, at constant pressure.
5
8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases?
•
The pressure dependence of the chemical potential of a substance depends on the molar volume of the phase. The lines show schematically the effect of increasing pressure on the chemical potentials of the solid and liquid phases and the corresponding effects on the freezing temperatures. (a) In this case the molar volume of the solid is less than that of the liquid and μ(s) increases less than μ(l). As a result, the freezing temperature rises.
From Atkins’ Chap 4 Physical transformations of pure substances
6
8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases?
• Here the molar volume is greater for the solid than the liquid (as for water), μ(s) increases more strongly than μ(1), and the freezing temperature is lowered. The chemical potential of ice rise more sharply than that of water; so if they are initially in equilibrium at 1 bar, there will be a tendency for the ice to melt at 2 bar.
From Atkins’ Chap 4 Physical transformations of pure substances
7
8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases?
•
For µ vs T plot at constant T, an increase in pressure leads to freezing point elevation.
V
•
liquid
m
V
solid
m
And to a freezing point depression for the case of water.
V
solid
m
V
liquid
m
8
Accessing the effect of pressure on the chemical potential Exercise : Calculate the effect on the chemical potentials of ice and water of increasing the pressure from 1.00 bar to 2.00 bar at 0 ℃. The density of ice is 0.917 g cm‐3 and that of liquid water is 0.999 g cm‐3
under these conditions. Answer:  μ 

 V m
 P T
The change in chemical potential of an incompressible substance when the pressure is changed by ∆p is ∆μ =Vm ∆p
The molar volume of two phases of water come from the mass density, ρ, and the molar mass, M, by using Vm = M/ρ. M  p 1.802x10
  ice  

A
kg mol -1  1.00x10
917 kg m-3 
-2
M  p 1.802x10


  water 

A
5
Pa 
kg mol -1  1.00x10
999 kg m-3 
-2
5
  1.97 J mol -1
Pa 
From Atkins’ Chap 4 Physical transformations of pure substances
  1.80 J mol -1
9
Exercise : Accessing the effect of an increase in pressure of 1.00 bar on the liquid and solid phases of carbon dioxide (of molar mass 44.0 g mol‐1) in equilibrium with densities 2.35 g cm‐3 and 2.50 g cm‐3, respectively. Answer: The molar volume of two phases of carbon dioxide come from the mass density, ρ, and the molar mass, M, by using Vm = M/ρ.  μ 

 V
 P T

M p
44.0x 10- 3k g mol -1 1.00x 105Pa 
-1
μ liquid  



1
.
87
J
mol
2.35 g cm -3 103k g g-1106 cm 3 m -3 
ρl

M p
44.0x 10- 3k g mol - 1 1.00x 105 Pa 
-1
μ solid  



1
.
76
J
mol
2.50 g cm - 3 10 3k g g-1106 cm 3m - 3 
ρl
From Atkins’ Chap 4 Physical transformations of pure substances
10
8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases?
•
•
For fixed value of pressure
(a) pressure lies below triple point pressure, and the solid sublimes. (b) The pressure corresponds to the triple point pressure. Triple point is where all 3 phases coexist in equilibrium at this point.
11
Terms:
相(Phase): 無論在整體的化學成分或物理狀態上，均勻的一種物質
形態。
• 相圖(Phase diagram):以內涵變數( p 對 T 作圖)或( 莫耳分率 對 T 作
圖)顯示的區塊圖，系統中每個區塊的相都是熱力學最穩定的形式
。
• 相邊界(phase boundary): 相圖上的線，表明在何種條件下，隔著
邊界這兩個相都是處在平衡狀態。
• 相轉換(Phase transition): 一種自發轉換由一相進入另一相。
• 相轉換溫度 (Transition temperature, Ttrs ): 在兩個相的化學勢相等
的溫度。
• 亞穩定相(metastable phase): 在熱力學上是不穩定的階段，但動力
學原因，阻止從一個相過渡到一個穩定的相。
‧ 蒸氣壓(Vapor pressure):在某一特定溫度下,聚在凝相上方的蒸氣
與其凝相達到動態平衡的壓力。
•
12
8.2 The Pressure–Temperature Phase Diagram
•
•
•
This usefulness of a phase diagram is that it displays this information graphically. A P–T phase diagram displays single phase regions, coexistence curves for which two phases coexist at equilibrium, and a triple point. The processes corresponding to paths a, b, c, and d.
13
8.2 The Pressure–Temperature Phase Diagram
• At the triple point, all 3 phases coexistent.
• All P,T points for which the same two phases coexist at equilibrium fall on a curve, call coexistence curve.
• The boiling point is the temperature at which vapor pressure is equal to external pressure. • Standard boiling temperature is when vapor pressure is 1 bar. • Normal boiling temperature when vapor pressure is 1 atm.
14
8.2 The Pressure–Temperature Phase Diagram
•
•
•
Values of the normal boiling and freezing temperatures
Liquid–gas line ends at the critical point, T=Tc and P=Pc
Substances for which T > Tc and P > Pc
are called supercritical fluids
15
8.2 The Pressure–Temperature Phase Diagram
•
•
•
•
•
The solid–gas and liquid–solid coexistence curves extend indefinitely, the liquid–gas line ends at the critical point, characterized by T=Tc and P=Pc
Substances for which T > Tc and P > Pc are called supercritical fluids.
臨界溫度, TC :是液相密度和上方蒸氣相密度相等，兩個相
之間的界面消失的溫度。
臨界壓力, PC :臨界溫度下的蒸氣壓
超臨界流體: 超臨界流體在溫度高於其臨界溫度，但其密度
是近似一個正常流體的狀態，性質介於液體和氣體之間。
16
8.2 The Pressure–Temperature Phase Diagram
•
For temperature versus heat curve, the temperature rises linearly with qP
in single‐phase regions, and remains constant along the two‐phase curves as the relative amounts of the two phases in equilibrium change.
17
Example 8.1
Draw a generic P–T phase diagram and pathways that correspond to the processes described here:
a. You hang wash out to dry at a temperature below the triple point. Initially, the water in the wet clothing has frozen. However, after a few hours in the sun, the clothing is warmer, dry, and soft.
18
Example 8.1
b. A small amount of ethanol is contained in a thermos bottle. A test tube is inserted into the neck of the thermos bottle through a rubber stopper. A few
minutes after filling the test tube with liquid nitrogen, the ethanol is no longer visible at the bottom of the bottle.
c. A transparent cylinder and piston assembly contains only a pure liquid in equilibrium with its vapor pressure. An interface is clearly visible between
the two phases. When you increase the temperature by a small amount, the interface disappears.
19
Solution
The phase diagram with the paths is shown here:
20
Solution
Paths a and b are not unique. Path a must occur at a pressure lower than the triple point pressure and process b must occur at a pressure greater than the triple point pressure. Path c will lie on the liquid–gas coexistence line up to the critical point, but can deviate once T > Tc and P > Pc.
21
8.2 The Pressure–Temperature Phase Diagram
•
A P–T phase diagram for H2O pressures up to 3.5 × 1010 bar shows that water has several solid phases that are stable in different pressure ranges because they have different densities.
22
8.3 The Phase Rule
•
If three phases, α, β and γ coexist in equilibrium, we have
 T , P     T , P    T , P 
23
8.4 The Pressure–Volume and Pressure–Volume–Temperature Phase Diagrams
•
A P–V phase diagram displays single‐ and two‐phase coexistence regions, a critical point, and a triple line for a substance.
solid
m
V
V
liquid
m
 V
gas
m
24
8.4 The Pressure–Volume and Pressure–Volume–Temperature Phase Diagrams
•
•
•
In process a, a solid is converted to a gas by increasing the temperature in an isobaric process.
In process b (freezing point) an isobaric transition from solid to gas.
In a process c shows a constant volume transition from a system consisting of solid and vapor in equilibrium to a supercritical fluid.
25
8.4 The Pressure–Volume and Pressure–Volume–Temperature Phase Diagrams
•
A 3D P–V–T diagram for an ideal gas shows constant pressure, constant volume, and constant temperature paths.
26
8.4 The Pressure–Volume and Pressure–Volume–Temperature Phase Diagrams
•
For a substance that contracts on freezing,
27
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
•
When pressure is applied to a system in which two phases are in equilibrium (at a), the equilibrium is disturbed. It can be restored by changing the temperature, so moving the state of the system to b. It follows that there is a relation between dP and dT that ensures that the system remains in equilibrium, ∆G = 0, as either variable is changed. From Atkins’ Chap 4 Physical transformations of pure substances
28
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
The slope of the phase boundaries
• For one component system with two phases, α and β
at equilibrium
dμ α = dμβ
• In each phase, dμ = dGӖ = VӖ dP ‐ SӖ dT , • Thus VӖα dP ‐ SӖα dT = VӖβ dP ‐ SӖβ dT , • At the phase boundary
dP S β- S α S
H



dT V β-V α V T V
• Here ∆S = SӖβ ‐ SӖα , dVӖ = VӖβ ‐ VӖα, 29
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
• The Clapeyron Equation is a relation between the change in P and T for two phases that remain in equilibrium when either variables is modified. It can therefore be regarded as an expression for the SLOPE of a phase boundary in a phase diagram:
dP
P Δtrs S
ΔtrsH



dT T ΔtrsV T ΔtrsV
30
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
• A typical solid‐liquid phase boundary slopes steeply upwards. This slope implies that, as the pressure is raised, the melting temperature rises. Most substances behave in this way.
From Atkins’ Chap 4 Physical transformations of pure substances
31
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
•
Clapeyron equation states that dP S m

dT Vm
•
•
This equation allows us to to calculate the slope of the coexistence curves in a P–T phase diagram if ΔSm and ΔVm for the transition are known.
From Table 8.2, the average is as follow:
S mfusion  22 Jmol 1K 1
Vmfusion  4 106m 3
32
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
•
fusion

V
We use the average values of and m
S mfusionto estimate the slope of the solid–liquid coexistence curve where average of 1 1
fusion
22
dP

S
Jmol
K


m




6
3
1
fusion
 4 10 m mol
 dT fusion Vm
 5.5 10 Pa K
6
1
 55 bar K
1
33
Trouton’s rule
• At constant pressure, q = ΔtrsH , and the change in molar entropy of the system is ΔvapS = ΔvapH /Tb = 88 J K‐1 mol‐1
• Predict the enthalpy of vaporization of methane from its boiling point, ‐88.6 ℃
ΔvapH º = Tb x (88 J K‐1mol‐1)
= (‐88.6 + 273.15 K) x (88 J K‐1)
= 16.24 kJ mol‐1
The experimental value is 8.18 kJ mol‐1
34
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
•
From Trouton’s rule, the average value is as follow:
vaporization
1 1
S m
 90 Jmol K
V
•
vaporization
m
 20 10 m mol
3
3
1
The slope of the liquid–gas coexistence line is given by
1 1
vaporization
dP

S
90
Jmol
K


m




vaporization
2 102m 3mol 1
 dT vaporization Vm
 4.8 103Pa K 1  5 102bar K 1
35
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
• The formula for the solid‐liquid boundary is obtained by integrating the Clapeyron
equation for fusion with approximation:
ΔtrsH
P *dP  ΔtrsV
P

T
T*
dT
ΔtrsH  T 
ln 

T
ΔtrsV T* 
ΔtrsH  T 
P  P* 
ln 
ΔtrsV T* 
 T T
T 
ln
  ln1
T*

T * 
*  T T *

T*

ΔfusH
T T * 
P  P* 
T * ΔtrsV
36
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
The liquid‐vapor boundaries
vap S
P
dP
trsH



dT T tvapV T trsV
37
Example Estimating the effect of p on the Tb
Example Estimating the effect of pressure on the boiling temperature
Estimate the typical size of the effect of increasing pressure on boiling point of a liquid. Answer: At the boiling point, the term ∆vapH / T is Trouton’s constant. Because the molar volume of a gas is so much greater than the molar volume of a liquid, we can take for Vm(g) the molar volume of a perfect gas (at low pressure, at least ):
Vm  Vm g- Vm l  Vm g
Trouton’s constant has the value 85 J K‐1 mol‐1. The molar volume of a perfect gas is about 25.1 L mol‐1 at 1 atm and near but above room temperature. Therefore,
dp
85 J K -1 mol -1
3
-1


3.4x10
Pa
K
dT
2.5x10 -2 m3 mol -1
or 0.034 atm K‐1
38
Example
Example Estimating dT/dp for water at its normal boiling point using the information in Table 3.2 and Vm(g) = RT/p
Answer: 28 K atm‐1 dp
85 J K -1 mol -1
3
-1


3.4x10
Pa
K
dT
2.5x10 -2 m3 mol -1
39
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
• An approximation form of the Clapeyron eq. applied when one phase is condensed and the other is a gas that may be treated as being perfect.
ΔvapV V g 
V g  nRT
P
• The Clausius‐Clapeyron equation for a liquid‐vapor or a solid‐vapor phase boundaries is based on approximations:
ΔvapH
ΔvapH Δvap H
dP



P
2
dT T V g-Vl  T V g
RT
ΔvapH
P
 d ln

dT
2
P
P  RT
dP
ΔvapH
P
ln
C
P
RT
40
ln(P/Pº)
Vapor pressure of water
T-1/103 K
• A plot of ln (p/p‫)ל‬
versus 1/T should be linear, and this is borne out by data on both vaporization and sublimation. • Std BP: standard Boilng Point.
• FP: Freezing Point
Chap 4 Physical transformations of pure substances
41
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
• By integrating between two limits in P, T:
 ΔvapH   1 1 
dP
P2
  
ln 

P
P1
R
T
T
P
/
P

1
 2
ΔvapH   1 1 
  
ln P2  ln P1 
R T2 T1 
P2 / P 
1
ΔvapH

R
T2
dT
T T 2
1
  ΔvapH   1 1 
  
P2  P1 exp
T2 T1 
 R
• The Clausius-Clapeyron equation can be regarded as an equation
for the temperature variation of the vapor pressure of liquid.
42
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
•
A typical liquid‐vapor phase boundary. The boundary can be regarded as a plot of the vapor pressure against the temperature. ΔvapH
P2
ln

P1
R
•
T2 T1 


 T2 T1 
Note that, if a logarithmic pressure scale is used, the phase boundary has the opposite curvature,
From Atkins’ Chap 4 Physical transformations of pure substances
43
8.6 Using the Clapeyron Equation to Calculate Vapor Pressure as a Function of T
•
Using ideal gas law and assuming  Η mvaporizati on
remain constant over the range of temperature of interest:
Pf
dP
P P 
i
 H mvaporizati on
R

Tf
dT
T T 2
i
 H mvaporizati on  1 1 
Pf

 
 
ln
Pi
R
T f T i 
•
This equation provides a way to determine the enthalpy of vaporization for a liquid by measuring its vapor pressure as a function of temperature.
44
Trouton’s rule
Example Estimation of the vapor pressure of benzene using Trouton’s
rule.
The boiling point of benzene is 80.1 ℃ at 1 atm. Estimate the vapor pressure of benzene at 25 ℃ using Trouton’s rule.
Ans: The vapor pressure at 353.3 K are 1.013 bar, and the estimated heat of vaporization is (88 J K‐1 mol‐1)(353.3 K) =31.1 kJ mol‐1
P2 vapH T2-T1
ln 
P1
R T1T2
ln
1.013 bar
P1

31.1x10  55.1
3
8.3145 298.2 353.3
P1 = 0.143 bar
45
8.5 Providing a Theoretical Basis for the P–T Phase Diagram
• The slope of the liquid‐vapor boundary is less steep than that of the solid‐liquid boundary.
• If the enthalpy of vaporization is assumed constant over the T and P ranges of interest, the Clausius‐Clapeyron
equation can be integrated to  ΔvapH   1 1 
 
P  P * exp

 R T * T 
• For the molar enthalpy of sublimation, fusion and vaporization at triple point, ∆subH = ∆fusH + ∆vapH
• For the slope of the solid‐vapor boundary: the enthalpy of sublimation is used in the Clapeyron equation giving a steeper boundary than for vapor. 46
The effect of temperature on the vapor pressure of a liquid.
Equation P  P * e-
ΔvapH   1 1 

  *
R T T 
can be used to estimate the vapor pressure of a liquid at any temperature from its normal boiling point, the temperature at which the vapor pressure is 1.00 atm (101 kPa). Thus, because the normal boiling point of benzene is 80℃(353 K) and from Table 2.3, ∆vapH º = 30.8 kJ mol‐1, to calculate the vapor pressure at 20℃(293 K), we write
ΔvapH   1 1  3.08x104 J mol-1  1
1


  *
-1
-1 
R T T  8.3145 J K mol  293 K 353 K

  2.15

• p = p* exp(‐χ ) = (101 kPa) e‐2.15 = 11.8 kPa
• Experimental value is 10 kPa.
47
Example 8.2
The normal boiling temperature of benzene is 353.24 K, and the vapor pressure of liquid benzene is 1.00 × 10‐4 Pa at 20.0°C. The enthalpy of fusion is 9.95 kJmol‐1, and the vapor pressure of solid benzene is 88.0 Pa at 44.3°C. Calculate the following:
a. Hmvaporization
b. Smvaporization
c. Triple point temperature and pressure
48
Solution
H mvaporization
a. We can calculate using the Clapeyron equation because we know the vapor pressure at two different temperatures
H mvaporization
Pf
ln  
Pi
R
H mvaporization  
 1 1
  
Tf Ti 
Pf
R ln
Pi
 1 1
  
Tf Ti 
101.325
4
1
.
00
10


 30.7kJ mol1
1
1





 353.24 273.15  20.0 
8.314 
49
Solution
 H mvaporizati
on
30 . 7  10 3

 86 . 9 Jmol
353 .24
K
b.
S
c.
At the point, the vapor pressures of the solid and liquid are equal :
vaporizati on
m

Tb
Ptpliquid
Pi liquid
 H mvaporizati
ln
 ln

P
P
R
on
Ptpsolid
Pi liquid
 H msub lim ation
ln
 ln

P
P
R
 1
1


liquid
T
 tp T i
 1
1


solid
T
T
tp
i

Ptpliquid
Pi solid
 H msub lim ation
 H mvaporizati
ln
 ln


solid
P
P
RT i
RT i liquid
T tp 
 H mvaporizati
on
 H msub lim ation
1
1








on
 H

vaporizati on
m
 Pi liquid
Pi solid
 H msub lim ation
 H mvaporizati on
R  ln
 ln


solid
P
P
RT i
RT i liquid

9 . 95  10 3


33 .2  10 3  9 . 95  10 3
8 .314   ln 10000  ln 88 . 0 
8 . 314  228 . 9

 H msub lim ation
RT tp






33 .2  10

8 .314  293 .15 
3
 278 K
50
Solution
We calculate the triple point pressure using the Clapeyron equation:
H mvaporization
Pf
ln  
Pi
R
 1 1
  
Tf Ti 
Ptp
30.7 103  1
1 
ln




101325
8.314
 278 353.24 
Ptp
ln
 8.6969
P
Ptp  5.98 103Pa
51
8.7 The Vapor Pressure of a Pure Substance Depends on the Applied Pressure
•
Consider a piston and cylinder assembly at 298 K is shown with the contents being pure water. (a) at a pressure greater than the vapor pressure, (b) at a pressure equal to the vapor pressure, and (c) at 1 bar for a mixture of argon and water.
52
8.7 The Vapor Pressure of a Pure Substance Depends on the Applied Pressure
• Pressure may be applied to a condensed phase either by (a) compressing the condensed phase or (b) subjecting it to an inert pressurizing gas. When pressure is applied, the vapor pressure of the condensed phase increases.
From Atkins’ Chap 4 Physical transformations of pure substances
53
8.7 The Vapor Pressure of a Pure Substance Depends on the Applied Pressure
• The vapor pressure of a pressurized liquid
• At equilibrium the chemical potentials of the liquid and its vapor are equal:
 g   l
• For any change that preserves equilibrium, the changes are the same: d g  d l 
• When increase the pressure on the liquid by dp,
d l  Vm l  dp  d g
Vm g  RT p d g  RT dp p
Replace
Now: RT dp
Vm l  dp
p
dp Vm l 
 
p
p*
p
RT
p*p
p dp
*
p
RT 
p*
dp
p
p*p
p*p
p*
*
 Vm l  dp Vm l 
p dp
p Vm l
ln 
p
p* RT
54
Example
Example Calculate the effect of an increase in pressure of 100 bar on the vapor pressure of benzene at 25℃ which has density 0.879 g cm‐3. Answer: The molar volume of two phases of benzene come from the mass density, ρ, and the molar mass, M, by using Vm = M /ρ. 
78.1 g mol-1x 100x105 Pa
p M P
ln * 

 0.3586
-3
6
3 -3
-1
-1
p RT 0.879 g cm 10 cm m 8.3145J K mol x 298 K 
p  p * exp0.3586  1.431p *
An increase of 43 % on the vapor pressure of benzene under 100 bar.
55
8.7 The Vapor Pressure of a Pure Substance Depends on the Applied Pressure
• The vapor pressure p increases when a pressure ∆p is applied and the vapor pressure, p*, of the liquid:
Vm p 
p  p exp 

RT


*
• For the temperature variation of the vapor pressure of liquid,
if Vm ∆P/RT « 1, where Vm is the molar volume of the liquid.
 Vmp 
p  p 1

RT 

*
which rearranges to
p - p* Vm p

RT
p
56
Example
• For water, which has density 0.997 g cm‐3 at 25℃ and therefore molar volume 18.1 cm3 mol‐1. when subjected to an increase in pressure of 10 bar.
Vm P 1.81x10-5 m3 mol-1x1.0x106 Pa 
-3


7.3x10
8.3145 J K-1 mol-1x298 K 
RT
where we have used 1 J
= 1 Pa‐m3. Because
Vm ΔP
 1,
RT
we obtain (p‐p*) / p* = 7.3x10‐3, an increase of 0.73 percent
57
8.7 The Vapor Pressure of a Pure Substance Depends on the Applied Pressure
•
To relate the pressures of the mixture, we have
P 
liquid
RT ln  Vm P  P 
P 
where p‫= ל‬ 1 bar 58
8.8 Surface Tension
• 表面張力（surface tension）在兩相(特別是氣‐液)界面上，處
處存在著一種張力，它垂直與表面的邊界，指向液體方向並與
表面相切。
• 把作用於單位邊界線上的這種力稱為表面張力，用γ表示，單位
是 N/m.
• 氣‐液界面的表面張力γ是製造單位界面(dAS) ，與所需功 dw 的
比例常數 dw = γ dAS.
• 廣義的表面自由能定義：保持對應量的特徵自然變數不變，每
增加單位表面積時，對應量熱力學函數的增加值。
• 狹義的表面自由能定義： 保持溫度、壓力和組成不變，每增加
單位表面積時，Gibbs自由能(以功的形式)的增加值，稱為表面
Gibbs自由能，或簡稱表面自由能或表面能，單位為 J/m2 。
59
8.8 Surface Tension
•
•
In liquid phase, we need to relate the energy of the droplet to its surface area, as they depend on each others.
The work associated with the creation of additional surface area on droplet at constant V and T is:
dA  γdσ
A = Helmholtz energy, γ = the surface tension σ = unit element of area. 60
8.8 Surface Tension
• The model used for calculating the work of forming a liquid film when a wire of length l is raised and pulls the surface with it through a height h. γ: either the force per unit length or energy per unit area.
F  2γl; AS  2h l
dw F dh 2γl


γ
dAS 2l dh 2l
61
From Atkins’ Chap 4 Physical transformations of pure substances
Surface Tensions of Some Liquids
Substance
t/℃
γ/mN m‐1
30
40
60
100
140
180
‐183
‐183
840
1102
453
472
138
106
13.1
6.2
Inorganic
Aluminum
Gold
Lead
Mercury
Lead chloride
Sodium chloride
Oxygen
Nitrogen
62
From Atkins’ Chap 4 Physical transformations of pure substances
8.8 Surface Tension
Values for the surface tension for a number of liquids are listed.
63
Surface Tensions of Some Liquids
Substance
t/℃
γ/mN m‐1
20
20
20
20
20
0
15
25
50
100
23.7
28.88
22.8
16.96
63.4
75.7
73.5
72.0
67.9
58.8
Organic
Acetone
Benzene
Ethanol
Diethyl ether
Glycerol
Water
64
From Atkins’ Chap 4 Physical transformations of pure substances
8.8 Surface Tension
The variation of the surface
tension of water with temperature.
65
From Atkins’ Chap 4 Physical transformations of pure substances
8.8 Surface Tension
• 溫度升高，表面張力下降，當達到臨界溫度TC 時，表面張力
趨向於零。這可用熱力學公式說明：
dG  S dT V dP   dA   B dnB
B
運用全微分的性質，可得：
S

( )T ,P ,nB  (
) A ,P ,n B
A
T
等式左方為正值，因為表面積增加，熵總是增加的。所以 γ
隨T 的增加而下降。
66
Curved surfaces
• 凸面(Convex)的曲率半徑取正值，凹面(Concave)的曲率半徑
取負值。
• A curved surface of a liquid, or a curved interface between phases, exerts a pressure so that the pressure is higher in the phase on the concave side of the interface. (在液體屈面或是
兩相的界面兩邊,施加於凹面的內部壓力較外部壓力大)
• The pressure inside of a rubber balloon is higher than the atmospheric pressure. (橡皮氣球的內部壓力比大氣壓力較高)
• Bubble is a region in which vapor is trapped by a thin film of liquid. (液膜泡是平衡蒸氣包在液體薄膜中)
• Cavity is a vapor‐filled hole in a liquid. (氣泡是在液體中的洞,
填充了平衡蒸氣壓)
• Droplet is a small volume of liquid at equilibrium with and surrounded by its vapor. (液滴是小體積的液體被周圍的平衡
蒸氣壓包圍)
67
Fig 6.6
• Spherical droplet of a pure liquid in contact with its vapor in a closed container at temperature T. Solid line for the position of the surface, and the dashed line shows the effect of an infinitesimal expansion of the droplet.
68
Surface Energy of droplet • The surface area AS is related to the volume VL of the liquid and to the amount nL of the liquid. 4 3
dVL  4 π r 2dr dAS  8 π r dr
VL  π r AS  4 π r 2
3
• At constant V and T, the work of surface formation is equal to the change in Helmholtz energy (dA) of the system.
• The fundamental equations at constant T for two phases are:
2
dAL T  -PL dVL  μL dnL  γ dAS
dA  dV
dAV T  -PV dVV  μV dnV
S
r
L
• At equilibrium: dA T  0  PV -PL  dVL  2 γ dVL or PL-PV   2 γ
r
r
where dA = dAL + dAV ; dVV = ‐ dVL ; dnL = ‐ dnV ; V =  L ;
69
Chap 6 Phase Equilibrium
• The Laplace equation for the vapor pressure at a curved surface show that, as bubble or cavity becomes larger, the pressure difference across its surface decreases. Pin ‐ Pout = 2 γ/ r
where γ is the surface tension and r is the radius, Thus the pressure inside the curvature is higher than the pressure outside the boundary, Pin > Pout .
70
8.8 Surface Tension
•
•
The surface tension has the units of energy/area or J m‐2, which is equivalent to N m‐1 (Newtons per meter).
At equilibrium, there is a balance between the inward and outward acting forces.
4r 2 Pouter  8r  4r 2 Pinner or
Pinner  Pouter
2

r
71
Surface Energy of cavity
• For a cavity, the outward force = Pressure x area = 4πr 2Pin
where the pressure inside is Pin and its radius is r.
• The inward force from the external pressure = 4πr 2Pout
• The change in surface area when the radius of sphere change from r to r + dr is 2
dAS  4 π r  dr  - 4 π r 2  8 π r dr
• As work = force x distance. The work done when the surface is stretched by dr is γdAS
dw  F x dr  8 πγr dr
• The force opposing stretching through a distance dr when the radius is r. At equilibrium, the outward and inward forces are balanced. The force inwards arises from both external pressure and the surface tension: 4πr 2Pin = 4πr 2Pout + 8πγr
• Now we have
Pin ‐ Pout = 2 γ/ r
72
8.8 Surface Tension
The dependence of the pressure inside
a curved surface on the radius of the
surface, for two different values of the
surface tension γ.
2γ
Pin  Pout 
r
73
From Atkins’ Chap 4 Physical transformations of pure substances
8.8 Surface Tension
•
•
Assume that a capillary of radius r is partially immersed in a liquid.
If the surface tension of the liquid is lower than that of the solid, the liquid will wet the surface, termed as capillary rise.
74
8.8 Surface Tension
•
If the surface tension of the liquid is higher than that of the solid, the liquid will avoid the surface, termed as capillary depression.
75
8.8 Surface Tension
When a capillary tube is first stood
in a liquid, the latter climbs up the
walls, so curving the surface. The
pressure just under the meniscus is
less than that arising from the
atmosphere by 2γ/r. The pressure is
equal at equal heights throughout the
liquid provided the hydrostatic
pressure (which is equal to ρgh)
cancels the pressure difference
arising from the curvature.
76
From Atkins’ Chap 4 Physical transformations of pure substances
8.8 Surface Tension
• Capillary action is the tendency of liquids to rise up capillary tubes of narrow bore. The height h to which a liquid of mass density ρ and surface tension γ will rise is 2γ
h
ρ gr
where g is the acceleration of free fall. 77
Illustration
If water at 25℃ rises through 7.36 cm in a capillary of radius 0.20 mm, its surface tension at that temperature is 1
γ  ρ g hr
2
= ½ (997.1 kg m‐3) (9.81 m s‐2) (7.36x10‐2 m) (2.0x10‐4 m)
= 72 mN m‐1
where we have used 1 kg m s‐2 = 1 N
78
8.8 Surface Tension: Contact Angle and Interfacial Tension
• There may be a nonzero angle between the edge of the meniscus and the wall called the contact angle θc, • The contact angle θc , is related to the surface tensions of the solid/gas; solid/liquid; and liquid/gas interfacial surface tension: γsg; γsl; γlg; by cos θc = (γsg‐ γsl) / γlg ;
• Capillary rise occurs for 0° < θc< 90°
• Capillary depress occurs for 90° < θc< 180°
79
8.8 Surface Tension
The balance of forces that results in a
contact angle, θc.
γsg = γsl + γlg cos θc.
θ = 0° θ < 90° θ = 90° θ > 90° θ = 180°
-
>0
-
=0
-
<0
80
8.8 Surface Tension
•
The difference in the pressure across the 2 / r
curved interface, , is balanced by the weight of the column in the gravitational 2
field, ρgh.
h
•
gr
For intermediate cases,
Pinner  Pouter
2
2
and h 

r cos 
gr cos 
81
Capillary rise and surface tension • If the contact angle between the tube and the liquid is θc
rather than zero, then the right hand side will multiply by cos(θc).
1
γ  ρ g hr cosC 
2
82
Example 8.3
The six‐legged water strider supports itself on the surface of a pond on four of its legs. Each of these legs causes a depression to be formed in the pond surface. Assume that each depression can be approximated as a hemisphere of radius 1.2 × 10‐4 m and that is 0°. Calculate the force that one of the insect’s legs exerts on the pond.
An Insect walking on the water
83
Solution
We have
2
2  71.99  10 3
3
 1.20  10 Pa
P 

4
r cos 
1.2  10
F  PA  P  r 2  1.20  103   1.2  10  4  5.4  10 5 N


84
Example 8.4
Water is transported upward in trees through channels in the trunk called xylem. Although the diameter of the xylem channels varies from species to species, a typical value is 2.0 × 10‐5 m. Is capillary rise sufficient to transport water to the top of a redwood tree that is 100 m high? Assume complete wetting of the xylem channels.
85
Solution
We have
2
2  71.99  10 3
h
 0.74m

5
pgr cos  997  9.81 2.0  10
Capillary rise is not sufficient to account for water supply to the top of a redwood tree.
86
Surface Wetting
• If we note that the work of adhesion of the liquid to the solid (per unit area of contact) is:
wad = γsg + γlg ‐ γsl
• The criteria for surface wetting is θc >0°
corresponds to the wetting of a surface, and it is possible when wad > 2 γlg
Wad
-1
• equation can be written as: cos C  
 lg
87
Surface Wetting
• The criteria for surface wetting is θc >0° corresponds to the wetting of a surface, and it is possible when wad > 2 γlg
(Source: Fonds der Chemischen Industrie, Germany; imageseries "Tenside")
88
Complete
wetting
• The variation of contact angle (shown by the semaphore‐like object) as the ratio wad / γlg changes. Poor
wetting
89
8.9 Chemistry in Supercritical Fluids
•
Supercritical fluids useful as solvents in chemical reactions as they are:
1. Exhibit favorable properties of liquids and gases.
2. Density is high.
• The properties of liquid crystals are intermediate between liquids and solids.
• They are generally formed from polar organic molecules with a rod‐like shape. 90
8.10 Liquid Crystals and LCD Displays
•
•
The properties of liquid crystals are intermediate between liquids and solids.
They are generally formed from polar organic molecules with a rod‐like shape.
91
8.10 Liquid Crystals and LCD Displays
•
The ordering in solid, liquid, and liquid crystal phases of such molecules are as follows:
Solid
Liquid
Liquid Crystal
92
8.10 Liquid Crystals and LCD Displays
•
The twisted nematic phase is important as liquid crystals with this structure are the basis of LCD industry.
93
8.10 Liquid Crystals and LCD Displays
• The way in which an LCD display functions is as follow:
a) Light passing through the first polarizer is transmitted by the second polarizer because the plane of polarization of the light is rotated by the crystal.
94
8.10 Liquid Crystals and LCD Displays
b) The orientational ordering of the twisted nematic phase is destroyed by application if the electric field. No light is passed.
95
8.10 Liquid Crystals and LCD Displays
c) Arrangement of electrodes in a LCD alphanumeric display.
96