Physical Chemistry

Lecturer: Liao-Ping Cheng

鄭 廖 平
Textbook 講義在iclass上，可下載

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2
 成績計算
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期末考: 34 分
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、其他
最高98 分

3
FOCUS 1 The properties of gases
Topic 1A The perfect gas
1A.1 Variables of state 狀態是由狀態
變數所表明
⚫ The variables needed to specify the state of a system:
n (no. of moles or mass), V (volume), p (pressure), T (temperature)
e.g., State 1: one mole of nitrogen gas at 1 atm and 298 K
State 2: one mole of nitrogen gas at 1atm and 1000 K
狀態變數改變，就是處於不同狀態!

(a) Pressure
⚫ The pressure of a fluid (gas or liquid) is defined as the forces the
fluid exert on the wall of the container per unit area; i.e., p = F/A.
⚫ Pressure units: Table 1A.1, pascal, bar, atmosphere, etc.
⚫ Mechanical equilibrium: Figure 1A.1
‘pask(ə)l’ 4
(b) Temperature
⚫ Thermal equilibrium: no transfer of heat can be observed, equality
of temperature.
⚫ The zeroth law of thermodynamics: If A is in thermal equilibrium
with B, and B is in thermal equilibrium with C, then C is also in
thermal equilibrium with A.
⚫ Thermodynamic temperature scale: temperatures are denoted T
and are normally reported in kelvins (K).
T(K) = θ(oC) + 273.15 (1A.1)
1A.2 Equation of state
⚫ General form of equation of state: relate the state variables p, T, V, n
For a pure substance, 4變數中取其3個即可完
p = f (T, V, n) (1A.2) 整表明氣體所處狀態

only 3 variables are needed to specify the state; the 4th variable is
fixed by the other three; e.g., knowing T, V, n of a particular substance,
p can be calculated by eq.(1A.2) 5
⚫ Each substance has its own equation of state.
Special case: for a perfect (ideal) gas, pV = nRT
體積減半時，撞擊器
(a) The empirical basis 壁倍增，壓力加倍

Boyle’s law: pV = constant, at constant n, T (1A.3a)

⚫ a limiting law, which holds only at sufficiently low pressure, p → 0,
where gas molecules are far apart and the there is no interaction
between molecules. 理想氣體
⚫ p-V plot, Fig. 1A.2: hyperbola, with n, T held constant 雙曲線 xy = k

Each curve represents the p-V relation at a specific temperature and

is called an isotherm. 溫度升高，氣體速度(能量)增加，
等溫線 碰撞頻率及力道增加 → 壓力增大

⚫ p vs. 1/V plot, Figure 1A.3: linear, extrapolated to origin, isotherm

1
p = constant  直線 y = k x
V 6
Charles’s law: V = constant  T, at constant n , p (1A.3b) pV = nRT
p = constant  T, at constant n ,V (1A.3c)

⚫ a limiting law and is good for p → 0

⚫ V-T plot, Figure 1A.4: linear, extrapolated to origin, each line 等壓線
represents the V-T relation at a specific pressure and is called an isobar.
⚫ p-T plot, Figure 1A.5: linear, extrapolated to origin, each line represents
the p-T relation at a specific volume and is called an isochore.
等容積線

Avogadro’s principle: V = constant  n, at constant p, T (1A.3d)

Summarize Boyle, Charles, Avogadro ideal gas law

pV= nRT (1A.4)
An ideal gas obeys this law under all conditions. A real gas?
7
A real gas approaches ideal as p → 0 pV = nRT
⚫ R: gas constant
R = NA ∙ k = 8.31447 J K-1mol-1
NA: Avogadro’s constant = 6.022141023 mol-1
k: Boltzmann’s constant = 1.3806510-23 J K-1

⚫ For a given amount of ideal gas, one cannot vary simultaneous all
of the three variables p, V, T only two degree of freedom
(須符合ideal gas law, 3擇2來改變)

Gas Constant R
dm3 = liter
dm = 0.1 m
注意單位!

8
⚫ p-V-T 3-D plot of an ideal gas: a surface, points on the surface are
Fig. 1A.6 states available to the ideal gas.
Fig. 1A.7 isothermal, isobaric, and isochoric sections. 剖面圖

Ex. 1A.1, p.8

⚫ standard temperature and pressure (STP): 0 oC and 1 atm

standard ambient T and p (SATP): 25 oC and 1 bar N
bar = 10 2 5

● molar volume: volume per mole of molecules, Vm m

1
V RT = atm
Vm = for ideal gas Vm = 1.01325
n p = 0.9869 atm
@ SATP Vm = 24.789 dm3 mol-1 = 0.02478 m3 mol-1
J=Nm
@ STP Vm = 22.414 dm3 mol-1
8.31447 J mol -1K -1  273.18 K
* p.9 左下角有一段文字用分子角度說明 1.01325  10 5 N m -2
Boyle’s law and Charles’s law，請自行參閱 9
(b) Mixtures of gases

A gas mixture: a homogeneous solution of different gases.

濃度一致，即使在局部區域亦復如是；
氣體最常用濃度單位是 mole fraction

nj
• mole fraction, x j = (1A.7)
n
j: component no.
n: total no. of moles, n = nA + nB + …
對總壓 p 的貢獻
⚫ partial pressure of j in a gas mixture, pj
any gas, real or ideal
p j = x j  p (1A.6)

p A + p B + ... = ( x A + x B + ...) p = p (1A.8)

x A + x B + ... = 1
10
Dalton’s law: the pressure exerted by a mixture of gases is the sum
of the pressures that each one would exert if it occupied
the container alone.
ideal gas only, no interaction

Ex. 1A.2, p.9

11
1B. The kinetic model
Three assumptions for the gas:
1. mass m, random motion, obeying laws of classical mechanics
2. negligible size: << distance between collision
3. interact through elastic collision internal modes of
motion are excited
移動動能不損失 轉動、震動
translational kinetic energy
elastic inelastic
(a) Pressure and molecular speeds
1 perfect gas (1B.2)
可推得 pV = nMv rms
2

3
M: molecular weight (molar mass) = mNA 將各個粒子的速
度平方後，取平
m: molecular mass 每個分子的質量 均值，再開根號
NA: Avogadro’s number 均方根
2 1/ 2
vrms: root-mean-square speed, v rms = v (1B.1) 12
How is that done? 1B.1, p.12
Fig. 1B.2
mvx
分子運動速
mv 度v，在x軸
之分量為vx
-mvx

碰撞之 momentum change = 2mvx

考慮Dt 期間，右方體積內之分子才
來得及碰撞器壁 (分子向右運動，離
牆壁太遠的碰不到)

nN A
Number density of molecules: (個/單位體積)
V
n: total number of moles
V: container volume
13
 nN A
右側方塊中之分子數 =  (v x DtA)
V
1 nN A
但半數朝右，半數朝左 → 碰撞器壁分子數 =  v x  Dt  A
2 V
每顆分子碰撞產生 2mvx 的動量變化
M = mNA
d (mv)
∴ total momentum change = 1 nN A  v  Dt  A  2mv F = ma =
x x dt
2 V
Newton’s 2nd law of motion: force = rate of momentum change
momemtum change nMAvx2
= =
Dt V
force nMvx2
Pressure = =
A V
Not all molecules travel with the same velocity use average velocity

nM  v x 
2

 p=
V 14
v =v +v +v
2 2 2 2 vrms =  v 2 1 / 2
x y z

speed of a molecule 2
vrms =  v x2  +  v y2  +  v z2 
random motion:  v x2  =  v y2  =  v z2 
nM  v x 
2

p=
2
vrms =  3v x2  帶入上頁壓力 p 之式子
V
1
pV = nMvrms
2
得證 (1.B.2)
3
2
vrms = f(T) only v
T → rms ∴ pV = constant
當n亦不動 即 Boyle’s law

1/ 2
又 1  3RT 
pV = nMvrms
2
= nRT ∴ v rms = 
3  M  BI 1B.1, p.16 15
 v v+dv
(b) The Maxwell-Boltzmann distribution of speeds

f(v)dv: fraction of molecules with speeds in the range v to v + dv

f(v): distribution function of speed.

Maxwell–Boltzmann distribution of speed: f (v) = Ke − / kT K: constant

k : Boltzmann constant;  : kinetic energy

𝑘NA = R → 1.38110-23 J K-1 ∙ 6.022 1023 mol-1 = 8.314 J K-1 mol-1

How is that done 1B.2

1 2 1 2 1 2
 (kinetic energy) = mvx + mv y + mvz
2 2 2
− ( mv x2 + mv 2y + mv z2 ) / 2 kT − mv y / 2 kT
f (v) = Ke = K x e − mvx / 2 kT K y e K z e − mv z / 2 kT
= f (v x ) f (v y ) f (v z )
f (v x ) = K x e − mvx2 / 2 kT
K = KxKyKz
16
  2kT 
1/ 2 1/ 2
  − mv  m 
= dv x = K x  =1 Kx = 
2

− x x − x  
/ 2 kT
f ( v ) dv K e x

 m   2kT 

total fraction = 1
1/ 2
 m  −mv
 f (v x ) =  (1B.3)
2

 e x / 2 kT

 2kT 

3/ 2
 m  −mv − mv y2 / 2 kT − mvz2 / 2 kT
f (v x ) f (v y ) f (v z )dv x dv y dv z = 
2

 e
/ 2 kT
x
e e dv x dv y dv z
 2kT 
− m ( vx2 + v y2 + vz2 ) / 2 kT − mv 2 / 2 kT
e =e
probability (fraction) that a molecule
has a velocity in the range:
vx to vx+dvx, vy to vy+dvy, vz to vz+dvz, i.e., f(v)dv

17
 3/ 2
 m  − mv 2 / 2 kT
f (v)dv =   e dv x dv y dv z
 2kT 

3/ 2
2 m 
直角坐標 → 球座標 f (v)dv = 4v   e − mv 2 / 2 kT
dv
 2kT 
Fig.1B.3
3/ 2
2 m 
f (v) = 4v   e − mv 2 / 2 kT

 2kT 
m mN A M
= =
k kN A R 3/ 2
 M 
= 4  
2
v e − Mv 2 / 2 RT
(1B.4)
 2RT 

18
⚫ Features of the Boltzmann-Maxwell distribution, f(v) Fig. 1B.4
 Gaussian, 有 e − Mv
2
/ 2 RT
∴ 當v ↑ → f(v) ↓↓

‚ Exponent 為 −Mv2/2RT ∴ 當M ↑ 時， 隨v上升， f(v)迅速下降，即具高速

率之大分子不多

ƒ Exponent 為 −Mv2/2RT, 高溫時，隨v上升 f(v)下降較緩慢， 即高溫下高速

率分子比率較高

„ 指數函數前有 v2 項 ∴當v → 0時， f(v) → 0 ，即小速率之分子，比率較低


…
- f (v)dv = 1

19
(c) Mean values
• The fraction of molecules with speed in the range v1 to v2: F(v1, v2)
(probability)
v2
F =  f ( v )dv (1B.5) Fig. 1B.5
v1


• Average value of vn  v 2  = v rms
2
=  v 2 f (v)dv
0
 n=2
 v  =  v n f (v)dv
n
3RT
0 = (1B.7)
(1B.6) M

1/ 2
 3RT 
v rms =  (1B.8)
 M 
3/ 2
n=1   M  
 v  = v mean =  vf (v)dv = 4  
2
 v 3 e -mv / 2 RT
dv
0
 2RT  0

3/ 2 2 1/ 2
 M  1  2 RT   8 RT 
= 4     =  
 2RT  2 M    M 
Ex. 1B.1, p.16 20
 1/ 2 1/ 2 1/ 2
vmean  8 RT   3RT   8 
⚫ =    =  = 0.921 (1B.9)
vrms  M   M   3 

⚫ most probable speed, vmp: v at the peak of f (v) vs. v

df (v)
= 0 at v = vmp
dv
1/ 2 1/ 2
 2 RT  vmp 2
It can be shown that vmp =  ⎯
⎯→ =  = 0.816 (1B.10)
 M  vrms 3

∴ vmp < vmean < vrms Fig. 1B.6

21
⚫ mean relative speed, vrel: mean speed with which one molecule
approaches another.
Fig. 1B.7
For molecules of the same kind: vrel = 2 (1B.11a)
1/ 2
vmean
1

For two molecules of masses mA and mB: vrel =  8kT  mA mB

2
=
   mA + mB
(1B.11b)
BI 1B.1 , p.17

§ 1B.2 Collisions
(a) The collision frequency, z
z: number of collisions made by one molecule in Dt
Dt: time interval

可推得 z = s vrel N (1B.12a)

N
N = : number density = no. molecules per unit volume
V 22
σ：collision cross-section of the molecules = π𝑑2 Fig 1B.8
d : collision diameter 碰撞截面
cf. Table 1B.2
in terms of pressure:
s rel p
z= (1B.12b) z = s vrel N
kT

How is that done 1B.3

Fig. 1B.8, 左方為運動粒子, 其他不動

Dt 期間分子行進距離 = vrel ∙ Dt
2 σ = π𝑑2
掃過柱狀體積 = vrel ∙Dt∙π𝑑
N N
體積內分子數 = vrel ∙Dt∙σ ∙ N=
V V
皆可參與碰撞 23
  rel DtsN (1B.12a)
 z= =  rel sN
Dt N nN A p
N = =
= v rel s V  nRT  kT
 
V  p 
v rel s p
=
kT (1B.12b)

T 固定， z 正比於 p (∵ number density 高) (1B.12b)

V 固定，當T↑時，z↑ (∵ vrel ↑) (1B.12a)

BI 1B.2 , p.17

24
(b) The mean free path, λ 平均自由路徑
λ : average distance a molecule travels between collision.
v v kT 1
 = rel = rel = T固定, λ正比於
z v relsp sp p
kT (1B.13), (1B.14)

 1/z = time between collision, Dt 兩次碰撞的時間間隔

∴ vrel∙Dt = distance between collision

BI 1B.3 , p.18

kT kV ∴ λ is independent of T for gases

• = =
sp s nRC in a container of fixed volume

pV= nRT·C (C: compression factor, for real gases)

• d << λ, gas → perfect 25

§1C. Real gases
• perfect gases: no interaction between molecules
real gas: with attractive & repulsive interactions, Fig. 1C.1

long-range short-range (molecular diameter)

(effective over serval
molecular diameters)
• low p: molecules are far apart → perfect gas

moderate p: average separation = a few molecular diameters

→ attractive force dominates
→ more compressible
high p: average separation is small → repulsive force dominates
→ less compressible

26
• Considering Fig. 1C.2(b): A →B →C →D →E →F
isothermal compression of a gas
sample initially in the state A
A: pV ≈ constant Boyle’s law
B: deviations from Boyle’s law, 分子靠近 CO2, p-V phase diagram
V C: first drop of liquid appears
下 CO2
C → E: pressure would not
降
build-up, i.e., constant p Force
D: more liquid appears
coexistence of Vapor-Liquid-Equilibrium
two phases
gas vapor pressure in
liquid a closed system

E: last trace of gas

F: reduction in volume of liquid ← requiring a great increase in p
27
(a) The compression factor, Z
• Z accounts for quantitatively the deviation of a real gas
from idealization.
V/n: molar volume (1C.1)
Vm
Z= o
Vm molar volume of an ideal gas at the same T and p
pV = nRT ideal → pVm = RT  pVm = ZRT (1C.2)
o

i.e., Z = 1 for an ideal gas

• Z = f (T, p)
• |Z ̶ 1|↑ 愈遠離理想

Figure 1C.3
(i) low pressure → Z ≈ 1 nearly perfect, with different slopes!
(ii) high pressure (圖中~400 atm) → Z > 1
28
 與理想氣體比較 ∴ Vm  Vmo → repulsive forces dominate
same T
ZRT attractive force
Vm = 擠開
p same high pressure, Z > 1
same T, p
Vm > Vmo
(iii) intermediate pressure → Z < 1
∴ Vm  Vmo → attractive force dominant

(b) Virial coefficients BI. 1C.1, p.20

pVm = RT (1 + Bp + C p 2 +   ) (1C.3a)
修正項
ideal
B C
pVm = RT (1 + + 2 +   ) (1C.3b)
Vm Vm
修正項
B C
• (1 + Bp + C p 2 +   ), (1 + + 2 +   ) : compression factor Z
Vm Vm
• B , C ,  B, C , are f (T ) : 2nd, 3rd,    virial coefficients 29
 C B
 高次項影響小 ignoring higher order terms
Vm2 Vm

BI 1C.2, p.21
dZ/dp

• dZ
=
d
dp T dp
(1 + Bp + C p 2 +   ) = B + 2 pC  +     (1C.4a)

dZ
as p → 0 → B ?
=0
dp As p → 0, Z = 1, but dZ/dp
𝑑𝑍 (slope) may be different for
然而 ideal gas, Z =1, 𝑑𝑝
=0
different gases. Fig. 1C.3 & 4

dZ
Similarly, 可推得，當 Vm →   =B (1C.4b)
d (1 / Vm )
B
另可證 B =
RT
30
• 𝐵′ = 𝑓(𝑇) ⇒
d𝑍
ቤ
d𝑝 p→0
depends on temperature

當 B' = 0 之溫度稱為 Boyle temperature, TB , cf. Table 1C.2

• 氣體在TB時接近理想氣體的範圍較大 Fig. 1C.4

B C C
又 pVm = RTB (1 + + 2 +   ), 其中B(TB) = 0, 而 V 2 很小
Vm Vm m

 pVm  RTB

31
(C) Critical constants p = vapor pressure
Fig. 1C.2 當T < Tc 恆溫壓縮氣體可生出液體，達成VLE
當T > Tc 恆溫壓縮氣體不生液相
當T = Tc 恆溫壓縮氣體，液相→ 0
Tc: a single point on the p-V diagram
critical point: 𝑇 = 𝑇𝑐 , 𝑝 = 𝑝𝑐 , 𝑉𝑚 = 𝑉𝑐

For CO2, Tc = 31.04 oC = 304.19 K

pc = 72.9 atm
Vc = 94.12 cm3 mol-1
亦即 T > 31.04 oC時恆溫壓縮CO2氣體
無法生成液體 CO2

BI 1C.3, p.23
32
§1C.2 The van der Waals equation
nRT n2
(a) Formulation p= −a 2 (1C.5a)
V − nb V

a, b: van der Waals coefficients, characteristic of each gas, ≠ f (T)

V − nb : assuming molecules are incompressible hard spheres
with molar volume b. 無法壓縮，ideal gas 則無此限制
repulsive
force, 迫使
→ nb: total volume of the hard spheres Vm >> b
分子遠離 → V − nb: actual free volume for travelling

an 2
2
: affecting the pressure by reducing the frequency & force
V
of collision with the wall. n n n2
  =a 2
attractive force V V V

nRT n2 RT a
p= −a 2 = − 2 (1C.5b) BI 1C.4, p.24
V − nb V Vm − b Vm
33
(b) The features of the equation 等溫線 isotherms
Calculated results of van der Waals equation → Fig. 1C.6, 1C.7
• Apart from oscillations for T < Tc , the isotherms resemble
experimental results.
• oscillation ← van der Waals’ loop
• a, b are found by fitting the
calculated curve to the experimental
curves. Table 1C.3

Features:
近
(1) High temperature and large molar volume → ideal gas law
low pressure
RT a RT
T↑ & Vm  : V − b  V 2 且 Vm  b p= ideal
m m Vm
34
(2) Liquid and gas coexists (VLE) when attractive and repulsive
terms have a similar magnitude. 2nd term 1st term
Van der Waals’ loops occur

(3) The critical constants (Vc Tc pc) are related to the

van der Waals coefficients a & b
RT a
p= − 2
Vm − b Vm

dp RT 2a
=− + =0 解得 Vc = 3b
dVm (Vm − b) Vm
2 3
(1C.5b) a
pc =
d2p 2 RT 6a Tc =
8a 27b 2
= − =0
dVm
2
(Vm − b) Vm
3 4
27 Rb
p cV c 3
Compression factor at critical condition: Zc = = = 0.375
RTc 8
obeying the van der Waals eq. 35
(c) The principle of corresponding state
對應狀態原理
X
• reduced properties: = reduced Xr
Xc
對比性質
T p V
Tr = , pr = , Vr = m
Tc pc Vc

Fig. 1C.9: Z vs. pr for different gases falls on the same curve.
一條線可代表不同氣體，亦即相同 pr , Tr 時，雖然是不同氣
體亦有相同 Z (相同Vr ) principle of corresponding states

• It is not exact !
pVm p pVV pV pV pV 必須有相同Zc 才
Z= = r c r c = c c r r = Zc r r 可能在相同 pr 及
RT RTr Tc RTc Tr Tr
Tr 時，有相同Z

Zc is not a real constant! (Table 1C.2)

≈ 0.3 BI. 1C.5
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 RT a
• van der Waals in terms of reduced properties p= − 2
Vm − b Vm
8a
27 Rb
RTrTc a 8Tr 3
p r pc = − 2 2 pr = − 2
VrVc − b Vr Vc 3Vr − 1 Vr
a material constants
27b 2 3b 3b 並無 a, b
∴ 與氣體種類無關

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