物理化学(偏摩尔量)

µ B = (∂G ∂n )T , p ,n ,L
B
c
as chemical potential of component B
dG = (∂G ∂T ) p,nB ,nc ,LdT + (∂G )T ,nB ,nc ,Ldp + ∑ µBdnB ∂p B
(∂G ) T , n B , nC = V
For closed system , dnB= 0
物理化学(Ⅰ) Ⅰ
PHYSICAL CHEMISTRY （9）
过程方向和限度判据：
孤立系统 封闭系统 绝热过程 等温无功
（等V, W’=0）
∆S，熵增加原理 ∆S，熵增加原理 ∆A，赫姆霍兹函数减小原 理 ∆G，吉布斯函数减小原理 ∆S，考虑环境、构建孤立系 ？（有物质交换发生）
等温等压 W’＝0 其它过程 开放系统
d U = T d S − p d V + ∑ µ B ,U d n B
B
dH = TdS + Vdp + ∑ µ B,H dnB
B
dA = − SdT − pdV + ∑ µ B, AdnB
B
d G = − S d T + V d p + ∑ µ B ,G d n B
B
At a given state,
V
xB
nB
偏摩尔体积也可看作是在等温等压 等温等压的条件 等温等压 无限大的系统中加入1 mol B所引起系统 下，在无限大 无限大 体积的变化量。 ③ Conditions: Constant T and p. ④ For a mono-component system,
VB = V (B)
* m
(2) Nonadditivity of extensive properties, such as volume。
V ≠ ∑nV
* i m ,i
(3) More than 3 independent variables are needed to define the state of a system.
Problem：集合公式和Gibbs - Duhem 方程的适 ： 用条件相同吗？
4. Determination of partial molar quantities (1) 切线法 (2) 解析法 如能从实验或理论得到方程
V = f (n2 )T , p ,n1
V2 = (∂V
∂n2
)T , p ,n1
B
c
Partial molar enthalpy Partial molar entropy
H B = ( ∂H
∂n B
)T , p , nc ,L
S B = ( ∂S
∂n B
)T , p ,nc ,L
Partial molar volumes of H2O and CH3CH2OH at 25oC
Discussion: ① Partial molar quantities are intensive properties of a system. （强度性质） ② Perspectives of physical significance
V = f (T, p, x1, x2 ,L)
Changing rate of V against nB at a given state (T, p, and concentration xB).
(∂G ∂T ) p , n B , nC = − S ∂p
dG = − SdT + Vdp + ∑ µ BdnB
B
Set :
U = f ( S , V , n1 , n2 , L) H = f ( S , p, n1 , n2 , L) A = f (T , V , n1 , n2 , L)
Define:
dV = (∂V
∂T ∂p + (∂V )T , p,n2 ,L dn1 + (∂V )T , p,n1 ,L dn2 + L ∂n1 ∂n2
) p,n1 ,n2 ,L dT + (∂V
Biblioteka Baidu
)T ,n1 ,n2 ,L dp
Define:
VB = (∂V
∂nB
)T , p ,nc ,L
as partial molar volume of substance B Partial molar internal energy U B = (∂U ∂n )T , p ,n ,L
µ B ,U = (∂U ∂n ) S ,V ,n ,L
B
c
µ B , H = (∂H ∂n ) S , p ,n ,L
B
c
µ B , A = (∂A ∂n )T ,V ,n ,L
B
c
Fundamental Equations of Chemical Thermodynamics for Open Systems (开放系统的热力学基本关系式 开放系统的热力学基本关系式) 开放系统的热力学基本关系式
For binary systems, (A+B),
Or,
∂VA ∂VB xA ( ) T , p + xB ( )T , p = 0 ∂xA ∂xA
Similarly, x ( ∂VA ) + x ( ∂VB ) = 0 A T,p B T,p
∂xB ∂xB
We may also have,
∂VA ∂VB xA ( ) T , p − xB ( )T , p = 0 ∂xA ∂xB
µ B ,U = µ B , H = µ B , A = µ B ,G
dU = TdS − pdV + ∑ µ BdnB
B
dH = TdS + Vdp + ∑ µ BdnB
B
dA = − SdT − pdV + ∑ µ BdnB
B
dG = − SdT + Vdp + ∑ µ BdnB
B
Discussion:
V = f (n1 , n2 , L)
V是nB的一次齐函数。根据齐函数的Euler定理
V = n1V1 + n2V2 + L
即：
V = ∑ nBVB
B
(2) Gibbs - Duhem equation
dV = (∂V ∂T ) p,n1 ,n2 ,L dT + (∂V ∂p )T ,n1 ,n2 ,L dp + ∑VBdnB
第三章
溶液热力学基础
Thermodynamics of Multi-component Systems (Solution Chemistry)
3.1 Partial molar quantities(偏摩尔量 偏摩尔量) 偏摩尔量
1. Characteristics of multi-component and heterogeneous systems (1) Composition and the amounts of substances are changeable in systems involving phase transitions and chemical reactions. (对其中的一 相而言,可看作开放系统)
V = f (T , p, n1 , n2 , L)
2. Partial molar quantities Taking volume as an example. For a system at thermal and mechanical equilibrium, we have
V = f (T , p, n1 , n2 , L)
V − n2V2 V1 = n1
(3) 截距法 定义:
V Vm = = f ( xB ) nA + nB
Vm
可以证明:
VA = Vm − xB ( ∂Vm ∂Vm
VB
∂xB ∂xB )T , p )T , p
VA 0 xB 1
VB = Vm + xA (
3.2 Chemical potential and fundamental equations of thermodynamics for open system 1. Chemical potential Define:
3. Two important equations of partial molar quantities (1) Additivity of partial molar properties (偏摩尔量的集合公式，或加和公式 偏摩尔量的集合公式 偏摩尔量的集合公式， 加和公式) 在指定T，p 均不变的条件下
Volume Change upon Mixing of C2H5OH(A) and H2O(B)
wA% VA*/cm3 VB*/cm3 (VA*+VB*)/cm3 20 25.34 80.32 105.66 40 50.68 60.24 110.92 60 76.02 40.16 116.18 80 101.36 20.08 121.44 V/cm3 ∆V/cm3 103.24 2.42 106.93 3.99 112.22 3.96 118.56 2.88
B
At constant T and p, From V = ∑nBVB
B
dV = ∑ VBdnB
B
dV = ∑VBdnB + ∑ nBdVB
B B
By comparison, we have
∑ n dV
B B
B
=0
∑ x dV
B B
B
=0
VA是nA的零次齐函数，根据齐函数的Euler定理
∂VA =0 ∑ nB ∂n B B T , p , nk ≠ B ∂VB =0 ∑ nB ∂n B A T , p , nk ≠ A
① µB是强度性质, µB= f ( T,p,xB,…) 热平衡和力平衡 没有非体 ② 适用条件: 处于热平衡 力平衡 热平衡 力平衡的,没有非体 积功的均相 均相系统 积功 均相
Assignments
Review (XC Fu): 4.1～4.3 Further Reading (PWA): 7.1～7.2 Homework: Fu: Chapter 4：1，4 Atkins: Problems 7.2