這是用戶在 2025-1-5 23:27 為 https://app.immersivetranslate.com/pdf-pro/860597f9-eaea-4278-baa3-468b52d4e459 保存的雙語快照頁面,由 沉浸式翻譯 提供雙語支持。了解如何保存?
(assume Δ H Δ H DeltaH^(@)-=\Delta H^{\circ} \equiv constant)
P = P e Δ H R ( 1 T 1 T ) P : initial P ad T P : final P at T P = P e Δ H R 1 T 1 T P :  initial  P  ad  T P :  final  P  at  T {:[P=P^(**)e^(-(Delta H)/(R))((1)/(T)-(1)/(T^(**)))],[P^(**):" initial "P" ad "T^(**)],[P:" final "P" at "T]:}\begin{aligned} & P=P^{*} e^{-\frac{\Delta H}{R}}\left(\frac{1}{T}-\frac{1}{T^{*}}\right) \\ & P^{*}: \text { initial } P \text { ad } T^{*} \\ & P: \text { final } P \text { at } T \end{aligned}
Ex: normal b.p. of benzene is 353 K , Δ vap H = 30.8 kJ 353 K , Δ vap H = 30.8 kJ 353K,Delta vap H=30.8kJ353 \mathrm{~K}, \Delta \operatorname{vap} H=30.8 \mathrm{~kJ}
苯的正常沸點是 353 K , Δ vap H = 30.8 kJ 353 K , Δ vap H = 30.8 kJ 353K,Delta vap H=30.8kJ353 \mathrm{~K}, \Delta \operatorname{vap} H=30.8 \mathrm{~kJ}

Plase find vapor prossure of benzene at 293 K
請查找在 293 K 下苯的蒸氣壓

Sol: Δ H R ( 1 T 2 1 T 1 ) = 30800 8.314 ( 1 293 1 353 ) = 2.14 Δ H R 1 T 2 1 T 1 = 30800 8.314 1 293 1 353 = 2.14 -(Delta H)/(R)((1)/(T_(2))-(1)/(T_(1)))=-(30800)/(8.314)((1)/(293)-(1)/(353))=-2.14-\frac{\Delta H}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)=-\frac{30800}{8.314}\left(\frac{1}{293}-\frac{1}{353}\right)=-2.14
P ( 293 k ) = ( 1 × 10 5 ) e 2.14 = 11877.6 Pa P ( 293 k ) = 1 × 10 5 e 2.14 = 11877.6 Pa P(293 k)=(1xx10^(5))e^(-2.14)=11877.6PaP(293 k)=\left(1 \times 10^{5}\right) e^{-2.14}=11877.6 \mathrm{~Pa}
chap 5 simple Mixture  第五章 簡單混合
molar concentration or molarity
摩爾濃度或摩爾性
[ J ] a C m o l d m 3 ( m o l L ) C θ = 1 m o l dm 3 at 1 bar molality, b , is mole (solute) kg (solvent) [ J ]  a  C m o l d m 3 m o l L C θ = 1 m o l dm 3  at  1  bar   molality,  b ,  is   mole (solute)  kg  (solvent)  {:[[J]" a "C(mol)/(dm^(3))((mol)/(L))],[C^(theta)=1(mol)/(dm^(3))" at "1" bar "],[" molality, "b","" is "(" mole (solute) ")/(kg" (solvent) ")]:}\begin{aligned} & {[J] \text { a } C \frac{m o l}{d m^{3}}\left(\frac{m o l}{L}\right)} \\ & C^{\theta}=1 \frac{m o l}{\mathrm{dm}^{3}} \text { at } 1 \text { bar } \\ & \text { molality, } b, \text { is } \frac{\text { mole (solute) }}{\mathrm{kg} \text { (solvent) }} \end{aligned}
Partial Molar Quantity  部分摩爾量
(a) PM volume  (a) PM 量
1 mole H 2 O ( ) H 2 O ( ) H_(2)O(ℓ)H_{2} \mathrm{O}(\ell) at 25 C , V m = 18 cm 3 mol 25 C , V m = 18 cm 3 mol 25^(@)C,V_(m)=(18cm^(3))/((mol))25^{\circ} \mathrm{C}, V_{m}=\frac{18 \mathrm{~cm}^{3}}{\mathrm{~mol}}
1 摩爾 H 2 O ( ) H 2 O ( ) H_(2)O(ℓ)H_{2} \mathrm{O}(\ell) 25 C , V m = 18 cm 3 mol 25 C , V m = 18 cm 3 mol 25^(@)C,V_(m)=(18cm^(3))/((mol))25^{\circ} \mathrm{C}, V_{m}=\frac{18 \mathrm{~cm}^{3}}{\mathrm{~mol}}

add I mole H 2 O ( ) H 2 O ( ) H_(2)O(ℓ)\mathrm{H}_{2} \mathrm{O}(\ell) to a large volume of ethanol,
將 I 摩爾 H 2 O ( ) H 2 O ( ) H_(2)O(ℓ)\mathrm{H}_{2} \mathrm{O}(\ell) 添加到大量乙醇中,
V eth + H 2 O V eth = 14 cm 3 V eth  + H 2 O V eth  = 14 cm 3 V_("eth ")+H_(2)O-V_("eth ")=14cm^(3)V_{\text {eth }}+\mathrm{H}_{2} \mathrm{O}-V_{\text {eth }}=14 \mathrm{~cm}^{3}
  • 14 cm 3 14 cm 3 14cm^(3)14 \mathrm{~cm}^{3} is the parthal molar volume in ethanol
    14 cm 3 14 cm 3 14cm^(3)14 \mathrm{~cm}^{3} 是乙醇中的部分摩爾體積
  • the volume occupied by molecules A depends on the identity of molecules B B BB surrounding A A AA
    分子 A 所佔的體積取決於圍繞 A A AA 的分子 B B BB 的身份
  • H 2 O H 2 O H_(2)O\mathrm{H}_{2} \mathrm{O} are held apart by H -bond in C 2 H 5 OH , H C 2 H 5 OH , H C_(2)H_(5)OH,H\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, \mathrm{H}-bond break S hence Vm ( H 2 O ) Vm H 2 O Vm(H_(2)O)darr\operatorname{Vm}\left(\mathrm{H}_{2} \mathrm{O}\right) \downarrow
    H 2 O H 2 O H_(2)O\mathrm{H}_{2} \mathrm{O} 由氫鍵分開,在 C 2 H 5 OH , H C 2 H 5 OH , H C_(2)H_(5)OH,H\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, \mathrm{H} 鍵斷裂 S 因此 Vm ( H 2 O ) Vm H 2 O Vm(H_(2)O)darr\operatorname{Vm}\left(\mathrm{H}_{2} \mathrm{O}\right) \downarrow
| Q A B | Q A B |(Q^(A))/(B)|\left|\frac{Q^{A}}{B}\right|
Parthal volume of A A AA in B B BB
= V A + B V B = V A = V A + B V B = V A =V_(A+B)-V_(B)=V_(A)=V_{A+B}-V_{B}=V_{A}
change in composition changes T.P. properties of A A AA and B B BB
成分的變化改變了 A A AA B B BB 的 T.P.性質
Definition of Parthal molar volume
Parthal 體積的定義

To find P.M.V. plot total volume verses composition the find slope V J = ( V n j ) P T , n V J = V n j P T , n V_(J)=((del V)/(deln_(j)))_(P*T,n^('))V_{J}=\left(\frac{\partial V}{\partial n_{j}}\right)_{P \cdot T, n^{\prime}}
要找到 P.M.V. 繪製總體積與組成的圖,找到斜率 V J = ( V n j ) P T , n V J = V n j P T , n V_(J)=((del V)/(deln_(j)))_(P*T,n^('))V_{J}=\left(\frac{\partial V}{\partial n_{j}}\right)_{P \cdot T, n^{\prime}}

n = n = n^(')=n^{\prime}= amount of other componemt
n = n = n^(')=n^{\prime}= 其他組件的數量


( n B n B n_(B)n_{B} )
vitotal volume of sample
樣本的總體積

n A n A n_(A)n_{A} : amount of A A AA
n A n A n_(A)n_{A} : A A AA 的數量

V A V A V_(A)V_{A} : slope of the plot of total V V VV as n A n A n_(A)n_{A} changes as n A n A n_(A)n_{A} changes
V A V A V_(A)V_{A} : 總 V V VV 隨著 n A n A n_(A)n_{A} 變化而變化時的斜率 n A n A n_(A)n_{A}

rarr\rightarrow could be NeGATIVE   rarr\rightarrow 可能是負面的
V j = ( d v d n j ) P , T , n V j = d v d n j P , T , n V_(j)=((dv)/(dn_(j)))P,T,n^(')V_{j}=\left(\frac{d v}{d n_{j}}\right) P, T, n^{\prime}
Partial molar dume of j j jj
部分摩爾體積 j j jj
d n A v d i x t u r e = V A d n A + V B d n B d n A v d i x t u r e = V A d n A + V B d n B {:[(dn_(A))/(∣v^(dixture))],[=V_(A)dn_(A)+V_(B)dn_(B)]:}\begin{aligned} & \frac{d n_{A}}{\mid v^{d i x t u r e}} \\ &=V_{A} d n_{A}+V_{B} d n_{B} \end{aligned}
if n A n B = n A n B = n_(A)-n_(B)=n_{A}-n_{B}= const γ A γ B = γ A γ B = =>gamma_(A)*gamma_(B)=\Rightarrow \gamma_{A} \cdot \gamma_{B}= const
rarr\rightarrow const slope
V = 0 n A γ A d n A + 0 n B γ B d n B V = 0 n A γ A d n A + 0 n B γ B d n B V=int_(0)^(n_(A))gamma_(A)dn_(A)+int_(0)^(n_(B))gamma_(B)dn_(B)V=\int_{0}^{n_{A}} \gamma_{A} d n_{A}+\int_{0}^{n_{B}} \gamma_{B} d n_{B}
total wlancl of mixture = A n A + γ B n B = A n A + γ B n B =sqrtAn_(A)+gamma_(B)n_(B)=\sqrt{A} n_{A}+\gamma_{B} n_{B}
(valid regardless of how solution is prepared)
(無論解決方案如何準備均有效)

Ex: at 25 C , ρ 25 C , ρ 25^(@)C,rho25^{\circ} \mathrm{C}, \rho of 50 wt % 50 wt % 50wt%50 \mathrm{wt} \% ethanol/water solution is
25 C , ρ 25 C , ρ 25^(@)C,rho25^{\circ} \mathrm{C}, \rho 50 wt % 50 wt % 50wt%50 \mathrm{wt} \% 乙醇/水溶液中是

0.914 g / Cm 3 0.914 g / Cm 3 0.914g//Cm^(3)0.914 \mathrm{~g} / \mathrm{Cm}^{3} given
V W = 17.4 cm 3 mol 1 γ E = ? V W = 17.4 cm 3 mol 1 γ E = ? V_(W)=17.4(cm^(3))/(mol^(1))quadgamma_(E)=?V_{W}=17.4 \frac{\mathrm{~cm}^{3}}{\mathrm{~mol}^{1}} \quad \gamma_{E}=?
sol:Assume ig water + ig ethanol
sol:假設 ig 水 + ig 乙醇
V = n W V r + n e V E 2 0.914 = 1 18 × 17.4 + 1 46 × V E V E = 56.2 cm 3 ( V m E = 58.4 cm 3 mol ) V = n W V r + n e V E 2 0.914 = 1 18 × 17.4 + 1 46 × V E V E = 56.2 cm 3 V m E = 58.4 cm 3 mol {:[V=n_(W)V_(r)+n_(e)V_(E)],[(2)/(0.914)=(1)/(18)xx17.4+(1)/(46)xxV_(E)],[V_(E)=56.2cm^(3)(V_(mE)=58.4(cm^(3))/((mol)))]:}\begin{aligned} & V=n_{W} V_{r}+n_{e} V_{E} \\ & \frac{2}{0.914}=\frac{1}{18} \times 17.4+\frac{1}{46} \times V_{E} \\ & V_{E}=56.2 \mathrm{~cm}^{3}\left(V_{m E}=58.4 \frac{\mathrm{~cm}^{3}}{\mathrm{~mol}}\right) \end{aligned}
Ex: Add MgSo to water
將 MgSo 加入水中
V M g 50 4 / H 2 O = 1.4 cm 3 / mol V M g 50 4 / H 2 O = 1.4 cm 3 / mol V_(Mg50_(4)//H_(2)O)=-1.4cm^(3)//molV_{M g 50_{4} / \mathrm{H}_{2} \mathrm{O}}=-1.4 \mathrm{~cm}^{3} / \mathrm{mol}
salt breaks H -bond in water
鹽打破水中的氫鍵
Ex: Add ethanol to 1 kg water at 25 C 25 C 25^(@)C25^{\circ} \mathrm{C}, a polynominal fit to the total volume of mixture is
25 C 25 C 25^(@)C25^{\circ} \mathrm{C} 時將乙醇添加到 1 公斤水中,混合物的總體積的多項式擬合為

V = 1002 + 55 x 0.36 x 2 + 0.03 x 3 V = 1002 + 55 x 0.36 x 2 + 0.03 x 3 V=1002+55 x-0.36x^(2)+0.03x^(3)V=1002+55 x-0.36 x^{2}+0.03 x^{3}
n n n n n_(n)n_{n} (mole)
total vol. of amount of ethanol
乙醇的總體積量

mixuture ( cm 3 ) cm 3 (cm^(3))\left(\mathrm{cm}^{3}\right)  混合物 ( cm 3 ) cm 3 (cm^(3))\left(\mathrm{cm}^{3}\right)
What is the P.M.V of ethanol
乙醇的 P.M.V 是多少

sol: r E = ( V n e ) = V x = 55 0.72 x + 0.09 X 2 r E = V n e = V x = 55 0.72 x + 0.09 X 2 r_(E)=((del V)/(del ne))=(del V)/(del x)=55-0.72 x+0.09X^(2)r_{E}=\left(\frac{\partial V}{\partial n e}\right)=\frac{\partial V}{\partial x}=55-0.72 x+0.09 X^{2}

for r w r w r_(w)r_{w} you need V = f ( x ) V = f ( x ) V=f(x)V=f(x)
對於 r w r w r_(w)r_{w} ,你需要 V = f ( x ) V = f ( x ) V=f(x)V=f(x)
n w 9 n w 9 n_(w)^(9)n_{w}^{9}
Partial molar M M M\mathcal{M} can be used in any extensive state function
部分摩爾 M M M\mathcal{M} 可以用於任何廣延狀態函數
Partial Molar any extensive state function
部分摩爾任何廣延狀態函數

Partial Molar Gibbs Engy
部分摩爾吉布斯能量

In a mixture, chemical potential ( μ ) ( μ ) (mu)(\mu) is used as P.M. Gibbs Engy.
在混合物中,化學勢能 ( μ ) ( μ ) (mu)(\mu) 被用作 P.M. 吉布斯能。
u j ( G n ) P , T , n u j G n P , T , n u_(j)-=((del G)/(del n))_(P,T,n^('))u_{j} \equiv\left(\frac{\partial G}{\partial n}\right)_{P, T, n^{\prime}}
slope of G G GG changes with n J n J n_(J)n_{J} at (P(T)(1)
pure substance M j = G J n j = G m  pure substance  M j = G J n j = G m " pure substance "M_(j)=(G_(J))/(n_(j))=G_(m)\text { pure substance } M_{j}=\frac{G_{J}}{n_{j}}=G_{m}
with the same argument with V V VV
V V VV 相同的論點
G = μ A n A + μ B n B G = μ A n A + μ B n B G=mu_(A)n_(A)+mu_(B)n_(B)G=\mu_{A} n_{A}+\mu_{B} n_{B}
M M M\mathcal{M} of a substance changes at the composition does
物質的成分改變時 M M M\mathcal{M} 會變化

General d G = V d P S d T + u A d n A + u B d n B + d G = V d P S d T + u A d n A + u B d n B + quad dG=VdP-SdT+u_(A)dn_(A)+u_(B)dn_(B)+dots dots\quad d G=V d P-S d T+u_{A} d n_{A}+u_{B} d n_{B}+\ldots \ldots  一般 d G = V d P S d T + u A d n A + u B d n B + d G = V d P S d T + u A d n A + u B d n B + quad dG=VdP-SdT+u_(A)dn_(A)+u_(B)dn_(B)+dots dots\quad d G=V d P-S d T+u_{A} d n_{A}+u_{B} d n_{B}+\ldots \ldots
‘physical’ (chemical composition) change
‘物理’(化學成分)變化

when § (T) d G = U A d n A + U B d n B + max non-expansion work d G = U A d n A + U B d n B + max  non-expansion work  {:[dG=U_(A)dn_(A)+U_(B)dn_(B)+dots],[max" non-expansion work "]:}\begin{aligned} d G & =U_{A} d n_{A}+U_{B} d n_{B}+\ldots \\ \max & \text { non-expansion work }\end{aligned}
During a chem reaction as reactants are converted to products, non-expansion work (dwadd) arises from changing composition. Mixing at (T)
在化學反應中,當反應物轉化為產品時,因成分變化而產生的非膨脹功(dwadd)會出現。在(T)下混合。

(1)
n A n A n_(A)n_{A} n B n B n_(B)n_{B}
T , P T , P T,PT, P T , P T , P T,PT, P
n_(A) n_(B) T,P T,P| $n_{A}$ | $n_{B}$ | | :--- | :--- | | $T, P$ | $T, P$ |
rarr\rightarrow
n = n A + n B n = n A + n B n=n_(A)+n_(B)n=n_{A}+n_{B}
P = P A + P B , T P = P A + P B , T P=P_(A)+P_(B),TP=P_{A}+P_{B}, T
n=n_(A)+n_(B) P=P_(A)+P_(B),T| $n=n_{A}+n_{B}$ | | :--- | | $P=P_{A}+P_{B}, T$ |
same P , T P , T P,TP, T
(1) chem pot = G m = G m =G_(m)=G_{m} (pure)
(1) 化學鍋 = G m = G m =G_(m)=G_{m} (純)
G m = G m θ + R T ln p p θ M = M θ + R T ln p p θ G m = G m θ + R T ln p p θ M = M θ + R T ln p p θ {:[G_(m)=G_(m)^(theta)+RT ln((p)/(p^(theta)))],[rarr M=M^(theta)+RT ln((p)/(p^(theta)))]:}\begin{aligned} G_{m} & =G_{m}^{\theta}+R T \ln \frac{p}{p^{\theta}} \\ \rightarrow M & =M^{\theta}+R T \ln \frac{p}{p^{\theta}} \end{aligned}
variation of u u uu with press change
u u uu 的變化與按壓變更
use bar p θ = 1 p p θ = p  use bar  p θ = 1 p p θ = p " use bar "rarrp^(theta)=1quad(p)/(p^(theta))=p\text { use bar } \rightarrow p^{\theta}=1 \quad \frac{p}{p^{\theta}}=p
i before mixing  我在混合之前
G i = n A ( μ A θ + R T ln P ) + n B ( μ B θ + R T ln P ) G i = n A μ A θ + R T ln P + n B μ B θ + R T ln P G_(i)=n_(A)(mu_(A)^(theta)+RT ln P)+n_(B)(mu_(B)^(theta)+RT ln P)G_{i}=n_{A}\left(\mu_{A}^{\theta}+R T \ln P\right)+n_{B}\left(\mu_{B}^{\theta}+R T \ln P\right)
(2) After mixing  (2) 混合後
G f = n A ( U A + R T ln P A ) + n B ( U B + R T ln P B ) Δ mix G = G A G A = n A R T ln P A P + n B R T ln P B P ( x = mole fraction n J = x J n , x J = P J P itotal mole P : total press ) . G f = n A U A + R T ln P A + n B U B + R T ln P B Δ mix G = G A G A = n A R T ln P A P + n B R T ln P B P x =  mole fraction  n J = x J n , x J = P J P  itotal mole  P : total  press  . {:[G_(f)=n_(A)(U_(A)^(⊖)+RT ln P_(A))+n_(B)(U_(B)^(⊖)+RT ln P_(B))],[Delta mix G=G_(A)-G_(A)=n_(A)RT ln((P_(A))/(P))+n_(B)RT ln((P_(B))/(P))],[([x=" mole fraction "],[n_(J)=x_(J)n],x_(J)=(P_(J))/(P)quad[" itotal mole "],[P:total" press "]:}]).\left.\begin{array}{l} G_{f}=n_{A}\left(U_{A}^{\ominus}+R T \ln P_{A}\right)+n_{B}\left(U_{B}^{\ominus}+R T \ln P_{B}\right) \\ \Delta \operatorname{mix} G=G_{A}-G_{A}=n_{A} R T \ln \frac{P_{A}}{P}+n_{B} R T \ln \frac{P_{B}}{P} \\ \left(\begin{array}{l} x=\text { mole fraction } \\ n_{J}=x_{J} n \end{array}, x_{J}=\frac{P_{J}}{P} \quad \begin{array}{l} \text { itotal mole } \\ P: \operatorname{total} \text { press } \end{array}\right. \end{array}\right) .
Δ m i x G Δ m i x G Delta_(mixG)\Delta_{m i x G} at § (1)
x A x B < 1 ln x A , ln x B < 0 x A x B < 1 ln x A , ln x B < 0 :'x_(A)x_(B) < 1quad:.ln x_(A),ln x_(B) < 0\because x_{A} x_{B}<1 \quad \therefore \ln x_{A}, \ln x_{B}<0
Δ Δ :.Delta\therefore \Delta mix G < 0 G < 0 G < 0G<0 for all compositions
Δ Δ :.Delta\therefore \Delta 混合 G < 0 G < 0 G < 0G<0 以適用於所有作品

=>\Rightarrow mixing of perfect gas is Always spontaneous in all compositions
=>\Rightarrow 完美氣體的混合在所有組成中總是自發的

Δ mix G = G f G A = n A R T ln P A P + n B R T ln P B P ( = n R T ( x A ln x A + x B ln x B ) B n A , P n B T , P Δ mix  G = G f G A = n A R T ln P A P + n B R T ln P B P ( = n R T x A ln x A + x B ln x B B n A , P n B T , P {:[Delta_("mix ")^(G)=G_(f)-G_(A)=n_(A)RT ln((P_(A))/(P))+n_(B)RT ln((P_(B))/(P))(],[=nRT(x_(A)ln x_(A)+x_(B)ln x_(B))B],[n_(A)","P(n_(B))/(T,P)]:}\begin{aligned} \Delta_{\text {mix }}^{G}=G_{f}-G_{A} & =n_{A} R T \ln \frac{P_{A}}{P}+n_{B} R T \ln \frac{P_{B}}{P}( \\ & =n R T\left(x_{A} \ln x_{A}+x_{B} \ln x_{B}\right) \mathbb{B} \\ n_{A}, P & \frac{n_{B}}{T, P} \end{aligned}
Ex: H 2 3 mol N 2 , 1 mol H 2 3 mol N 2 , 1 mol quadH_(2)3molN_(2),1mol\quad H_{2} 3 \mathrm{~mol} N_{2}, 1 \mathrm{~mol}

Δ mix = Δ mix = Delta_(mix)=\Delta_{\operatorname{mix}}= ?
Δ mix G = 3 R T ln 3 2 p 3 p + 1 R T ln 1 2 p p = 8.314 × 298 ( 3 ln 1 2 + ln 1 2 ) = 6869 J Δ mix G = 3 R T ln 3 2 p 3 p + 1 R T ln 1 2 p p = 8.314 × 298 3 ln 1 2 + ln 1 2 = 6869 J {:[Delta mix G=3RT ln(((3)/(2)p)/(3p))+1RT ln(((1)/(2)p)/(p))],[=8.314 xx298(3ln((1)/(2))+ln((1)/(2)))],[=-6869J]:}\begin{aligned} \Delta \operatorname{mix} G & =3 R T \ln \frac{\frac{3}{2} p}{3 p}+1 R T \ln \frac{\frac{1}{2} p}{p} \\ & =8.314 \times 298\left(3 \ln \frac{1}{2}+\ln \frac{1}{2}\right) \\ & =-6869 \mathrm{~J} \end{aligned}
Δ G < 0 Δ G < 0 Delta G < 0\Delta G<0 doesn’t mean spontaneous d G 0 d G 0 :'dG!=0\because d G \neq 0
Δ G < 0 Δ G < 0 Delta G < 0\Delta G<0 並不意味著自發的 d G 0 d G 0 :'dG!=0\because d G \neq 0
Δ G = V d P S d T + Δ mix G Δ G = V d P S d T + Δ mix G Delta G=int VdP-int SdT+Delta_(mix)G\Delta G=\int V d P-\int S d T+\Delta_{\operatorname{mix}} G
sol: must know volume to know how press chorge
sol: 必須知道體積才能知道如何施加壓力
2 V 1 = 2 R T H 2 3 V 2 = 4 R T N 2 V V 2 = 3 4 V 1 = 3 V V 2 = 4 V 2 V 1 = 2 R T H 2 3 V 2 = 4 R T N 2 V V 2 = 3 4 V 1 = 3 V V 2 = 4 V {:[2V_(1)=2RT,H_(2)],[3V_(2)=4RT,N_(2)],[(V)/(V_(2))=(3)/(4),V_(1)=3V],[V_(2)=4V]:}\begin{array}{ll} 2 V_{1}=2 R T & H_{2} \\ 3 V_{2}=4 R T & N_{2} \\ \frac{V}{V_{2}}=\frac{3}{4} & V_{1}=3 \mathrm{~V} \\ V_{2}=4 \mathrm{~V} \end{array}
final P H 2 = 2 × 3 7 = 6 7 atm P H 2 = 2 × 3 7 = 6 7 atm P_(H_(2))=2xx(3)/(7)=(6)/(7)atmP_{H_{2}}=2 \times \frac{3}{7}=\frac{6}{7} \mathrm{~atm}
P N 2 = 3 × 4 7 = 12 7 P N 2 = 3 × 4 7 = 12 7 P_(N_(2))=3xx(4)/(7)=(12)/(7)P_{N_{2}}=3 \times \frac{4}{7}=\frac{12}{7}
Δ mix G = 2 R T ln 6 / 7 2 + 4 R T ln 12 / 7 3 = 8.314 × 298 ( 2 ln 12 7 + 4 ln 36 7 ) = 9144.4 J Δ mix  G = 2 R T ln 6 / 7 2 + 4 R T ln 12 / 7 3 = 8.314 × 298 2 ln 12 7 + 4 ln 36 7 = 9144.4 J {:[Delta_("mix "G)=2RT ln((6//7)/(2))+4RT ln((12//7)/(3))],[=8.314 xx298(2ln((12)/(7))+4ln((36)/(7)))],[=-9144.4J]:}\begin{aligned} \Delta_{\text {mix } G} & =2 R T \ln \frac{6 / 7}{2}+4 R T \ln \frac{12 / 7}{3} \\ & =8.314 \times 298\left(2 \ln \frac{12}{7}+4 \ln \frac{36}{7}\right) \\ & =-9144.4 \mathrm{~J} \end{aligned}
Other Mixing Functions  其他混合功能
( G T ) P 1 n = S , Δ mix S = ( Δ mix G ) T G T P 1 n = S , Δ mix  S = ( Δ mix G ) T ((del G)/(del T))_(P_(1)n)=-S,quadDelta_("mix ")S=-(del(Delta mix G))/(del T)\left(\frac{\partial G}{\partial T}\right)_{P_{1} n}=-S, \quad \Delta_{\text {mix }} S=-\frac{\partial(\Delta \operatorname{mix} G)}{\partial T}
mixing perfect gas initially at same press & temp Δ mix S = n R ( X A ln X A + X B ln X B ) > 0 Δ mix  S = n R X A ln X A + X B ln X B > 0 Delta_("mix ")S=-nR(X_(A)ln X_(A)+X_(B)ln X_(B)) > 0\Delta_{\text {mix }} S=-n R\left(X_{A} \ln X_{A}+X_{B} \ln X_{B}\right)>0 for all composition
在相同壓力和溫度下混合完美氣體 Δ mix S = n R ( X A ln X A + X B ln X B ) > 0 Δ mix  S = n R X A ln X A + X B ln X B > 0 Delta_("mix ")S=-nR(X_(A)ln X_(A)+X_(B)ln X_(B)) > 0\Delta_{\text {mix }} S=-n R\left(X_{A} \ln X_{A}+X_{B} \ln X_{B}\right)>0 以獲得所有成分

Ex: For equal amount of pertect gases mixed at P P PP with total amount of n n nn moles
例如:在 P P PP 的條件下混合相等量的保護氣體,總量為 n n nn 摩爾
sol: Δ mix S = n R ( 1 2 ln 1 2 + 1 2 ln 1 2 ) = n R ln 2 > 0 E x : 25 C 3 mm Imol H 2 N 2  sol:  Δ mix S = n R 1 2 ln 1 2 + 1 2 ln 1 2 = n R ln 2 > 0 E x : 25 C 3 mm Imol H 2 N 2 {:[" sol: "Delta mix S=-nR((1)/(2)ln((1)/(2))+(1)/(2)ln((1)/(2)))],[=nR ln 2 > 0],[Ex:25^(@)C[[3mm,∣Imol],[H_(2),N_(2)]]rarr◻]:}\begin{aligned} \text { sol: } \Delta \operatorname{mix} S & =-n R\left(\frac{1}{2} \ln \frac{1}{2}+\frac{1}{2} \ln \frac{1}{2}\right) \\ & =n R \ln 2>0 \\ E x: 25^{\circ} \mathrm{C} & \begin{array}{|c|c|} 3 \mathrm{~mm} & \mid \mathrm{Imol} \\ H_{2} & N_{2} \end{array} \rightarrow \square \end{aligned}
  • press changes  按壓變更
Δ mix G = 6869 J Δ mix S = Δ mix G T = + 6869 298 = + 23.2 J / K Δ mix G = 6869 J Δ  mix  S = Δ mix G T = + 6869 298 = + 23.2 J / K {:[Delta mix G=-6869J],[:.Delta" mix "S=-(Delta mix G)/(T)=(+6869)/(298)=+23.2J//K]:}\begin{aligned} \Delta \operatorname{mix} G & =-6869 \mathrm{~J} \\ \therefore \Delta \text { mix } S & =-\frac{\Delta \operatorname{mix} G}{T}=\frac{+6869}{298}=+23.2 \mathrm{~J} / \mathrm{K} \end{aligned}
The chemical potential of Liquid
液體的化學勢
chem pot of A in vapor [ u p : pure j [ chem pot vapor p  chem pot of   A in vapor  u p :  pure  j  chem pot   vapor  p {:{:[" chem pot of "],[" A in vapor "]:}quad[[u^(**)],[p^(**):" pure "j][[" chem pot "],[" vapor "p]:}:}\begin{aligned} & \begin{array}{l} \text { chem pot of } \\ \text { A in vapor } \end{array} \quad\left[\begin{array} { l } { u ^ { * } } \\ { p ^ { * } : \text { pure } j } \end{array} \left[\begin{array}{l} \text { chem pot } \\ \text { vapor } p \end{array}\right.\right. \end{aligned}
Ideal soln  理想溶液
μ A = μ A + R T ln p A p θ ( pure A ) μ A = μ A + R T ln p A p θ (  pure  A ) mu_(A)^(**)=mu_(A)^(**)+RT ln((p_(A)^(**))/(p^(theta)))(" pure "A)\mu_{A}^{*}=\mu_{A}^{*}+R T \ln \frac{p_{A}^{*}}{p^{\theta}}(\text { pure } A)
add B
M A = M A θ + R T ln P A P ( B A ) Δ M A = M A M A θ = R T ln P A P A M A = M A θ + R T ln P A P ( B A ) Δ M A = M A M A θ = R T ln P A P A {:[M_(A)=M_(A)^(theta)+RT ln((P_(A))/(P^(**)))(B!=A)],[DeltaM_(A)=M_(A)-M_(A)^(theta)=RT ln((P_(A))/(P_(A)^(**)))]:}\begin{aligned} & M_{A}=M_{A}^{\theta}+R T \ln \frac{P_{A}}{P^{*}}(B \neq A) \\ & \Delta M_{A}=M_{A}-M_{A}^{\theta}=R T \ln \frac{P_{A}}{P_{A}^{*}} \end{aligned}
u A = u A + R T ln x A u A = u A + R T ln x A u_(A)=u_(A)^(**)+RT ln x_(A)u_{A}=u_{A}^{*}+R T \ln x_{A} chem pot of A A AA in an ideal soln or real soln
u A = u A + R T ln x A u A = u A + R T ln x A u_(A)=u_(A)^(**)+RT ln x_(A)u_{A}=u_{A}^{*}+R T \ln x_{A} 化學容器的 A A AA 在理想溶液或實際溶液中

with [ B ] 0 [ B ] 0 [B]rarr0[B] \rightarrow 0
Raoult’s Law  拉烏爾定律
P A = X A P A P A = X A P A P_(A)=X_(A)P_(A)^(**)P_{A}=X_{A} P_{A}^{*}
mole froction of A in liquid phase
液相中 A 的摩爾分率

Ex: Rcoult’s Law
at T = 20 C P benz = 75 T = 20 C P benz  = 75 T=20^(@)CquadP_("benz ")^(**)=75T=20^{\circ} \mathrm{C} \quad P_{\text {benz }}^{*}=75 torr   T = 20 C P benz = 75 T = 20 C P benz  = 75 T=20^(@)CquadP_("benz ")^(**)=75T=20^{\circ} \mathrm{C} \quad P_{\text {benz }}^{*}=75 托爾
P methylbenz = 21 torr P methylbenz  = 21  torr  P_("methylbenz ")=21" torr "P_{\text {methylbenz }}=21 \text { torr }
after mixing x ben = x mbenz = 1 2 x ben  = x mbenz  = 1 2 x_("ben ")=x_("mbenz ")=(1)/(2)x_{\text {ben }}=x_{\text {mbenz }}=\frac{1}{2} what’s the composition in vapor?
混合 x ben = x mbenz = 1 2 x ben  = x mbenz  = 1 2 x_("ben ")=x_("mbenz ")=(1)/(2)x_{\text {ben }}=x_{\text {mbenz }}=\frac{1}{2} 後,蒸氣中的成分是什麼?

sol: Assume ideal soln
假設理想溶液
P benz = 1 2 × 75 = 38 P mbenz = 1 2 × 21 = 11 > P total = 49 P benz  = 1 2 × 75 = 38 P mbenz  = 1 2 × 21 = 11 > P total  = 49 {:[P_("benz ")=(1)/(2)xx75=38],[P_("mbenz ")=(1)/(2)xx21=11]:} > P_("total ")=49\begin{aligned} & P_{\text {benz }}=\frac{1}{2} \times 75=38 \\ & P_{\text {mbenz }}=\frac{1}{2} \times 21=11 \end{aligned}>P_{\text {total }}=49
in vapor (assume P.G.)
在蒸氣中(假設為 P.G.)
y b e n z = 38 49 = 0.78 , y m b = 21 49 = 0.22 y b e n z = 38 49 = 0.78 , y m b = 21 49 = 0.22 y_(benz)=(38)/(49)=0.78,y_(m*b)=(21)/(49)=0.22y_{b e n z}=\frac{38}{49}=0.78, y_{m \cdot b}=\frac{21}{49}=0.22
y y yy : molar froction in gas phase
y y yy : 氣相中的摩爾分數

rarr\rightarrow more volatile specie becomes richer in gas phase
rarr\rightarrow 更揮發的物種在氣相中變得更富含氣體
Ideal Dilute solution  理想稀釋溶液
= = == Real solutions with low solute conc
= = == 低溶質濃度的實際解決方案

[solvent obeys Raoult’s Law
溶劑遵循拉烏爾定律

( B P B = X B K B B P B = X B K B ^(B)P_(B)=X_(B)K_(B){ }^{B} P_{B}=X_{B} K_{B}
A K B ( K B ( K_(B)(K_{B}( press unit) : Henry’s Law constant
A K B ( K B ( K_(B)(K_{B}( press unit) : 亨利定律常數

(Empirical)  (經驗的)
other Expressions of Henry’s Law
亨利定律的其他表達式
P B = C K B C ( m o l L ) K B ( P a L m o l ) P B = S K B S ( g L ) K B ( P a L g ) P B = C K B C m o l L K B P a L m o l P B = S K B S g L K B P a L g {:[P_(B)=CK_(B)^(')quad C((mol)/(L))quadK_(B)^(')((Pa)/(L*mol))],[P_(B)=SK_(B)^('')quad S((g)/(L))quadK_(B)^('')((Pa)/(L*g))]:}\begin{aligned} & P_{B}=C K_{B}^{\prime} \quad C\left(\frac{m o l}{L}\right) \quad K_{B}^{\prime}\left(\frac{P a}{L \cdot m o l}\right) \\ & P_{B}=S K_{B}{ }^{\prime \prime} \quad S\left(\frac{g}{L}\right) \quad K_{B}^{\prime \prime}\left(\frac{P a}{L \cdot g}\right) \end{aligned}
P B > P A P B > P A :'P_(B)^(**) > P_(A)\because P_{B}^{*}>P_{A}
:.\therefore ideal soln fulls betwen P A & P B P A & P B P_(A)^(**)&P_(B)^(**)P_{A}^{*} \& P_{B}^{*}
:.\therefore 理想解滿足 P A & P B P A & P B P_(A)^(**)&P_(B)^(**)P_{A}^{*} \& P_{B}^{*} 之間

(1) at low [ B ] ( X B 0 ) [ B ] X B 0 [B](X_(B)rarr0)[B]\left(X_{B} \rightarrow 0\right)
(1) 在低 [ B ] ( X B 0 ) [ B ] X B 0 [B](X_(B)rarr0)[B]\left(X_{B} \rightarrow 0\right)

solvent molecules (A) are in an environment similar to pure solnet
溶劑分子 (A) 位於類似於純溶劑的環境中
P A P A P A P A P_(A)∼P_(A)^(**)P_{A} \sim P_{A}^{*}
(2) at low [B]  (2) 在低 [B]
solute molecules in a complete different environment
在完全不同的環境中的溶質分子

P B P B :.P_(B)\therefore P_{B} largely deviates form P B P B P_(B)^(**)P_{B}^{*}
P B P B :.P_(B)\therefore P_{B} 在很大程度上偏離了 P B P B P_(B)^(**)P_{B}^{*}

(3) if A B A B A∼BA \sim B molecular structure similar them A + B A + B A+BA+B froms ideal soln that fllows Raoult’s Law
(3) 如果 A B A B A∼BA \sim B 的分子結構類似它們 A + B A + B A+BA+B 形成理想溶液,則遵循拉烏爾定律

Ex: The vapor P P PP of chloromethane at various mole fractions in a mixture at 25 C 25 C 25^(@)C25^{\circ} \mathrm{C} was found as:
例:在 25 C 25 C 25^(@)C25^{\circ} \mathrm{C} 的混合物中,氯甲烷的蒸氣 P P PP 在不同的摩爾分數下被發現為:
x x xx 0.005 0.009 0.019 0.024
P ( k P a ) P ( k P a ) P(kPa)P(k P a) 27.3 48.4 101 126
x 0.005 0.009 0.019 0.024 P(kPa) 27.3 48.4 101 126| $x$ | 0.005 | 0.009 | 0.019 | 0.024 | | :---: | :---: | :---: | :---: | :---: | | $P(k P a)$ | 27.3 | 48.4 | 101 | 126 |
Estimate Henry’s Law constant for chloromethane
估算氯甲烷的亨利定律常數

sol: Regression  回歸
P = 5207.5 x + 1.4685 R 2 = 0.99 ( P = x B K B ) x = 1 P = k B = 5209 kPa P = 5207.5 x + 1.4685 R 2 = 0.99 P = x B K B x = 1 P = k B = 5209 kPa {:[P=5207.5 x+1.4685],[R^(2)=0.99quad(P=x_(B)K_(B))],[x=1quad P=k_(B)=5209kPa]:}\begin{aligned} & P=5207.5 x+1.4685 \\ & R^{2}=0.99 \quad\left(P=x_{B} K_{B}\right) \\ & x=1 \quad P=k_{B}=5209 \mathrm{kPa} \end{aligned}
The Properties of solns
溶液的性質
  1. liquid mixtures of ideal solutions in equilibrium
    理想溶液的液體混合物處於平衡狀態
M j = M j + R T ln X j chem pot for ideal soln M j = M j + R T ln X j  chem pot for ideal soln  M_(j)=M_(j)^(**)+RT ln X_(j)quad" chem pot for ideal soln "M_{j}=M_{j}^{*}+R T \ln X_{j} \quad \text { chem pot for ideal soln }
Before mixing G i = n A U A + n B U B G i = n A U A + n B U B G_(i)=n_(A)U_(A)^(**)+n_(B)U_(B)^(**)G_{i}=n_{A} U_{A}^{*}+n_{B} U_{B}^{*}  在混合之前 G i = n A U A + n B U B G i = n A U A + n B U B G_(i)=n_(A)U_(A)^(**)+n_(B)U_(B)^(**)G_{i}=n_{A} U_{A}^{*}+n_{B} U_{B}^{*}
After G f = n A ( M A + R T ln x A ) + n B ( M B + R T ln X B ) G f = n A M A + R T ln x A + n B M B + R T ln X B G_(f)=n_(A)(M_(A)^(**)+RT ln x_(A))+n_(B)(M_(B)^(**)+RT ln X_(B))G_{f}=n_{A}\left(M_{A}^{*}+R T \ln x_{A}\right)+n_{B}\left(M_{B}^{*}+R T \ln X_{B}\right)   G f = n A ( M A + R T ln x A ) + n B ( M B + R T ln X B ) G f = n A M A + R T ln x A + n B M B + R T ln X B G_(f)=n_(A)(M_(A)^(**)+RT ln x_(A))+n_(B)(M_(B)^(**)+RT ln X_(B))G_{f}=n_{A}\left(M_{A}^{*}+R T \ln x_{A}\right)+n_{B}\left(M_{B}^{*}+R T \ln X_{B}\right) 之後
Δ mix G = n A R R ln x A + n B R T ln x B = n R T ( x A ln x A + x B ln x B ) n A = x A n , n B = x B n , n = n A + n B Δ mix  G = n A R R ln x A + n B R T ln x B = n R T x A ln x A + x B ln x B n A = x A n , n B = x B n , n = n A + n B {:[Delta_("mix "G)=n_(AR)R ln x_(A)+n_(B)RT ln x_(B)],[=nRT(x_(A)ln x_(A)+x_(B)ln x_(B))],[n_(A)=x_(A)*n","quadn_(B)=x_(B)*n","n=n_(A)+n_(B)]:}\begin{aligned} \Delta_{\text {mix } G} & =n_{A R} R \ln x_{A}+n_{B} R T \ln x_{B} \\ & =n R T\left(x_{A} \ln x_{A}+x_{B} \ln x_{B}\right) \\ & n_{A}=x_{A} \cdot n, \quad n_{B}=x_{B} \cdot n, n=n_{A}+n_{B} \end{aligned}
Δ mix S = Δ mix G T Δ mix  S = Δ mix  G T Delta_("mix ")S=-(Delta_("mix "G))/(T)\Delta_{\text {mix }} S=-\frac{\Delta_{\text {mix } G}}{T}
= n R ( x A ln x A + x B ln x B ) = n R x A ln x A + x B ln x B =-nR(x_(A)ln x_(A)+x_(B)ln x_(B))=-n R\left(x_{A} \ln x_{A}+x_{B} \ln x_{B}\right)
Δ mix H = Δ mix G + T Δ mix = 0 Δ mix V = ( Δ mix G ) P | T = 0 Δ mix  H = Δ mix  G + T Δ mix  = 0 _ Δ mix  V = Δ mix  G P T = 0 _ {:[Delta_("mix "H)=Delta_("mix "G)+TDelta_("mix ")=0_],[Delta_("mix "V)=(del(Delta_("mix "G)))/(del P)|_(T)=0_]:}\begin{aligned} & \Delta_{\text {mix } H}=\Delta_{\text {mix } G}+T \Delta_{\text {mix }}=\underline{0} \\ & \Delta_{\text {mix } V}=\left.\frac{\partial\left(\Delta_{\text {mix } G}\right)}{\partial P}\right|_{T}=\underline{0} \end{aligned}
( Δ mix Δ mix  Delta_("mix ")\Delta_{\text {mix }} is not a function of P P PP )
( Δ mix Δ mix  Delta_("mix ")\Delta_{\text {mix }} 不是 P P PP 的函數 )

Δ mix , Δ mix V Δ mix  , Δ mix  V Delta_("mix "),Delta_("mix ")V\Delta_{\text {mix }}, \Delta_{\text {mix }} V are Zero for mixing ideal solutions
Δ mix , Δ mix V Δ mix  , Δ mix  V Delta_("mix "),Delta_("mix ")V\Delta_{\text {mix }}, \Delta_{\text {mix }} V 是混合理想溶液的零值

Ex: Mixing two ideal solns at T = 298 K T = 298 K T=298KT=298 \mathrm{~K} I mole benzene +2 mol methyl benzene
Ex: 混合兩個理想溶液在 T = 298 K T = 298 K T=298KT=298 \mathrm{~K} 1 摩爾苯 + 2 摩爾甲苯
Δ mix G m = 1 R T ln 1 3 + 2 R T ln 2 3 3 or λ T ( x A ln x A + x B ln x B ) = 1.6 kJ / mol 8.314 × 298 ( 1 3 ln 1 3 + 2 3 ln 2 3 ) Δ mix S m = Δ mix n or Δ mix G m T = ( 1600 ) 298 = 8.314 ( 1 3 ln 1 3 + 2 3 ln 2 3 ) Δ mix 2 H m = 0 = + 5.3 J / mol k Δ mix V m = 0 Δ mix  G m = 1 R T ln 1 3 + 2 R T ln 2 3 3  or  λ T x A ln x A + x B ln x B = 1.6 kJ / mol 8.314 × 298 1 3 ln 1 3 + 2 3 ln 2 3 Δ mix  S m = Δ mix  n  or  Δ mix  G m T = ( 1600 ) 298 = 8.314 1 3 ln 1 3 + 2 3 ln 2 3 Δ  mix  2 H m = 0 = + 5.3 J / mol k Δ  mix  V m = 0 {:[Delta_("mix ")G_(m)=(1RT ln((1)/(3))+2RT ln((2)/(3)))/(3)" or "lambda T(x_(A)ln x_(A)+x_(B)ln x_(B))],[=-1.6kJ//molquad8.314 xx298((1)/(3)ln((1)/(3))+(2)/(3)ln((2)/(3)))],[Delta_("mix ")S_(m)=(Delta_("mix "))/(n)" or "(-Delta_("mix ")G_(m))/(T)=(-(-1600))/(298)],[=-8.314((1)/(3)ln((1)/(3))+(2)/(3)ln((2)/(3)))quad Delta" mix "^(2)H_(m)=0],[=+5.3J//mol*kquad Delta" mix "V_(m)=0]:}\begin{aligned} & \Delta_{\text {mix }} G_{m}=\frac{1 R T \ln \frac{1}{3}+2 R T \ln \frac{2}{3}}{3} \text { or } \lambda T\left(x_{A} \ln x_{A}+x_{B} \ln x_{B}\right) \\ & =-1.6 \mathrm{~kJ} / \mathrm{mol} \quad 8.314 \times 298\left(\frac{1}{3} \ln \frac{1}{3}+\frac{2}{3} \ln \frac{2}{3}\right) \\ & \Delta_{\text {mix }} S_{m}=\frac{\Delta_{\text {mix }}}{n} \text { or } \frac{-\Delta_{\text {mix }} G_{m}}{T}=\frac{-(-1600)}{298} \\ & =-8.314\left(\frac{1}{3} \ln \frac{1}{3}+\frac{2}{3} \ln \frac{2}{3}\right) \quad \Delta \text { mix }^{2} H_{m}=0 \\ & =+5.3 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{k} \quad \Delta \text { mix } V_{m}=0 \end{aligned}
Real solutions  實際解決方案
Interactions between A A AA and B B BB molecules A A A B B B A A A B B B A-A!=A-B!=B-BA-A \neq A-B \neq B-B
A A AA B B BB 分子 A A A B B B A A A B B B A-A!=A-B!=B-BA-A \neq A-B \neq B-B 之間的相互作用
Δ mix H 0 Δ mix V 0 Δ mix G = Δ mix H T Δ mix S Δ mix  H 0 Δ mix  V 0 Δ mix  G = Δ mix  H T Δ mix  S {:[:.Delta_("mix ")H!=0quadDelta_("mix ")V!=0],[=>Delta_("mix ")G=Delta_("mix ")H-TDelta_("mix ")S]:}\begin{aligned} & \therefore \Delta_{\text {mix }} H \neq 0 \quad \Delta_{\text {mix }} V \neq 0 \\ \Rightarrow & \Delta_{\text {mix }} G=\Delta_{\text {mix }} H-T \Delta_{\text {mix }} S \end{aligned}
if Δ mixH Δ mixH  Delta_("mixH ")\Delta_{\text {mixH }} is large and postive Δ mix Δ mix  Delta_("mix ")\Delta_{\text {mix }} is large and nasative
如果 Δ mixH Δ mixH  Delta_("mixH ")\Delta_{\text {mixH }} 很大且正數 Δ mix Δ mix  Delta_("mix ")\Delta_{\text {mix }} 很大且為負數

Δ mix G > 0 Δ mix  G > 0 =>Delta_("mix ")G > 0\Rightarrow \Delta_{\text {mix }} G>0 spontaneous seperation
Δ mix G > 0 Δ mix  G > 0 =>Delta_("mix ")G > 0\Rightarrow \Delta_{\text {mix }} G>0 自發分離

Fimmiscible or partially miscible Heating T T T uarrT \uparrow increase solnbility
不相容或部分相容 加熱 T T T uarrT \uparrow 增加溶解度

Driving force for mixing is the increase of S ( Δ S > 0 ) S ( Δ S > 0 ) S(Delta S > 0)S(\Delta S>0)
混合的驅動力是 S ( Δ S > 0 ) S ( Δ S > 0 ) S(Delta S > 0)S(\Delta S>0) 的增加

Colligative Properties 依數
依數的性質

The propertios that depend only on the number of solute purticlos present NOT their identity normally happens in dilute solutions
依賴於溶質粒子數量而非其身份的性質通常發生在稀溶液中

=>\Rightarrow vapor pree , b p , b p darr,bp uarr\downarrow, b p \uparrow, osmosis P P P uarrP \uparrow from the presence of solnte m p m p mp darrm p \downarrow
=>\Rightarrow 蒸氣 pree , b p , b p darr,bp uarr\downarrow, b p \uparrow ,滲透 P P P uarrP \uparrow 來自 solnte 的存在 m p m p mp darrm p \downarrow

Colligative property is based on 2 assumptions
共濟性質基於兩個假設

(1) solute is not volatile ( p = 0 p = 0 p^(**)=0p^{*}=0 )
(1) 溶質不揮發 ( p = 0 p = 0 p^(**)=0p^{*}=0 )

(2) Solute does not dissolve in solid solvent rarr\rightarrow solvent and solute seperate when solution freezes
(2) 溶質不溶解於固體溶劑 rarr\rightarrow 溶劑和溶質在溶液凍結時分開
The 2 assumptions  這兩個假設
rarr\rightarrow solute has no influenu on vapor and solid
rarr\rightarrow 溶質對蒸氣和固體沒有影響

rarr\rightarrow only liq line shifts when solute is added
rarr\rightarrow 只有在添加溶質時,液體線會移動

rarr\rightarrow the presence of solute results in the decrease of chemical potential of liquid solvent
溶質的存在導致液體溶劑的化學勢能降低
μ A = μ A + R T ln x A x A < 1 μ A < μ A μ A = μ A + R T ln x A x A < 1 μ A < μ A {:[mu_(A)=mu_(A)^(**)+RT ln x_(A)],[:'x_(A) < 1quad:.mu_(A) < mu_(A)^(**)]:}\begin{aligned} & \mu_{A}=\mu_{A}^{*}+R T \ln x_{A} \\ & \because x_{A}<1 \quad \therefore \mu_{A}<\mu_{A}^{*} \end{aligned}
Decrease of M A M A M_(A)M_{A} comes from increase of S S SS
M A M A M_(A)M_{A} 的減少來自 S S SS 的增加
Δ mix S = S mix S pure > 0 Δ mix  S = S mix  S pure  > 0 Delta_("mix ")S=S_("mix ")-S_("pure ") > 0\Delta_{\text {mix }} S=S_{\text {mix }}-S_{\text {pure }}>0
spontaneous mixing  自發混合
rarr\rightarrow Solvent energy becomes more disperse after mixing, hence there’s less driving force to vaporize & freeze
溶劑能量在混合後變得更加分散,因此蒸發和冷凍的驅動力減少

Phase diagiam of binany systems
二元系統的相圖
  1. Vapor press diagiams
    P A P B P A P B P_(A)P_(B)P_{A} P_{B}
    Raoult’s P A = X A P A , P B = X B P B P A = X A P A , P B = X B P B P_(A)=X_(A)P_(A)^(**),P_(B)=X_(B)P_(B)^(**)P_{A}=X_{A} P_{A}^{*}, P_{B}=X_{B} P_{B}^{*}
    x A + x B = 1 x A + x B = 1 x_(A)+x_(B)=1x_{A}+x_{B}=1
    Total Press  總按壓
P = P A + P B = X A P A + ( 1 x A ) P B = P B + ( P A P B ) X A P = P A + P B = X A P A + 1 x A P B = P B + P A P B X A {:[P=P_(A)+P_(B)],[=X_(A)P_(A)^(**)+(1-x_(A))P_(B)^(**)],[=P_(B)^(**)+(P_(A)^(**)-P_(B)^(**))X_(A)]:}\begin{aligned} P & =P_{A}+P_{B} \\ & =X_{A} P_{A}^{*}+\left(1-x_{A}\right) P_{B}^{*} \\ & =P_{B}^{*}+\left(P_{A}^{*}-P_{B}^{*}\right) X_{A} \end{aligned}
Aa X A X A X_(A)X_{A} goes from 0 1 , P 0 1 , P 0rarr1,P0 \rightarrow 1, P goes P B P A P B P A P_(B)^(**)rarrP_(A)^(**)P_{B}^{*} \rightarrow P_{A}^{*} composition in vapor
Aa X A X A X_(A)X_{A} 0 1 , P 0 1 , P 0rarr1,P0 \rightarrow 1, P P B P A P B P A P_(B)^(**)rarrP_(A)^(**)P_{B}^{*} \rightarrow P_{A}^{*} 的蒸氣組成
y A = P A P , y B = P B P y A = x A P A P B + ( P A P B ) x A , y B = 1 y A y A = P A P , y B = P B P y A = x A P A P B + P A P B x A , y B = 1 y A {:[y_(A)=(P_(A))/(P)","y_(B)=(P_(B))/(P)],[y_(A)=(x_(A)P_(A)^(**))/(P_(B)^(**)+(P_(A)^(**)-P_(B)^(**))x_(A),y_(B)=1-y_(A))]:}\begin{aligned} & y_{A}=\frac{P_{A}}{P}, y_{B}=\frac{P_{B}}{P} \\ & y_{A}=\frac{x_{A} P_{A}^{*}}{P_{B}^{*}+\left(P_{A}^{*}-P_{B}^{*}\right) x_{A}, y_{B}=1-y_{A}} \end{aligned}
composition in vapor varies with composition in liquid
氣相中的組成隨液相中的組成而變化

for P A > P B P A > P B P_(A)^(**) > P_(B)^(**)P_{A}^{*}>P_{B}^{*} (A more volatile)
對於 P A > P B P A > P B P_(A)^(**) > P_(B)^(**)P_{A}^{*}>P_{B}^{*} (更具波動性)
y A > x A y A > x A y_(A) > x_(A)y_{A}>x_{A}
if B B BB not volatile
P B = 0 Y A = 1 , y B = 0 P B = 0 Y A = 1 , y B = 0 {:[ rarrP_(B)^(**)=0],[ rarrY_(A)=1","y_(B)=0]:}\begin{aligned} & \rightarrow P_{B}^{*}=0 \\ & \rightarrow Y_{A}=1, y_{B}=0 \end{aligned}
Ex: at 25 C P B = 75 25 C P B = 75 25^(@)CquadP_(B)^(**)=7525^{\circ} \mathrm{C} \quad P_{B}^{*}=75 torr
D M B B = 21 tarr D M B B = 21 tarr D_(MB)B^(**)=21 tarrD_{M B} B^{*}=21 \operatorname{tarr}
(A) a equimolar mixture what’s P B P B P_(B)P_{B} and P M B P M B P_(MB)P_{M B} in eam?
(A) 一個等摩爾混合物是什麼 P B P B P_(B)P_{B} P M B P M B P_(MB)P_{M B} 在 eam 中?

sol:
P B = X B P B = 1 2 × 15 = 38 torr P M B = 1 2 21 = 10 torr y B = 38 38 + 10 = 0.19 , y m B = 0.21 P B = X B P B = 1 2 × 15 = 38  torr  P M B = 1 2 21 = 10  torr  y B = 38 38 + 10 = 0.19 , y m B = 0.21 {:[P_(B)=X_(B)P_(B)^(**)=(1)/(2)xx15=38" torr "],[P_(MB)=(1)/(2)*21=10" torr "],[y_(B)=(38)/(38+10)=0.19","y_(mB)=0.21]:}\begin{aligned} & P_{B}=X_{B} P_{B}^{*}=\frac{1}{2} \times 15=38 \text { torr } \\ & P_{M B}=\frac{1}{2} \cdot 21=10 \text { torr } \\ & y_{B}=\frac{38}{38+10}=0.19, y_{m B}=0.21 \end{aligned}
( B ) ( B ) (B)(B) with X B = 0.79 , X M B = 0.21 X B = 0.79 , X M B = 0.21 X_(B)=0.79,X_(MB)=0.21X_{B}=0.79, X_{M B}=0.21 what’s the composition in vapor ϕ ϕ phi\phi ?
( B ) ( B ) (B)(B) X B = 0.79 , X M B = 0.21 X B = 0.79 , X M B = 0.21 X_(B)=0.79,X_(MB)=0.21X_{B}=0.79, X_{M B}=0.21 在蒸氣中成分是什麼 ϕ ϕ phi\phi ?

Sol: P B = 0.79 × 75 = 59.25 P B = 0.79 × 75 = 59.25 P_(B)=0.79 xx75=59.25P_{B}=0.79 \times 75=59.25
P m B = 0.21 × 21 = 44.1 P = 63.66 y B = 59.25 63.66 = 0.93 y m B = 0.07 P m B = 0.21 × 21 = 44.1 P = 63.66 y B = 59.25 63.66 = 0.93 y m B = 0.07 {:[P_(mB)=0.21 xx21=44.1quad P=63.66],[y_(B)=(59.25)/(63.66)=0.93quady_(mB)=0.07]:}\begin{aligned} & P_{m B}=0.21 \times 21=44.1 \quad P=63.66 \\ & y_{B}=\frac{59.25}{63.66}=0.93 \quad y_{m B}=0.07 \end{aligned}
Plot x A , y A x A , y A x_(A),y_(A)x_{A}, y_{A} V.S. vapor P P PP
x A , y A x A , y A x_(A),y_(A)x_{A}, y_{A} 與蒸氣 P P PP


for points betwee Lan V Lines, the system has two phase present
對於 Lan V 線之間的點,系統有兩個階段

mole fraction  摩爾分率
P A = V a p o r p r e s s o f a m i x t u r e w i t h c o m p o s i t i o n X A P A = V a p o r p r e s s o f a m i x t u r e w i t h c o m p o s i t i o n X A P_(A)=VaporpressofamixturewithcompositionX_(A)P_{A}=V a p o r ~ p r e s s ~ o f ~ a ~ m i x t u r e ~ w i t h ~ c o m p o s i t i o n ~ X_{A}
y A y A y_(A)y_{A} : comp of vapor in egm with the lig at press P A P A P_(A)P_{A}
y A y A y_(A)y_{A} : 在壓力下,與液體的蒸氣組成 P A P A P_(A)P_{A}


(molar fraction of A A AA )
a 1 ( p 1 ) a 1 p 1 a_(1)(p_(1))a_{1}\left(p_{1}\right) : liq phase,   a 1 ( p 1 ) a 1 p 1 a_(1)(p_(1))a_{1}\left(p_{1}\right) : 液相,
a 1 a 1 a_(1)^(')a_{1}^{\prime} : vapor phase in eqm with a 1 a 1 a_(1)a_{1} vapor pressure p 1 p 1 p_(1)p_{1}
a 1 a 1 a_(1)^(')a_{1}^{\prime} : 與 a 1 a 1 a_(1)a_{1} 的蒸氣壓 p 1 p 1 p_(1)p_{1} 處於平衡的蒸氣相

(tiny amount of vapor present)
(存在微量蒸氣)

P 2 P 2 P_(2)P_{2} : composition of l i q = a 2 , V a p o r = a 2 l i q = a 2 , V a p o r = a 2 l_(iq)=a_(2),Vapor=a_(2)^(')l_{i q}=a_{2}, ~ V a p o r=a_{2}^{\prime}
P 2 P 2 P_(2)P_{2} : l i q = a 2 , V a p o r = a 2 l i q = a 2 , V a p o r = a 2 l_(iq)=a_(2),Vapor=a_(2)^(')l_{i q}=a_{2}, ~ V a p o r=a_{2}^{\prime} 的組成

P 3 : 11 l i q = a 3 P 3 : 11 l i q = a 3 P_(3):11quadl_(iq)=a_(3)P_{3}: 11 \quad l_{i q}=a_{3}, Vapor = a 3 = a 3 =a_(3)^(')=a_{3}{ }^{\prime}
(tiny amount of liquid prsent)
(存在微量液體)

P4: vapor phase only
P4:僅蒸氣相

from a a aa, lower the pressure while overall composition stays the same
a a aa ,降低壓力,同時整體成分保持不變

Ex:  例:
b p ( M ) = 110.6 C b p ( 0 ) = 125.6 C b p ( M ) = 110.6 C b p ( 0 ) = 125.6 C {:[b_(p)(M)=110.6^(@)C],[b_(p)(0)=125.6^(@)C]:}\begin{aligned} & b_{p}(M)=110.6^{\circ} \mathrm{C} \\ & b_{p}(0)=125.6^{\circ} \mathrm{C} \end{aligned}
(Table of T V.S. Z A Z A Z_(A)Z_{A} )
What is the composition of lig and vapor phase when X M = 0.25 X M = 0.25 X_(M)=0.25X_{M}=0.25
X M = 0.25 X M = 0.25 X_(M)=0.25X_{M}=0.25 時,液相和氣相的組成是什麼
x 0 = 0.25 x 0 = 0.25 x_(0)=0.25x_{0}=0.25
The Levev Rule  Levev 規則
The two-phase region between α α alpha\alpha and β β beta\beta represent the amount of α α alpha\alpha and β β beta\beta guautitatualy
α α alpha\alpha β β beta\beta 之間的兩相區域代表了 α α alpha\alpha β β beta\beta 的量

α , β α , β alpha,beta\alpha, \beta in equilibrium   α , β α , β alpha,beta\alpha, \beta 在平衡中
n A l a = n B l B n A l a = n B l B n_(A)l_(a)=n_(B)l_(B)n_{A} l_{a}=n_{B} l_{B}
l l ll horizontal distance
l l ll 水平距離

n α , n β n α , n β n_(alpha),n_(beta)n_{\alpha}, n_{\beta} : two amount of α α alpha\alpha and β β beta\beta phase
n α , n β n α , n β n_(alpha),n_(beta)n_{\alpha}, n_{\beta} : 兩個 α α alpha\alpha β β beta\beta 相位的數量

Ex: a sample was prepared to be x A = 0.4 x A = 0.4 x_(A)=0.4x_{A}=0.4 with two phase
例如:樣本已準備好以 x A = 0.4 x A = 0.4 x_(A)=0.4x_{A}=0.4 進行兩相處理

in eqm: α α alpha\alpha and B x A , α = 0.6 B x A , α = 0.6 B quadx_(A,alpha)=0.6B \quad x_{A, \alpha}=0.6
x A , B = 0 , 2 x A , B = 0 , 2 x_(A,B)=0,2x_{A, B}=0,2
the ration of the amount of two phase = ?
兩相的量的比率 = ?

sol:
l α = 0.6 0.4 = 0.2 l β = 0.4 0.2 = 0.2 n α n β = l β l α = 0.2 0.2 = 1 l α = 0.6 0.4 = 0.2 l β = 0.4 0.2 = 0.2 n α n β = l β l α = 0.2 0.2 = 1 {:[l_(alpha)=0.6-0.4=0.2],[l_(beta)=0.4-0.2=0.2],[(n_(alpha))/(n_(beta))=(l_(beta))/(l_(alpha))=(0.2)/(0.2)=1]:}\begin{aligned} & l_{\alpha}=0.6-0.4=0.2 \\ & l_{\beta}=0.4-0.2=0.2 \\ & \frac{n_{\alpha}}{n_{\beta}}=\frac{l_{\beta}}{l_{\alpha}}=\frac{0.2}{0.2}=1 \end{aligned}
X A B = 0 , 2 χ A = 0.4 X A B = 0 , 2 χ A = 0.4 X_(AB)=0,2chi_(A)=0.4X_{A B}=0,2 \chi_{A}=0.4
Temperature - composition diagram
溫度 - 成分圖

rarr\rightarrow for seperation or distillation of two compounds with difterout b.P.
rarr\rightarrow 用於分離或蒸餾兩種具有不同沸點的化合物。

(a) Distillation of A + B A + B A+BA+B
(a) A + B A + B A+BA+B 的蒸餾


a 1 a 1 a_(1)a_{1} iheated to a 2 a 2 a_(2)a_{2}   a 1 a 1 a_(1)a_{1} 加熱至 a 2 a 2 a_(2)a_{2}
a 1 a 1 a_(1)a_{1} : liq composition   a 1 a 1 a_(1)a_{1} : 液體成分
a 2 a 2 a_(2)^(')a_{2}^{\prime} : vapor   a 2 a 2 a_(2)^(')a_{2}^{\prime} : 蒸氣
T 2 T 2 T_(2)T_{2} : b.P. of original mixture
T 2 T 2 T_(2)T_{2} : 原始混合物的 b.P.

a 2 a 2 a 2 a 2 a_(2)rarra_(2)^(')a_{2} \rightarrow a_{2}^{\prime} : continue heatira T 2 T 2 T_(2)T_{2}
a 2 a 2 a 2 a 2 a_(2)rarra_(2)^(')a_{2} \rightarrow a_{2}^{\prime} : 繼續 heatira T 2 T 2 T_(2)T_{2}

until L/v egm  直到 L/v egm
i y i > x i A y i > x i A y_(i) > x_(i)=>Ay_{i}>x_{i} \Rightarrow A more volatile then B B BB, i.e. b p ( A ) < b p ( B ) b p ( A ) < b p ( B ) bp(A) < bp(B)b p(A)<b p(B)
y i > x i A y i > x i A y_(i) > x_(i)=>Ay_{i}>x_{i} \Rightarrow A B B BB 更不穩定,即 b p ( A ) < b p ( B ) b p ( A ) < b p ( B ) bp(A) < bp(B)b p(A)<b p(B)

Take vapor a 2 a 2 a_(2)^(')a_{2}^{\prime} and condense to a 3 ( x a 3 = y a 2 ) a 3 a 3 x a 3 = y a 2 a 3 a_(3)(xa_(3)=ya_(2)^('))a_(3)a_{3}\left(x a_{3}=y a_{2}^{\prime}\right) a_{3} heat at T 3 T 3 T_(3)T_{3} to reach L / v L / v L//vL / v eqm a 3 ( y a 3 > x a 3 ) a 3 y a 3 > x a 3 a_(3)^(')(ya_(3)^(') > x_(a_(3)))a_{3}^{\prime}\left(y a_{3}^{\prime}>x_{a_{3}}\right)
將蒸氣 a 2 a 2 a_(2)^(')a_{2}^{\prime} 凝結至 a 3 ( x a 3 = y a 2 ) a 3 a 3 x a 3 = y a 2 a 3 a_(3)(xa_(3)=ya_(2)^('))a_(3)a_{3}\left(x a_{3}=y a_{2}^{\prime}\right) a_{3} ,在 T 3 T 3 T_(3)T_{3} 加熱以達到 L / v L / v L//vL / v 平衡 a 3 ( y a 3 > x a 3 ) a 3 y a 3 > x a 3 a_(3)^(')(ya_(3)^(') > x_(a_(3)))a_{3}^{\prime}\left(y a_{3}^{\prime}>x_{a_{3}}\right)

Take vapor a 3 a 3 a_(3)^(')a_{3}^{\prime} and condense
將蒸氣 a 3 a 3 a_(3)^(')a_{3}^{\prime} 凝結

pure A (vapor) and pure B (liqq)
純 A (蒸氣) 和純 B (液體)

=>\Rightarrow frational distillation
分餾

simple distillation  簡單蒸餾
rarr\rightarrow seperate volatile solvent from non-volatile solute
rarr\rightarrow 分離揮發性溶劑與非揮發性溶質

rarr\rightarrow with draw vapor and condense
rarr\rightarrow 以抽取蒸氣並凝結

Theoretical plate number
理論塔板數

= number used to expross efficiency of a fractioning column
= 用於表達分餾塔效率的數字

2 plates from  2 盤來自
x = 0.2 x = 0.9 x = 0.2 x = 0.9 x=0.2 rarr x=0.9x=0.2 \rightarrow x=0.9
4 plates (less efficient)
4 片(效率較低)

(b) Azeotropes  (b) 共沸物
Sometimes minimum occurs in phase diagrams ( T X ) ( T X ) (T-X)(T-X) when unfavorable interaction between A A AA and B B BB increase the vapor of the mixture and lower Tbp
有時在相圖中,當 A A AA B B BB 之間的不利相互作用增加混合物的蒸氣並降低 Tbp 時,會出現最小值 ( T X ) ( T X ) (T-X)(T-X)


A + B A + B A+BA+B : A less volatile heating at b.P. shifts to less A (more B) a 1 a 1 a_(1)a_{1} heats and boils at a 2 , a 2 a 2 a 2 , a 2 a 2 a_(2),a_(2)-a_(2)^(')a_{2}, a_{2}-a_{2}^{\prime} egm, remove a 2 a 2 a_(2)^(')a_{2}^{\prime} remaining mixture x A a 2 x A a 2 x_(A)uarra_(2)^(')x_{A} \uparrow a_{2}^{\prime} condenses to a 3 , a 3 a 3 a 3 , a 3 a 3 a_(3),a_(3)-a_(3)^(')a_{3}, a_{3}-a_{3}^{\prime} eqm, a 3 a 3 a_(3)^(')a_{3}^{\prime} condense a 4 a 4 a_(4)a_{4} composition eventnally shifts to b b bb and NOT beyond b: composition of azeotrope
A + B A + B A+BA+B : 較不揮發的加熱在 b.P. 轉向較少 A (更多 B) a 1 a 1 a_(1)a_{1} 加熱並在 a 2 , a 2 a 2 a 2 , a 2 a 2 a_(2),a_(2)-a_(2)^(')a_{2}, a_{2}-a_{2}^{\prime} egm 沸騰,去除 a 2 a 2 a_(2)^(')a_{2}^{\prime} 剩餘混合物 x A a 2 x A a 2 x_(A)uarra_(2)^(')x_{A} \uparrow a_{2}^{\prime} 凝結為 a 3 , a 3 a 3 a 3 , a 3 a 3 a_(3),a_(3)-a_(3)^(')a_{3}, a_{3}-a_{3}^{\prime} eqm, a 3 a 3 a_(3)^(')a_{3}^{\prime} 凝結 a 4 a 4 a_(4)a_{4} 成分最終轉向 b b bb 而不超過 b:共沸物的成分

Tb: b.P. of azeotrope

at b b bb, distillation can no longer be used to seporate A A AA and B B BB
b b bb 時,蒸餾不再能用來分離 A A AA B B BB

A + B A + B A+BA+B A more volatile   A + B A + B A+BA+B 更具波動性
a 1 a 2 eqm a 2 discard a 2 a 1 a 2 eqm a 2  discard  a 2 a_(1)longrightarrowa_(2)rarr_("eqm")^("")a_(2)^(')longrightarrow" discard "a_(2)^(')a_{1} \longrightarrow a_{2} \xrightarrow[\mathrm{eqm}]{ } a_{2}^{\prime} \longrightarrow \text { discard } a_{2}^{\prime}
=>\Rightarrow remaining liquid richer in B and reaches a 3 a 3 a 3 a 3 a_(3)rarra_(3)^(')a_{3} \rightarrow a_{3}^{\prime} (discard)
=>\Rightarrow 剩餘液體富含 B 並達到 a 3 a 3 a 3 a 3 a_(3)rarra_(3)^(')a_{3} \rightarrow a_{3}^{\prime} (丟棄)

=>\Rightarrow remaining liq eventually shifts to word b (azeotrope x A = y A x A = y A x_(A)=y_(A)x_{A}=y_{A} )
=>\Rightarrow 剩餘的液體最終轉移到字 b(共沸物 x A = y A x A = y A x_(A)=y_(A)x_{A}=y_{A}

© Immiscible liquids for distllation
© 不相容液體用於蒸餾
Immiscible: both liquids are saturated with the other only in tiny amount
不相容:兩種液體僅在微量中彼此飽和
total vapor P = P A + P B  total vapor  P = P A + P B :." total vapor "P=P_(A)^(**)+P_(B)^(**)\therefore \text { total vapor } P=P_{A}^{*}+P_{B}^{*}
when P = P = P=P= erternal pressure (1atm)
boiling starts : the two will not boil at the same
煮沸開始:這兩者不會同時煮沸

temperature  溫度
Thay will boil at a temperature lower the A A AA or B B BB alone
Thay 將在低於 A A AA B B BB 單獨的溫度下煮沸
n e w T b . p < T b p ( A ) < T b p ( B ) p = 1 = p A + p B P A < 1 lower b.p. P B < 1 n e w T b . p < T b p ( A ) < T b p ( B ) p = 1 = p A + p B P A < 1  lower b.p.  P B < 1 {:[newT_(b.p) < T_(bp)(A)],[ < T_(bp)(B)],[p=1=p_(A)^(**)+p_(B)^(**)],[:'P_(A)^(**) < 1rarr" lower b.p. "],[P_(B)^(**) < 1]:}\begin{aligned} & n e w T_{b . p}<T_{b p}(A) \\ &<T_{b p}(B) \\ & p=1=p_{A}^{*}+p_{B}^{*} \\ & \because P_{A}^{*}<1 \rightarrow \text { lower b.p. } \\ & P_{B}^{*}<1 \end{aligned}
Steam Distillation  蒸汽蒸餾
water + insoluble organic solvent to lowor b.p. to protect temp-sensitive compound
水 + 不溶性有機溶劑以降低沸點以保護對溫度敏感的化合物

Ex: phenylamin + water 98 C P = 7.07 kPa P = 94.3 kPa 98 C P = 7.07 kPa P = 94.3 kPa 98^(@)CquadP^(**)=7.07kPaquadP^(**)=94.3kPa98^{\circ} \mathrm{C} \quad P^{*}=7.07 \mathrm{kPa} \quad P^{*}=94.3 \mathrm{kPa}
苯胺 + 水 98 C P = 7.07 kPa P = 94.3 kPa 98 C P = 7.07 kPa P = 94.3 kPa 98^(@)CquadP^(**)=7.07kPaquadP^(**)=94.3kPa98^{\circ} \mathrm{C} \quad P^{*}=7.07 \mathrm{kPa} \quad P^{*}=94.3 \mathrm{kPa}
7.07 + 94.3 > 101.325 kPa 7.07 + 94.3 > 101.325 kPa 7.07+94.3 > 101.325kPa7.07+94.3>101.325 \mathrm{kPa}
mixture boils at 98 C 98 C 98^(@)C98^{\circ} \mathrm{C} (bp water = 100 C = 100 C =100^(@)C=100^{\circ} \mathrm{C}, bp phenylamin = 184 C = 184 C =184^(@)C=184^{\circ} \mathrm{C} )
混合物在 98 C 98 C 98^(@)C98^{\circ} \mathrm{C} 沸騰(水的沸點 = 100 C = 100 C =100^(@)C=100^{\circ} \mathrm{C} ,苯胺的沸點 = 184 C = 184 C =184^(@)C=184^{\circ} \mathrm{C}

(d) Partially miscible  (d) 部分可混溶
L-L phase diagrom  L-L 相位圖

(1) B completely dissolve in A 1 A 1 A rarr1A \rightarrow 1 phase
(1) B 完全溶解於 A 1 A 1 A rarr1A \rightarrow 1

(8) B stops dissolving in A 2 A 2 A rarr2A \rightarrow 2 phase in eqm
(8) B 在平衡中停止在 A 2 A 2 A rarr2A \rightarrow 2 相中溶解
A staturated in B
B B BB staturated in A A AA
B B BB 飽和於 A A AA

when T T T uarrT \uparrow miscibility , 2 , 2 uarr,2\uparrow, 2 phase region narrows
T T T uarrT \uparrow 混溶性 , 2 , 2 uarr,2\uparrow, 2 相區域變窄

when T = T = T=T= Tuc, thermal energy >potential energy rarr\rightarrow I phase
T = T = T=T= Tuc,熱能 > 潛能 rarr\rightarrow I 相

(3) Seperation of 2 phases
(3) 兩相分離

A A AA-rich phase, composition a a a^(')a^{\prime}
A A AA -富相,成分 a a a^(')a^{\prime}

B" " a " a "quada^(')" \quad a^{\prime}
m n = A rich φ B rich φ m n = A  rich  φ B  rich  φ (m)/(n)=(A-" rich "varphi)/(B-" rich "varphi)\frac{m}{n}=\frac{A-\text { rich } \varphi}{B-\text { rich } \varphi}
(4) large excess of B B BB dissolves all A 1 A 1 A rarr1A \rightarrow 1 phase
(4) 大量的 B B BB 溶解所有 A 1 A 1 A rarr1A \rightarrow 1
Tuci upper critical temp
Tuci 上臨界溫度
= highest + temp + hat =  highest  +  temp  +  hat  =" highest "+" temp "+" hat "=\text { highest }+ \text { temp }+ \text { hat }
phase seperation occurs  相分離發生
Tlc: Lower critical temp
Tlc: 低臨界溫度
= loweos + temp + hat =  loweos  +  temp  +  hat  =" loweos "+" temp "+" hat "=\text { loweos }+ \text { temp }+ \text { hat }
phase seperation occurs  相分離發生

when P = 1 P = 1 P=1P=1
low energy molecules form weak complex as T , P = 2 T , P = 2 T uarr,P=2T \uparrow, P=2, high energy breaks up complex (ex: double cream)
低能量分子形成弱複合物如 T , P = 2 T , P = 2 T uarr,P=2T \uparrow, P=2 ,高能量則會破壞複合物(例如:雙重奶油)

some systems have both
一些系統同時擁有兩者

Tuc and TLC  Tuc 和 TLC
Distillation of Partially Miscible Liquids
部分混溶液體的蒸餾

Common combinations: Partially miscible liquids form low-boiling azeotope that becomes fully miscible
常見組合:部分可混溶的液體形成低沸點的共沸物,隨後變為完全可混溶

-before  -之前
after boil.  煮沸後。
a 1 ( x B = 0.95 ) boils at a 2 eqm vapor b 1 ( y B < x B ) cools b 2 (1 phase) cools b 3 (2 phase) a 1 x B = 0.95  boils at  a 2  eqm   vapor  b 1 y B < x B  cools  b 2  (1 phase)   cools  b 3  (2 phase)  {:[a_(1)(x_(B)=0.95)" boils at "a_(2)],[ vec(" eqm ")" vapor "b_(1)(y_(B) < x_(B))],[ vec(" cools ")b_(2)" (1 phase) " vec(" cools ")b_(3)" (2 phase) "]:}\begin{gathered} a_{1}\left(x_{B}=0.95\right) \text { boils at } a_{2} \\ \overrightarrow{\text { eqm }} \text { vapor } b_{1}\left(y_{B}<x_{B}\right) \\ \overrightarrow{\text { cools }} b_{2} \text { (1 phase) } \overrightarrow{\text { cools }} b_{3} \text { (2 phase) } \end{gathered}
when we remove b 1 b 1 b_(1)b_{1} to condense to b 2 b 2 b_(2)b_{2} remaining liq becomes richer in B
當我們去除 b 1 b 1 b_(1)b_{1} 以濃縮至 b 2 b 2 b_(2)b_{2} 時,剩餘的液體變得更富含 B

vapor gradually becomes richer in A until the formation of azeotrope
蒸氣逐漸在 A 中變得更濃,直到形成共沸物

II Fully Miscible after boil (no TLC)
II 完全混溶於煮沸後(無薄層色譜)

a 1 heat a 1 a 1 eq. ) a 1  heat  a 1 a 1  eq.  ¯  )  {:a_(1) vec(" heat ")a_(1) vec(a_(1)) bar(" eq. ")" ) ":}\begin{aligned} & a_{1} \overrightarrow{\text { heat }} a_{1} \overrightarrow{a_{1}} \overline{\text { eq. }} \text { ) } \end{aligned}
condense b 1 b 3 b 1 b 3 b_(1)rarrb_(3)b_{1} \rightarrow b_{3}, remaining (liquid richer in B B BB )
濃縮 b 1 b 3 b 1 b 3 b_(1)rarrb_(3)b_{1} \rightarrow b_{3} ,剩餘(液體更富含 B B BB
b 2 : 3 phase < V Liquid, seperates ( e 2 , e 2 ) b 3 : 2 phase ( b 3 , b 3 ) e 1 ( 2 ϕ ) ( b . P . ) heat e 2 2 e 3 ( v ) e 2 : vapor + liq ( 2 ϕ ) = 3 ϕ b 2 : 3  phase  < V  Liquid, seperates   (  e 2 , e 2 b 3 : 2  phase  b 3 , b 3 e 1 ( 2 ϕ ) ( b . P . )  heat  e 2 2 e 3 ( v ) e 2 : vapor + liq ( 2 ϕ ) = 3 ϕ {:[b_(2):3" phase " < V" Liquid, seperates "],[" ( "{:e_(2)^('),e_(2)^(''))],[b_(3):2" phase "(b_(3)^('),b_(3)^(''))],[e_(1)(2phi) vec((b.P.))_(" heat ")e_(2)_(2)longleftarrowe_(3)(v)],[e_(2):vapor+liq(2phi)=3phi]:}\begin{aligned} & b_{2}: 3 \text { phase }<\mathrm{V} \text { Liquid, seperates } \\ & \text { ( } \left.e_{2}^{\prime}, e_{2}^{\prime \prime}\right) \\ & b_{3}: 2 \text { phase }\left(b_{3}^{\prime}, b_{3}^{\prime \prime}\right) \\ & e_{1}(2 \phi) \underset{\text { heat }}{\overrightarrow{(b . P .)}} \underset{2}{e_{2}} \longleftarrow e_{3}(v) \\ & e_{2}: \operatorname{vapor}+\operatorname{liq}(2 \phi)=3 \phi \end{aligned}
vapor and liq have same composition (trace) ( X e 2 = y e 2 ) X e 2 = y e 2 quad(X_(e_(2))=y_(e_(2)))\quad\left(X_{e_{2}}=y_{e_{2}}\right) rarr\rightarrow azeotrope
蒸氣和液體具有相同的成分(微量) ( X e 2 = y e 2 ) X e 2 = y e 2 quad(X_(e_(2))=y_(e_(2)))\quad\left(X_{e_{2}}=y_{e_{2}}\right) rarr\rightarrow 共沸物
Ex: T ( k ) T ( k ) T(k)T(k)

a 1 a 2 b 1 ( a 1 a 2 b 1 ( a_(1)longrightarrowa_(2)longleftrightarrowb_(1)(a_{1} \longrightarrow a_{2} \longleftrightarrow b_{1}( less in A ) A ) A)A)   a 1 a 2 b 1 ( a 1 a 2 b 1 ( a_(1)longrightarrowa_(2)longleftrightarrowb_(1)(a_{1} \longrightarrow a_{2} \longleftrightarrow b_{1}( A ) A ) A)A) 中更少
b p = 350 K b p = 350 K b_(p)=350Kb_{p}=350 \mathrm{~K}
remove b 1 b 1 b_(1)b_{1} vapor rarr\rightarrow remaining liq richer in A λ A λ A lambdaA \lambda
去除 b 1 b 1 b_(1)b_{1} 蒸氣 rarr\rightarrow 剩餘液體更富含 A λ A λ A lambdaA \lambda

(1) The boiling range of the original liq ( 350 k 390 k ) liq ( 350 k 390 k ) liq((350k)/(390k))\operatorname{liq}\binom{350 \mathrm{k}}{390 \mathrm{k}}
(1) 原始 liq ( 350 k 390 k ) liq ( 350 k 390 k ) liq((350k)/(390k))\operatorname{liq}\binom{350 \mathrm{k}}{390 \mathrm{k}} 的沸點範圍

(2) The composition of vapor during boil Y A = 0.66 0.95 Y A = 0.66 0.95 Y_(A)=0.66∼0.95Y_{A}=0.66 \sim 0.95
(2) 沸騰時的蒸氣組成 Y A = 0.66 0.95 Y A = 0.66 0.95 Y_(A)=0.66∼0.95Y_{A}=0.66 \sim 0.95

11 11 '11\prime 11
liq
X A = 0.95 0.99 X A = 0.95 0.99 X_(A)=0.95∼0.99X_{A}=0.95 \sim 0.99
(3) b b bb, condeses ( 1 st 1 st  1^("st ")1^{\text {st }} distillate)
(3) b b bb , 凝縮 ( 1 st 1 st  1^("st ")1^{\text {st }} 蒸餾液)
at 330 K ( x , y ) = ( 0.87 , 0.49 ) , L V = 0.66 0.49 0.87 0.66 = 0.17 0.21 at 320 K , 3 , V L = 0.80 0.66 0.66 0.45 = 0.14 0.21  at  330 K ( x , y ) = ( 0.87 , 0.49 ) , L V = 0.66 0.49 0.87 0.66 = 0.17 0.21  at  320 K , 3 , V L = 0.80 0.66 0.66 0.45 = 0.14 0.21 {:[" at "330K(x","y)=(0.87","0.49)","(L)/(V)=(0.66-0.49)/(0.87-0.66)=(0.17)/(0.21)],[" at "320K","3O/","(V)/(L)=(0.80-0.66)/(0.66-0.45)=(0.14)/(0.21)]:}\begin{aligned} & \text { at } 330 \mathrm{~K}(x, y)=(0.87,0.49), \frac{L}{V}=\frac{0.66-0.49}{0.87-0.66}=\frac{0.17}{0.21} \\ & \text { at } 320 \mathrm{~K}, 3 \varnothing, \frac{V}{L}=\frac{0.80-0.66}{0.66-0.45}=\frac{0.14}{0.21} \end{aligned}
at 298 K B rich ( x A = 0.2 , x B = 0.8 ) A r i c h ( x A = 0.9 , x B = 0.1 ) = 0.9 0.66 0.66 0.2 A r i c h ϕ ( x A = 0.8 , X B = 0.2 ) , B r i c h ϕ ( x A = 0.3 298 K B rich x A = 0.2 , x B = 0.8 A r i c h x A = 0.9 , x B = 0.1 = 0.9 0.66 0.66 0.2 A r i c h ϕ x A = 0.8 , X B = 0.2 , B r i c h ϕ x A = 0.3 298K(B-rich(x_(A)=0.2,x_(B)=0.8))/(A-rich(x_(A)=0.9,x_(B)=0.1))=(0.9-0.66)/(0.66-0.2)A-rich phi(x_(A)=0.8,X_(B)=0.2),B-richphi(x_(A)=0.3:}298 \mathrm{~K} \frac{B-\operatorname{rich}\left(x_{A}=0.2, x_{B}=0.8\right)}{A-r i c h\left(x_{A}=0.9, x_{B}=0.1\right)}=\frac{0.9-0.66}{0.66-0.2} A-r i c h \phi\left(x_{A}=0.8, X_{B}=0.2\right), B-r i c h ~ \phi\left(x_{A}=0.3\right.   298 K B rich ( x A = 0.2 , x B = 0.8 ) A r i c h ( x A = 0.9 , x B = 0.1 ) = 0.9 0.66 0.66 0.2 A r i c h ϕ ( x A = 0.8 , X B = 0.2 ) , B r i c h ϕ ( x A = 0.3 298 K B rich x A = 0.2 , x B = 0.8 A r i c h x A = 0.9 , x B = 0.1 = 0.9 0.66 0.66 0.2 A r i c h ϕ x A = 0.8 , X B = 0.2 , B r i c h ϕ x A = 0.3 298K(B-rich(x_(A)=0.2,x_(B)=0.8))/(A-rich(x_(A)=0.9,x_(B)=0.1))=(0.9-0.66)/(0.66-0.2)A-rich phi(x_(A)=0.8,X_(B)=0.2),B-richphi(x_(A)=0.3:}298 \mathrm{~K} \frac{B-\operatorname{rich}\left(x_{A}=0.2, x_{B}=0.8\right)}{A-r i c h\left(x_{A}=0.9, x_{B}=0.1\right)}=\frac{0.9-0.66}{0.66-0.2} A-r i c h \phi\left(x_{A}=0.8, X_{B}=0.2\right), B-r i c h ~ \phi\left(x_{A}=0.3\right.
= 0.24 0.46 = 0.24 0.46 =(0.24)/(0.46)=\frac{0.24}{0.46}
L A r i c h L B rich = 0.66 0.3 0.8 0.66 = 0.36 0.14 L A r i c h L B  rich  = 0.66 0.3 0.8 0.66 = 0.36 0.14 (L_(A)-rich)/(L_(B)-" rich ")=(0.66-0.3)/(0.8-0.66)=(0.36)/(0.14)\frac{L_{A}-r i c h}{L_{B}-\text { rich }}=\frac{0.66-0.3}{0.8-0.66}=\frac{0.36}{0.14}
Ex: when C 1 ( X A = 0 , 4 ) C 1 X A = 0 , 4 C_(1)(X_(A)=0,4)C_{1}\left(X_{A}=0,4\right) boils to C 4 C 4 C_(4)C_{4}
C 1 , 2 liq , B r i c h A r i c h = 0.9 0.4 0.4 0.2 = 5 2 C 1 , 2  liq  , B r i c h A r i c h = 0.9 0.4 0.4 0.2 = 5 2 C_(1,2" liq ")O/,(B-r_(ich))/(A-r_(ich))=(0.9-0.4)/(0.4-0.2)=(5)/(2)C_{1,2 \text { liq }} \varnothing, \frac{B-r_{i c h}}{A-r_{i c h}}=\frac{0.9-0.4}{0.4-0.2}=\frac{5}{2}
C 2 , 2 C 2 , 2 C_(2),2C_{2}, 2 liq ϕ , 1 ϕ , 1 phi,1\phi, 1 vapor ϕ ( b . P 1 = 320 K ) ϕ b . P 1 = 320 K phi(b.P_(1)=320(K))\phi\left(b . P_{1}=320 \mathrm{~K}\right)
C 2 , 2 C 2 , 2 C_(2),2C_{2}, 2 液體 ϕ , 1 ϕ , 1 phi,1\phi, 1 蒸氣 ϕ ( b . P 1 = 320 K ) ϕ b . P 1 = 320 K phi(b.P_(1)=320(K))\phi\left(b . P_{1}=320 \mathrm{~K}\right)
B-rich A r i c h = 0.8 0.4 0.4 0.3 = 4 1 L V = 0.45 0.4 0.4 0.3 = 1 2  B-rich  A r i c h = 0.8 0.4 0.4 0.3 = 4 1 L V = 0.45 0.4 0.4 0.3 = 1 2 {:[(" B-rich ")/(A-rich)=(0.8-0.4)/(0.4-0.3)=(4)/(1)],[(L)/(V)=(0.45-0.4)/(0.4-0.3)=(1)/(2)]:}\begin{aligned} & \frac{\text { B-rich }}{A-r i c h}=\frac{0.8-0.4}{0.4-0.3}=\frac{4}{1} \\ & \frac{L}{V}=\frac{0.45-0.4}{0.4-0.3}=\frac{1}{2} \end{aligned}
C 3 = 1 C 3 = 1 C_(3)=1C_{3}=1 liq ϕ , 1 ϕ , 1 phi,1\phi, 1 vapor ϕ L V = ϕ L V = phi(L)/(V)=\phi \frac{L}{V}=
C 3 = 1 C 3 = 1 C_(3)=1C_{3}=1 液體 ϕ , 1 ϕ , 1 phi,1\phi, 1 蒸氣 ϕ L V = ϕ L V = phi(L)/(V)=\phi \frac{L}{V}=

C 4 C 4 C_(4)C_{4} : 1 1 quad1\quad 1 vapor O/\varnothing
C 4 C 4 C_(4)C_{4} : 1 1 quad1\quad 1 蒸氣 O/\varnothing
Phase Diagram of Ternary System
三元系相圖

phase rule F = C P + 2 F = C P + 2 F=C-P+2F=C-P+2  相位規則 F = C P + 2 F = C P + 2 F=C-P+2F=C-P+2
C = 3 F = 3 p + 2 = 5 p C = 3 F = 3 p + 2 = 5 p C=3quad F=3-p+2=5-pC=3 \quad F=3-p+2=5-p
if (1)(D) F = 5 P 2 = 3 P F = 5 P 2 = 3 P F^(')=5-P-2=3-PF^{\prime}=5-P-2=3-P
on a 3 ϕ 3 ϕ 3phi3 \phi diogram
在一個 3 ϕ 3 ϕ 3phi3 \phi 圖表上

P = 1 F = 2 P = 1 F = 2 P=1quadF^(')=2P=1 \quad F^{\prime}=2 area means single ϕ ϕ phi\phi in eam
P = 1 F = 2 P = 1 F = 2 P=1quadF^(')=2P=1 \quad F^{\prime}=2 區域表示單一 ϕ ϕ phi\phi 在 eam

P = 2 F = 1 P = 2 F = 1 P=2quadF^(')=1P=2 \quad F^{\prime}=1 line means 2 ϕ 2 ϕ 2phi2 \phi in eqm
P = 2 F = 1 P = 2 F = 1 P=2quadF^(')=1P=2 \quad F^{\prime}=1 行表示 2 ϕ 2 ϕ 2phi2 \phi 在 eqm 中

P = 3 F = 0 P = 3 F = 0 P=3quadF^(')=0P=3 \quad F^{\prime}=0 a point means 3 ϕ 3 ϕ 3phi3 \phi in egm
P = 3 F = 0 P = 3 F = 0 P=3quadF^(')=0P=3 \quad F^{\prime}=0 一個點意味著 3 ϕ 3 ϕ 3phi3 \phi 在 egm 中

Triangular Phase Diagram
三角相圖

mole fraction x A + x B + x C = 1 x A + x B + x C = 1 x_(A)+x_(B)+x_(C)=1x_{A}+x_{B}+x_{C}=1  摩爾分率 x A + x B + x C = 1 x A + x B + x C = 1 x_(A)+x_(B)+x_(C)=1x_{A}+x_{B}+x_{C}=1

equilateral /_\\triangle at any point P P PP
等邊 /_\\triangle 在任何點 P P PP
x A + x B + x ¯ C = 1 A B : x C = 0 , A B binary system A C : x B = 0 A C B C : X A = 0 B C x A ¯ + x B ¯ + x ¯ C = 1 A B ¯ : x C = 0 , A B  binary system  A C ¯ : x B = 0 A C B C ¯ : X A = 0 B C {:[ bar(x_(A))+ bar(x_(B))+ bar(x)_(C)=1],[ bar(AB):x_(C)=0","quad A-B" binary system "],[ bar(AC):x_(B)=0quad A-C],[ bar(BC):X_(A)=0quad B-C]:}\begin{aligned} & \overline{x_{A}}+\overline{x_{B}}+\bar{x}_{C}=1 \\ & \overline{A B}: x_{C}=0, \quad A-B \text { binary system } \\ & \overline{A C}: x_{B}=0 \quad A-C \\ & \overline{B C}: X_{A}=0 \quad B-C \end{aligned}
B : C = constant B : C =  constant  B:C=" constant "B: C=\text { constant }
Adding A A AA to B C B C B-CB-C mixture
A A AA 添加到 B C B C B-CB-C 混合物中
Partially miscible 3 liquids
部分可混溶的三種液體

water, autic acid. chloroform
水,醋酸。氯仿

(W)
(A)
©
W-c : partially miscible  W-c : 部分可混溶
W A , A C W A , A C W-A,A-CW-A, A-C ifully miscible   W A , A C W A , A C W-A,A-CW-A, A-C 完全混溶
when T T T uarrT \uparrow, miscibility uarr\uparrow
T T T uarrT \uparrow ,混溶性 uarr\uparrow

§
A ^ Δ Δ Δ A ^ Δ Δ Δ hat(A)rarr Delta rarr Delta rarr Delta\hat{A} \rightarrow \Delta \rightarrow \Delta \rightarrow \Delta

a 1 : 2 ϕ 3 a 1 : 2 ϕ 3 a_(1):2phi_(3)a_{1}: 2 \phi_{3}, lever rule W l = 0.95 0.4 0.4 0.2 W l = 0.95 0.4 0.4 0.2 (W)/(l)=(0.95-0.4)/(0.4-0.2)\frac{W}{l}=\frac{0.95-0.4}{0.4-0.2}
a 1 : 2 ϕ 3 a 1 : 2 ϕ 3 a_(1):2phi_(3)a_{1}: 2 \phi_{3} , 槓桿法則 W l = 0.95 0.4 0.4 0.2 W l = 0.95 0.4 0.4 0.2 (W)/(l)=(0.95-0.4)/(0.4-0.2)\frac{W}{l}=\frac{0.95-0.4}{0.4-0.2}


0.4
( w + C w + C w+Cw+C only) ( x c = 0.4 x c = 0.4 x_(c)=0.4x_{c}=0.4 )
a 1 a 2 = add A a 1 a 2 = add A a_(1)rarra_(2)=add Aa_{1} \rightarrow a_{2}=\operatorname{add} A to help W / C W / C W//CW / C dissolve in each other ( more W in C-rich ϕ ϕ phi\phi )
a 1 a 2 = add A a 1 a 2 = add A a_(1)rarra_(2)=add Aa_{1} \rightarrow a_{2}=\operatorname{add} A 以幫助 W / C W / C W//CW / C 彼此溶解(在富含 C 的 ϕ ϕ phi\phi 中更多 W)

a 3 : L ϕ s a 3 : L ϕ s a_(3):Lphi_(s)a_{3}: L \phi_{s}, but c r i c h ϕ c r i c h ϕ c-rich phic-r i c h \phi in trace amount (lever rule)
a 3 : L ϕ s a 3 : L ϕ s a_(3):Lphi_(s)a_{3}: L \phi_{s} ,但 c r i c h ϕ c r i c h ϕ c-rich phic-r i c h \phi 以微量存在(杠杆法則)

a 4 a 4 a_(4)a_{4} : single O/\varnothing   a 4 a 4 a_(4)a_{4} : 單一 O/\varnothing

p: plaitpoint 褶點
: the composition of C - rich are identīcal
C - rich 的成分是相同的

tie-line ( a 2 a 2 ) a 2 a 2 (a_(2)^(')-a_(2)^(''))\left(a_{2}^{\prime}-a_{2}^{\prime \prime}\right) :the two ends of the line vepresent the composition of the two ϕ s ϕ s phi_(s)\phi_{s} in egm(empirical)
tie-line ( a 2 a 2 ) a 2 a 2 (a_(2)^(')-a_(2)^(''))\left(a_{2}^{\prime}-a_{2}^{\prime \prime}\right) :該線的兩端代表兩個 ϕ s ϕ s phi_(s)\phi_{s} 在 egm(經驗)中的組成

a 2 a 2 a_(2)^('')a_{2}^{\prime \prime} closer to A A AA-pealc than a a a^('')a^{\prime \prime} :more A A AA in w w ww-rich ϕ ϕ phi\phi
a 2 a 2 a_(2)^('')a_{2}^{\prime \prime} 更接近 A A AA -pealc 比 a a a^('')a^{\prime \prime} :更多 A A AA w w ww -豐富 ϕ ϕ phi\phi
w -rich ϕ c-rich ϕ = a 2 a 2 a 1 a 2 w -rich  ϕ  c-rich  ϕ = a 2 a 2 ¯ a 1 a 2 ¯ (w-"-rich "phi)/(" c-rich "phi)=( bar(a_(2)-a_(2))^(''))/( bar(a_(1)-a_(2)^(')))\frac{w-\text {-rich } \phi}{\text { c-rich } \phi}=\frac{\overline{a_{2}-a_{2}}{ }^{\prime \prime}}{\overline{a_{1}-a_{2}^{\prime}}}
Ex:point Z ( X A = 0.18 ) Z X A = 0.18 Z(X_(A)=0.18)Z\left(X_{A}=0.18\right) composition of two ϕ S ϕ S phi_(S)\phi_{S} and the ratio?
Ex:點 Z ( X A = 0.18 ) Z X A = 0.18 Z(X_(A)=0.18)Z\left(X_{A}=0.18\right) 由兩個 ϕ S ϕ S phi_(S)\phi_{S} 的組成及比例?
a 2 ( x c , x w , x A ) = ( 0.2 , 0.57 , 0.23 ) a 2 ( x c , x w , x A ) = ( 0.82 , 0.08 , 0.1 ) a 2 z ( c r i c h ) z a 2 ( w r i c h ) = R ( 測量 ) R ( ) a 2 x c , x w , x A = ( 0.2 , 0.57 , 0.23 ) a 2 x c , x w , x A = ( 0.82 , 0.08 , 0.1 ) a 2 z ( c r i c h ) z a 2 ( w r i c h ) = R (  測量  ) R ( 測   星  {:[a_(2)^(')(x_(c),x_(w),x_(A))=(0.2","0.57","0.23)],[a_(2)^('')(x_(c),x_(w),x_(A))=(0.82","0.08","0.1)],[(a_(2)^(')-z(c-rich))/(z-a_(2)^('')(w-rich))=(R(" 測量 "))/(R_(("測 ")" 星 "))]:}\begin{aligned} & a_{2}^{\prime}\left(x_{c}, x_{w}, x_{A}\right)=(0.2,0.57,0.23) \\ & a_{2}^{\prime \prime}\left(x_{c}, x_{w}, x_{A}\right)=(0.82,0.08,0.1) \\ & \frac{a_{2}^{\prime}-z(c-r i c h)}{z-a_{2}^{\prime \prime}(w-r i c h)}=\frac{R(\text { 測量 })}{\left.R_{(\text {測 }} \text { 星 }\right)} \end{aligned}
E x : Q ( X A = 0.34 ) E x : Q X A = 0.34 Ex:Q(X_(A)=0.34)E x: Q\left(X_{A}=0.34\right)
trace ϕ = c ϕ = c phi=c\phi=c-rich x c , x w , x A = 0 , 61 , 0 , 21 , 0 , 27 x c , x w , x A = 0 , 61 , 0 , 21 , 0 , 27 x_(c),x_(w),x_(A)=0,61,0,21,0,27x_{c}, x_{w}, x_{A}=0,61,0,21,0,27
dominant ϕ = w ϕ = w phi=w\phi=w-rich x c , x w , x A = 0 , 37 , 0 , 28 , 0 , 35 x c , x w , x A = 0 , 37 , 0 , 28 , 0 , 35 x_(c),x_(w),x_(A)=0,37,0,28,0,35x_{c}, x_{w}, x_{A}=0,37,0,28,0,35
主導 ϕ = w ϕ = w phi=w\phi=w -富裕 x c , x w , x A = 0 , 37 , 0 , 28 , 0 , 35 x c , x w , x A = 0 , 37 , 0 , 28 , 0 , 35 x_(c),x_(w),x_(A)=0,37,0,28,0,35x_{c}, x_{w}, x_{A}=0,37,0,28,0,35
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