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 ( ℓ ) H 2 O ( ℓ ) 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 ( ℓ ) H 2 O ( ℓ ) H_(2)O(ℓ)H_{2} \mathrm{O}(\ell) 在 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 ( ℓ ) H 2 O ( ℓ ) H_(2)O(ℓ)\mathrm{H}_{2} \mathrm{O}(\ell) to a large volume of ethanol,
將 I 摩爾 H 2 O ( ℓ ) H 2 O ( ℓ ) 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 和 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
在 25 ∘ C , ρ 25 ∘ C , ρ 25^(@)C,rho25^{\circ} \mathrm{C}, \rho 的 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
在 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}
へ 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
與 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