{:[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 353K,Delta vap H=30.8kJ353 \mathrm{~K}, \Delta \operatorname{vap} H=30.8 \mathrm{~kJ} 苯的正常沸點是 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: -(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
1 mole H_(2)O(ℓ)H_{2} \mathrm{O}(\ell) at 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} \mathrm{O}(\ell) 在 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)\mathrm{H}_{2} \mathrm{O}(\ell) to a large volume of ethanol, 將 I 摩爾 H_(2)O(ℓ)\mathrm{H}_{2} \mathrm{O}(\ell) 添加到大量乙醇中,
14cm^(3)14 \mathrm{~cm}^{3} is the parthal molar volume in ethanol 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 BB surrounding AA 分子 A 所佔的體積取決於圍繞 AA 的分子 BB 的身份
H_(2)O\mathrm{H}_{2} \mathrm{O} are held apart by H -bond in C_(2)H_(5)OH,H\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, \mathrm{H}-bond break S hence Vm(H_(2)O)darr\operatorname{Vm}\left(\mathrm{H}_{2} \mathrm{O}\right) \downarrow H_(2)O\mathrm{H}_{2} \mathrm{O} 由氫鍵分開,在 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)darr\operatorname{Vm}\left(\mathrm{H}_{2} \mathrm{O}\right) \downarrow
|(Q^(A))/(B)|\left|\frac{Q^{A}}{B}\right|
Parthal volume of AA in BB
=V_(A+B)-V_(B)=V_(A)=V_{A+B}-V_{B}=V_{A}
change in composition changes T.P. properties of AA and BB 成分的變化改變了 AA 和 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)=((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)=((del V)/(deln_(j)))_(P*T,n^('))V_{J}=\left(\frac{\partial V}{\partial n_{j}}\right)_{P \cdot T, n^{\prime}} n^(')=n^{\prime}= amount of other componemt n^(')=n^{\prime}= 其他組件的數量
( n_(B)n_{B} )
vitotal volume of sample 樣本的總體積 n_(A)n_{A} : amount of AA n_(A)n_{A} : AA 的數量 V_(A)V_{A} : slope of the plot of total VV as n_(A)n_{A} changes as n_(A)n_{A} changes V_(A)V_{A} : 總 VV 隨著 n_(A)n_{A} 變化而變化時的斜率 n_(A)n_{A} rarr\rightarrow could be NeGATIVE rarr\rightarrow 可能是負面的
{:[(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}
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 =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,rho25^{\circ} \mathrm{C}, \rho of 50wt%50 \mathrm{wt} \% ethanol/water solution is 在 25^(@)C,rho25^{\circ} \mathrm{C}, \rho 的 50wt%50 \mathrm{wt} \% 乙醇/水溶液中是 0.914g//Cm^(3)0.914 \mathrm{~g} / \mathrm{Cm}^{3} given
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^(@)C25^{\circ} \mathrm{C}, a polynominal fit to the total volume of mixture is 在 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
へ n_(n)n_{n} (mole)
total vol. of amount of ethanol 乙醇的總體積量
mixuture (cm^(3))\left(\mathrm{cm}^{3}\right) 混合物 (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)=((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} you need V=f(x)V=f(x) 對於 r_(w)r_{w} ,你需要 V=f(x)V=f(x)
n_(w)^(9)n_{w}^{9}
Partial molar M\mathcal{M} can be used in any extensive state function 部分摩爾 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 (mu)(\mu) is used as P.M. Gibbs Engy. 在混合物中,化學勢能 (mu)(\mu) 被用作 P.M. 吉布斯能。