Microsoft Word - 熱力-第六章_99_.doc

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1 Chapter 6 hermodynamic roperties of Fluids 6-. roperty relations for homogeneous phases. he first law for a closed system of n moles is: du = dq + dw (For the special case of a reversible process, du = dq rev + dw rev dq rev = ds nd law (For - work only: dw rev = -d du = ds d. Define: ( H U + A U S H S du = ds d dh = ds + d da = -Sd d d = -Sd + d 不但要會證明, 且先決條件要清楚! (For a function of two variables, F(x, y, = xy hen, df = d(xy = x dy + y dx he total derivative of a function of multiple variables, F(x, y, z,... is defined as: df = ( F/ x y,z---- dx + ( F/ y x,z---- dy + ( F/ z x,y ---- dz d(x 3 y = 而 du = ds d; dh = ds + d; da = -Sd d; d = -Sd + d Southern 所以 : aiwan University du = ( --- ds ( --- d ( --- = ; ( --- = - du = ds d dh = ( --- ds + ( --- d dh = ds + d ( --- = ; ( --- = 6-

2 da = ( --- d + ( --- d da = -Sd - d ( --- = -S ; ( --- = - d = ( --- d + ( --- d d = -Sd + d ( --- = -S ; ( --- = (3he Maxwell Relationships: 任何一函數 F(x, y, z 都可應用 df = ( F/ x y,z---- dx + ( F/ y x,z---- dy + ( F/ z x,y ---- dz 而變成 : df = A dx + B dy + C dz 但不是隨便一個 M dx + N dy + Q dz 都可變成 df 的形式 d(3xy + x 3 y = A dx + B dy A = B = x y dx + 4xy 3 dy = d = 若 M dx + N dy = df 是成立的稱之為 exact differential ( M/ y x = ( N/ x y 6x y 5 dx + 0x 3 y 4 dy 3x y 4 dx + 5x 4 y dy he Maxwell Relationships: he four equations, du = ds d; dh = ds + d; da = -Sd d; d = -Sd + d, are all exact differentials. du = ds d is a exact differential ( S = -( Southern aiwan University S dh = ds + d is a exact differential ( --- = ( --- da = -Sd d is a exact differential ( --- = ( --- d = -Sd + d is a exact differential ( --- = (

3 3. Enthalpy and entropy as functions of and : ( 在均勻相中, 把 enthalpy and entropy 表示成的函數得到最有用的物性關係式 ( 以 mol 為基量 : H ( = Cp S C ( = p H S dh = ds + d ( = ( but S ( ( = - H S dh = ds + d ( = ( + = - ( + = - ( 重要的熱力學狀態方程式 he functional relations chosen here for H and S are H =H(, and S = S(, H H S S dh = ( d + ( d ds = ( d + ( d dh = C p d + ds = C p d - ( d ( d 4.Alternative Forms for Liquids: (Chapter 3: olume expansivity: β ( Isothermal compressibility: κ - ( Southern aiwan University S ( ( = - ( = -β H ( = H ( = S ( + = (-β + = (-β S ( + = - ( - ( = (-β 6-3

4 dh = C p d + ds = C p d - ( d = C p d + (-βd d ( d = Cp -βd 5.Internal energy and entropy as functions of and : 以 mol 為基量 : ( du = ds d U = S U S = - (However, U S = C = One of Maxwell s equations: = S U C = U = 重要的熱力學狀態方程式 S - = - (3 he functional relations chosen here for U and S are U = U(, and S = S(, U U S S du = ( d + ( d ; ds = ( d + ( d du = C d + [ -] d Southern aiwan University ds = C d + d Chapter 3 ( = 詳見 Example 3- du = C d + [ -] d ds = C d + d 6-4

5 Example 6-: Show the Joule-homson coefficientμ J = Sol: H H dh = d + d Joule-homson 實驗過程是 H 不變 0 = 0 = H d + H H d H + H H = C [ - ] H H H μ J = H = - H = - = - Cp Cp = - [ - ] = [ - ] C C Example 6-: 證明 ideal gas 的 internal energy 只和溫度成 du = nĉ v d 的關係而與體積無關 U U Sol: du = ( d + ( d U 而 ( = C du = ds d U S ( = - = - Southern du = C d + [ aiwan -] d University 6-5

6 Example 6-3: Show thermodynamicaly that Π = for a van der Waals gas. U = 0 for a perfect gas, and derive its value Example 6-4: Derive U = - where β= he expansion coefficient of a gas = κ= he isothermal compressibility of a gas = - Sol: A B C B C A C A B Southern aiwan University Example 6-5: rove that U = C - 6-6

7 Example 6-6: Derive the Joule coefficient for a van der Waals gas. U 6. he Helmholtz and ibbs free energy: ( 由 nd law 知 : dq ds For a spontaneous process or irreversible process: dq ds > dq<ds or dq-ds<0 only - work st law: du = dq + dw = dq-d, S 若保持不變 (du,s <0 du+ d-ds<0 自發程序若發生在系統的體積與 entropy 不變時, 則內能要減少 dq 把 ds irrev. 大於 rev. 等於轉換成另一種表示 (du,s 0 若, 為定值 du+ d-ds<0 可變成 du-d(s<0 此式被稱為 Clausius inequality or d(u-s<0 irrev. 小於 rev. 等於但 A = U-S (da, <0 Southern 自發程序若發生在系統的溫度與體積不變時 aiwan University, 則 Helmholtz free energy 要減少 dq irrev. 小於把 ds 轉換成另一種表示 (da, 0 rev. 等於若, 為定值 du+ d-ds<0 可變成 du+ d(-d(s<0 or d(u+-d(s<0 即 d(h-s<0 但 = H-S (d, <0 6-7

8 自發程序若發生在系統的溫度與壓力不變時, 則 ibbs free energy 要減少 dq 把 ds 轉換成另一種表示 (d, 0 irrev. 小於 rev. 等於由於一般的反應或程序最常在恆溫恆壓下進行, 所以, 以值來判斷該反應 ( 程序是否為自發為最重要 (Maximum work and Helmholtz free energy : 一個系統對外作可逆 work 時最大的 work isothermal & reversible 的條件下 : 即 W rev. = W max ds = dq rev dqrev = ds da = du-d(s = du-ds = dq rev + dw rev -ds = ds+ dw rev -ds = dw rev = dw max ΔA = W max ΔA 是一個程序進行時, 所能做的最大 work Helmholtz free energy 又被稱為 work function (3 reversible 的條件下 : du = dq rev + dw( 各種功 = ds + (-d + dw net W net = 除了 - work 之外的所有 work d = dh d(s = d(u + d(s = du + d( - d(s = [ds + (-d + dw net ] + [d +d] - [ds+sd] =-Sd + d + dw net 恆溫恆壓可逆程序時 : d = dw net or Δ = W net 恆溫恆壓可逆程序時,ibbs free energy 的變化等於 - work 之外的所有 work Southern (4Basic calculating ofδ: aiwan University 一般從 = H S 的定義即可求得 Δ 在 isothermal 下, 有時需考慮以下的 case: d = -Sd + d = d Δ = d = d nr for ideal gases: Δ = d = d = nr ln for liquid or solid :Δ =( 6-8

9 Example 6-7: Calculate q, W,ΔU,ΔH,ΔS, andδ for the conversion of 3 moles of H O from liquid water to water vapor at atm, 00? Example 6-8: wo moles of an ideal gas expands from 3 atm, 49. liters to atm isothermally and reversibly. Find q, W,ΔU,ΔH,ΔS, andδ for this process? Example 6-9: 計算下列程序的自由能變化? H O(l, -0 H O(l, -0 水在 0 之 vapor pressure:.49 mm-hg; 冰在 0 之 vapor pressure 為.950 mm-hg Southern ibbs free energies of Chemical aiwan Reactions University We define standardfree energies of formation Δ f o similarly to the enthalpies of formation, Δ f H o. Δ f o 和 Δ f H o 具有類似的特色且求 Δ rxn o 與求 Δ rxn H o 的計算方式相同 rxn H o =Σγ j x f H o (products-σγ i x f H o (reactants rxn o =Σγ j x f o (products-σγ i x f o (reactants 若 chemical reaction 是在恆溫下進行, 若可以求出此反應的 rxn S o 與 rxn H o, 則 rxn o = rxn H o rxn S o 6-9

10 6-0 Example 6-0: Determine rxn (98K for the following chemical reaction: H [g] + O [g] H O [l] H [g] O [g] H O [l] f H o, kj/ml S o, J/mol K f o, kj/mol Sol: Focus on ---he variation of the ibbs free energy with temperature ibbs-helmholtz equation: = H S or H = + S = ( ( = ( = -S = ( = S = H = H or = H or ( = H is called the ibbs-helmholtz equation. = H ( d = - d H ( = - d H ( Southern aiwan University

11 Example 6-: 知標準生成自由能,Δ o NO(g=0.7 kcal/mol;δ o NO(g=.4 cal/mol 標準生成焓, ΔH o NO(g=.6 kcal/mol;δh o NO(g=8. cal/mol 且 O 的 Cp =6.5;NO 的 C p =7.0; NO 的 C p =9.0[ 單位皆是 (cal/mol K], 求 atm, 00 時, NO (g + O (g NO (g 之 Δ? 標準生成自由能標準生成焓 --- 等, 若沒特別說明就當成 atm 5 7. he ibbs energy as a generating function: ( d = - Sd + d = (, 的函數關係是存在的 he special [or canonical( 正則 ] variables for the ibbs energy are temperature and pressure. (d( = R H S d - R d = (d Sd- R R d R H = d- R R d Southern aiwan University 6-. Residual properties. Define: (the residual ibbs energy R - ig :actual gas 的 ibbs free energy ig :ideal-gas 的 ibbs energy 全都是在 mol 且同溫同壓的條件下 (the residual volume R - ig =-(R/ 6-

12 For real gas: Z =(/R = ZR/ R - ig R =-(R/ = (Z- (3he definition for the generic residual property is: M R M-M ig M 與 M ig 分別為 mol 的 actual gas 與 ideal-gas 的 extensive property, 如 U H S 與且必需為同溫同壓下的數值. = H S R = H R S R H d( = d- R R R at constant temperature real gas: d( = R R ig ideal gas: d( R d d ig = d R R R R d( = d R R R = R R =( Z R d= [ (Z-]d R d 求 Residual properties, 除了知道定義之外, 先會求 real gas 與 ideal gas 的性質, 再互相減扣即可所以, 6-3 不再介紹 6-3. wo-phase systems. oint X-- two phases α and β of a pure +d α β Southern species coexist at, in equilibrium, aiwan University 力( 蒸 = 0 氣 ( 取 mole 為基量當溫度變成 + d 時, α phase 的 ibbs energy d β phase 的 ibbs energy d 壓溫度壓 +d 6-

13 但仍要保持 α 與 βphase 的向平衡 d d 即 d d 但 d Sd d S d d S d d d S S = d = S 另一方面, H S 定溫時, H tr S 但既然任何 phase line 的點都代表相平衡 0 H tr S d = d H tr d = d d = d d = d H vap ( g H fus ( s H sub ( g s liquid-vapor equilibrium solid-liquid equilibrium solid-vapor equilibrium Clapeyron eq. Example 6-: 已知水和冰的密度分別是 0.97 g/cm 3 與 g/cm 3, 而水的凝固熱為 J/mol, Southern aiwan University 則壓力為 00 atm 時, 水的 melting point 為多少? 6-3

14 . 考慮 liquid-vapor equilibrium: d d H vap ( g ( g d d H g vap ( 把 vapor 視為 ideal gas g = R/ d d H g vap H vap R or H vap d = R d d H R p vap d 同理 H vap ln = ( R H sub ln = ( R Clausius-Clapeyron equation (3 d H vap d = d R = H vap ln = ( + C R y = mx + c H vap d + C R ln Slope = -ΔH/R Southern aiwan University vap H 以ln 對 / 作圖得 slope = R 之直線 / 6-4

15 Example 6-3: Find the boiling point of water at 750 mm-hg if the heat of the vaporization of water is 539 cal/g at 00, atm. Example 6-4: he normal boiling point of toluene is 0.6. Find the vapor pressure of toluene at 77 if the heat of vaporization is kj/mol. Example 6-5: 已知固體苯在 43K 及 73K 的飽和蒸氣壓為.4 與 mm-hg; 液體苯在 303K 及 83K 的飽和蒸氣壓為 9 與 46.4 mm-hg 請求出苯的(A 熔融熱 (Btriple point? (A487cal/mol (B39.6 mm-hg; 80K Example 6-6: Southern aiwan University he vapor pressure of water at varius temperature are listed in the following table. Calculate the heat of vaporization of water. (K (atm sol: 將上面的 able 改成 / 與 ln /.74x0-3.86x x x0-3 ln

16 再以 ln 對 / 作圖, 求其 slope 0 - ln y = x E E E-03 / vap H slope = = H vap = x 8.34 = 446 J/mol R 不可以從 able 中, 任意找兩組數據代入 H vap ln = ( R 來求 H vap Southern aiwan University 6-6

hap First Law of hermodynamics Isochoric process: Isochoric rocess: hange in the state at constant volume,, ƌq /d (U / ) ; m mol J K - mol - kg J K -

hap First Law of hermodynamics.6 hange in State at onstant olume : : heat capacity : ƌq / d (ƌ: inexact differentials, ƌq ) U,, du,, :,, U f (, ) : du (U/ ) d + (U/) d, du ƌq ex d (for -work only, ƌw -