1、2018/9/23,1,提示: 放映时按鼠标右键,从快捷菜单中选择“定位”“以前查看过的”, 便可返回到调用对应例题的原文稿处; 选择“定位”“按标题” “例？”，便可选择例题.,1molH2由 p1 = 101.325kPa, t1 = 0,分别经(a)向真空膨胀; (b)反抗恒定外压 p(环) = 50.663kPa, 恒温膨胀至p2 = 50.663kPa, 试求两种不同途径中系统与环境交换的体积功W(a)及W(b).,两种途径均为恒外压膨胀, 且都是恒温. (1) p(环, a) = 0, W(a) = p(环, a) (V2V1) = 0,(2) p(环, b) = 50.663kP
2、a ,例1,功并非状态函数的增量, 而是途径函数.,p(环) = 50.663kPa ,1molH2 p1 = 101.325kPa T1 = 273.15K,1molH2 p2 = 50.663kPa T2T1 = 273.15K,p(环) = 0,2018/9/23,2,3mol理想气体于恒温298.15K条件下由始态 V1 = 20.0dm3可逆膨胀到末态V2 = 50.0dm3 . 求始, 末态气体的压力p1 , p2 以及膨胀过程的可逆功Wr .,p1 = nRT/V1 = 3 8.314 298.15/(20 103) Pa = 371.8 kPa,p2 = nRT/V2 = 3
3、8.314 298.15/(50 103) Pa = 148.7 kPa,因是理想气体, 恒温, 可逆过程, 故,例2,2018/9/23,3,n = 50 g / 28 gmol1 = 1.79 mol,逆向反抗恒外压膨胀回到始态, p(环) = p1 = 0.1 Mpa,正向恒温可逆压缩,298 K时, 将0.05 kg的N2由0.1 MPa恒温可逆压缩到 2MPa, 试计算此过程的功. 若压缩后的气体再反抗0.1MPa的外压力 进行恒温膨胀回到始态, 问此过程的功又是多少?,例3,2018/9/23,4,理想气体恒温可逆膨胀, 从V1至10V1, 对外作功41.85 kJ, p1 = 2
4、02.6 kPa, 求V1; 又若气体为2 mol时, 温度为多少?,由 W = nRTln(V2 / V1)= nRTln(10V1 /V1) = 2.303nRT = 41850 J 求得 nT = 2185.7 molK 故 V1 = nRT / p1 = (8.3142185.7 / 202600)m3= 0.0897 m3 = 89.7dm3 因 n = 2 mol 则 T = (2185.7 / 2)K = 1092.85 K,例4,2018/9/23,5,求5 mol H2 (视为理想气体)在下列各过程中的体积功. (1)由300 K, 100 kPa恒压下加热到800 K; (
5、2)5 mol H2由300 K, 100 kPa恒容下加热到800 K; (3)5 mol H2由300 K, 1.0 MPa恒温可逆膨胀到1.0 kPa; (4)5 mol H2由300 K, 1.0 MPa自由膨胀到1.0 kPa.,(1) 恒压过程 p (环) = p1 = p2 = p W = p(V2 V1) = nR (T2 T1) = 20.78 kJ (2) 恒容过程, W = 0. (3) 理想气体恒温可逆过程W1 = nRTln(p1 / p2) = 86.15 kJ (4) 理想气体自由膨胀, W = 0.,例5,2018/9/23,6,10 mol的理想气体, 压力1
6、013 kPa, 温度300 K, 分别求出恒温时下列过程的功: (1)向真空中膨胀; (2)在外压力101.3 kPa下体积胀大1 dm3; (3)在外压力101.3 kPa下膨胀到该气体压力也是101.3 kPa; (4)恒温可逆膨胀至气体的压力为101.3 kPa.,(1) W = 0 p(环) = 0 (2) W = 101.31 J = 101.3 J (3) W = p(环)(nRT /p2 nRT/p1)= 101.3108.314300 (1/101.3 1/1013)J = 22.45 kJ (4) W = nRTln(p2 / p1)= 108.314300ln(101.3
7、 / 1013) J = 57.43 kJ,例6,2018/9/23,7,计算2 mol理想气体在以下过程中所作的功: (1)25时, 从10.0 dm3恒温可逆膨胀到30.0 dm3; (2)使外压力保持为101.3 kPa , 从10.0 dm3恒温膨胀到30.0 dm3; (3)在气体压力与外压保持相等并恒定的条件下, 将气体加热, 温度从T1 = 298 K升到T2, 体积从10.0 dm3膨胀到30.0 dm3.,(1)理想气体恒温可逆过程= 2 8.314 298 ln(30.0 / 10.0) J = 5.44 kJ,(2)恒外压过程, p(环) = 101.3 kPaW2 =
8、p(环)(V2 V1) = 101.3(30.010.0)103 kJ = 2.03 kJ,(3)恒压过程, p(环) = p1 = nRT/V1 = (28.314298)/(10.0103)Pa = 4.96 105 Pa W3 = p(V2 V1) = 4.96105 (30.0 10.0)103 J = 9.91kJ,例7,2018/9/23,8,(1) W = p(环)(V2 V1)= 101.3103 (1.6771.043103 )18.02103 J = 3059 J,(2) W p(环)V2 = 101.3103 1.67718.02103 J = 3061 J,W = p(
9、环)(V2V1) 101.3(30.6318.021.043103)J =3101J 或 W= p(环)(V2 V1) p(环)V2 = nRT= 1 8.314 373.15 J = 3102 J,计算: (1)1 mol水在100, 101.3 kPa下气化时的体积功. 已知在100, 101.3 kPa时水的比体积为1.043103dm3g1, 水蒸气的比体积为1.677dm3g1, H2O的摩尔质量为18.02 gmol1; (2)忽略液体体积计算体积功; (3)把水蒸气看做理想气体, 计算体积功.,例8,2018/9/23,9,101.3 kPa下, 冰在0的体积质量(密度)为0.9
10、168 106 gm3, 水在100时的体积质量为0.9584 106 gm3, 试求将1mol 0的冰变成100的水的过程的功及变成100的水蒸气过程的功. 设水蒸气服从理想气体行为. H2O的摩尔质量为18.02 gmol1.,冰变成水: W = p(环)V(水) V(冰)= 101.3 103(18.02 / 0.9584 18.02 / 0.9168) 106 J = 0.0864 J,冰变成水蒸气: W = p(环)V(气)V(冰) = 101.3103 8.314373 /101.310318.02 /(0.9168106) J = 3099 J,例9,2018/9/23,10,求
11、100 g, 25的饱和水蒸气恒温可逆膨胀到0.101 kPa时的功. 已知25时水的饱和蒸气压为3.17 kPa. 设水蒸气符合理想气体行为. (已知H2O的摩尔质量为18.02 gmol1.),例10,2018/9/23,11,计算25时50 g 铁溶解在盐酸中所做的功: (1)在密闭容器中, (2)在敞开的烧杯中. 设外压力恒定, 产生的H2视作理想气体, 液体体积可忽略. (已知Fe的摩尔质量为55.8 gmol1.),Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g) (1) V = 0, W = p(环)V = 0,(2) H2反抗恒外压作功: W = p(环)(
12、V2 V1)= p(环)V2 = pV2 = nRT,例11,2018/9/23,12,物质的量为 n的理想气体由始态 p1, V1 , T 恒温变化压力, 体积, 到达末态 p2, V2 , T, 求过程的焓变 H.,H = U + (pV) = U + (nRT),理想气体恒温的单纯 pVT 变化中, U = 0, 所以 H = U + nR (T) = 0,计算表明, 理想气体作单纯 pVT 变化时其焓不随系统的压力和体积等变化, 或者说这种情况下系统的焓只是温度的函数.,例12,2018/9/23,13,已知在101.3 kPa下, 18时1 mol Zn溶于稀盐酸时放出151.5 k
13、J的热, 反应析出1 mol H2气. 求反应过程的W, U, H.,W =p(V2V1) =n(g)RT =n(H2)RT= 18.314291.15J= 242 kJ H = Qp =151.5 kJ U = QW =151.5 kJ2.421 kJ =153.9 kJ,例13,2018/9/23,14,3 mol单原子理想气体, 从始态T1 = 300 K, p1 = 100 kPa, 反抗恒外压50 kPa作不可逆膨胀, 至终态T2 = 300 K, p2 = 50 kPa, 求这一过程的Q, W, U, H.,T2 = T1 U = 0, H = 0 W = p(环) (V2V1)=
14、 p(环) (nRT/ p2nRT/ p1)= 3.741 kJ Q = W = 3.741 kJ,例14,p(环) =50 kPa,2018/9/23,15,某压缩机气缸吸入101.325kPa, 25的空气, 经压缩后压力提高至192.5kPa, 相应使温度上升到79. 已知该温度范围内空气的CV,m近似为25.29Jmol1 K1, 试求1mol空气压缩过程的W, Q及系统的U.,W = U = 1366J,U = U1 = nCV,m(T2T1) = 1 25.29(352.15298.15)J = 1366J,因是非恒容过程, Q不是恒容热, 故Q U, 也不能用CV,m直接求非恒容
15、过程的热.,(空气视为理想气体),例15,2018/9/23,16,2 mol H2从400 K, 100 kPa恒压加热到1000 K, 已知Cp,m(H2) = 29.2 Jmol1K1, 求U, H, Q, W各为多少?,Qp = 2 mol29.2 Jmol1K1(1000400)K= 35.04 kJ U = H (pV) = HnR(T2T1) = 25.06 kJ W = UQ =9.98 kJ,例16,2molH2 p1 = 100kPa T1 = 400K,2molH2 p2 p1 = 100kPa T2= 1000K,dp=0,2018/9/23,17,0.2 mol某理想
16、气体, 从273 K, 1 MPa恒压加热到523 K, 计算该过程的Q, W, U, H. 已知该气体的Cp, m = (207103 T / K)Jmol1K1.,Q = = na(T2T1)b(T22T12)/2 = 1.14 kJW = pV = nRT = 0.42 kJ U = QW = 0.72 kJ H = Q = 1.14 kJ,例17,2018/9/23,18,3 mol某理想气体, 在101.3 kPa下恒压由20加热到80, 计算此过程的Q, W, U, H. 已知该气体 CV, m = (31.3813.4103 T / K)Jmol1K1.,U = QW = (7.
17、921.50)kJ = 6.42 kJ,W = p(环)(V2V1)= nR(T2T1)= 38.314(8020)J = 1497 J = 1.50 kJ,例18,2018/9/23,19,3 mol某理想气体由409 K, 0.15 MPa经恒容变化到 p2 = 0.10 MPa, 求过程的Q, W, U及H. 该气体Cp, m = 29.4 Jmol1K1.,T2 = p2T1 / p1 = (0.10409 / 0.15)K = 273 KQV = U = n CV, m T= 3 mol(29.48.31)Jmol1K1(409273)K= 8.635 kJ W = 0,H = U
18、+ (pV) = UnRT= 8635 J3 mol8.314 Jmol1K1(409273)K= 12.040 kJ,例19,2018/9/23,20,(1) dV = 0, W = 0QV = U = nCV, m (T2T1) = (1020283)J = 56.6 kJH = nCp, m(T2T1) = 10(208.314)283J =80.129 kJ,(2) Q = 0, U2 = U1 = 56.6 kJ, H2 = H1 = 80.129 kJ W = U2 = 56.6 kJ,(3) dp = 0, U3 = U1 = 56.6 kJQp = H3 = H1 = 80.1
19、29 kJ,某理想气体, 其CV, m = 20 Jmol1K1, 现有10 mol该气体处于283 K, 采取下列不同途径升温至566 K. 试计算各个过程的Q, W, U, H, 并比较之. (1)体积保持不变; (2)系统与环境无热交换; (3)压力保持不变.,例20,2018/9/23,21,已知液体A的正常沸点为350 K, 此时A的气化焓 vapHm = 38 kJmol1. A蒸气视为理想气体, 其平均恒压摩尔热容为30 JK1mol1. 试求2mol A从400K, 50.663kPa的气态变为350K, 101.325kPa的液态的U和H.,H1 = nCp, m(T2T1)
20、= 2 mol 30 JK1mol1 (50)K =3.00 kJ H2 = n liqHm =2 mol 38 kJmol1 =76 kJ H = H1 + H2 = (763.0)kJ =79 kJ U = H(pV) H(pVg) = H + nRT =79 kJ + 2 8.314 400 103 kJ =72.35 kJ,A(蒸气)n = 2molT1 = 400Kp1 = 50663Pa,A(液体)n = 2molT3 = 350Kp3 = 101325Pa,A(蒸气)n = 2molT2 = 350Kp2 = 101325Pa,例21,2018/9/23,22,已知苯在101.3
21、 kPa下的熔点为5. 在5时, fusHm = 9916 Jmol1, , , 计算在101.3 kPa, 5下的fusHm.,例22,2018/9/23,23,水在101.3 kPa, 100时, vapHm= 40.59 kJmol1. 求10 mol 水蒸气与水的内能之差. (设水蒸气为理想气体, 液态水的体积可忽略不计.),U = Hp(VgVl ) HpVg = HnRT= 10mol 40.95kJmol110mol 8.314Jmol1K1 373.15K= 400.95 kJ31.024 kJ = 369.9 kJ,例23,2018/9/23,24,已知 25时乙炔C2H2(
22、g)的标准摩尔生成焓fHm(C2H2, g) =2267 kJmol1, 标准摩尔燃烧焓cHm(C2H2 g) =12996 kJmol1, 及苯C6H6(l)的标准摩尔燃烧焓cHm(C6H6, l ) =32675kJmol1. 求25时苯的标准摩尔生成焓 fHm(C6H6, l)., fHm(C6H6 , l)= 3 fHm(C2H2, g) + 3cHm(C2H2, g)cHm(C6H6, l)= 3 2267 + 3 (12996)(32675) kJmol1= 488 kJmol1,例24,2018/9/23,25,先求f Hm (环丙烷, 298 K): 3C(石墨) + 3H2(
23、g) = C3H6(环丙烷, g )f Hm(环丙烷, 298 K) =3cHm(石墨, 298 K)+3c Hm (H2, g, 298K) cHm(环丙烷, 298 K)= 5363 kJmol1 则 r Hm(298 K) = f Hm(丙烯, g, 298 K) f Hm(环丙烷, g, 298 K) =3323 kJmol1.,已知某些物质的标准摩尔燃烧焓与标准摩尔生成焓的数据: 物 质 cHm(298 K)/ kJmol1 f Hm(298 K)/ kJmol1H2(g) 28584 0C(石墨) 39351 0 C3H6 (环丙烷, g) 209168 C3H6 (丙烯, g)
24、2040 计算由环丙烷(g)异构化制丙烯(g)时在298 K的rHm .,例25,2018/9/23,26,求下列反应在393 K 的rHm(393 K): C2H5OH(g) + HCl(g) = C2H5Cl(g) + H2O(g)物质 f Hm (298 K)/ kJmol1 Cp, m / JK1mol1 C2H5Cl(g) 105.0 13.07 +188.5103(T /K)H2O(g) 241.84 30.00 +10.71103(T /K)C2H5OH(g) 235.3 19.07 +212.7103(T /K)HCl(g) 92.31 26.53 +4.62103(T /K)
25、,rHm(393K) = rHm(298K) += 19.23103Jmol1 + 2.5318.09103 (T/K)JK1 mol1 dT=20.06 kJmol1,例26,r Hm(298K) =f Hm(C2H5OH,g, 298K)f Hm(HCl, g,298K)+ f Hm(C2H5Cl, g, 298K) + f Hm(H2O,g, 298K)= 19.23 kJmol1B Cp, m(B) = 2.5318.09103 (T/K) JK1mol1,2018/9/23,27,气相反应 A(g) + B(g) = Y(g) 在500, 100 kPa 进行时, Q, W, r H
26、m, rUm各为多少, 并写出计算过程. 已知数据(Cp, m的适用范围为 25 800) : 物质 f Hm(298 K)/ kJmol1 Cp, m / JK1mol1 A(g) 235 19.1 B(g) 52 4.2 Y(g) 241 30.0,r Hm(298 K) =58 kJmol1 BCp, m (B) = 6.7 JK1mol1r Hm(773 K) = r Hm(298 K) + =54.82 kJmol1 Q = r Hm(773 K) =54.82 kJmol1,例27, rUm = rHm B (g)RT =48.39 kJmol1 W =pV =B(g)RT = 6
27、.43 kJmol1, rUm = rHm B (g)RT =48.39 kJmol1 W =pV =B(g)RT = 6.43 kJmol1,2018/9/23,28,求反应2H2(g)+ O2(g) = 2H2O(l)的标准摩尔焓rHm(T)随温度T的函数关系式. 已知: 2H2(g) + O2(g) = 2H2O(l) rHm(298 K) =573.208 kJmol1Cp, m(H2, g) = ( 27.20 + 3.8103(T / K) JK1mol1; Cp, m(O2, g) = ( 27.20 + 4.0103(T / K) JK1mol1; Cp,m(H2O, g) =
28、 36.867.9103(T/K) + 9.28106(T/K)2 JK1mol1.,rHm(T) = H0 + aT +(1/2)bT 2 +(1/3)cT 3 因 a = 236.86 227.2027.20 = 7.88 JK1mol1b / 2 = 2(7.9103)23.81034103 /2= 13.7103 JK1mol1c / 3 = (29.28 106 ) / 3 = 6.19106 JK1mol1 得 rHm(T)/ Jmol1 = H0 / Jmol17.88 (T / K)13.7103(T / K)2 + 6.19106 (T / K)3,例28,2018/9/23
29、,29,将 rHm(298 K) =573 208 Jmol1 代入上式, 得H0 = 573 208 + 7.88(298)+ 13.7103 (298)26.19106 (298)3 Jmol1= 570 902 Jmol1,可得题给反应的标准摩尔焓rHm(T)随温度T的函数关系式: rHm(T) = 570 9027.88(T / K)13.7103(T / K)2 + 6.19106 (T / K)3 Jmol1,2018/9/23,30,试将甲烷的标准摩尔生成焓表述为温度的函数, 并计算在500 K时, 从单质生成甲烷的标准摩尔焓变. 已知 C(石墨) +2H2(g) = CH4(g
30、) rHm(298 K) =75. 25 kJmol1; Cp,m(C,石墨,s) = 4.662+20.1103(T/K)5.02106 (T/K)2 JK1mol1; Cp, m(H2, g) = 27.196 + 3.8103(T/K) JK1mol1; Cp, m(CH4, g) = 12.552 + 95.4103(T/K)20.08106 (T/K)2 JK1mol1.,B Cp, m(B) =Cp, m(C,石墨, s)2Cp, m(H2, g) + Cp, m(CH4, g)= 46.5 + 67.7103(T / K)15.06106(T / K)2 JK1mol1,例29,
31、2018/9/23,31,将T = 298 K及fHm(CH4 , g , 298K) =75.25 JK1mol1 代入 得 H0 =64 270 Jmol1 故 fHm(CH4 , g , T)/ Jmol1 =64 27046.5(T/K)+ 33.9103(T/K)2 +5.02106 (T/K)3 将 T = 500 K 代入 得 f Hm(CH4 , g , 500 K) =79.67 JK1mol1, f Hm(CH4, g, T)/Jmol1 = H0/Jmol146.5(T/K) +33.9103(T/K)2 + 5.02106(T/K)3,2018/9/23,32,已知下列
32、反应在298K下的标准摩尔反应焓: (1) C2H6(g) +(7/2)O2(g) = 2CO2(g) + 3H2O(l), Hl =1426.8 kJmol1; (2) H2(g) + (1/2)O2(g) = H2O(g), H2 =241.84 kJmol1; (3) (1/2)H2(g) + (1/2)Cl2(g) = HCl(g), H3 =92.3 kJmol1 ; (4) C2H6(g) + Cl2(g) = C2H5Cl(g) + HCl(g), H4 =112.0 kJmol1 ; 求反应 C2H5Cl(g) + (13/4)O2 = (1/2)Cl2(g) + 2CO2(g
33、) + (5/2)H2O(g) 的rHm(298 K), rUm(298 K).,由 (1)式(4)式 + (3)式(2)式(1/2), 得: C2H5Cl(g) + (13/4)O2 = (1/2)Cl2(g) + 2CO2(g) + (5/2)H2O(g)rHm(298 K) = Hl H4 + H3H2(1/2) = 1 286 kJmol1rUm(298 K) = rHm (298 K)BB(g)RT= 1286 kJmol1 0.758.314103298 K= 1288 kJmol1,例30,2018/9/23,33,试从H = f (T, p)出发, 证明: 若一恒量某种气体从2
34、98K, 100kPa 恒温压缩时, 系统的焓增加, 则气体在298K, 100kPa下的节流膨胀系数(即J-T系数)J -T 0.,由H = f (T, p) 得,例31,2018/9/23,34,试证: 理想气体的 J -T = 0 .,例32,2018/9/23,35,例33,2018/9/23,36,实际气体分子间力不可忽略, 节流为降压过程, 亦即体积增大, 拉开分子间的距离, 克服引力需要吸收能量, 而使内势能增加, 即 (U/p)T 0 , (T/p)H 0 说明节流过程 (dp 0 , 则当|(U/p)T | | (pV)/pT | 时, uH 为正值; 当 |(U/p)T |
35、 |(pV)/pT | 时为负值. 究竟是正是负要看气体的性质及所处温度, 压力而定.,2018/9/23,37,1 mol氧气由0, 106Pa，经过（1）绝热可逆膨胀；（2）对抗恒定外压Psu= 105Pa绝热不可逆膨胀，使气体最后压力为105Pa，求此两种情况的最后温度及环境对系统所作的功。 已知：已知该气体 CV, m = 21.05Jmol1K1.,例34,已知该气体 CV, m = 21.05Jmol1K1.,解（1）绝热可逆膨胀,1molO2 p1 = 106Pa T1 = 273.15K,1molO2 p2 = 105Pa T2= ?K,绝热可逆膨胀,2018/9/23,38,绝热过程： W1 = U = nCV, m(T2- T1) = 1mol21.05 Jmol1K1 (142.4-273.15)K =-2753J,(2)绝热恒外压逆膨胀,1molO2 p1 = 106Pa T1 = 273.15K,1molO2 p2 = 105Pa T2= ?K,绝热恒外压膨胀,因不可逆,Psu= 105Pa,2018/9/23,39,由此可见，由同一始态经过可逆与不可逆两种绝热变化不可能达到同一终态，即 ; 。,
