分享第一章 热力学第一定律 习题答案.doc

2011精品第一章 热力学第一定律练习参考答案1. 一隔板将一刚性绝热容器分成左右两侧，左室气体的压力大于右室气体的压力。现将隔板抽去，左、右气体的压力达到平衡。若以全部气体作为体系，则U、Q、W为正？为负？或为零？解: p2p1 刚性绝热容器 Q=0，W=0，U=02. 试证明1mol理想气体在恒后下升温1K时，气体与环境交换的功等于摩尔气体常数R 。解: 恒压下，W= - p外V= - p外= - R (p外= p，n=1mol，T=1 )3. 已知冰和水的密度分别为0.92103 kgm-3和1.0103 kgm-3，现有1mol的水发生如下变化:(1) 在100、101.325kPa下蒸发为水蒸气，且水蒸气可视为理想气体；(2) 在0、101.325kPa下变为冰。试求上述过程体系所作的体积功。解: 恒压、相变过程，(1) W= -p外(V2 V1) = - 101.325103=-3100 ( J )(2) W= - p外(V2 V1) = - 101.325103= -0.16 ( J )4. 若一封闭体系从某一始态变化到某一终态。(1) Q、W、QW、U是否已完全确定；(2) 若在绝热条件下，使体系从某一始态变化到某一终态，则(1)中的各量是否已完全确定？为什么？解:(1) Q+W、U完全确定。( Q+W=U；Q、W与过程有关)(2) Q、W、Q+W、U完全确定。(Q=0，W = U)5. 1mol理想气体从100、0.025m3经下述四个过程变为100、0.1m3:(1) 恒温可逆膨胀；(2) 向真空膨胀；(3) 恒外压为终态压力下膨胀；(4) 恒温下先以恒外压等于0.05m3的压力膨胀至0.05m3，再以恒外压等于终态压力下膨胀至0.1m3。求诸过程体系所作的体积功。解: 纯p V T 变化。(1) W= - p外dV= - = - nRT ln= - 18.314373ln = - 4300 ( J )(2) W= -p外dV = 0(3) 恒压，W= - p外(V2 V1) = - (V2 V1) = - (0.1-0.025) = - 2326 ( J )(4) 恒温、恒外压，W= - p外1(V2 V1)-p外2(V3 V2)= -(V2 V1)-(V3 V2)= - (0.05-0.025) - (0.1-0.05)=-3101 ( J )6. 在一个带有无重量无摩擦活塞的绝热圆筒内充入理想气体，圆筒内壁上绕有电炉丝。通电时气体缓慢膨胀，设为等压过程。若(1)选理想气体为体系；(2)选电阻丝和理想气体为体系。两过程的Q、H分别是等于、小于还是大于零？解:(1) 理想气体体系: Qp = H 0(2) （理想气体电阻丝）体系: Qp = 0；H = W电 0( 由 U2 U1 = Qp +(p2V2 p1V1)+ W电 得 H= Qp+ W电 )7. 在373K和101.325kPa的条件下，1mol体积为18.80 cm3的液态水变为30200 cm3的水蒸气，已知水的蒸发热为4.067104 Jmol -1。求此过程体系的H及U。解: 定T、p下的相变。H= Qp =14.067104=4.067104 ( J )W= - p外(V2 V1) = -101.325103(3020018.80)10-6 = - 3058 ( J )U= Q+W=3.761104 ( J )8. 分别判断下到各过程中的Q、W、U和H为正、为负还是为零？(1) 理想气体自由膨胀。(2) 理想气体恒温可逆膨胀。(3) 理想气体节流膨胀。(4) 理想气体绝热、反抗恒外压膨胀。(5) 水蒸气通过蒸气机对外做出一定量的功之后恢复原态，以水蒸气为体系。(6) 水(101325 Pa，273.15K) 冰(101325 Pa，273.15K)。(7) 在充满氧的定容绝热反应器中，在墨剧烈燃烧，以反应器及其中所有物质为体系。解:(1) W= - p外V=0，Q=0，U=H=0(2) U= 0，H=0，W= - Q= -p外Dv = - nRT ln 0(3) U= 0，H=0，W= - Q=0(4) Q=0，W= - p外V 0，U 0，H0 (降温)(5) U=H=0，Q0，W0(6) 定T、p相变。Q0，W0，U 0，H0 (H= Qp)(7) Q=0，W= 0，U =0，H=U + Vp09. 已知H2(g)的Cp,m(29.070.83610-3T2.0110-6T2) JK-1mol -1，现将1mol的H2(g)从300K升至1000K，试求:(1) 恒压升温吸收的热及H2(g)的H；(2) 恒容升温吸收的热及H2(g)的U。解: 纯p V T 变化。(1) H= Qp =nCp,mdT =129.07(T2 T1)0.83610-3(1/2)(T2 2T12)2.0110-6(1/3)(T2 3T13)=20620 ( J )(2) U= QV =nCV,mdT =nCp,mdTnR dT=2062018.314(1000300) = 14800 ( J )10. 在0和506.6kPa条件下，2 dm3的双原子理想气体体系以下述二个过程恒温膨胀至压力为101325 Pa，求Q、W、U和H。(1) 可逆膨胀；(2) 对抗恒外压101.325 kPa膨胀。解: 恒温、纯p V T 变化。(1) U=H=0，W1 = -Q1 = - pdV = - nRT ln = -RT ln= - p1V1 ln= - 506.6103210-3ln= - 1631 ( J )(2) U=H=0，W2 = - Q2 = - p外(V2 V1) = - p外()= - p外()= -101.32510-3()= - 811 ( J )11. (1)在373K、101.325 kPa下，1mol水全部蒸发为水蒸气，求此过程的Q、W、U和H。已知水的汽化热为40.7 kJmol -1。(2)若在373K、101.325kPa下的1mol水向真空蒸发，变成同温同压的水蒸气，上述各量又如何？(假设水蒸气可视为理想气体)。解: (1) 恒T、p下的相变。水的密度为1 cm3/g。H= Qp =140.7103= 40.7103 ( J )W = - p外(V2 V1) =- p外(V1)= - 101.32510-3(1810-6) = -3099 ( J )U= Q+W= 40.71033099 =37601 ( J )(2) 与(1)的始、终态相同， H = 40.7103 ( J ) ， U=37601 ( J )W = - p外(V2 V1) =0 ( p外 = 0 )Q =U = 37601 ( J ) (HQ)12. 1mol单原子理想气体，始态压力为202.65 kPa，体积为11.2 dm3，经过pT为常数的可逆压缩过程至终态压力为405.3 kPa，求:(1) 终态的体积与温度；(2) 体系的U及H；(3) 该过程体系所作的功。解: 纯p V T 可逆变化。(1) T1 = = = 273 ( K )T2 = = = 136.5 ( K )V2 =2.810-3 ( m3 ) =2.8 (dm3 )(2) U = nCV,mdT = 1(3/2)8.314(136.5 -273) = -1702 ( J )H = nCp,mdT =n(CV,m+R )dT= 1(5/2)8.314(136.5 -273) = -2837 ( J )(3) p = =， p = ( 常数= p1T1 = p2T2 )W = -pdV = -dV = -2(2.810-3)0.5 (11.210-3)0.5 = 2269.7 ( J ) = 2.27 ( kJ )13. 某理想气体的CV,m20.92 JK-1mol -1，现将1mol的该理想气体于27、101.325kPa时受某恒外压恒温压缩至平衡态，再将此平衡态恒容升温至97，此时压力为1013.25kPa。求整个过程的Q、W、U和H。解: 理想气体，纯p V T变化。V1 =0.0246 ( m3 )V2 = V3 =3.03410-3 ( m3 ) 恒温时，T2 = T1 p外= p2 =821554(Pa )恒外压恒温压缩: U1 =H1=0Q1 = - W1 = p外(V2 V1) =821554(3.03410-3 -0.0246) = -17717.6 ( J )恒容升温: U2 = QV2 =nCV,mdT =120.92(370300) = 1464.4 ( J )W2= - p外V=0H2 =n(CV,m+R )dT=1(20.92 +8.314)(370300)= 2046.4 ( J )整个过程:Q = Q1 + Q2= -16253.2 ( J )， W = W1 + W2 = -17717.6 ( J )U =U1 +U2 = 1464.4 ( J )，H=H1 +H2= 2046.4 ( J )14. 1mol单原子分子理想气体，在273.2K、1.0105 kPa时发生一变化过程，体积增大一倍，Q1674 J。H2092 J。(1) 计算终态的温度、压力和此过程的W、U。(2) 若该气体经恒温和恒容两步可逆过程到达上述终态，试计算Q、W、U和H。解: 理想气体，纯p V T变化。(1) H = nCp,mdT = n(5/2)R(T2 T1)2092=1(5/2)8.314(T2 273.2) 得: T2 = 373.8 ( K )V1 =0.0227 ( m3 )V2 = 2V1 = 0.0454 ( m3 )p2 = = = 68453.2 ( J )U = nCV,mdT = 1(3/2)8.314(373.8 273.2) = 1254.6 ( J )W =U-Q =1254.6-1674 = - 419.4 ( J )(2) 可逆过程，与(1)的始、终态相同。U = 1254.6 ( J ) ，H =2092 ( J )恒温可逆: U1 =H1=0Q1 = - W1 = nRT1 ln =18.314273.2ln=1574.4 ( J )恒容可逆: QV2 =U2 =nCV,mdT =1254.6 ( J )W2= - p外V=0Q = Q1 + Q2= 1574.4 + 1254.6=2829 ( J )， W = W1 + W2 = -1574.4 ( J )15. 1mol双原子理想气体在0和101.325 kPa时经绝热可逆膨胀至50.65 kPa，求该过程的W及U。解:= Cp,m/CV,m = (7/2)R / (5/2)R =1.4， T1p11 -= T2p21 -T2= T1 (p1/ p2) (1) / =273(101.325/ 50.65) (11.4) / 1.4 =224 ( K )U = W = nCV,mdT = 1(5/2)8.314(224 273) = 1018.5 ( J )W = -1018.5 ( J )16. 某理想气体的Cp,m28.8 JK-1mol -1，其起始状态为p 1303.99 kPa，V11.43 dm3，T1298 K。经一可逆绝热膨胀至2.86 dm3。求:(1) 终态的温度与压力；(2) 该过程的U及H。解:(1) n =0.175 (mol )= Cp,m/CV,m = 28.8 / (28.8-8.314) =1.4T1V1-1 = T2V2-1T2= T1 (V1/ V2)-1 =298(1.43/ 2.86) 1.41 =226 ( K )p1V1= p2V2p2= p1 (V1/ V2) =303.99103(1.43/ 2.86) 1.4 =115.19103 (Pa )(2)H = nCp,mdT = 0.17528.8(226298) = 363 ( J )U = nCV,mdT = 0.175(28.88.314)(226298) = 258 ( J )17. 今有10 dm3 O2从2.0105 kPa经绝热可逆膨胀到30 dm3，试计算此过程的Q、W、U及H。(假设O2可视为理想气体)。解: 双原子理想气体。= Cp,m/CV,m = (7/2)R / (5/2)R =1.4p1V1= p2V2p2= p1 (V1/ V2) =2.0105(10/ 30) 1.4 = 42959.6 (Pa )Q=0，U = W = nCV,mdT = n(5/2)R(T2 T1) = (5/2)(p2V2 p1V1)= (5/2)(42959.63010-3 2.01051010-3) = 1778 ( J )W = -1778 ( J )H = nCp,mdT = n(7/2)R(T2 T1) = (7/2)(p2V2 p1V1)=2489.2 ( J )18. 证明解: H = U + pV， 定p下，两边对T求微分 ，又 ， 得 19. 证明解: 又 dH定V下，两边对T求微分代入式，得20. 25的0.5克正庚烷在恒容条件下完全燃烧使热容为8175.5 JK-1的量热计温度上升了2.94，求正庚烷在25完全燃烧的H。解: 正庚烷(C7H16)的摩尔质量为100 g/mol。反应进度=1 的QV为QV =(100/0.5)2.948175.5 = 4807149 ( J ) = 4807.2 ( kJmol -1 )25: C7H16 (l)+11O2 (g) = 7CO2 (g) +8H2O (l)rHm = Qp = QV +nRT =4807.2103 +(4)8.314298 =4817.1 ( kJmol -1 )21. 试求下到反应在298K、101.325 kPa时的恒压热效应。(1) 2H2S (g)+SO2 (g) = 3H2O (l)+3S (斜方) QV = 223.8 kJ(2) 2C (石墨)+O2 (g) = 2CO (g) QV = 231.3 kJ(3) H2 (g)+Cl2 (g) = 2HCl (g) QV = 184 kJ解:(1) Qp = QV +nRT =223.8103 +(3)8.314298 =231.2 ( kJ )(2) Qp = QV +nRT =231.3103 +(+1)8.314298 =228.8 ( kJ )(3) Qp = QV +nRT =184103 +(0)8.314298 =184 ( kJ )22. 某反应体系，起始时含10mol H2和20mol O2，在反应进行的t时刻，生成了4mol的H2O。请计算下述反应方程式的反应进度:(1) H2 + O2 = H2O(2) 2H2 + O2 = 2H2O(3) H2 + O2 = H2O解: 各物质的摩尔数的变化量为:n(H2)=4 moln(O2)=2 moln(H2O)= 4 mol(1) =(4)/(1)=(2)/(0.5)= 4/1= 4 mol(2) =(4)/(2)=(2)/(1)= 4/2= 2 mol(3) =(4)/(0.5)=(2)/(0.25)= 4/0.5= 8 mol23. 已知下列反应在298 K时的热效应。(1) Na (s) +Cl2 (g) = NaCl (s) rHm = 411 kJ(2) H2 (g) +S (s) +2O2 (g) = H2SO4 (l) rHm = 811.3 kJ(3) 2Na (s) +S (s) +2O2 (g) = Na2SO4 (s) rHm = 1383 kJ(4) H2 (g)+ Cl2 (g) = HCl (g) rHm = 92.3 kJ求反应2NaCl (s) + H2SO4 (l) = Na2SO4 (s) + 2HCl (g)在298 K时的rHm和rUm。解: 因为反应(3)+2(4)-2(1)-(2) 即得反应2NaCl (s) + H2SO4 (l) = Na2SO4 (s) + 2HCl (g)所以rHm=rHm(3)+2rHm(4)-2rHm(1)-rHm(2)= 1383+2(92.3)-2(411 )-(811.3)=65.7( kJmol -1 )rUm =rHm-nRT = 65.7-28.31429810-3 = 60.7 ( kJmol -1 )24. 已知下述反应在298 K时的热效应:(1) C6H5COOH (l) +O2 (g) = 7CO2 (g) + 3 H2O (l) rHm = 3230 kJ(2) C (石墨)+O2 (g) = CO2 (g) rHm = 394 kJ(3) H2 (g) + O2 (g) = H2O (l) rHm = 286 kJ求C6H5COOH (l)的标准生成热fHmo 。解: 因为反应(2)7+(3)3 -(1) 即得C6H5COOH (l) 的生成反应7C (石墨) + 3H2 (g) +O2 (g) = C6H5COOH (l)所以fHmo=rHm(2)7+rHm(3)3-rHm(1)= (394)7+(286)3 -(3230)= 386 ( kJmol -1 )25. 已知下列反应在298 K时的热效应:(1) C (金刚石)+O2 (g) = CO2 (g) rHmo = 395.4 kJ(2) C (石墨)+O2 (g) = CO2 (g) rHmo = 393.5 kJ求C (石墨) = C (金刚石) 在298 K时的rHmo 。解: 因为反应(2)-(1) 即得反应C (石墨) = C (金刚石)所以rHmo=rHmo(2)-rHmo(1)= (393.5) -(395.4)= 1.9 ( kJmol -1 )26. 试分别由生成焓和燃烧焓计算下列反应:3C2H2 (g) = C6H6 (l)在101.325 kPa和298.15 K时的rHm和rUm。解: 查附录得生成焓和燃烧焓数据:C2H2 (g) C6H6 (l)fHmo/ kJmol -1 226.73 49.04cHmo/ kJmol 1 -1300 -3268由生成焓计算:rHm = 149.04-3226.73 = -631.15 ( kJmol -1 )rUm =rHm-nRT = -631.15-(3)8.314298.1510-3 = -623.7 ( kJmol -1 )由燃烧焓计算:rHm =3(-1300) -1(-3268) = -632 ( kJmol -1 )rUm =rHm-nRT = -632 -(3)8.314298.1510-3 = -624.6 ( kJmol -1 )27. KCl (s) 298.15 K时的溶解过程:KCl (s) = K+ (aq ,) + Cl - (aq ,) Hm = 17.18 kJmol -1已知Cl - (aq ,)和KCl (s)的摩尔生成焓分别为167.44 kJmol -1和435.87 kJmol -1，求K+ (aq ,)的摩尔生成焓。解: Hm = 17.18 = 1(-167.44)+1fHmo,K+ 1(-435.87)fHmo,K+ = -251.25 ( kJmol -1 )28. 在298 K时H2O (l) 的标准摩尔生成焓为285.8 kJmol -1，已知在25至100的温度范围内H2 (g)、O2 (g)及H2O (l)的Cp,m分别为28.83 JK -1mol -1、29.16 JK -1mol -1及75.31 JK -1mol -1。求100时H2O (l) 的标准摩尔生成焓。解: H2O (l) 的生成反应为:H2 (g) + O2 (g) = H2O (l)Cp =175.31128.83(1/2)29.16= 31.9 (JK -1mol -1)fHmo(T2) =fHmo(T1) + Cp dT = -285.8 + 31.9(373-298)10-3 = -283.4 ( kJmol -1 )29. 反应N2 (g) + 3H2 (g) = 2NH3 (g) 在298 K时的rHmo = 92.88 kJmol -1，求此反应在398 K时的rHmo 。已知:Cp,m(N2 , g) = (26.98+5.91210-3T3.37610-7T2) JK-1mol -1Cp,m(H2 , g) = (29.07+0.83710-3T +20.1210-7T2) JK-1mol -1Cp,m(NH3 , g) = (25.89+33.0010-3T30.4610-7T2) JK-1mol -1解:a =225.89329.0712