物理化学(复旦大学药学院） 第一章习题答案.doc

第一章习题解答第 6 页 共 6 页第一章习题解答1. (1) Q=DU-W=200-160=40 kJ(2) DU=Q+W=260-100=160 kJ2. W= -pDV= -R3.(1) W= -pDV= -p(VgVl) -pVg= -nRT= -18.314373.15= -3102 J(2) W= -pDV= -p(VsVl)4. 在压力p和房间容积V恒定时，提高温度，部分空气溢出室外，因此室内气体量n是温度的函数，。5.(1)恒温可逆膨胀(2)真空膨胀W = 0(3)恒外压膨胀W = -p外(V2V3) = = -2327 J(4)二次膨胀W=W1 + W2 以上结果表明，功与具体过程有关，不是状态函数。6.(1) 理想气体为系统，等压过程， Q=DH0(2) 电热丝+理想气体为系统，等压过程，Q=0，DU=W0，DH=DU+D(pV)07.DH=nDHm,汽化=40670 JDU=DHD(pV)=DHp(Vg-Vl)=40670101325(302001880)10-6 =406703058=37611 J8. 过程QWDUDH(1)理想气体自由膨胀0000(2)理想气体恒温可逆膨胀+00(3)理想气体节流膨胀0000(4)理想气体绝热恒外压膨胀0(5)水蒸气通过蒸气机对外做功后复原，以水蒸气为系统+00(6)水(1 atm，0C) 冰(1atm，0C)或0(7)刚性容器中石磨燃烧，整体为系统000+9.Cp,m=29.070.836103T+2.0110-6T2(1) Qp=DH=20349380+625=20.62 kJ(2) QV=DU=DHD(pV)=DH(p2V2p1V1) =DHnR(T2T1)=20.62R(1000-300)10-3 =14.80 kJ10（1）等温可逆膨胀DU =DH = 0Q = -W（2）等温恒外压膨胀DU =DH =0Q = -W = p2 (V2V1) = p2V2p2V1= p1V1p2V1= (p1p2)V1=(506.6-101.3)103210-3 = 810 J11. （1）常压蒸发：Q=DH= 40.7 kJW= -p(VgVl) -pVg -RT= -8.314373= -3.1 kJDU=Q+W=37.6 kJ(2) 真空蒸发：DH=40.7 kJ W=0DU =Q=37.6 kJ12.(1) p1T1=p2T2(2) DU=nCV,m(T2T1)= DH=nCp,m(T2T1)=(3) 以T为积分变量求算：pT=C(常数)也可以用p或V为积分变量进行求算。13DU=nCV,m(T2T1)=20.92(370300)=1464 JDH=nCp,m(T2T1)=(20.92+R)(370300)=2046 J始态体积体积变化：压力 W=W1+W2= -p2(V2V1)+0= -821554(0.0030260.0246)= 17724 JQ=DU -W=146417724=16260 J14. (1) DH=Cp,m(T2T1)T2=373.8 Kp2=0.684105 PaDU= CV,m(T2T1) JW=DUQ=12551674=419 J(2) 状态函数变化同上 DU=1255 JDH=2092 JW=W1=8.314273.2ln2=1574 JQ=DUW=1255+1574 = 2829 J15.双原子分子W=DU=nCV,m(T2T1)16. （1）等温可逆膨胀DU =DH = 0Q = W（2）绝热可逆膨胀Cp.m=20.79CV.m= Cp.m-R=12.476TVg-1=常数Q =0DU=W=nCV,m(T2T1)=12.476(160.06273.15)= -1411 JDH=nCp,m(T2T1)= 20.79(160.06273.15)= -2351 J17.(1) (2) DU=nCV,m(T2T1)=0.1755(28.8R) (224.9298)= -263 JDH=nCp,m(T2T1)=0.175528.8 (224.9298)= -369 J18. 设物质量为1mol始态温度： K求终态温度： K JJQ=0W=DU= -1779 J若设物质量为n mol，可如下计算：始态温度，终态温度在计算DU、DH、W时n可消去，得相同的结果。19. 证明U=HpV20.证明(1)H=f(T,p)V不变，对T求导代入(1)21.发酵反应C6H12O6(s) 2C2H5OH(l) + 2CO2(g)Dn=2DU=DH-DnRT= -67.8 -2R29810-3 = -72.76 kJmol-122nQV+CDT=0QV =4807200 JC7H16(l) + 11O2(g) = 8H2O(l) + 7CO2(g)Dn =4DcHm = QV + DnRT =48072004R298 = 4817100 Jmol1 =4817.1 kJmol-123(1)2H2S(g)+SO2(g) = 8H2O(l) + 3S(斜方)Dn =3QV =223.8 kJDrHm = QV + DnRT = 223.8 + (3)RT10-3 = 231.2 kJ (2)2C(石墨) + O2(g) = 2CO(g)Dn = 1QV =231.3 kJDrHm = QV + DnRT = 231.3 +RT103 = 228.8 kJ(3)H2(g)+Cl2(g) = 2HCl (g)Dn =0QV =184 kJDrHm = QV =184 kJ24.(1) x=4 mol(2) x=2 mol(3) x=8 mol252NaCl(s) + H2SO4(l) = Na2SO4(s) + 2HCl(g)DfHOm(kJmol-1)411811.3138392.3DrHOm=S(nDfHOm)产物S(nDfHOm)反应物= (1383292.3)(811.32411) = 65.7 kJmol-1DrUOm =DrHmDnRT=65.72RT10-3=60.7 kJmol-126生成反应7C(s) + 3H2(g) + O2 (g) = C6H5COOH(l)DcHOm(kJ mol-1)3942863230DrHOm=S(nDcHOm) 反应物S(nDcHOm) 产物= 7(394) + 3(286) (3230)= 386 kJmol-127.反应C(石墨) C(金刚石)DcHOm(kJmol-1)393.5395.4DrHOm=DcHOm，石墨DcHOm，金刚石 =393.5(395.4)=1.9 kJmol-128. 3C2H2 (g) = C6H6 (l)DfHOm (kJmol-1)226.7349.04DcHOm (kJmol-1)1299.63267.5由生成热数据计算：DrHOm=49.043226.73=631.15 kJmol-1DrUOm=DrHOmDnRT=631.15+38.314298.210-3= -623.71 kJmol-1由燃烧热数据计算：DrHOm=3(1299.6)(3267.5) =631.3 kJmol-1DrUOm=DrHOmDnRT=631.3+38.314298.210-3= -623.86 kJmol-129.反应KCl(s) K+(aq, ) + Cl-(aq, )DfHOm(kJmol-1)435.87?167.44DrHOm=17.18 kJmol-1DrHOm=S(nDfHOm)产物S(nDfHOm)反应物17.18=DfHOm (K+,aq, )167.44(435.87)DfHOm (K+,aq, )=251.25 kJmol-130.生成反应H2(g) + 0.5O2(g) = H2O(g)DrHOm,298=285.83 kJmol-1Cp,m(JK-1mol-1)28.82429.35575.291DCp=75.291(28.824+0.529.355)=31.79 JK-1=285.83+31.79(373298)10-3 =283.45 kJmol-1 31.反应N2(g) + 3H2(g) = 2NH3(g)DrHOm,298=92.88 kJmol-1ab103c107N2(g)26.985.9123.376H2(g)29.070.83720.12NH3(g)25.8933.0030.46D62.4162.599117.904=92880+6241+2178144= 97086 Jmol-1H2(g) + I2(s)H2(g) + I2(g)DrHOm,291= 49.455 kJ mol-12HI(g)2HI(g)Cp,m=55.64 JK1mol1T=386.7K D熔HOm=16736 Jmol-1Cp,m=62.76 JK1mol1Cp,m=7R/2 JK1mol1T=457.5K D蒸HOm=42677 Jmol-1sll gDrHOm，473=？DHHICp.m=7R/2DHI2DHH2Cp.m=7R/232.按图示过程计算：DHH2=nCp,mDT=3.5R(473291)=5296 JDHHI=nCp,mDT=23.5R(473291)=10592 JDHI2=DH1(s,291386.7K) + DH2(sl) + DH3(