物理化学(复旦大学药学院） 第一章习题答案

1、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=-1´8.314´373.15=-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

2、.(1) 理想气体为系统，等压过程， Q=DH>0(2) 电热丝+理想气体为系统，等压过程，Q=0，DU=W¢>0，DH=DU+D(pV)>07.DH=n×DHm,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，0

3、6;C)® 冰(1atm，0°C)或»0(7)刚性容器中石磨燃烧，整体为系统000+9.Cp,m=29.070.836´103T+2.01´10-6T2(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= (p1p

4、2)V1=(506.6-101.3)´103´2´10-3 = 810 J11. （1）常压蒸发：Q=DH=40.7 kJW=-p(VgVl)»-pVg»-RT=-8.314´373=-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´(370

5、300)=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=0.684´105 PaDU= CV,m(T2T1) JW=DUQ=12551674=419 J(2) 状态函数变化同上 DU=1255 JDH=2092 JW=W1=8.314´273.2×ln2=157

6、4 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=nC

7、p,m(T2T1)=0.1755´28.8´(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) ¾® 2C2H5OH(l) + 2CO2(g)Dn=2DU=DH-DnRT=-67.8-2R´298´10-3 = -72.76 kJ×mol

8、-122nQV+CDT=0QV=4807200 JC7H16(l) + 11O2(g) = 8H2O(l) + 7CO2(g)Dn =4DcHm = QV+ DnRT =48072004R´298 = 4817100 J×mol1 =4817.1 kJ×mol-123(1)2H2S(g)+SO2(g) = 8H2O(l) + 3S(斜方)Dn =3QV =223.8 kJDrHm = QV + DnRT = 223.8 + (3)RT´10-3 = 231.2 kJ (2)2C(石墨) + O2(g) = 2CO(g)Dn = 1QV =231.3 kJ

9、DrHm = QV + DnRT = 231.3 +RT´103 = 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(kJ×mol-1)411811.3138392.3DrHOm=S(nDfHOm)产物S(nDfHOm)反应物= (13832´92.3)(811.32´411) = 65.7 kJ×

10、;mol-1DrUOm =DrH°mDnRT=65.72RT´10-3=60.7 kJ×mol-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 kJ×mol-127.反应C(石墨)®C(金刚石)DcHOm(kJ×mol-1)393.5395.4DrHOm=DcHOm，石墨DcHOm，金刚石

11、=393.5(395.4)=1.9 kJ×mol-128. 3C2H2 (g) = C6H6 (l)DfHOm (kJ×mol-1)226.7349.04DcHOm (kJ×mol-1)1299.63267.5由生成热数据计算：DrHOm=49.043´226.73=631.15 kJ×mol-1DrUOm=DrHOmDnRT=631.15+3´8.314´298.2´10-3=-623.71 kJ×mol-1由燃烧热数据计算：DrHOm=3´(1299.6)(3267.5) =631.3 kJ

12、×mol-1DrUOm=DrHOmDnRT=631.3+3´8.314´298.2´10-3=-623.86 kJ×mol-129.反应KCl(s)® K+(aq, ¥) + Cl-(aq, ¥)DfHOm(kJ×mol-1)435.87?167.44DrHOm=17.18 kJ×mol-1DrHOm=S(nDfHOm)产物S(nDfHOm)反应物17.18=DfHOm (K+,aq, ¥)167.44(435.87)DfHOm (K+,aq, ¥)=251.25 kJ

13、15;mol-130.生成反应H2(g) + 0.5O2(g) = H2O(g)DrHOm,298=285.83 kJ×mol-1Cp,m(J×K-1×mol-1)28.82429.35575.291DCp=75.291(28.824+0.5´29.355)=31.79J×K-1=285.83+31.79´(373298)´10-3=283.45 kJ×mol-131.反应N2(g) + 3H2(g) = 2NH3(g)DrHOm,298=92.88 kJ×mol-1ab´103c´1

14、07N2(g)26.985.9123.376H2(g)29.070.83720.12NH3(g)25.8933.0030.46D62.4162.599117.904=92880+6241+2178144= 97086 J×mol-1H2(g) + I2(s)H2(g) + I2(g)DrHOm,291= 49.455 kJ ×mol-12HI(g)2HI(g)Cp,m=55.64 J×K1×mol1T=386.7K D熔HOm=16736 J×mol-1Cp,m=62.76 J×K1×mol1Cp,m=7R/2 J×K1×mol1T=457.5K D蒸HOm=42677 J×mol-1s®ll ®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=2´3.5R(473291)=10592 JDHI2=DH1(s,291®386.7K) + DH2(s®l) +DH3(l,1386.7&#