复旦 物理化学 第一章 习题答案.doc

第一章习题解答1. 体系为隔离体系, DU=0W=Q=02.(1) W=pDV=p(Vg-Vl)pVg=nRT=18.314373.15=3102 J(2) W=pDV=p(VsVl)3.(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 4.DH=nDHm,汽化=40670 JDU=DHD(pV)=DHp(Vg-Vl)=40670101325(302001880)106 =406703058=37611 J5.Cp,m=29.070.836103T+2.01106T2(1) Qp=DH=20349380+625=20.62 kJ(2) QV=DU=DHD(pV)=DH(p2V1p1V1)V2=V1 QV=DHnR(T2T1)=20.62R(1000-300)103=14.80 kJ(3) 6（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)1032103 = 810 J7.(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。8DU=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 J9.双原子分子W=DU=nCV,m(T2T1)10.(1) (2) DU=nCV.m(T2T1)=n (28.8R)(224.9298)= 263 JDH=nCp.m(T2T1)=n 28.8(224.9298)= 369 J11.证明U=HpV12.证明（1）H=f(T,p)V不变，对T求导代入(1)13nQV+CDT=0QV=4807200 JC7H16(l) + 11O2(g) = 8H2O(l) + 7CO2(g)Dn =4DcHm = QV + DnRT =48072004R298 = 4817100 Jmol1 =4817.1 kJmol114(1)2H2S(g)+SO2(g) = 8H2O(l) + 3S(斜方)Dn =3QV =223.8 kJDrHm = QV + DnRT = 223.8 + (3)RT103 = 231.2 kJ (2)2C(石墨) + O2(g) = 2CO2(g)Dn = 1QV =231.3 kJDrHm = QV + DnRT = 228.8 +RT103 = 228.8 kJ(3)2H2(g)+Cl2(g) = HCl (g)Dn =0QV =184 kJDrHm = QV =184 kJ15.(1) x=4 mol(2) x=2 mol(3) x=8 mol162NaCl(s) + H2SO4(l) = Na2SO4(s) + 2HCl(g)DfHm(kJmol1)411811.3138392.3DrHm=S(nDfHm)产物S(nDfHm)反应物= (1383292.3)(811.32411) = 65.7 kJDrUm =DrHmDnRT=65.72RT103=60.7 kJ17.DrHm=S(nDcHm) 反应物S(nDcHm) 产物= (22834285.8)(01370)=339.2 kJ18生成反应7C(s) + 3H2(g) + O2 (g) = C6H5COOH(l)DcHm(kJmol1)3942863230DrHm=S(nDcHm) 反应物S(nDcHm) 产物= 7(394) + 3(286) (3230)= 386 kJ19.反应C(石墨) C(金刚石)DcHm(kJmol1)393.5395.4DrHm=DcHm，石墨DcHm，金刚石 =393.5(395.4)=1.9 kJ20.反应CH4(g)+2O2(g) = CO2(g) + 2H2O(l)DfHm(kJmol1)74.8393.5285.8DrHm=S(nDfHm)产物S(nDfHm)反应物=393.5+2(285.5)(74.8)=890.3 kJ21.反应(COOH)2(s)+2CH3OH(l) = (COOCH3)2(l) + 2H2O(l)DcHm(kJmol1)251.5726.61677.80DrHm=S(nDcHm) 反应物S(nDcHm) 产物=251.5+2(726.6)(1677.8)=26.9 kJ22.反应KCl(s) K+(aq, ) + Cl(aq, )DfHm(kJmol1)435.87?167.44DrHm=17.18 kJDrHm=S(nDfHm)产物S(nDfHm)反应物17.18=DfHm (K+,aq, )167.44(435.87)DfHm (K+,aq, )=251.25 kJmol123.生成反应H2(g) + 0.5O2(g) = H2O(g)DrH298=285.8 kJmol1Cp,m(JK1mol1)28.8329.1675.31DCp=75.31(28.83+0.529.16)=31.9JK1=285.8+31.9(373298)103=283.4 kJmol1 24.反应N2(g) + 3H2(g) = 2NH3(g)DrH298=92.888 kJmol1ab103c107N2(g)26.985.9123.376H2(g)29.070.83720.12NH3(g)25.8933.0030.46D62.4162.599117.904=92880+6241+2178144=97086 JH2(g) + I2(s)H2(g) + I2(g)DrH291=49.455 kJ2HI(g)2HI(g)Cp,m=55.64 JK1mol1T=386.7K D熔Hm=16736 JCp,m=62.76 JK1mol1Cp,m=7R/2 JK1mol1T=457.5K D蒸Hm=42677 Jsll gDrH473=？DHHICp.m=7R/2DHI2DHH2Cp.m=7R/225.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(l,1386.7457.5K) + DH4(lg)+ DH5(g,457.5473K)=55.64(386.7291)+16736+62.76(457.5386.7)+42677+3.5R(473457.5)=69632 JDHH2+DHI2+DrH473=DrH291+DHHI5296+69632+DrH473=49455+10592DrH473=14881 J