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1、.第一章习题解答第 5 页 共 5 页第一章习题解答1. 体系为隔离体系, DU=0W=Q=02.(1) W=pDV=p(Vg-Vl)»pVg=nRT=1´8.314´373.15=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=n×DHm,汽化=40670 JDU=DHD(pV)=DHp(Vg-Vl)=40670101325(302001880)´106 =406703058=37611 J5.
2、Cp,m=29.070.836´103T+2.01´106T2(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)´103´2´103 = 810 J
3、7.(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)
4、 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 =48072004R´298 = 4817100 J×mol1 =4817.1 kJ×mol114(1)2H2S(g)+SO2(g
5、) = 8H2O(l) + 3S(斜方)Dn =3QV =223.8 kJDrHm = QV + DnRT = 223.8 + (3)RT´103 = 231.2 kJ (2)2C(石墨) + O2(g) = 2CO2(g)Dn = 1QV =231.3 kJDrHm = QV + DnRT = 228.8 +RT´103 = 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) =
6、 Na2SO4(s) + 2HCl(g)DfH°m(kJ×mol1)411811.3138392.3DrH°m=S(nDfH°m)产物S(nDfH°m)反应物= (13832´92.3)(811.32´411) = 65.7 kJDrU°m =DrH°mDnRT=65.72RT´103=60.7 kJ17.DrH°m=S(nDcH°m) 反应物S(nDcH°m) 产物= (2´2834´285.8)(01370)=339.2 kJ18生成反应7C
7、(s) + 3H2(g) + O2 (g) = C6H5COOH(l)DcH°m(kJ×mol1)3942863230DrH°m=S(nDcH°m) 反应物S(nDcH°m) 产物= 7´(394) + 3´(286) (3230)= 386 kJ19.反应C(石墨) ® C(金刚石)DcH°m(kJ×mol1)393.5395.4DrH°m=DcH°m，石墨DcH°m，金刚石 =393.5(395.4)=1.9 kJ20.反应CH4(g)+2O2(g) = CO2
8、(g) + 2H2O(l)DfH°m(kJ×mol1)74.8393.5285.8DrH°m=S(nDfH°m)产物S(nDfH°m)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)DcH°m(kJ×mol1)251.5726.61677.80DrH°m=S(nDcH°m) 反应物S(nDcH°m) 产物=251.5+2´(726.6)(1677.8)=26
9、.9 kJ22.反应KCl(s)® K+(aq, ¥) + Cl(aq, ¥)DfH°m(kJ×mol1)435.87?167.44DrH°m=17.18 kJDrH°m=S(nDfH°m)产物S(nDfH°m)反应物17.18=DfH°m (K+,aq, ¥)167.44(435.87)DfH°m (K+,aq, ¥)=251.25 kJ×mol123.生成反应H2(g) + 0.5O2(g) = H2O(g)DrH298=285.8 kJ×m
10、ol1Cp,m(J×K1×mol1)28.8329.1675.31DCp=75.31(28.83+0.5´29.16)=31.9J×K1=285.8+31.9´(373298)´103=283.4 kJ×mol1 24.反应N2(g) + 3H2(g) = 2NH3(g)DrH298=92.888 kJ×mol1ab´103c´107N2(g)26.985.9123.376H2(g)29.070.83720.12NH3(g)25.8933.0030.46D62.4162.599117.904=9
11、2880+6241+2178144=97086 JH2(g) + I2(s)H2(g) + I2(g)DrH291=49.455 kJ2HI(g)2HI(g)Cp,m=55.64 J×K1×mol1T=386.7K D熔H°m=16736 JCp,m=62.76 J×K1×mol1Cp,m=7R/2 J×K1×mol1T=457.5K D蒸H°m=42677 Js®ll ®gDrH473=？DHHICp.m=7R/2DHI2DHH2Cp.m=7R/225.按图示过程计算：DHH2=nCp,mDT
12、=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®457.5K) + DH4(l®g)+ DH5(g,457.5®473K)=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;