第一章 習(xí)題解答 1. (1) Q=DU-W=200-160=40 kJ (2) DU=Q+W=260-100=160 kJ 2. \W= -pDV= -R 3. (1) W= -pDV= -p(Vg–Vl) -pVg= -nRT= -18.314373.15= -3102 J (2) W= -pDV= -p(Vs–Vl) 4. 在壓力p和房間容積V恒定時(shí)，提高溫度，部分空氣溢出室外，因此室內(nèi)氣體量n是溫度的函數(shù)，。 5. (1) 恒溫可逆膨脹 (2) 真空膨脹 W = 0 (3) 恒外壓膨脹 W = -p外(V2–V3) = = -2327 J (4) 二次膨脹 W=W1 + W2 以上結(jié)果表明，功與具體過(guò)程有關(guān)，不是狀態(tài)函數(shù)。 6. (1) 理想氣體為系統(tǒng)，等壓過(guò)程， Q=DH>0 (2) 電熱絲+理想氣體為系統(tǒng)，等壓過(guò)程，Q=0，DU=W>0，DH=DU+D(pV)>0 7. DH=nDHm,汽化=40670 J DU=DH–D(pV)=DH–p(Vg-Vl)=40670–101325(30200–1880)10-6 =40670–3058=37611 J 8. 過(guò)程 Q W DU DH (1)理想氣體自由膨脹 0 0 0 0 (2)理想氣體恒溫可逆膨脹 + – 0 0 (3)理想氣體節(jié)流膨脹 0 0 0 0 (4)理想氣體絕熱恒外壓膨脹 0 – – – (5)水蒸氣通過(guò)蒸氣機(jī)對(duì)外做功后復(fù)原，以水蒸氣為系統(tǒng) + – 0 0 (6)水(1 atm，0C) 冰(1atm，0C) – –或0 – – (7)剛性容器中石磨燃燒，整體為系統(tǒng) 0 0 0 + 9. Cp,m=29.07–0.836103T+2.0110-6T2 (1) Qp=DH =20349–380+625=20.62 kJ (2) QV=DU=DH–D(pV)=DH–(p2V2–p1V1) =DH–nR(T2–T1) =20.62–R(1000-300)10-3 =14.80 kJ 10．（1）等溫可逆膨脹 DU =DH = 0 Q = -W （2）等溫恒外壓膨脹 DU =DH =0 Q = -W = p2 (V2–V1) = p2V2–p2V1= p1V1–p2V1= (p1–p2)V1 =(506.6-101.3)103210-3 = 810 J 11. （1）常壓蒸發(fā)：Q=DH= 40.7 kJ W= -p(Vg–Vl) -pVg -RT= -8.314373= -3.1 kJ DU=Q+W=37.6 kJ (2) 真空蒸發(fā)：DH=40.7 kJ W=0 DU =Q=37.6 kJ 12. (1) p1T1=p2T2 (2) DU=nCV,m(T2–T1)= DH=nCp,m(T2–T1)= (3) 以T為積分變量求算： pT=C(常數(shù)) 也可以用p或V為積分變量進(jìn)行求算。 13．DU=nCV,m(T2–T1)=20.92(370–300)=1464 J DH=nCp,m(T2–T1)=(20.92+R)(370–300)=2046 J 始態(tài)體積 體積變化： 壓力 W=W1+W2= -p2(V2–V1)+0= -821554(0.003026–0.0246)= 17724 J Q=DU -W=1464–17724=–16260 J 14. (1) DH=Cp,m(T2–T1) T2=373.8 K p2=0.684105 Pa DU= CV,m(T2–T1) J W=DU–Q=1255–1674=–419 J (2) 狀態(tài)函數(shù)變化同上 DU=1255 J DH=2092 J W=W1=–8.314273.2ln2=–1574 J Q=DU–W=1255+1574 = 2829 J 15. 雙原子分子 W=DU=nCV,m(T2–T1) 16. （1）等溫可逆膨脹 DU =DH = 0 Q = W （2）絕熱可逆膨脹 Cp.m=20.79 CV.m= Cp.m-R=12.476 TVg-1=常數(shù) Q =0 DU=W=nCV,m(T2–T1)=12.476(160.06–273.15)= -1411 J DH=nCp,m(T2–T1)= 20.79(160.06–273.15)= -2351 J 17. (1) (2) DU=nCV,m(T2–T1)=0.1755(28.8–R) (224.9–298)= -263 J DH=nCp,m(T2–T1)=0.175528.8 (224.9–298))= -369 J 18. 設(shè)物質(zhì)量為1mol 始態(tài)溫度： K 求終態(tài)溫度： K J J Q=0 W=DU= -1779 J 若設(shè)物質(zhì)量為n mol，可如下計(jì)算： 始態(tài)溫度， 終態(tài)溫度 在計(jì)算DU、DH、W時(shí)n可消去，得相同的結(jié)果。 19. 證明 U=H–pV 20. 證明 (1) H=f(T,p) V不變，對(duì)T求導(dǎo) 代入(1) 21. 發(fā)酵反應(yīng) C6H12O6(s) 2C2H5OH(l) + 2CO2(g) Dn=2 DU=DH-DnRT= -67.8 -2R29810-3 = -72.76 kJmol-1 22． nQV+CDT=0 QV =–4807200 J C7H16(l) + 11O2(g) = 8H2O(l) + 7CO2(g) Dn =–4 DcHm = QV + DnRT =–4807200–4R298 = –4817100 Jmol–1 =–4817.1 kJmol-1 23．(1) 2H2S(g)+SO2(g) = 8H2O(l) + 3S(斜方) Dn =–3 QV =–223.8 kJ DrHm = QV + DnRT = –223.8 + (–3)RT10-3 = –231.2 kJ (2) 2C(石墨) + O2(g) = 2CO(g) Dn = 1 QV =–231.3 kJ DrHm = QV + DnRT = –231.3 +RT10–3 = –228.8 kJ (3) H2(g)+Cl2(g) = 2HCl (g) Dn =0 QV =–184 kJ DrHm = QV =–184 kJ 24. (1) x=4 mol (2) x=2 mol (3) x=8 mol 25． 2NaCl(s) + H2SO4(l) = Na2SO4(s) + 2HCl(g) DfHOm(kJmol-1) –411 –811.3 –1383 –92.3 DrHOm=S(nDfHOm)產(chǎn)物–S(nDfHOm)反應(yīng)物 = (–1383–292.3)–(–811.3–2411) = 65.7 kJmol-1 DrUOm =DrHm–DnRT=65.7–2RT10-3=60.7 kJmol-1 26．生成反應(yīng) 7C(s) + 3H2(g) + O2 (g) = C6H5COOH(l) DcHOm(kJ mol-1) –394 –286 –3230 DrHOm=S(nDcHOm) 反應(yīng)物–S(nDcHOm) 產(chǎn)物 = [7(–394) + 3(–286)] – (–3230)= –386 kJmol-1 27. 反應(yīng) C(石墨) C(金剛石) DcHOm(kJmol-1) –393.5 –395.4 DrHOm=DcHOm，石墨–DcHOm，金剛石 =–393.5–(–395.4)=1.9 kJmol-1 28. 3C2H2 (g) = C6H6 (l) DfHOm (kJmol-1) 226.73 49.04 DcHOm (kJmol-1) –1299.6 –3267.5 由生成熱數(shù)據(jù)計(jì)算： DrHOm=49.04–3226.73=–631.15 kJmol-1 DrUOm=DrHOm–DnRT=–631.15+38.314298.210-3= -623.71 kJmol-1 由燃燒熱數(shù)據(jù)計(jì)算： DrHOm=3(–1299.6)–(–3267.5) =–631.3 kJmol-1 DrUOm=DrHOm–DnRT=–631.3+38.314298.210-3= -623.86 kJmol-1 29. 反應(yīng) KCl(s) K+(aq, ) + Cl-(aq, ) DfHOm(kJmol-1) –435.87 ? –167.44 DrHOm=17.18 kJmol-1 DrHOm=S(nDfHOm)產(chǎn)物–S(nDfHOm)反應(yīng)物 17.18=[DfHOm (K+,aq, )–167.44]–(–435.87) DfHOm (K+,aq, )=–251.25 kJmol-1 30. 生成反應(yīng) H2(g) + 0.5O2(g) = H2O(g) DrHOm,298=–285.83 kJmol-1 Cp,m(JK-1mol-1) 28.824 29.355 75.291 DCp=75.291–(28.824+0.529.355)=31.79 JK-1 =–285.83+31.79(373–298)10-3 =–283.45 kJmol-1 31. 反應(yīng)N2(g) + 3H2(g) = 2NH3(g) DrHOm,298=–92.88 kJmol-1 a b103 c107 N2(g) 26.98 5.912 –3.376 H2(g) 29.07 –0.837 20.12 NH3(g) 25.89 33.00 –30.46 D –62.41 62.599 –117.904 =–92880+[–6241+2178–144]= –97086 Jmol-1 H2(g) + I2(s) H2(g) + I2(g) DrHOm,291= 49.455 kJ mol-1 2HI(g) 2HI(g) Cp,m=55.64 JK–1mol–1 T=386.7K D熔HOm=16736 Jmol-1 Cp,m=62.76 JK–1mol–1 Cp,m=7R/2 JK–1mol–1 T=457.5K D蒸HOm=42677 Jmol-1 sl l g DrHOm，473=？ DHHI Cp.m=7R/2 DHI2 DHH2 Cp.m=7R/2 32. 按圖示過(guò)程計(jì)算： DHH2=nCp,mDT=3.5R(473–291)=5296 J DHHI=nCp,mDT=23.5R(473–291)=10592 J DHI2=DH1(s,291386.7K) + DH2(sl) + DH3(l,1386.7457.5K) + DH4(lg) + DH5(g,457.5473K) =55.64(386.7–291)+16736+62.76(457.5–386.7)+42677+3.5R(473–457.5) =69632 J DHH2+DHI2+DrHOm,473=DrHOm,291+DHHI 5296+69632+DrHOm,473=49455+10592 DrHOm,473=–14.881 kJ mol-1