物理化學(xué)(復(fù)旦大學(xué)藥學(xué)院) 第一章習(xí)題答案

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1、第一章習(xí)題解答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. 在壓力p和房間容積V恒定時，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(V2V3) = = -2327 J(4)二次膨脹W=W1 + W2 以上結(jié)果表明，功與具體過程有關(guān)，不是狀態(tài)函數(shù)。6.(1)

2、 理想氣體為系統(tǒng)，等壓過程， Q=DH0(2) 電熱絲+理想氣體為系統(tǒng)，等壓過程，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. 過程QWDUDH(1)理想氣體自由膨脹0000(2)理想氣體恒溫可逆膨脹+00(3)理想氣體節(jié)流膨脹0000(4)理想氣體絕熱恒外壓膨脹0(5)水蒸氣通過蒸氣機對外做功后復(fù)原，以水蒸氣為系統(tǒng)+00(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)000+

3、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）等溫可逆膨脹DU =DH = 0Q = -W（2）等溫恒外壓膨脹DU =DH =0Q = -W = p2 (V2V1) = p2V2p2V1= p1V1p2V1= (p1p2)V1=(506.6-101.3)103210-3 = 810 J11. （1）常壓蒸發(fā)：Q=DH= 40.7 kJW= -p(VgVl) -p

4、Vg -RT= -8.314373= -3.1 kJDU=Q+W=37.6 kJ(2) 真空蒸發(fā)：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為積分變量進行求算。13DU=nCV,m(T2T1)=20.92(370300)=1464 JDH=nCp,m(T2T1)=(20.92+R)(370300)=2046 J始態(tài)體積體積變化：壓力 W=W1+W2= -p2(V2V1)+0= -821554(0.0030260.0246)=

5、 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) 狀態(tài)函數(shù)變化同上 DU=1255 JDH=2092 JW=W1=8.314273.2ln2=1574 JQ=DUW=1255+1574 = 2829 J15.雙原子分子W=DU=nCV,m(T2T1)16. （1）等溫可逆膨脹DU =DH = 0Q = W（2）絕熱可逆膨脹Cp.m=20.79CV.m= Cp.m-R=12.476TVg-1=常數(shù)Q =0DU=W

6、=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. 設(shè)物質(zhì)量為1mol始態(tài)溫度： K求終態(tài)溫度： K JJQ=0W=DU= -1779 J若設(shè)物質(zhì)量為n mol，可如下計算：始態(tài)溫度，終態(tài)溫度在計算DU、DH、W時n可消去，得相同的結(jié)果。19. 證明U=HpV20.證

7、明(1)H=f(T,p)V不變，對T求導(dǎo)代入(1)21.發(fā)酵反應(yīng)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

8、 = 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)411811.3138392.3DrH

9、Om=S(nDfHOm)產(chǎn)物S(nDfHOm)反應(yīng)物= (1383292.3)(811.32411) = 65.7 kJmol-1DrUOm =DrHmDnRT=65.72RT10-3=60.7 kJmol-126生成反應(yīng)7C(s) + 3H2(g) + O2 (g) = C6H5COOH(l)DcHOm(kJ mol-1)3942863230DrHOm=S(nDcHOm) 反應(yīng)物S(nDcHOm) 產(chǎn)物= 7(394) + 3(286) (3230)= 386 kJmol-127.反應(yīng)C(石墨) C(金剛石)DcHOm(kJmol-1)393.5395.4DrHOm=DcHOm，石墨DcHO

10、m，金剛石 =393.5(395.4)=1.9 kJmol-128. 3C2H2 (g) = C6H6 (l)DfHOm (kJmol-1)226.7349.04DcHOm (kJmol-1)1299.63267.5由生成熱數(shù)據(jù)計算：DrHOm=49.043226.73=631.15 kJmol-1DrUOm=DrHOmDnRT=631.15+38.314298.210-3= -623.71 kJmol-1由燃燒熱數(shù)據(jù)計算：DrHOm=3(1299.6)(3267.5) =631.3 kJmol-1DrUOm=DrHOmDnRT=631.3+38.314298.210-3= -623.86 k

11、Jmol-129.反應(yīng)KCl(s) K+(aq, ) + Cl-(aq, )DfHOm(kJmol-1)435.87?167.44DrHOm=17.18 kJmol-1DrHOm=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-130.生成反應(yīng)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)=31.7

12、9 JK-1=285.83+31.79(373298)10-3 =283.45 kJmol-1 31.反應(yīng)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 JK1

13、mol1T=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.按圖示過程計算：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+DrHOm,473=DrHOm,291+DHHI5296+69632+DrHOm,473=49455+10592DrHOm,473=14.881 kJ mol-1