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

1、第一章習(xí)題解答1. (1) Q=U-W=200-160=40 kJ(2) U=Q+W=260-100=160 kJ2. W= -pV= -R3.(1) W= -pV= -p(VgVl) -pVg= -nRT= -18.314373.15= -3102 J(2) W= -pV= -p(VsVl)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(V2V3) = = -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

2、為系統(tǒng)，等壓過(guò)程， Q=H0(2) 電熱絲+理想氣體為系統(tǒng)，等壓過(guò)程，Q=0，U=W0，H=U+(pV)07.H=nHm,汽化=40670 JU=H(pV)=Hp(Vg-Vl)=40670101325(302001880)106 =406703058=37611 J8. 過(guò)程QWUH(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(guò)蒸氣機(jī)對(duì)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+9.Cp,m=29.070.83610

3、3T+2.01106T2(1) Qp=H=20349380+625=20.62 kJ(2) QV=U=H(pV)=H(p2V2p1V1) =HnR(T2T1)=20.62R(1000-300)103 =14.80 kJ10（1）等溫可逆膨脹U =H = 0Q = -W（2）等溫恒外壓膨脹U =H =0Q = -W = p2 (V2V1) = p2V2p2V1= p1V1p2V1= (p1p2)V1=(506.6-101.3)1032103 = 810 J11. （1）常壓蒸發(fā)：Q=H= 40.7 kJW= -p(VgVl) -pVg -RT= -8.314373= -3.1 kJU=Q+W=3

4、7.6 kJ(2) 真空蒸發(fā)：H=40.7 kJ W=0U =Q=37.6 kJ12.(1) p1T1=p2T2(2) U=nCV,m(T2T1)= H=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(jìn)行求算。13U=nCV,m(T2T1)=20.92(370300)=1464 JH=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)= 17724 JQ=U -W=146417724=16260 J14. (1)

5、H=Cp,m(T2T1)T2=373.8 Kp2=0.684105 PaU= CV,m(T2T1) JW=UQ=12551674=419 J(2) 狀態(tài)函數(shù)變化同上 U=1255 JH=2092 JW=W1=8.314273.2ln2=1574 JQ=UW=1255+1574 = 2829 J15.雙原子分子W=U=nCV,m(T2T1)16. （1）等溫可逆膨脹U =H = 0Q = W（2）絕熱可逆膨脹Cp.m=20.79CV.m= Cp.m-R=12.476TV-1=常數(shù)Q =0U=W=nCV,m(T2T1)=12.476(160.06273.15)= -1411 JH=nCp,m(T2

6、T1)= 20.79(160.06273.15)= -2351 J17.(1) (2) U=nCV,m(T2T1)=0.1755(28.8R) (224.9298)= -263 JH=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=U= -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U、H、W時(shí)n可消去，得相同的結(jié)果。19. 證明U=HpV20.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) + 2C

7、O2(g)n=2U=H-nRT= -67.8 -2R298103 = -72.76 kJmol122nQV+CT=0QV =4807200 JC7H16(l) + 11O2(g) = 8H2O(l) + 7CO2(g)n =4cHm = QV + nRT =48072004R298 = 4817100 Jmol1 =4817.1 kJmol123(1)2H2S(g)+SO2(g) = 8H2O(l) + 3S(斜方)n =3QV =223.8 kJrHm = QV + nRT = 223.8 + (3)RT103 = 231.2 kJ (2)2C(石墨) + O2(g) = 2CO(g)n =

8、 1QV =231.3 kJrHm = QV + nRT = 231.3 +RT103 = 228.8 kJ(3)H2(g)+Cl2(g) = 2HCl (g)n =0QV =184 kJrHm = QV =184 kJ24.(1) =4 mol(2) =2 mol(3) =8 mol252NaCl(s) + H2SO4(l) = Na2SO4(s) + 2HCl(g)fHOm(kJmol1)411811.3138392.3rHOm=(nfHOm)產(chǎn)物(nfHOm)反應(yīng)物= (1383292.3)(811.32411) = 65.7 kJmol1rUOm =rHmnRT=65.72RT103=

9、60.7 kJmol126生成反應(yīng)7C(s) + 3H2(g) + O2 (g) = C6H5COOH(l)cHOm(kJ mol1)3942863230rHOm=(ncHOm) 反應(yīng)物(ncHOm) 產(chǎn)物= 7(394) + 3(286) (3230)= 386 kJmol127.反應(yīng)C(石墨) C(金剛石)cHOm(kJmol1)393.5395.4rHOm=cHOm，石墨cHOm，金剛石 =393.5(395.4)=1.9 kJmol128. 3C2H2 (g) = C6H6 (l)fHOm (kJmol1)226.7349.04cHOm (kJmol1)1299.63267.5由生成熱

10、數(shù)據(jù)計(jì)算：rHOm=49.043226.73=631.15 kJmol1rUOm=rHOmnRT=631.15+38.314298.2103= -623.71 kJmol1由燃燒熱數(shù)據(jù)計(jì)算：rHOm=3(1299.6)(3267.5) =631.3 kJmol1rUOm=rHOmnRT=631.3+38.314298.2103= -623.86 kJmol129.反應(yīng)KCl(s) K+(aq, ) + Cl(aq, )fHOm(kJmol1)435.87?167.44rHOm=17.18 kJmol1rHOm=(nfHOm)產(chǎn)物(nfHOm)反應(yīng)物17.18=fHOm (K+,aq, )167

11、.44(435.87)fHOm (K+,aq, )=251.25 kJmol130.生成反應(yīng)H2(g) + 0.5O2(g) = H2O(g)rHOm,298=285.83 kJmol1Cp,m(JK1mol1)28.82429.35575.291Cp=75.291(28.824+0.529.355)=31.79 JK1=285.83+31.79(373298)103 =283.45 kJmol1 31.反應(yīng)N2(g) + 3H2(g) = 2NH3(g)rHOm,298=92.88 kJmol1ab103c107N2(g)26.985.9123.376H2(g)29.070.83720.12

12、NH3(g)25.8933.0030.4662.4162.599117.904=92880+6241+2178144= 97086 Jmol1H2(g) + I2(s)H2(g) + I2(g)rHOm,291= 49.455 kJ mol12HI(g)2HI(g)Cp,m=55.64 JK1mol1T=386.7K 熔HOm=16736 Jmol1Cp,m=62.76 JK1mol1Cp,m=7R/2 JK1mol1T=457.5K 蒸HOm=42677 Jmol1sll grHOm，473=？HHICp.m=7R/2HI2HH2Cp.m=7R/232.按圖示過(guò)程計(jì)算：HH2=nCp,mT=3.5R(473291)=5296 JHHI=nCp,mT=23.5R(473291)=10592 JHI2=H1(s,291386.7K) + H2(sl) + H3(