[化学]物理化学总习题解答.doc
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1、物理化学习题解答(一)习题 p606210解: (1) 平均自由程:,未知数n怎么求?其物理意义是什么?由公式pV=NkBT, n=N/V=p/(kBT)n=100103/p(1.38110-23298)=2.431025m3=0.707/3.14(0.310-9)22.431025= 1.0310-7m (2) Z/=Vad2n, 未知数Va怎么求?其物理意义是什么?Va=15.01m.s-1Z/=15.013.14(0.310-9)22.431025=1.47108/s(3) Z=n Z/=2.4310251.47108=1.771033/s13解: (1) 理想气体: pV=nRT, ,
2、=nR/pV=nR/nRT=1/T(2) 范德华气体:(p+n2a/V2)(V-nb)=nRT,展开方程式得,pV-nbp+n2a/V-n3ab/V2=nRT =(RV3-nbRV2)/(RTV3- 2anV2+4abn2V-2ab2n3)21解: 2C+O2=2CO C+O2=CO2(1) VO2=10.21=0.21单位体积;VCO=2VO20.92=0.3864单位体积;VCO2=VO20.08=0.0168单位体积;V总=V空+VCO+VCO2-VO2=1+0.3864+0.0168-0.21=1.1932单位体积.(2) xN2=VN2/V总=10.78/1.1932=0.654;x
3、Ar=VAr/V总=10.0094/1.1932=0.00788xCO=VCO/V总=0.3864/1.1932=0.324;xCO2= VCO2/V总=(10.0003+0.0168)/1.1932=0.0143(3) 2C + O2 = 2CO C + O2 = CO2 24g 1mol 2mol 12/g 1mol 1mol xg x/24mol x/12mol y/g y/12mol y/12mol x+y=1000 x+y=1000 x=958.33/g x/24:y/12=92:8 x=23y y=41.67/gnO2=x/24+y/12=958.33/24+41.67/12=43
4、.40mol;nCO=x/12=958.33/12=79.86mol;nCO2=y/12=41.67/12=3.47mol;n空=nO2/0.21=206.68moln总=n空+nCO+nCO2-nO2=206.68+79.86+3.47-43.40=246.61molV总=n总RT/p=246.618.314293/105=6.00m325解:(1) =23000/60=100/s-1, V=r=40/m.s-1 EH2=mH2V2=2/6.0231023(40)2=2.621810-20JEO2=mO2V2=32/6.0231023(40)2=4.19510-19J(2) n/n0(H2)
5、=exp(-EH2/kBT)=exp-2.621810-20/(1.38110-23293)= 1.534710-3n/n0(O2)=exp(-EO2/kBT)=exp-4.19510-19/(1.38110-23293)= 9.47710-46(3) n(H2)/n(O2)= 1.534710-3 /9.47710-46=1.61941042物理化学习题解答(二)习题 p1291333解:(1) V2=V1,W=0,U= 1.5R(T2-T1)=1.58.314(546-273)=3404.58JU =Q+W,Q=U=3404.58Jp2=nRT2/V2=18.314546/(22.410-
6、3)=202.65kPa(2) T2=T1,U=0W= -nRTln(V2/V1)= -18.314546ln(44.8/22.4)= -3146.50J U =Q+W,Q= -W= 3146.50Jp3=nRT3/V3=18.314546/(44.810-3)=101.33kPa(3) U= 1.5R/(T1-T3)=1.58.314(273-546)= -3404.58JQ= 2.5R(T1-T3)=2.58.314(273-546)= -5674.31JW=U-Q= -3404.58-(-5674.31)=2269.73Jp1=nRT3/V3=18.314273/(22.410-3)=1
7、01.33kPa8解:(1) V1=nRT1/p1=18.314423/(100103)=35.1710-3m3W= -nRTln(V2/V1)= -18.314423ln(10/35.17)=4422.78J(2) p1Vm,1=RT1+bp1-a/Vm,1+ab/Vm,12 100103Vm,1=8.314423+3.7110-5100103-0.417/Vm,1+0.4173.7110-5/Vm,12105Vm,13=3520.532Vm,12-0.417Vm,1+1.5470710-5,Vm,1=35.08710-3m3W= =-RTln(Vm,2-b)/(Vm,1-b) -a(1/V
8、m,2-1/Vm,1)= -8.314423ln(10-0.0371)/(35.087-0.0371)- 0.417103 (1/10-1/35.087)= 4423.826-29.815= 4394.01J9解:(1) W= -pe(V2-V1)= -100103(1677-1.043)1810-6= -3016.72J(2) W= -pe(V2-V1)-peV2= -10010316771810-6= -3018.60J;W%=(3018.6-3016.72)/3016.72100%=0.063%(3) V2=nRT2/p2=18.314373/(100103)=31.0112dm3W=
9、-pe(V2-V1)-peV2= -10010331.01110-3= -3101.12J(4) vapHm= 40.69 kJ.mol-1;vapUm =vapHm +W=40.69-3.02=37.67kJ.mol-1(5) vapUm 0(实际上是T、P的函数),vapHm-W 由于体积膨胀,分子间的平均距离增大,必须克服分子间引力做功,热力学能也增大,故蒸发的焓变大于系统所做的功。23解:(1) T2=T1,U=0W= =-nRTln(V2/V1)= -18.314298ln2=-1717.32JU =Q+W,Q= -W=1717.32J(2) 绝热,Q=0,T2=T1(V1/V2)-
10、1=298(1/2)5/3-1=187.73KU=nCv,m(T2-T1)=1.58.314(187.73-298)= -1375.2JU =Q+W,W=U= -1375.2J(3) V1=nRT1/p1=18.314298/(200103)=12.38786dm3p1=104V1+b,b=200103-10412.38786=76121.4p2=104V1+p1=10412.38786+200103=323.8786kPaT2=P2V2/nR=323.8786212.38786/8.314=965.16K =-5106(V22-V12)-b(V2-V1)= -15106V12-bV1= -1
11、5106 (12.3878610-3)2-76121.412.3878610-3= -3244.87J-W3-W1-W2;U3U1U228解:Ag(s)+1/2Cl2(g)=AgCl(s) fHm (AgCl,s,298.15K)=?由1/2(1)+1/2(2)得: (5) Ag(s)+HCl(g)+1/4O2(g)=AgCl(s)+1/2H2O(l)由(5)+(3)得: (6) Ag(s)+1/4O2(g)+1/2H2(g)=AgCl(s)+1/2H2O(l) 由(6)-1/2(4)得: (7) Ag(s)+1/2Cl2(g)=AgCl(s)即由1/2(1)+1/2(2)+(3)-1/2(4
12、)可得:Ag(s)+1/2Cl2(g)=AgCl(s)故:fHm (AgCl,s,298.15K)=1/2(rHm,1+rHm,2-rHm,4 )+rHm,3=1/2(-324.9-30.57+285.84)-92.31= -127.125 kJ.mol-131解:C(g)+4H(g)=CH4(g) rHm(298.15K)=?(1) C(石墨)=C(g) subHm,1(298.15K)=711.1 kJ.mol-1(2) C(石墨)+2H2(g)=CH4(g) fHm,2(298.15K)=-74.78 kJ.mol-1(3) H2(g)=2H(g) divHm,3(298.15K)=43
13、1.7 kJ.mol-1由(2)-(1)-2(3)可得:C(g)+4H(g)=CH4(g) 故:rHm(298.15K)= fHm,2-subHm,1 -2divHm,3=-74.78-711.1-2431.7= -1649.28 kJ.mol-135解: 可近似看作绝热反应 r1Hm=0(298K,100kPa) 2C2H2(g)+5O2(g) +20N2(g)4CO2(g)+2H2O(g) +20N2(g) (T=?,100kPa)H1 r2Hm H2(298K,100kPa) 2C2H2(g)+5O2(g) +20N2(g)4CO2(g)+2H2O(g) +20N2(g) (298K,1
14、00kPa)H1=0;n(N2)=80/205=20molr2Hm=4fHm(CO2)+2fHm(H2O)-2fHm(C2H2)=4(-393.51)+2(-241.82)-2226.7= -2511.08 kJ.mol-1H2= 4Cp,m(CO2)+2 Cp,m(H2O)+20 Cp,m(N2)dT= (437.1+233.58+2029.12)dT +(48.7910-3+210.2910-3 +203.7610-3)TdT= 797.96(T-298)+ 65.4710-3(T 2-2982)=(817.47+ 0.06547T)(T-298)r1Hm=H1+r2Hm+H2,H2= -
15、r2Hm(817.47+ 0.06547T)(T-298)=2511.08103T=2806.1K或者:H2= 4Cp,m(CO2)+2 Cp,m(H2O)+20 Cp,m(N2)dT=(437.1+233.58+2029.12) (T2-T1)= 797.96(T-298) r1Hm=H1+r2Hm+H2,H2= -r2Hm797.96(T-298)=2511.08103T=3444.87K物理化学习题解答(三)习题 p2002034解:(1) p总(VO2+VN2)=n总RT,pO2VO2=nO2RT,pN2VN2=nN2RTp总= n总RT/(nO2RT1/pO2+ nN2RT/pN2)
16、=n总/(nO2/pO2+ nN2/pN2) =1/(0.2/20+0.8/80)=50kPa(2) 等温,U=0,H=0,故:Q= -WV=0,Q= -W=0为了计算状态函数,设计如下可逆途径:(a) O2(298K、V 、20kPa)O2(298K、2V 、10kPa),等温可逆WO2,r= -nO2RTln2= -0.28.314298= -343.46JQO2,r= -WO2=343.46JSO2= QO2/T =1.1526J/KGO2= -nO2RTln2= -343.46J(b) (N2(298K、V 、80kPa)O2(298K、2V 、40kPa)WN2,r= -nN2RTl
17、n2= -0.88.314298= -1373.85JQN2,r= -WN2= 1373.85JSN2= QN2,r/T =4.6103J/KGN2= -nN2RTln2= -1373.85J S=SO2+SN2=5.763J/KG=GO2+GN2= -1717.3J(2) 等温可逆,U=0,H=0,S= -5.7627J/K,G=1717.3J故:W=Q= TS= -5.7627298= -1717.3J 6解:(1) 等温可逆,U=0,H=0,故:Q= -W W= -nRTlnV2/V1=28.314300ln50/20= -4570.82JS=Q/T=4570.82/300=15.324
18、J/K(2) 真空膨胀,U=0,H=0,Q= -W=0,S=15.324J/K(3) 等温,U=0,H=0W= -pe(V2-V1)= -100103(50-20)10-3= -3000JQ= -W=3000J,S=15.324J/K8解:(a) (0.15kg,273K) H2O(S)H2O(l)(0.15kg,273K) 等温等压可逆相变Q1=H1=mfusH =0.15103g333.4J.g-1=50.01kJS1=Q1/T=50010/273=183.19J.K-1 (b) m1Cp(T-273)=m2Cp(298-T),0.15(T-273)=0.1(298-T),T=294.74
19、K(0.15kg,273K) H2O(l)H2O(l)(0.15kg,TK) 等压可逆S2= mCp lnT2/T1=0.151034.184ln(294.74/273)=48.09J/K(c) (1.0kg,298K) H2O(l)H2O(l)(1.0kg,TK) 等压可逆S3= mCp lnT2/T1=1.01034.184ln(294.74/298)= -46.02J.K-1故:S=S1+S2+S3=183.19+48.09-46.02=185.26J.K-111解:(1) 等温可逆,U=0,H=0W=-nRTlnV2/V1 = -nRTlnP1/P2=18.314298ln(100/6
20、00)= -4439.21JQ= -W=4439.21JSsys=Q/T=4439.21/298=14.90J.K-1A=U-TS = 0-4439.21= -4439.21JG=H-TS = 0-4439.21= -4439.21JSsub=-Q/T= -4439.21/298=14.90J.K-1Siso=Ssys+Ssub =0(2) 等温,U=0,H=0等外压,W= -pe(V2-V1)= -pe (nRT/p2-nRT/p1)= -60010318.314298(1/600-1/100)10-3=12.39kJQ= -W= -12.39kJS=14.90 J.K-1A= -4439.
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