化学原理2习题答案1

第一章习题解答1.CH4在25℃、95.9kPa时所占的体积为2L，求在标准状况下的体积（0℃、100kPa)。假设符合理想气体状态定律。已知：p1=95.9kPa，t1=25℃，V1=2p2=100kPa，t2=0℃求：V2＝？解：按照理想气体定律pV=nRTp1V1/T1=p2V2/T2V2=p1V1T2/(T1p2)=95.9kPa×2L×(25℃/℃+273.2)K/[(0℃/℃+273.2)K×100kPa]=1.76L2.0.016kgN2在273.15K、50.6kPa时的体积。假设符合理想气体状态定律。已知：T＝273.15K，p=50.6kPa，m=0.016kg求：V＝？解：按照理想气体定律pV=nRTV＝nRT/p=0.016kg/0.028(kg/mol)×273.15K×8.315J·mol-1·K-1/50.6×103Pa＝25.6×10－3m3＝25.6L3.用理想气体状态方程求5.07MPa、－40℃时甲烷的密度。已知：p=5.07×106Pa，T＝(－40℃/℃＋273.15)K=233.15K求：密度＝？解：由理想气体状态方程pV=nRT=m/MRT=m/V=pM/RT=5.07×106Pa×0.016kg/mol/[8.315J/(mol·K)×233.15K]=41.84kg/m34.已知：T＝（0℃/℃+273.15)K解：=m/V=pM/RTM＝RT/pp/kPa101.367.5350.6533.77/(gL-1)1.97681.31490.98510.6560M(g/mol)44.321744.224144.173744.1200用线性回归方法得到，p=0，M=44.02206绘得下图：M－p图4444.0544.144.1544.244.2544.344.35020406080100120p/kPaM/(g/mol)5.解：由Amagat分体积定律得x(CO2)=0.015由Dalton分压定律得p(CO2)=p×x(CO2)=101.3kPa×0.015=1.52kPa6.已知：V(O2)＝0.5L，p(O2)＝100kPaV(N2)＝1.6L，p(N2)＝50kPat＝25℃求：（1）p＝？（2）混合后，p(O2)，p(N2)解：n(O2)＝p(O2)V(O2)/RTn(N2)＝p(N2)V(N2)/RT由于气体服从理想气体状态定律n＝n(O2)＋n(N2)＝pV/RT（1）pV＝p(O2)V(O2)＋p(N2)V(N2)p＝(100kPa×0.5L＋50kPa×1.6L)/2.1L＝61.9kPa（2）混合后p(O2)＝p×n(O2)/n＝p×p(O2)V(O2)/pV＝p(O2)V(O2)/V＝100kPa×0.5L/2.1L＝23.8kPap(N2)＝p－p(O2)＝61.9kPa-23.8kPa=38.1kPa7.解：（1）体积分数＝摩尔分数体积分数不等于质量分数p(O2)＝21kPa，p(N2)=79kPaV(O2)＝0.21L，V(N2)=0.79L（2）升高温度，分压力都增加，而分体积和质量分数都不变（3）等温下，将体积增加1倍，则分压力皆减少一半，p(O2)＝10.5kPa，p(N2)=39.5kPa，而分体积增加一倍，V(O2)＝0.42L，V(N2)=1.58L（4）充入N2，使总压达到150kPa，则p(O2)不变，p(N2)＝129kPa，V(O2)＝21kPa/150kPa×1L=0.14L，V(N2)＝0.86L（5）维持温度、压力不变充入N2，则O2分压减少，分体积不变，N2分压与分体积皆增加。8.证明：由Dalton分压定律得p(N2)＝0.3pp(O2)＝0.5pp(CO2)＝0.2pp(N2):p(O2):p(CO2)＝3：5：29.已知：p(A)＝100kPa，T(A)＝300K，V(A)＝1Lp(B)＝200kPa，T(B)＝300K，V(B)＝4.5LT＝T(A)，p＝100kPa求：V、体积分数及物质的量分数解：ni＝piVi/RTn＝ini＝ipiVi/RT＝pV/RTxi＝ni/n＝piVi/pVpV=pAVA+pBVBV＝(pAVA+pBVB)/p＝(100kPa×1L＋200kPa×4.5L)/100kPa＝10LxA＝nA/n＝pAVA/pV=0.1xB＝0.9体积分数分别为10%、90%。10.已知：m=1kg，M＝0.016kg/mol，p=40MPa，T＝273.15Ka=2.29×1011Pa·cm6·mol-2=0.229Pa·m6·mol-2b=42.8cm3·mol-1=42.8×10-6m3·mol-1求：V解：此题用迭代法求解(p+n2a/V2)(V-nb)=nRTV=nRT/(p+n2a/V2)+nb(*)n=1kg/0.016kg·mol-1=62.5mol迭代初始值用理想气体状态定律计算V0＝nRT/p=62.5mol×8.315J·mol-1·K-1×273.15K/40×106kPa＝3.55×10-3m3代入(*)式V1＝62.5×8.315×273.15/(40×106＋62.5×62.5×0.229/(3.55×10-3)2)+62.5×42.8×10－6＝3.95×10－3m3V2=4.13×10－3m3V3=4.21×10－3m3V4=4.24×10－3m3V5=4.26×10－3m3V6=4.26×10－3m311.解：若为一定量的理想气体，pV/RT=常数，故不为理想气体TbVVapmmR2TnnbVVanpR2212.解：n=1mol，T＝(100℃/℃+273.15)K=373.15K，V＝10L＝0.010m3a=2.29×1011Pa·cm6·mol-2=0.229Pa·m6·mol-2b=42.8cm3·mol-1=42.8×10-6m3·mol-1按理想气体计算：p＝RT/V＝8.315×373.15/0.010＝3.103×105Pa按范德华方程计算：p＝RT/(Vm-b)-a/Vm2＝8.315×373.15/(0.010-42.8×10-6)－0.229/0.0102＝3.093×105Pa13.解：T=(-88℃/℃+273.15)K=185.15K，p=4.53MPa,n=1molTc=(-118.4℃/℃+273.15)K=154.75K，pc=5.04MpaTr=T/Tc=1.20，pr=p/pc=0.90查图得到，Z=0.84V=ZRT/p=0.84×8.315×185.15/4.53×106=2.85×10-4m314.解：T＝298.15K，V=40L=0.040m3，p=20MPaTc=(-118.4℃/℃+273.15)K=154.75K，pc=5.04MpaTr=T/Tc=1.93，pr=p/pc=3.97查图得到，Z=0.98n=ZpV/RT=0.98×20×106×0.040/(8.315×298.15)=316.24molm=316.24×0.032=10.12kg15.解：p=14.0MPa，=9.63×10-2kg·L-1=96.3kg/m3Tc=(-82.1℃/℃+273.15)K=191.05K，pc=4.64MpaPr=3.02pM＝ZRTZT＝pM/R=14.0×106×0.016/(96.3×8.315)=279.74KZ=279.74/T=279.74/TcTr=279.74/(191.05·Tr)=1.464/TrTr0.91.01.11.21.41.61.71.82.0Z1.631.461.331.221.050.9150.860.810.732查图得：Tr1.01.11.21.31.41.61.82.0Z0.420.470.560.660.720.840.910.95由两条曲线交点得到：Tr=1.72T=Tr×Tc=191.05×1.72=328.61K=55.46℃1.15答案0.60.70.80.911.11.41.61.82TrZ计算结果查图结果16.解：n=348/48=6mol，V=10L=0.010m3，T=373.15Ka=13.9×1011Pa·cm6·mol-2=1.39Pa·m6·mol-2b=116.4cm3·mol-1=116.4×10-6m3·mol-1Tc=(152.01℃/℃+273.15)K=425.16K，pc=3.8MPaTr=0.878按理想气体计算：p=nRT/V=6×8.315×373.15/0.010=1.86×106Pa按范德华方程计算：p=nRT/(V-nb)-n2a/V2=6×8.315×373.15/(0.010-6×116.4×10-6)-6×6×1.39/0.0102=1.50×106Pa按压缩因子图计算：Z=pV/(nRT)=PrPcV/(nRT)=Pr×3.8×106×0.010/(6×8.315×373.15)=2.041PrPr0.20.30.40.5Z0.410.610.821.02查图：Pr0.20.30.40.50.6Z0.890.820.740.660.58由图查得pr=0.372p=0.372×3.8=1.42MPa1.16答案00.20.40.60.811.20.150.250.350.450.55PrZ计算结果查图结果