2010-2011学年春季《热学》习题3(答案)

2010-2011春《热学》习题邢玉梅2011-05-011第二章分子动理学理论的平衡态理论习题三：1.质量为10Kg的氮气，当压强为1.0atm,体积为7700cm3时，其分子的平均平动能是多少？解:∵mRPVTmM而T23k∴J104.5)10022.6()10(2)028.0()107700()10013.1(32323241231365molKgmolKgmPamNPVMmRkPVMAmm2.求常温下质量为m1=3.00g的水蒸气与m2=3.00g的氢气组成的混合理想气体的定体摩尔热容.设：水蒸汽的定容比热为1Vc，氢气的定容比热为2Vc，则混合气体的定容比热可由量热学,21QQQ即221121)(McMccMMVVV，得)1()(212211MMMcMccVVV，定容摩尔热容量)2(VmolVcMC，对混合气体，其平均摩尔质量满足关系：)3(1112211MMMMMMMmolmolmol（注：（3）式可由理想气体状态方程及道尔顿分压定律得出。）综合上述三式，其中21MMM，可得MMMMMMMcMcMCmolmolVVV22112211111）（，令：水蒸汽和氢气的定容摩尔热容量分别为111molVVMcC、222molVVMcC，则上式变为2211222111molmolmolVmolVVMMMMMMCMMCC，对水蒸汽，;3261RkNCAV对氢气，RkNCAV25252，故有2010-2011春《热学》习题邢玉梅2011-05-012)KJ/mol(21.2120512318323251833RRRCV3.在等温大气模式中，设气温为5oC，同时测得海平面的大气压和山顶的气压分别为aP1000.15和aP1078.05，试问山顶的海拔为多少？解：根据玻尔兹曼分布律，可得等温高度公式，代入数据，得z=RTMmglnp0p=(8.31J×mol-1×K-1)´(278K)(29´10-3Kg×mol-1)´(9.8m×s-2)ln1.00´105Pa0.78´105Pa=2019.7(m)。4.当液体与其饱和蒸气共存时，气化率与凝结率相等。设所有碰到液面上的蒸气分子都能凝结为液体，并假定当把液面上的蒸气迅速抽去时，液体的气化率与存在饱和蒸气时的气化率相同。已知水银在0oC时的饱和蒸气压为2N0246.0m，问每秒通过每平方厘米液面有多少克水银向真空中气化。解：以饱和蒸汽为研究对象，设所有碰到液面的蒸汽分子都凝结为液体，则t内碰到A液面的气体质量为：tAMRTkTpNMtAvnNMtAmMmAmAm84141代入数据：Mm=200.5´10-3kg×mol-1,p=0.0246pa,DA=10-4m2,T=273KM=(200.5´10-3kg×mol-1)´(0.0246Pa)´(10-4m2)4´(8.31J×mol-1×K-1)´(273K)´4´8´(8.31J×mol-1×K-1)´(273K)3.1416´(200.5Kg×mol-1)得)(1023.99kgM。按题意，假设凝结率等于汽化率，则该结果也就是每秒通过每平方厘米液面逸出的水银分子的质量。5.某种气体的分子由四个原子组成，它们分别处在正四面体的四个顶点上。（1）求这种分子的平动、转动和振动自由度数；（2）根据能量均分定理求这种气体的定容摩尔热容。解：（1）因n个原子组成的分子最多有3n个自由度。其中3个平动自由度，3个转动自由度，3n-6个是振动自由度，这里n=4，12433ni，故有12个自由度。（2）作刚性近似，自由度数6i，)KJ/mol(93.243RCV6.Asealedcubicalcontainer20.0cmonasidecontainsthreetimesAvogadro’snumberofmolecules2010-2011春《热学》习题邢玉梅2011-05-013atatemperatureof20.0℃.Findtheforceexertedbythegasononeofthewallsofthecontainer.[Answer]Now12310023.6molNA,1131.8KmolJR,KT15.293themoleofthegasmolNNNNnAA33,thesidelengthofthecontainermL2.0One-sidedareaofthecontainer22204.0)2.0(mmLAThevolumeofthecontainer333008.0)2.0(mmLVAssumingthegasisideal,accordingtotheEquationofStateforanidealgas,nRTpVSop=nRTV=(3mol)×(8.31J×mol-1×K-1)×(293.15K)0.008m3=9.135´105J×m-3=9.135´105PaThen,theforceexertedbythegasononeofthewallsofthecontainerNmPamPapAF422510654.33654004.0)10135.9(7.Inaperiodof1.00s,5.00×1023nitrogenmoleculesstrikeawallwithanareaof8.00cm2.Ifthemoleculesmovewithaspeedof300m/sandstrikethewallheadoninperfectlyelasticcollisions,whatisthepressureexertedonthewall?(ThemassofoneN2moleculeis4.68×10-26kg.)[Answer]themoleofnitrogenmoleculesmolmoln830.010022.61000.512323ThetotalmassofnitrogenmoleculesKgKgNmmsum0234.0)1068.4()1000.5(2623Assumingthevelocityattitudeofnitrogenmoleculesbeforecollisionisthepositivedirection,sothespeedsofnitrogenmoleculesbeforecollisionandaftercollisionaresmv/300andsmv/300',respectively.AccordingtotheMomentumtheorem,theimpulseofsinglenitrogenmoleculecausedbythewallis12326'/10808.2)/300/300()1068.4()(smKgsmsmKgvvmmvmvpTheminussignrepresentsthedirectionoftheimpulseofnitrogenmoleculeisoppositetothatofthevelocity.Thetotalimpulseofallnitrogenmoleculesis11232304.14)10808.2()1000.5(smKgsmKgNppsumTheforceofthenitrogenmoleculescausedbywallis2010-2011春《热学》习题邢玉梅2011-05-014NsmKgssmKgtpFsumN04.1404.14)00.1()04.14(21AccordingtoTheNewtonthirdlaw,theforceexertedbythenitrogenmoleculesonthewallisNFFNw04.14Sincetheareaofthewallstrikebynitrogenmolecules241000.8mAThepressureexertedonthewallPamNmNAFPw4242410755.110755.11000.804.148.Inaconstant-volumeprocess,209Jofenergyistransferredbyheatto1.00molofanidealmonatomicgasinitiallyat300K.Find(1)theincreaseininternalenergyofthegas.(2)Theworkitdoes.(3)Itsfinaltemperature.[Answer](1)Intheconstant-volumeprocess,theenergywastransformedintointernalenergy,sotheincreaseoftheinternalenergyis209J.(2)Intheconstant-volumeprocess,thesystemdoesnotdowork,sotheworkitdoesis0J.(3)1131.8,20923UKmolJRJTRmSo,KKmolJmol77.16)31.8(231J209T11K77.316K77.16K300TSothefinaltemperatureis316.77K9.Onemoleofanidealdiatomicgaswith2/5RCvoccupiesavolumeiVatapressureiP.Thegasundergoesaprocessinwhichthepressureisproportionaltothevolume.Attheendoftheprocess,itisfoundthatrmsvofthegasmoleculeshasdoubledfromitsinitialvalue.Determinetheamountofenergytransferredtothegasbyheat.[Answer]TTTTMRTvmrms3431,21525EnnRTTRnTnCvtransAccordingtotheequationofstateforanidealgas,nRTViiPSoiitransVP215E2010-2011春《热学》习题邢玉梅2011-05-01510.Thedimensionsofaroomare4.20m×3.00m×2.50m.(1)Findthenumberofmoleculesofairinitatatmosphericpressureand20.0℃.(2)Findthemassofthisair,assumingthattheairconsistsofdiatomicmoleculeswithamolarmassof28.9g/mol.(3)Findtheaveragekineticenergyofamolecule.(4)Findtheroot-mean-squaremolecularspeed.(5)Ontheassumptionthatthespecificheatisaconstantindependentoftemperature,wehave2/5intnRTE.Findtheinternalenergyintheair.(6)Findtheinternalenergyoftheairintheroomat25.0℃.[Answer](1)P=1.013´105Pa,R=8.31J×mol-1×K-1,T=293.15K350.3150.200.320.4VmmmmAccordingt