化工热力学(第三版)陈钟秀课后习题答案.docx

其次章21.运用下述方法计匏IknXU甲烷贮存在体枳为0.1246m'、温度为Soe的容器中产生的压力：(1)志向气体方程：(2)RK方程：(3)行因化关系式.挈：甲烷的摩尔体积V=O.124611vV1.knx>1.=J24.6cm,11w1.ff1.和甲烷的临界参物T<=I9O.6KPt=4.0MPaV.=99cmVnw1.=Q<XJ8(1)志向气体方程P=RT,'V=8.314x3”1*124.6x16f21.S6MPaR-K方程,=0.42748现=。42体业*1=3.222布儿心FGP14.6XIO6,=0.08664=0.08664=2.985×10smj11w/.P=旦«"V-hT''3V(V+b)8.314x323.153.222(12.46-2.985)×105323.1505×12.46×105(12.46+2.985)×105=19.O4MPu(3)普/化关系式Tr=TTt.=323.15/19().6=1.695V,=V/Vt=124.6/99=1.259<2.利川件仄法计算,Z=Zc+Z'Z等6盘黑打=皿死迭代：令ZHI-P“16S7乂TIn.69S.fiW.三W:ZJ=O.X938ZXbI623Z=Z''+Z'=0.8938+0.0(BX0.4M3=0.8975虻时.P=RP产4.6x4.687=21.56MPa同理.!iZ1=0.8975依上述过程计算.一至计算出的相邻的西个Zifi相差很小.送代结束.-Z和PWtf1.：.P=1.9.22MPa22分别运用志向气体方株和Pitztf密遇化关心式计算5IOK.2.5MPa正丁烷的摩尔体枳.已知国魁值为14M>.7CmVnN-»?：RI倒正丁烷的推界替数：T<=425.2KP.=3.80()MPaV,=99cm"mn1<o-O.I9341管(片+<w1.11.)三8；：就：；9(_().02989+().()49×0.1336)=-7.378×1060422=0.083JZ=-0.3417(3OY3O4.2)S1.=0.139-?=0.1390"2=403588(303/3()4.2)4-B22=$(B；+(a2B)=Wm：2(-0.3417-0.2250.03588)=-1.1.9.931.6乂T”=(rZj”=(132.9x304.2)"=201.068K93.1依+94.(W'=93.55cmimo1.Z钎平幽詈，蚪他+他0.295+0.225=-=0.137P1=Ziv¾vw=0.2845×8.314×201.068/(93.55×106)=5.O838MP4.Trij=v=303/201.068=1.507P=PP,=0.1013/5.0838=0.0199成=。.083-詈=。83-僚=136()|72B11,=0.139-=0.)39-20.1721.5074y=0.1083B12=(堤+明国J=*器;"渭(-0.136+0.137×0.1083)=-39.84×105BZ=y；BN+2yy21.j+£BN=0.24,×(-7.3780)+2×0.24X0.76×(-39.84K1O,)+O.76,(-1.19.93×10*)=-S4.27x1.%nu>iBPPV:.Z1=I+-s-V=O.O2486m'mo1.RTRT.,.V=nV=100xI1.×8138%-'12×0.02486=168.58m,(2)P=y1P-=O.24×O.IOI3-=O.O25Pa,Z.,0.2845P、=Y、P虫=0.76×0.1013-=0.074.WPa2-Zm0.28452，.格压力为2.0WP;，、2攵为。TK条(1吓的2.83mNHa压缗到O.I42m',七压缩£海度44K6K,则其压力为若I?分别用内述方法计皿(I)VandcrWaaJs方程；(2)Rcd1.ich-Kwang方程t<3)PEg-RObinsOn方程I<4)忤篇化关系式.解：ftRI/NH'的舱i界参数：r=405.6KP1.=H.2KMPaVv=72.5cmVno1.-O.25O(I)求取气体的廉尔体积财干状班1|P=2,O3MP».T=447K.V=2.83mj7；=7/7；=477/405.6=1.176.=2.03/11.28=0.18Af维法B''=0.083-=-S3-0-42?,=-0.24267,j61.176,6*=0139-詈=0139-器y=005194器26+025-96Z=+筹普小黑与M.n=2.83mV1.885×IO'mVm<M=15(>1.mo1.对于状名Ih除尔体税V=O.14211V1.501.mo1.=9.458×1.05mVmo1.T=448.6K(2) Wmderwaak力利=()A253Pam,mo227W27x8.314?X4056MP.64×11.28IO6bRT8.314×405.6'=玄=81.1.28"=3.737×10-smjmo/-1r,KTa8.314×448.60.4253,一r.P=T=7r=17.65MPa3V(9.458-3.737)X10s(3.737x10)(3) Rai1.ich-Kwang方程a=0.42748-=0.427488314x4()5；6,-=8.679Pa,tto2P11.28×106b=0.086649=0.086648314x406=2.59×10Wmo/1P,11.28XIO68.679=I834PS.3I4×4486V-ftTwV(V÷b)(9.458-2.59)×10'5448.6,×9.458×IO'5(9.458+2.59)×IOT(I)PCngRobinson方程VTr=TTc=448.6/405.6=1.106.k=0.3746+.5422(uo-0.26992,J=0.3746+1.54226×0.25-0.26992×O.252=0.7433()=1.+fc(1.-7,)y=1.+0.7433X(I-1.106"S)T=0.9247a(T)=a,a(T)0.45724-(7')=0.45724×胃皆:需X0.9247=0.4262mft-rm,I'b=0.07780以=0.07780X心火4。芋=2.32610sm-moiP.11.28×Kf.P=旦必V-bVy+b)+bV-b)8.314x448.60.4262(9.458-2.326)×IO,-9.458×(9.458+2,326)101°+2,326×(9.458+2.326)×IOm=19.00Ma(5)普遍化关系大V匕=VV=9.458×10s7.25×105=1.305<2适用普质法.迭代进行计算.方法网Ii(3>26试计券台行30%(摩尔分数)氯气(1)和70%1.隼尔分散)正丁烷气体混介物7g.1.S8C.6.88SM!条件下的体枳,已知BiI=I4c11ie°IB*="265cmY11>1.B2=-9.5cmVmo1.棒Bm=A+2y,M1.j+=0.32×14+2×0.3×0.7×(-9.5)+O.72×(-265)=-132.5&毋Inv>1BI1PVZ=1+-=TV(I米尔体积)=4.24x1.(尸m',no1.RTRT假设气体混合物总的摩尔数为1.则0.3nx28÷0.7nx58-7n-0.1429mo1.V=廿丫(摩尔体枳E.I429×4.24xOyO.57CmS28.试用R-K方和和SRK力积计算273K,101.3MP*F氟的原缩因子.C1.知试验也为2685W：IdHIEoS的普遍化形式长阳泰二RNH，的的界落%T1=I26.2KPE.394MR1.-0.(M(1) R-K方程的普造化。=0.42748-=0.42748314%I262：=i5571pa.n.Zwo尸P13.394×IO6b=0.08664旦=0.0866483七*=2678IOSm'.P13.394XIOh4aPnbPAa1.5577A=-;o=-=;rRT2sRTBhRT's2.678XIO5X8.314X27315,BhbP2.678×105×101.3×10ft1.1952/?=ZVZ1<TZx8314×273ZZ=-=-1.551.f-1.-t1.+1-U+J、两式联；£,迭代求解缩因子Z(2) SRK方程的砰道化7；=7/7；=273/126.2=2.163m=0.480+1.574«-0.176行=0.48()+1.574x0.(M-O.I76x0.(M2=0.5427a(7')=1.1.+n(-7:"s)=1+0.5427(1-2.163")T=0.2563a=0.42748×a(T=0.4274«''4R”：X.2563=0.3992RJ-M6osmoiP',3.394×IO6b=0.08664区=0.08664'”小段?=X)OTwJJ/03992=0.39751UP3.394X1.O62.678×IO5X8.314×273'"一二=旦=2.678XK)TXm.3x0=KZVZRTZ8314273ZZ=-1.-ZI-AB三(mmm.=-'两式联立，迭代求解压缩因子Z第三章S-I.物炕的体积防张系数Q和碎羽田缩笔皎人的定义分别为J：更.试导山听从VundcrWaa1.s状态方程的”和A的表达火W：Vim1.IerWaab方程P=±V-hV'由Z=fg)的性质像).闺,傕)=T存所以RT-WnV-b.(叫RV'(V-b)(万厂tm,-2a(V-1.)y故z7.m_八)，7V,-2(V-Jt一JK)v(v-)3=V8p)1.ftTy,-2a(V-b)23-2.某走向气体借活塞之助装于用瓶中.压力为34.4SMPa.淋原为93t.抗拒恒定的外1.k力3*MPa而等海65朕.直到两倍丁K初始容积为止.试计力此过程之A1.1.Mi、AS、M.G.,TdS、iV.Q.和W.解：老向气体辞湘过程.AU=O.W=0。=.心pdV=vpdV=J；与dVr=7In2=2109.2j.n>H÷-2109.2J'11u1.走向'i体等海膨朕过程dr=。、R<IS=-dPPAS=JjaS=-Aj,"InP=-RIn/'，=1.n2=5.763M11>KjA=(-TS=-3665.763=-2IO9.26M114K)AG=AHTNS-AA-2iw.26j'<11M>!KiTdS=TAS=A=-2IO9.261.'<mo1.K)"=J,"=J：专1.1.,=Rr1.n2=2IO9,2Jmo1.3-3.试求算Ikam在JQJmNIK,度为”3K下的内能*培.Mi.Cy.Ci,ft,I1.1.K假设氯气听从志向气体定律.已知:'I'.0.1013MPa：认的Cj,；女忆天系刀G=27.22+0.00418711(nx)!K)：(2)假定在SC及0.1013MP时氟的均为零I(3) ft.298K及0.1013MPa时班的崎为191.76JxmO1.K).3-4.收乳在27U、0.1MP;.协力为零,试求227X?、IOMRI卜双的培、W(ft.已知1.i志向气体状态F的定压摩尔热容为C=31.696+10.1.44×10'T-4.038X106T2J/(mo1.K)»?：分析就力学过程300KA1.MPaXI实气体H三三0,S=OIF-4R',3(X)KA1MPa理想气体5K.1.()MPa真实气体5(X)K.I0MPa理想气体钝因求二甜家的临界参数hT,=4I7K.Pk7.701MPa、=0.073(I>3（三）K,OjMPa的我实气体转换为志向气体的剁余培和婀余场肥=0.083-詈=-0.6324弟=0.675/7=I.592*=。/39-k=5485等=0.722r"=401.4TI=TdT,=300217=0.719P,=PP,=0.M7.701=0.013利用件靠法评理好卜-嘿+也喘"Ae修+噎)t<<itWW,'=-91.4Mm<,s,"=-O.2()37">fK1.志向气体由300K、0.1MPa到5<XHGIoMPa过程的均变和城变MI=J：/ZT=''31.696+10.144XIO3T-4.038X106T2(=7.O2kJnx)1.M1=r-ein-=,31.69(+10.1.44×10*,-4.038077-en-IJ1.1.TPi0.1=-20.39Amo1.K)(3)5OOK.IOMPa的志向气体找换为“女气体的剁余端和剥氽俄Tr=T=5(W,4I7=I.I99P,=PdR=Ia7.701=1.299利用铃雉法计。Bu=0.083-上半=-0.2326=0.675/T26=0.42I1.fd/，Q172oi,=0.139-yf=-005874常=0.722/刀"=0.281xAk喏化喇AM野啕代人数如计算科”fn-3.4IK>"人S"=U.768JMNK).h=Hj-Hi=Hj=-H,+“，+”：=91.41+7020-3410=3.701KJ,mo1.s=S1.SI=S-s,tv'+S：=0.2037-20.39-4.76g=-24.95J4w>1.K>3-5.W1.Mffifi化方法让酰：班化碳在473.2K、30MPn卜的恰与麻已知在相同条件下,二班化建处于志向状态的端为8377JMwI.慵为-2S.86MmMK).W：杳附用:将二氧化碳的够界参数为:TV=3O4.2K.P<三Z376MPa,«-0.225：.T/T,=473.23<M.2=1.556P,=P，'P<=3O7.376=4.067利用普压法计算i!t表，由炫性内插法计算得出I吗1.=T,741等=。.042C“1-O.85I7!1.4296K1.K1.RR上二”+,此1乜回。.由明"7：RjRReR计算储/严=4.377KJmo1.S*r=-7.635JA>m>1.K)：.H=M+2=4377+8.377=4KJjmo1.S=S*+W=2635-25.86=-33.5m,>1.K)36.UWI走21C时.ImoI乙焕的悒和疏汽与饱和液体的U,V.H和S的近似值.乙炊在0.1013MPa,Ot的志向气体状6的H、S定为零,乙快的正常潴怠为84C2C时的，汽JK为4.459MPa.3-7.将UIkg相373.5e3Mpar.七.试计算北过程中AU、AH、AS、AAMAG之情.3-8.试估算纯笨由0.1013MPa.801C的憎和淞体受为1.013MPa、180一的饱和薇汽时该过程的AVHJi和S.已知纯装在正常沸点时的汽化港热为3.733J/mo1.：饱和液体在正常沸点下的体枳为的.7Cmo1.:定笈摩尔热容C=I6036+O2357n(mo1.K):M：次维里泰数b=T8(1.KIO1.)"s'mOrW：1.iS的物性般数：1>562.1K.P1.=4.894MPa>=0.2712.3jV由两项推里方程Z,=生"=H二_78(1.k)，门*RTRT矶(7)J1.1.01.3×10f,8.3I4×10×4532.4'I=0.8597V=型J8597x83I4x453=396QM加/P1.013AV=Vi-V1AV=V2-V1=3196.16-95.7=31()0.5an3mo1.H=例,+(-)+H+Hf+(H/)必=5r+(-S1.')+5+S+(57)分析:黄和液体笨1013'11.3"K,y=95.-fmj¼o具体过程.WS但和W汽IOBMP-43KV,tH,S,SHr风口饱和重汽.IO1.JMPa.353K3.计算终一过程右史和编空（I）饱和液体（忸T、P汽化1一饱和蒸汽H*30733KJKmo1.5=/A=3O733/353=«7.1KJ.Kmo1.K（2）饱和蒸汽（353K>OIOBMPii）一志向气体Jq=O.Tc562.1P0.1013Pc-4.894=0.0207点<T,.Pr）落在图28B1.的找左上方,所以,用件JS化箧咀系效法进行计究,山式（3-61）.（3-62）计。然同翁a像制=-().()207X0.628X(2.2626+1.2824)+0.271(8.1124+1.7112)=-0.0807>1w'=W)三62,卜图啕=-0.0207(2.2626+0.271x8.1124)=-0.09234S1.*=-0.09234X8.314=0.7677KKmHK(3)志向气体(3S3K.OJOBMPa)一态向气体<453K.I.OI3MPa>邮=J：c：打r455/=(16.036+O.235T)tf=16.036(453-353)+(4532-3532)=I1.IO2.31K2。/Srf=,f-/?/»jATPi尸(16.036,cr1."1.013=+0.2357'(-8.314加JkITJ0.1013453=16.036/n+0.2357(453-353)-19.1=KJ/Kmo1.K(4)志向气体(453K.I.OI3MPa)-其实气体(4S3K,1.OI3MPa)4537；=-=0.806562.1P”摆二02。7。点TPr)落在图28图曲线左上方.所以，用普电化维里原政法进行计算.由式(361.).(3.62)计W玩=-0.806X0.207()1.1826+0.5129+0.271(2.2161+0.2863)=-0.3961'<1H'(IB'=”+dT1.dTt=-0.2070(1.1826+0.27I×2.2I6=-0.3691=3.0631KJ/Kmo1.KH：=1850.73KJ/Kmo1.1求AH27=Hv+(-W1k)+AHB+ht+(”2")=40361.7KJjKmo1.5=Sv+(-S1R)+AS；7+S÷=932给KJjKmo1.K39行A和B四个容戕A容戕充沿抱和灌态水.B霭戏无足饱和舞气.阚个容揩的体枳均为I1.压力5为IM1.Sn加这两个容器埠炸.试“噂个容;S被破坏的史严峻？仪定A、B容器内物货辕可逆绝热影张.快速地热秘胀到OJMPa3-10.一容睛内的液体水和蒸汽在IMPa压力下处于平衡状悠，帧敏为Ikg.材如容：内液体和焦汽汁占一半体校.试求杏器内的液体水利蒸汽的总培.解：森按压力拌列的饱和水蒸汽衣1MI时,/，=762.8W/依=2778.W/依V1=IA273CMfgV=I94.4cm,Ig侬据咫意液体和盛汽各占一半体枳.设千度为X则-V=(1.-)½A×194.4=(I-X)XI.1273解之得I=0.577%所以Z、H=xH+(1.-x)H1.=0.00577X2778.1+(1-O.577)X672.81=774.441/依31).过热蒸汽的状态为533KbCIO336MPi.好过喷嘴膨彩，M1.I压力为0.2067MPa.假M过程为UJ逆绝热I1.达到Tfti.试何蒸汽在嗓啸出11的状态加何？3-12.试求算366K.2026MRI卜Imo1.乙燎的体枳.ft.i!i255K.0.KM3MPa时乙烷的培*场为零.已知/.烷在志向气体状态1；的厚尔恒压热容C=10.038+239.3(MXIo77一73.358X10«尸J/(11>1.K)3-13.试采第RK方程求N在227C、5MPa卜气相正丁烷的利余埼和利余储.tt：杳附用用正丁烷的临界整航Tt三42S2KR1.M3.800MPa、=0.193nZR-Kfi½P=-一一V-bT''-V(V+h)a=0.42748=0.42748S3I4-422-=2904P4-MK<"mo1.'P3.8×IO6b=0.086643=0.08664二=8.06×10sw'-mo'P3.8×IO65xIb=8.314x500.1529.(>4XV-8.O6×IO's5(X).15,i3v(V+8.06×I0,)试能求得:V=5.61×IO,m,mn1.力,=理吗=。/438V56.1X1.Os=3.874Aa29.041.ihRT'-8.()6×IOjx8.314×5(X).I5ijz=t4,t=0.681工Z-1-3"RTbRT'7n1.+=Z-I-1.5-1.n(1.+)=-1-0997*=-1.0997x8.314×5(X).15=-4513JItno1.父=h"叫.2bRT,7,n,+7=-O.8O9S=-0.809×8.3I4=-6.726J/(“，/K)3M.假设二氧化碳听从RK状态方积.试计口MrC、10.13MPa时二粒化球的逸度.W：爽阳戏和怎化碳的格界给数：T1.川4.22K、P门7.376MPa.”0.42748誓=0,42748嘴穿=6.466吐淄N5RT力=0.08664一=0.086648.314x304.27.376X此RTv-brv(v+ft).,.IO,Bx1.O68.314x323.I56.4661V-29.71XIO*6323.1/V(V+29.71×IO*)迭代求得：1.294.9cm>mo1.,b29.71八一V294.9Aa6.466BbK'-=29.71×1(×8.314x323.15m=4.506z=t4(t)=!-4,506°-10071-0.1(X)7U+0.1(X)7=0.6997哈Z-一冷bRT'=-0.73264.M9M1.,a315.试i1.电液态水在3QC下，压力分别为3)饱和蒸汽压、<b)100XIo'PaF的逸女和逸慢系数，已知：(1)水在309时Ifi和蒸汽压沪0.(M24x0fPat(2)30,O-KX>05Pa是出内将液态水的摩尔体枳视为常数.共做为0.01809n的kmoh(3)INOSPa以下的水蕤气可以视为志向K体.W：(a>30r.DjM24x1.3IhY汽灌平衡时.乂Ix1.伊Pa以下的水薇气可以视为志向气体.>=0.(M24×10,Pa<x0'Pa.,30,C.O.(M24×1.(k'1.下的水蒸气可以衩为志向气体.又走向气体的=P.,工$=吊=0.0424*10'Pas=z7=(b)30C,100xIO5Pa.乃=EWexpJ；露P£=f1.sP1.s=%=Yip1.=。.。即1。X100-0324)O=007174£SRTRT8.3I4×3O3.15.£=1.074f.Sf：=1.074xs=1.074X0.0424xIO5=4.554×IO*Pa3-16.Iff1.A和B两股水蒸汽通过绝热混合获得0.5MPn的饱和蒸汽，其中A股是I度为98%的湿蒸汽,1代力为05MPa.近卡为IkgsiIfUB股是473.15K0.5MPa的过热蒸汽.试求B股过热蒸汽的流量该为多少？解：A股：会按压力揖列的性和水废汽表<),5MPi1.(I5I.9C)时，H1=640.23U!ksH=2748.7fc/fkgHi=0.98X2748.7+0.02X640.23=2706.53H/依BK:473.I5K.0.5MPa的过热蒸汽Hf1.=2H55AU/kg依据题点，为等压过程,Zi=QF的肛混合过程中的散羯损失.绝热混合Qp-O.所以/=Ot后培位不变设B股过热蒸汽的流量为XkgE以1秒为Ht?基欣列能成淅宽式2706.53X1+2855.4a=2748.7(1+x)=0.3952依/s2748.7-2706.532855.4-2748.7读混合过程为不行逆抱姓混合,所以S0混合M后的燧值不相等.只行可逆沦热过程.AS=O闪为是等压过程，该国也不成"HAU=Oj?.第四章4-1.在20在、I01.3MPaW.乙1（1）与H2O<2）所形成的溶液其体枳可用下式衣示:V=58.36-32.46x,-42.98.v?+58.77x123.45.r；.试将乙醉和水的伪摩尔体枳彳、¼去示为浓度X?的由数.解：由二元溶液的偏摩尔性般与瞟尔件质问的关系:=jW+(1-T.PXMI闵=-32.46-85.96.v,+176.31.d-93.8工；所以V=58.36-32.46a、-42.98.Y+58.77J23.45.-x,-32.46-85.96a+176.3It；-93.8i*S1.，=58.36+42.98.4-117.54+70.35KV!mo1.¼58.36-32.46m42.98E+58.77W-23.45.vt+1.x,)-32.4685.,176.31.x；-93.8Y一>II=25.9-85.96.v,+219.294-211.34+7().35imo1.4-2.某二元姐分液体混合物在胤定T&P下的始可用下式表示：H=4.v1+6(X).v,+.vix,(40.vi+20.r,).式中,H冷位为防叫试确定在该出以压力状态F(I)H1.n我东的耳和瓦I(2)纯炮分培H,和出的数值：3无限稀出卜液体的保家尔后H；和H*的数值.W>(I)已知=4(X).t1.+600E+X1X2(40i+20)（八）用m1.x,俗入（八）,并化简得：H=4(X).v,+6(1-i)+x,(-x1)40xi+20(1-切=600-180XI-20x：由二元溶液的恰摩尔性班与摩尔性Hi间的关系：TT,1.()-rr-(M1=M+(I-X1).M2=.W-,神卜”+(If偎1.瓦=Hf僵1.由式(B)(三1.-,8o6ox'所以77；=60()-180.r1-20i,+(I-X1)-180-6O.v1.2=420-60+4().7/n/(c)T=6(X)-1.80i-20xi,-.V1-18()-60a；=600+40.r1.,mo1.(o)将M=I及x,=0分别代入式(B>得纯斑分於”,和心H,=400Jmo1.H,=6Q0J)io1.(3)，：和是指在Xr=Ox,=1时的耳和用.将XK)代入式(C)中得，1.x=420W?/.将4=1代入式(D)中行：=640，/”".4-1成验室笏要用制120Oan)防拣落液.它由30%的甲Bf和70%fiH2O(2)摩尔比)祖成.试求东安多少体枳的25P的甲醉与水泥令，己知甲静和水在25匕、30%(廉尔分数)的甲的溶液的米摩尔体积:V1=38.632CFMSmo，V2=17.765CHJmO1.251C下炖物质的体积:V1=4O.727cm'/mo1.Vi=1.8.067n,w解：由M=Z(X,福)罔：V=x1.½+x2代人敢他将：V=0.3×3X.6524<).7i7.765=24.03<n/，00配M1»徐溶液由物抵的“hn=49.95"1”24.03所需甲醇、水的物境的M分别为:F=0.3x49.95=14.985，。/n2=0.7X49.95=34.965，”“/则所焉甲静.水的体积为IV1,=14.985x40.727=610.29zw/V21=34.965X18.068=63USmoi将四种组分的体校简沽加和：%+%=610.29+631.75=1242.04，。/124204-12(X)则混合石生成的溶液体粒要缩小：:=3.503%I2(M)4-4.有人提出用下列方程组表示恒温、忸压下商沽二元体察的储摩尔体积：Vi-Vt=a+(b-a)xi-bx,V2-V2a+(b-a)x2-b.x式中,V1.和V：是纯凯分的庠尔体枳,*b只是T,P的沿数.试从热力学角僮分析这些方程是否合理？W:依据Gibbs-Duhcm方程Z(IMH),“=°得恒温、忸压下xydVx+xzdVz=OBH(IV.J(IV,芭-/=_与7=4丁axiaxiax2由5S冶方程都xi-=(b-a)xx-2bx必（八）x2=(b-)x2-2bx(B)比较上述结5K.式（八）式(B).即所洽出的方程组在融状况下不SiitGibWDuhem方程.故不合理.45试计算甲乙用“)和甲米(2)的等分子混合物在323K和23X101%下的j,4和F46试推外听从M"d">Mt"s方程的气体的推S表达式,4-9.344.75K时.由氢和闪烧沮成的：元气体混令物,此中内烧的摩尔分数为0.792.混合物的儿，！3.7974MI.试川RK万科和相应的混介观则计口混合物中乳的逸度系呢己知氯-内烷系的"0.07.设的试验值为1.439.解：已知混育气体的T=M4.7SK1.>=5.TO74MI.ftR1.:得两组分的他界参数乳(1>：y=O.2O8T,=33.2KP1.=1.297MPi丙烷(23y1.=0.792T<=369.8KPt=4.246MPuVc=65.Ocm'1.mn1.(»-().22Vv=2O3cm'/mo1.o>=O.I52。42748管=。卅8嚼需30.42748篁1.42748骑篝二&3。哂，”“尸W(IJax2=(W2)*IS(I-,2)=(0.1447×18.30)QS(1-0.07)=1.513PM-KasWW尸«=RI+2y,y：a1:+j=0.208*X0.1447+2O.2O8X0.792×I.5I3+0.792?18.30=11.98RJ-mt,Kas-尸h.=0.08664口=0.086648314x33f=1.844×105m,mo,P1.1.1.297X1.O6A=0.08664必=0.08664&可.少岁=6274X1。小球P24.246X10f,bn=Zyibi=0.208×1.844×IO5+0.792×6.274×10,=5.3526XIormW=aa11.98bnRT't5.3526×1(5×8.314×344.75,5=4.206B_bP_5.3526XI(T'X3.7974X10"_0.07091ZZRTZX8.314X344.75-IT联1.2两式.迭代求娇竹：Z0.7575hj(6s所以.混合气体的摩尔体积为：V=组=皿5x2./"567X炉加疝P3.7974XIO6EdMiM急H¾2号卜爵号)H祟啮刑自卜(自卜n(竿上给PuN加卜(第分别代入数据计算科：M0.某二元液体混合物在固定T和P下其超K!均可用下列方程於衣示：M=11xX40"+20E.其中足的旗位为Jn>1.试求后：和/7；(用*表示).4-I2.473K.5MPaF两气体混合物的逸度系数可表示为IIn¢=.四外。+工),式中“和六M1.1.分1和沮分2的摩尔分率.试求/.八的花送式.并求出当F1.FV)5时.£、£各为多少？M3.在侦定.P下,测得某.元体系的活度来数值可用1列方耳.表示"n%=ax1÷.(3.v1-x,)（八）GK试求出而In/,=ax：+/7X(.r1.-3.v,)(b>的表达犬I并同（八）,(b)方程应是否满JftGibbS-DUhem方程？若用(ch<d>方程式&示该:无体察的活度数值时.则是否也淌有Gibte-Duhcm力打?Iny1.=X式a+bxjn2=x1.(a+bxi)(d)4I7测将乙胞”)-6ft?<2>体系在50乙到WOiC的其次幅里系数可近似地用下式底示:1.1.=-8.551.×10,jSSZ.产/.7”4=-21.5(y×10j2=-1.74×103J式中,T的机位是K,B的冷位是cm'mo1.,试计比乙拓和乙(K两组分的等分子蒸气泡介物在0.8*1.0'Pa和80'C时的，与，；.M1.某二元混合物在肖定t、P下熠可用下式表示：H=X1（«,-bixi）+.V,（,+b2x2）,其中a、b为常St.试求凯分】的恒烁尔物H1的表示式-1.vH网町=1-2b1.x1.+Z