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=-2I