化工热力学(第三版)陈钟秀课后习题答案.docx

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1、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.knx1.=J24.6cm,11w1.ff1.和甲烷的临界参物T=I9O.6KPt=4.0MPaV.=99cmVnw1.=QXJ8(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=2.98510smj11w/.P=旦V-hT3V(V+b)

2、8.314x323.153.222(12.46-2.985)105323.150512.46105(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=1.259.7CmVnN-?：RI倒正丁烷的推界替数：T=425.2KP.=3.80()MPaV,=99cmmn1o-O.I9341管(片+iBPPV:.Z1=I+-s-V=O.O2486mmo1.RTRT.,.V=nV=100xI1.8138%-120.02486=168.58m,(2)P=y1P-=O.24O.IOI3-=O.O25Pa,Z

3、.,0.2845P、=Y、P虫=0.760.1013-=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t3)PEg-RObinsOn方程I4)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=1.1

4、76.=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=2.83mV1.885IOmVm1.mo1.对于状名Ih除尔体税V=O.14211V1.501.mo1.=9.4581.05mVmo1.T=448.6K(2) Wmderwaak力利=()A253Pam,mo227W27x8.314?X4056MP.6411.28IO6bRT8.314405.6=玄=81.1.28=3.73710-smjmo/-1r,KTa8.314448.60.42

5、53,一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,tto2P11.28106b=0.086649=0.086648314x406=2.5910Wmo/1P,11.28XIO68.679=I834PS.3I44486V-ftTwV(Vb)(9.458-2.59)105448.6,9.458IO5(9.458+2.59)IOT(I)PCngRobinson方程VTr=TTc=448.6/405.6=1.106.k=0.3746+.

6、5422(uo-0.26992,J=0.3746+1.542260.25-0.26992O.252=0.7433()=1.+fc(1.-7,)y=1.+0.7433X(I-1.106S)T=0.9247a(T)=a,a(T)0.45724-(7)=0.45724胃皆:需X0.9247=0.4262mft-rm,Ib=0.07780以=0.07780X心火4。芋=2.32610sm-moiP.11.28Kf.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

7、=19.00Ma(5)普遍化关系大V匕=VV=9.45810s7.25105=1.30526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=I4c11ieIB*=265cmY111.B2=-9.5cmVmo1.棒Bm=A+2y,M1.j+=0.3214+20.30.7(-9.5)+O.72(-265)=-132.5&毋Inv1BI1PVZ=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0.3nx280.7nx58-7n-0.1429mo1.V=廿丫(摩尔体枳

8、E.I4294.24xOyO.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P13.394IO6b=0.08664旦=0.0866483七*=2678IOSm.P13.394XIOh4aPnbPAa1.5577A=-;o=-=;rRT2sRTBhRTs2.678XIO5X8.314X27315,BhbP2.678105101.310ft1.1952/?=ZVZ1H-2109.2J11u1.走向i体等海膨朕过程dr=。、RKjA=(-TS=-3665.763=-2IO9.26M114K)AG=AHTNS-AA-2iw.26j!KiTdS=TAS=A=-2I