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1、本科生期末考试试卷统一格式（16 开）：20 20 学年第学期期末考试试卷化工热力学（A 或 B 卷共页）（考试时间：20 年月日）学院专业班年级学号姓名题号一二三四五六七八九十成绩得分一、简答题（共 8 题，共 40 分，每题 5 分）1.写出封闭系统和稳定流动系统的热力学第一定律。答：封闭系统的热力学第一定律：WQU稳流系统的热力学第一定律：sWQZguH2212.写出维里方程中维里系数B、C 的物理意义，并写出舍项维里方程的混合规则。答：第二维里系数 B 代表两分子间的相互作用，第三维里系数 C 代表三分子间相互作用，B 和 C 的数值都仅仅与温度T 有关；舍项维里方程的混合规则为：ni
2、njijjiMByyB11，10ijijijcijcijijBBpRTB，6.10422.0083.0prijTB，2.41172.0139.0prijTB，cijprTTT，5.01cjciijcijTTkT，cijcijcijcijVRTZp，331315.0CjcicijVVV，cjcicijZZZ5.0，jiij5.03.写出混合物中 i 组元逸度和逸度系数的定义式。答：逸度定义：iiifRTdypTd?ln,（T 恒定）1?l i m0iippyf逸度系数的定义：iiipyf?4.请写出剩余性质及超额性质的定义及定义式。答：剩余性质：是指同温同压下的理想气体与真实流体的摩尔广度性质之
3、差，即：pTMpTMMid,；超额性质：是指真实混合物与同温同压和相同组成的理想混合物的摩尔广度性质之差，即：idmmMMEM5.为什么 K 值法可以用于烃类混合物的汽液平衡计算？答：烃类混合物可以近似看作是理想混合物，于是在汽液平衡基本表达式中的1i，ivi?，在压力不高的情况下，Ponding因子近似为1，于是，汽液平衡表达式化简为：visisiiiidippxyK。由该式可以看出，K 值仅仅与温度和压力有关，而与组成无关，因此，可以永K值法计算烃类系统的汽液平衡。6.汽相和液相均用逸度系数计算的困难是什么？答：根据逸度系数的计算方程，需要选择一个同时适用于汽相和液相的状态方程，且计算精度
4、相当。这种方程的形式复杂，参数较多，计算比较困难。文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3
5、HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3
6、 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N
7、3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6
8、N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B
9、6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4
10、B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3二、推导题（共 2 题
11、，共 15 分，其中第一题 8 分，第二题 7 分）1.请推导出汽液相平衡的关系式。（其中：液相用活度系数表示，以Lewis-Randell 规则为基准；汽相用逸度系数表示。）答：根据相平衡准则，有liviff?；其中，等式左边项可以根据逸度系数的定义式变形为：viivipyf?；等式的右边项可以根据活度系数的定义式变形为：iiilixff?，而标准态取为同温同压下的纯液体，于是有RTppVpfsilisisiiexp，带入相平衡准则，得到：RTppVxppysiliiisisiviiexp?2.从汽液相平衡的关系式出发，进行适当的假设和简化，推导出拉乌尔定律。答：1)压力远离临界区和近临界区
12、时，指数项1expRTppVsili。2)若体系中各组元是同分异构体、顺反异构体、光学异构体或碳数相近的同系物，那么，汽液两相均可视为理想化合物，根据Lewis-Randall 规则，有ivi?；同时，1i。3)低压下，汽相可视为理想气体，于是有：1?vi，1si。综上所述，汽液平衡体系若满足1），2），3），则：isiixppy，即为拉乌尔定律。三、计算题（共 4 题，共 45 分，其中第一题 15 分，第二题 15分，第三题 5 分，第四题 10分）1.求某气体在 473 K，30 105 Pa时，Hm=?文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I
13、3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8
14、I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F
15、8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1
16、F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q
17、1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10
18、Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X1
19、0Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3已知：pVm=RT+10-5p，其中：p 单位 Pa，Vm单位 m3 mol-1，Cpid=7.0+1.0 10-3T(J mol-1 K-1)。设：该饱和
20、液体在 273 K 时，Hm=0(J mol-1)。其中安托尼常数 A=20.7，B=2219.2，C=-31.13。（安托尼方程中压力p：Pa，T：K，压力用自然对数表示）（设z=1）答：首先涉及路径，273 K 饱和蒸汽压下的气体可近似视为理想气体。13.31/2.22197.20/lnKtPapsi2732227313.312.22191lnTTsiTTTdpd-2824 48.231lnTdpdZRHsivkJmol-16.150101.07.04732733-473273dTTdTCHigpigkJ mol-1Vm=RT/p+10-5510TVTV301052103005dpHRJ
21、mol-111.17402RigvmmHHHHHkJ mol-12.有人提出用下列方程组来表示恒温恒压下简单二元体系的偏摩尔体积：2222221111)()(bxxabaVVbxxabaVV273K,pis,l473K,3MPa,g273K,pis,ig473K,3MPa,i gHvRH2igHH文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B
22、3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10
23、B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT1
24、0B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT
25、10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 H
26、T10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3
27、HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3
28、 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3其中：V1和 V2是纯组分的摩尔体积，a、b 只是 T、p 的函数，试从热力学角度分析这些方程是否合理？答：由于该方程涉及到偏摩尔性质和温度压力等参数，因此如果该方程合理，必须要满足 Gibbs-Duhem 方程。首先，衡量等温等压下的Gibbs-Duhem是否满足：即：02211VdxVdx。对二元体系，做衡等变形，得：0222111dxVdxdxVdx由已知得：111
29、2bxabdxVd，2222bxabdxVd。于是，有：021222111xxbadxVdxdxVdx，因此，该表达式不合理。文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B
30、3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10
31、B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT1
32、0B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT
33、10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 H
34、T10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3
35、HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3文档编码:CR1J3Z4B6N3 HT10B3P10K2Z4 ZW1X10Q1F8I3