物理化学(三、四)1

班级                 姓名                               分数                

一、选择题 ( 共10题  18分 )

1. 298 K时,HCl(g,Mr=36.5)溶解在甲苯中的亨利常数为245 kPakgmol-1,当HCl(g)在甲苯溶液中的浓度达2%时,HCl(g)的平衡压力为：                                       (    )     

   (A)  138 kPa     (B)  11.99 kPa     (C)  4.9 kPa     (D)  49 kPa                 

2. 已知 I2(g)的基本振动频率=21 420 m-1, kB=1.38×10-23 JK-1, h=6.627×10-34 Js, c=3×108 ms-1, 则 I2(g) 的振动特征温度v为：                                        (    )     

      (A)  2.13×10-14 K                                                          

      (B)  1.03×10-8 K                                                          

      (C)  308.5 K                                                             

      (D)  3.23×10-3 K                                                          

3. 在288 K时，H2O(l)的饱和蒸气压为1702 Pa,当0.6 mol的不挥发溶质B溶于0.540 kg H2O时,蒸气压下降42 Pa,溶液中H2O的活度系数γx应该为：                        (    )      

      (A)  0.9804    (B)  0.9753                                                

      (C)  1.005     (D)  0.9948                                                

4.近定域粒子体系和经典极限下的非定域粒子体系的                (    )     

      (A) 最概然分布公式不同                                                    

      (B) 最概然分布公式相同                                                    

      (C) 某一能量分布类型的微观状态数相同                                      

      (D) 以粒子配分函数表示的热力学函数的统计表达示相同                        

5.  1molHCl溶于20 dm3水中形成稀溶液。在该温度下，该溶液上方HCl的蒸气分压pHCl      [式中xHCl=1mol/(n+1mol)]（填 >; =; <）

6. 在平动、转动、振动运动对热力学函数的贡献中，下述关系式中哪一个是错误的?  

                                                                    (    )     

     (A)  Fr= Gr                                                            

     (B)  Uv= Hv                                                            

     (C)  CV,v= Cp,v                                                      

     (D)  Cp,t= CV,                                                      

7. 在 400 K，液体A的蒸气压为4×104 Nm-2，液体B的蒸气压为6×104 Nm-2，两者组成理想液体混合物。当气-液平衡时，在溶液中A的摩尔分数为0.6，则在气相中B的摩尔分数应为：                                                       (    )     

       (A)  0.31            (B)  0.40                                          

       (C)  0.50            (D)  0.60                                          

8. 根据逸度-组成图,下面表达不正确的是：                              (    )     

    (A)  Ba虚线符合fA=fA*xA                                                

    (B)  Ab'虚线符合fB=kBxB                                             

    (C)  ab虚线符合f=kBxB+fA*xA                                       

    (D)  Adb实线符合fB=fB*aB                                         

9. 298 K, p下,两瓶含萘的苯溶液,第一瓶为2 dm3(溶有0.5 mol萘),第二瓶为1 dm3(溶有0.25 mol萘),若以μ1和μ2分别表示两瓶中萘的化学势,则                               (    )    

     (A)  μ1= 10μ2       (B)  μ1= 2μ2                                     

     (C)  μ1= (1/2) μ2     (D)  μ1=μ2                                      

10.  某双原子分子 AB 取振动基态能量为零，在 T 时的振动配分函数为 1.02，则粒 子分布在 v = 0 的基态上的分布数N0/N应为：                                       (    )    

     (A)  1.02            (B)  0                                                

     (C)  1               (D)  1/1.02                                           

二、填空题 ( 共 9题  17分 )

11. 所谓正偏差溶液是指在不同分子间的引力弱于同类分子间的引力,故分子逸出液面的倾向增加,此时pA____pA*xA, ΔmixV____0 , ΔG____0。( 填 > , =, < )

12. 已知N2分子的振动特征温度=3340 K，在298.15 K时，N2的标准摩尔振动熵     (298.15 K)=                    。

13. 25 C 下,1g葡萄糖(C6H12O6)溶于1kg水中，此液的渗透压为                       。

已知Mr(C6H12O6)=180.16。

14. 当两能级差ε2-ε1= kT时,则两能级上最概然分布时分子数之比N2*/N1* 为:__________________；当两能级差为ε2-ε1= kT,且其简并度g1=1,g2=3,则N2*/N1*为_________________。                                                  

15. 313 K时,液体A的饱和蒸气压是液体B的饱和蒸气压的21倍,A,B两液体形成理想溶液。若气相中A和B的摩尔分数相等,则液相中A和B的摩尔分数比xA/xB为______。          

16. 已知N2分子的=2. K，则N2在25C，101 325 Pa压力下的标准摩尔振动吉布斯自由能=                   。（假定最低振动能级的能量为零。）

17. 实际溶液的非理想性程度可用超额函数表示，它与 ______ 有关。例如，某溶液的超额自由能 GE为 _________ ，当 GE为正值时，表示该溶液的 G 对理想情况产生了_________ 偏差。                                                                

18.热力学函数与分子配分函数的关系式对于定域粒子体系和离域粒子体系都相同的是               。

19. 理想气体分子间作用力_______, 理想溶液分子间作用力________。                 

三、计算题 ( 共 5题  50分 )

20.在598 K时,含铊的汞齐中汞的活度系数γ1在x2为1～0.2范围内服从公式

                     lnγ1=-0.096(1+0.263x1/x2)-2                                       

    试用 (1) 亨利定律为基准：当x1→0, γ2→1                                  

         (2) 拉乌尔定律为基准：x2→1时, γ2→1                                  

    求 x2=0.5时铊的活度系数。                                                     

21. 在铅和金刚石中，Pb原子和金刚石原子的基本振动频率分别为21012和41013s-1，试计算它们的振动特征温度和振动配分函数在300 K下的数值。（选取振动基态为能量零点。）

(k=1.3805, h=6.626)

22. 两个10 dm3容器用一活塞连接,抽空后一个注入0.1 kg H2O；另一个压入202 650 Pa的干燥CO2气,两者温度均为298.15 K。恒温下旋开活塞,待达到平衡后,问体系内的总压力为多少?已知CO2在298.15 K时在水中的溶解度为2 mol/(1 kg H2O)，纯水在298.15 K时的蒸气压为3199.7 Pa。                                                                      

23. I2分子的振动基态能量选为零，在激发态的振动波数为：213.30，425.39，636.27，845.93 和 1054.38 cm-1。试求：                                          

     (1) 用直接求和的方法计算298 K时的振动配分函数；                             

     (2) 在298 K时，基态和第一激发态I2分子占总分子数的比例是多少？           

     (3) 在298 K时，I2的平均振动能。                                            

24. 混合气体A,B,C的组成为：yA=0.4，yB=0.2。在一定温度下等温压缩此混合气体,问： 

    (1)  加压至多大时开始有液相析出?液相组成是多少?                             

    (2)  继续加压至多大压力,此混合气体刚好全部凝结为液体?                       

    已知在此温度下,各纯物质的饱和蒸气压分别为 pA*=53 329 Pa; pB*=26 6 Pa; pc*=13 332 Pa (设混合气体为理想气体,构成的溶液为理想溶液)。             

四、问答题 ( 共 3题  15分 )

25. 苯甲酸在水中和苯中分配时,在水中呈单分子,在苯中缔合为(COOH)n,单分子浓度很小。证明在定温定压平衡时,c水/c苯1/n =K。c水和c苯 分别为苯甲酸在水层和苯层中的总浓度(以单分子计算)。      

26. 若取双原子分子的转动惯量I=10，则其第三和第四转动能级的能量间隔等于多少？

27. 在恒温恒压下,从纯水中取出1 mol纯溶剂(l)（蒸气压为 p1*）加入大量的、溶剂摩尔分数为 x1的溶液中（溶剂蒸气分压为 p1）。                                    

      (1)  设蒸气为理想气体，溶剂遵守拉乌尔定律，计算该1 mol纯水(l)的ΔGm（以 x1表示）；                                                       

      (2)  设蒸气不是理想气体，但溶剂仍遵守拉乌尔定律，结果是否相同？           

      (3)  若蒸气是理想气体，但溶剂不严格遵守拉乌尔定律，ΔGm又如何表示？     

一、选择题 ( 共10题  18分 )

1.  [答]    (A)                                                                     

        在 1 kg 溶剂中：                                                        

          mB=nB/mA=0.5591 mol·kg-1                                      

          pB=kB,m× mB                                                  

            =(245 kPa·kg·mol-1)×(0.5591 mol·kg-1)                          

            =138 kPa                                                         

2.  [答]  (C)  v= hc/k = 308.5 K                                                                       

3.  [答]   (D)                                                                      

         x=n/(n+nB)

=(0.54 kg/0.018 kg·mol-1)/[(0.540 kg/0.018 kg·mol-1)+0.6]         

              = 0.9804             

4.  [答]  (B)                                                                       

5. [答]< 因为HCl在水中解离成H+和Cl- ，xHCl将下降。                                                        

6.  [答]  (D)                                                                       

        Fr= Gr= -NkTlnqr                                                

        UV = HV = NkT×[x/(ex-1)]                                         

        CV,V = Cp,V = Nk×[x2ex/(ex-1)2]     x = v/T                

        Cp,t= (5/2)Nk     CV,t= (3/2)Nk                                

    所以 Cp,t≠ CV,t                                                     

7.  [答]  (C) yB=pB/p总= pB*xB/(pA*xA+ pB*xB) = 0.50                  

8. [答]     (C)  ab线应符合 f=fB* xB+fA* xA                                                                  

9.D     10.D                                                             

二、填空题 ( 共 9题  17分 )

11. [答] >, >, >, >.

12.  [答]  (1分)

                      =                           (1分)

13.[答]                                                (2分) 

14. [答]      0.368；           1.104                                          

      N2*/N1*= exp[-(U2-U1)/ kT] = e-1= 0.368                         

      N2*/N1*= (g2/ g1) exp[-(U2-U1)/kT] = 1.104                  

15.  [答]    1/21                                                                    

        pA= pxA=21 pxA=pyA                                         

        pB= pxB= pxB=pyB                                           

        因为 yA=yB                                                            

        所以 xA/xB=1/21                                            (2分)     

16. [答]  

          =                              (1分)

          =-0.0324                                            (1分)

17.  [答]    温度,压力和组分的活度系数；                             (1分)     

      mixG(re)- mixG(ig)；                                     (0.5分)     

      正                                                       (0.5分)     

18.[答]  U, H, CV, Cp                                                     (2分)

19.  [答]     (1) 为零   (2) 相等                                                             

三、计算题 ( 共 5题  50分 )

20. [答]    根据Gibbs-Duhem公式：                                                   

            x1dlnγ1+x2dlnγ2=0                                           

            dlnγ2=-(x1/x2)dlnγ1                                         

                   =-(x1/x2)d[(-2.303×0.096)/(1+0.263x1/x2)2]         

                   =-(x1/x2)[(2×2.303×0.096)/(1+0.263x1/x2)3]        

                    ×0.263(x1/x2)                              (2分)     

        用积分公式：                                                             

         dx/(b+ax)3=(1/a2)[b/2(b+ax)2-1/(b+ax)]+k                    

        lnγ2=(-0.1163/0.2632)[1/2(1+0.263x1/x2)2                         

                -1/(1+0.263x1/x2)] +k ------------------- (1)            (2分)     

        (1) 以 Henry 定律为基准：x2→0, γ→1,代入(1)式得 k=0     (1分)     

           当x2= 0.5时,解得：                                                    

            γ2=2.23                                           (2分)     

        (2) 以Raoult's定律为基准：x2→1, γ2→1,代入(1)式得 k= -0.84(1分)     

            γ2=0.92                                         (2分)     

21. [答]  对铅   =96 K                                         (2分)

            =3.65                           (3分)

     对金刚石   =1920 K                                   (2分)

                =1                         (3分)

22. [答]    CO2的总量：                                                            

        n(CO2)=pV/RT                                                      

              = 202 650 Pa×10×10-3 m3/(8.314 J·K-1·mol-1)(298.15 K)           

              = 0.8175 mol                                       (2分)     

       设服从亨利定律,溶在水中的CO2的物质的量为n(CO2,l)                     

           101 325 Pa=k×0.2 mol                                                     

           p(CO2)=kn(CO2,l)                                                   

           n(CO2,l)=(0.2mol)p(CO2)/101325Pa                                    

                  =1.9738×10-6 p(CO2)    ---------------- (1)         (2分)     

          p(CO2)=n(CO2,g)RT/V(g)={n(CO2)-n(CO2,l)}RT/V(g)           

           ={0.8175 mol-n(CO2,l)}×(8.314 J·K-1·mol-1)×(298.15K)        

             /[10+(10-0.1)]×10-3 m3  ----------------- (2)               

     联立(1)、(2)式,解得：                                                      

          p(CO2)=81 734 Pa                                        (3分)     

          n(CO2,l)=0.1613 mol                                                   

        p=p*x                                                      

           =3199.7 Pa×5.55/(5.55+0.1613)=3103.7 Pa                 (2分)     

       p总= p+p(CO2)=84 838 Pa                                 (1分)     

23.[答]  (1) q v=∑qiexp( i/kT)                                             

          =1+exp(-hc1/kT)+exp(-hc2/kT)+exp(-hc3/kT)    

            +exp(-hc4/kT)+exp(-hc5/kT)                          

          =1+0.357+0.128+0.046+0.017+0.006=1.554                            (3分)      

      (2) n0/N=exp(- 0/kT)/qv=1/1.554=0.                                   (2分)      

          n1/N=exp(- 1/kT)/qv=0.357/1.554=0.23                                (2分)      

      (3) U=∑Pi i=(1/1.554)×0+(0.357/1.54)×hc1+···              

         =1.39 kJ·mol-1                                                      (3分)      

24.[答]    (1) 当加压至p时开始有液相析出,成气液平衡,设液相组成为                  

            xA,xB,xC,此时气相组成视为不变                                 

            yA=pA/p=pA*   xA/p=0.4                                        

            yB=pB/p=pB*  xB/p=0.2                                        

            yC=pC/p=pC*  xC/p=0.4         ------------- (1)                   

            xA+xB+xC=1                ------------- (2)                   

        联立(1),(2)式解得：                                                     

            p=22220 Pa                                         (3分)     

            xA=xB=0.1667                                       (3分)     

            xC=0.6666                                          (1分)     

        (2) 当气体全部凝结为液相时,液相组成为：                                   

            xA=0.4；xB=0.2；xC=0.4,                                         

            此时平衡蒸气的气相分压为：                                           

            pA= pA*   xA= 21 332 Pa                                   (1分)     

            pB= pB*  xB= 5333 Pa                                     (1分)     

            pC= pC*  xC= 5333 Pa                                     (1分)     

            p=pA+pB+pC= 31 998 Pa                                 (1分)     

四、问答题 ( 共 3题  15分 )

25.  [答]          (ΦCOOH)n nΦCOOH                          

        在苯中：c苯-m            m  (设m为单分子浓度)                       

        K1=mn/( c苯-m)≈mn/c苯     (因在苯中m很小)                     

        m= K1c苯1/ n                                            (3分)   

        ΦCOOH(在苯中)    ΦCOOH(在水中)                              

        K2=c水/m = c水/(c苯K1)1/ n                               

        令 K=(K1)1/ nK2                                              

        则 K=c水/c苯1/n                                          (2分)   

26. [答]  ,  J=0, 1, 2,                                   (2分)

     第3能级J=2, 第4能级J=3

                                       (1分)

          = J

          =J                                           (2分)

27. [答]      (1)  ΔGm=μ1(sln) -μ1*=RTlnx1                           (2分)     

         (2)  因为溶剂遵守拉乌尔定律，所以结果与 (1) 相同     (1分)     

         (3)  ΔGm=μ1(sln) -μ1*= RTlna1= RTln(p1/p*)                (2分)