学而不思则惘，思而不学则殆普化无机试卷（气体一）答案一、选择题1. (0101) (C)2. (0102) (B)3. (0103) (D)4. (0104) (B)5. (0105) (B)6. (0106) (C)7. (0108) (A)8. (0123) (A)9. (0124) (B)10. (0125) (C)11. (0126) (B)12. (0127) (C)13. (0130) (C)14. (0131) (B)15. (0132) (B)16. (0133) (C)17. (0134) (B)18. (0135) (B)19. (0136) (C)20. (0144) (C)21. (0145) (D)22. (0146) (C)23. (0147) (D)24. (0148) (D)25. (0150) (B)26. (0151) (D)27. (0153) (C)28. (0154) (A)29. (0155) (C)30. (0156) (D)31. (0157) (B)32. (0160) (B)33. (0164) (D)34. (0165) (C)35. (0167) (B)36. (0168) (C)37. (0169) (A)38. (0170) (D)39. (0171) (C)40. (0172) (A)学而不思则惘，思而不学则殆41. (0173) (D)42. (0174) (B)43. (0175) (C)44. (0176) (B)45. (0177) (C)46. (0178) (D)47. (0179) (C)48. (0180) (C)49. (0181) (A)50. (0182) (B)51. (0183) (B)52. (0184) (A)53. (0185) (B)54. (0186) (B)55. (0187) (C)56. (0188) (C)57. (0189) (D)58. (0190) (A)59. (0191) (A)60. (0192) (B)61. (0193) (D)62. (0194) (A)63. (0195) (A)64. (0196) (C)65. (0197) (A)66. (0198) (B)67. (0199) (B)二、填空题 ( 共7题 13分 )68. (0107) 低压高温69. (0128) 分子间的作用力 , 分子的体积70. (0137) 3NH p : 3.9 MPa 2N p : 1.5 MPa 2H p : 4.6 MPa71. (0149) 0.5872. (0161) 46773. (0162) (1) H 2＞ He ＞ Ne ＞ CO 2; (2) 分子平均能量都相等 。

74. (0163) 1.004三、计算题 ( 共3题 155分 )75. (0109)pV 96.9×150×10-32X n = ── = ──────── = 0.00250 (mol) RT 8.31×(427+273)2MX 2(g) = 2MX(g) + X 2(g)1.120 g 0.720 g 0.400 gX 2的摩尔质量为: 0.400/0.00250 = 160 (g ·mol -1) X 的相对原子质量为 80n MX = 2×0.00250 = 5.00×10-3 (mol)学而不思则惘，思而不学则殆MX 的摩尔质量为: 0.720/(5.00×10-3)=144 (g ·mol -1)M 的相对原子质量为 6476. (0110)n 1.0×109p =── RT =────── ×8.31×1200=1.7×10-11 (kPa)V 6.0×102377. (0111)pV =nRT 当 p ,V 一定时, n 随 T 变,pV 1 1Δn =───×（──── ─ ──── ）R T (冬) T (夏)1.00×106 1 1=103×─────（ ─── ─ ─── ）= 10.5×103(mol)8.31 248 3144CH m =10.5×16.0=168 (kg)78. (0112)2H 2O ＋CaC 2= Ca(OH)2＋C 2H 2↑设反应后得到了 n mol 的 C 2H 2,即含 2n mol H 2O ,21.0×10-3所以 n = 1.00×102×────── = 8.62×10-4 (mol)8.31×2932×8.62×10-4×18.0H 2O 的质量分数=────────────×100% = 2.06 %1.50879. (0113)T 1 p 2 273 207d 2= d 1·──·── = 1.43 ──·── = 2.76 (g ·dm -3 )T 2 p 1 290 10180. (0114)p 1V 1 p 2V 2(1) 膨胀前为状态１, 膨胀后为状态２, ───── = ─────T 1 T 2202×V 101×4V───── = ────── T 2= 600 (K)300 T 2(2) 22H H x p p 总= Ne H H H 222n n n x +=25.0 75.02H n = ─── = 12.5(mol) n Ne = ─── = 3.71(mol)学而不思则惘，思而不学则殆2.020.212.5 2H x =────── = 0.77212.5＋3.712H p = 202×0.772 = 156 (kPa)81. (0115)(1) p 1V 1= p 2V 2101×1.0101×1.0 = 50.5×V 2 V 2= ───── = 2.0 (dm 3)50.5V 1 V 2 1.0 V 373(2) ── = ── ── = ─── V 2 = ───×1.0 = 1.1 (dm 3)T 1 T 2 330 373 33082. (0116)(1) p 空V =p’空V’ (不考虑水)(101.3-17.3)×1.0=(202.6-17.3)×V’ V’= 0.45 (dm 3)pV p 'V’(2) ─── = ─── (不考虑水)T T '(101.3-17.3)×1.0 (101.3-1.2)×V’───────── = ───────── ∴ V’= 0.72(dm 3)330 28383. (0117)设球的体积为 V , 当两球都在沸水中时 101(2V ) =nR ×373 ---------------------- (1)若一球在沸水中 , 另一球在冰水中 , 两球平衡时压力为 p , 在沸水中有n 1mol N 2, 而在冰水中有 n 2mol N 2则: n 1＋ n 2= n n 1373 = n 2273p (2V ) = n 1R ×373 ＋ n 2R ×273= n 1R ×373 ＋ n 1R ×373 = 2n 1R ×373 ------------------------------------- (2)(1) 101 n 1 n 2 1 373── = ── = ── = ── (1＋── )= ──(1＋ ── ) = 1.18(2) p 2n 1 2 n 1 2 273p = 85.6 kPa84. (0118)(1) 冷却前为状态１，冷却后为状态２，冷却前后空气质量不变。

p 1V 1 p 2V 2──── = ────T 1 T 2(101.3-29.3)×10.0 (101.3－6.66) V 2─────────── = ──────────学而不思则惘，思而不学则殆373 293293 72.0冷却后气体体积 V 2= 10.0×───×─── = 5.98 (m 3)373 94.6pV 29.3×10.0×103(2) 冷却前乙醇含量 n 1= ─── = ──────── = 94.5 (mol)RT 8.31×3736.66×5.98×103冷却后乙醇含量 n 2= ──────── = 16.4 (mol)8.31×293乙醇凝聚量 n = 94.5－16.4 = 78.1 (mol)85. (0119)01194.62初态时: n (Cl 2)= ───── = 0.0651 (mol)35.5×24.16n (SO 2)= ───── = 0.0650 (mol)64.0终态时, 设有 y mol SO 2起反应则: SO 2 ＋ Cl 2 = SO 2Cl 2初态 n/mol 0.0650 0.0651 0终态 n/mol 0.0650-y 0.0651-y y在终态 气体总量 n = (0.130－y )mol用 pV =nRT 式求 y 202×2.00 = (0.130－y )×8.31×463202×2.00y = 0.130 － ───── = 0.02508.31×4630.0650－0.02502SO x = ──────── = 0.3810.130－0.02502SO p = 202×0.381 = 77.0 (kPa)0.0651－0.02502Cl x = ──────── = 0.3810.130－0.02502Cl p = 77.0 kPa0.0250学而不思则惘，思而不学则殆22Cl SO x = ──────── = 0.2380.130－0.025022Cl SO p = 202×0.238 = 48.1 (kPa)86. (0120)设混合前氢压力为 p22H N n n n +=总 n = pV/RT400×V 总 500×2N V 2H p 2H V────── = ───── ＋ ─────R ×400 R ×500 R ×300V 总 2N V 2H p 2H V─── = ─── ＋ ─────1 1 300按定义 V 总 = 2N V +2H V2H p 2H VV 总－2N V =2H V = ─────300∴ 2H p = 300 kPa87. (0121)燃烧反应方程式: 2C 2H 6(g)+ 7O 2(g)= 6H 2O + 4CO 2(g)pV 50.02O n = ─── = 99.2×────── = 2.00 (mol)RT 8.31×29862H C n = 6.00/30 = 0.200 mol7C 2H 6完全燃烧时需 O 2为: ─×0.200 = 0.700 (mol)2剩余 O 2为: 2.00 - 0.70 = 1.30 (mol)生成的水: O H 2n = 0.600 mol生成的 CO 2: 2CO n = 0.400 mol学而不思则惘，思而不学则殆(1) T = 573 K, H 2O 为气体,∑n i RTp = 2O p +O H 2p + 2CO p = ──────V(0.600+0.400+1.30)×8.31×573= ──────────────── = 219 (kPa)50.0(2) T = 363 K 时, 如 H 2O 为气体,O H 2n RTO H 2p = ────── = 36.2(kPa) 比90℃ 时饱和水蒸气压要小H 2O 全部气化。