Tiêu đề 32. Kinetic Theory of Gases Easy Ans.pdf ✅

1. A 2. D 3. C 4. B 5. C 6. B 7. C 8. A 9. A 10. C 11. C 12. A 13. D 14. D 15. A 16. B 17. B 18. C 19. C 20. D 21. D 22. B 23. A 24. A 25. A 26. C 27. C 28. C 29. B 30. B 31. A 32. B 33. C 34. C 35. B 36. B 37. B 38. A 39. A 40. B 41. A 42. C 43. D 44. C 45. D 46. B 47. B 48. B 49. A 50. B 51. A 52. b 53. C 54. A 55. A 56. D 57. C 58. D 59. C 60. A 61. A 62. C
63. A 64. D 65. C 66. A 67. B 68. B 69. A 70. C 71. C 72. B 73. D 74. A 75. D 76. A 77. A 78. C 79. B 80. A 81. B 82. A 83. C 84. A 85. A 86. D 87. B 88. A 89. D 90. A 91. C 92. C 93. C 94. A 95. A 96. C 97. B 98. A 99. C 100. A 101. B 102. C 103. D 104. B 105. A 106. C 107. A 108. B 109. D 110. A 111. B 112. A 113. D 114. B 115. B 116. B 117. C 118. B 119. B 120. D 121. D 122. D 123. C
124.(d) As P v (8.99 10 ) (3180 ) 3.03 10 N/m 3.0 atm 3 1 3 1 2 2 2 5 2 = rms =   =  = −  125.(c) Due to increase in temperature root mean square velocity of gas molecules increases. So they strike the wall more often with higher velocity. Hence the pressure exerted by a gas on the walls of the container increases. 126.(d) Kinetic energy E = J 5 1.5  10 , volume V = 20 litre = 3 3 20 10 m −  Pressure V E 3 2 = 6 2 3 5 5 10 / 20 10 1.5 10 3 2 =  N m           = − . 127.(d) Mean square velocity of molecule m 3kT = For gas A, x component of mean square velocity of molecule 2 = w  Mean square velocity m kT w 3 3 2 = = .....(i) For B gas mean square velocity m kT v 2 2 3 = = .....(ii) From (i) and (ii) 1 3 2 2 2 = v w so 3 2 2 2 = v w . 128.(a) 3 3 V 10 m − = , 22 N = 3.0 10 , m kg 26 5.3 10 − =  , v m s rms = 400 / 2 4 2 3 26 22 2 (400 ) 8.48 10 / 10 5.3 10 3.0 10 3 1 3 1 v N m V mN P rms =     = =  − − . 129.(c) 2 3 1 rms v V M P =  V MT P  At constant volume and temperature, if the mass of the gas is doubled then pressure will become twice. 130.(c) From ideal gas equation PV = RT we get 6 2 3 1 1 1 1 1 2 1 2 1 2 =                 =                 = V V P P V V P P T T  T2 = 6T1 = 6  300 = 1800 K = 1527 C. 131.(c) From PV = RT we get 5 9 0.5 1 300 270 2 1 1 2 1 2  =            =                 = P P T T V V  3 2 900 5 9 V = 500  = m 132.(c) From PV = RT we get 4 1 2 / 2 1 1 1 1 2 1 1 2 1 2 =                 =                 = V V T T V V T T P P  4 1 2 P P = 133.(b) As 32 gm O2 means 1 mole therefore 8 gm O2 means 1 / 4mole i.e. 4 1  = So from PV = RT we get PV RT 4 1 = or 4 RT PV = 134.(d) PV  Mass of gas  Temperature In this problem pressure and volume remains constant so M1T1 = M2T2 = constant  13 12 325 300 (52 273) (27 273) 2 1 1 2 = = + + = = T T M M  M M gm 12 gm 13 12 13 13 12 2 = 1  =  = i.e. the mass of gas released from the flask = 13 gm – 12 gm = 1 gm. 135.(d) M1 = M , T1 = 60 + 273 = 333 K , 4 3 4 2 M M M = M − = [As 1 / 4th part of air escapes] If pressure and volume of gas remains constant then MT = constant  2 1 1 2 M M T T = 3 4 3 / 4  =      = M M  T =  T =  333 = 444 K = 171 C 3 4 3 4 2 1 136.(a) 250 18 10 4.5 Molecular wt. of water Mass of water 3 =  = = − kg kg  , T = 273 K and 5 2 P = 10 N/m (STP) From PV = RT  3 5 5.66 10 250 8.3 273 m P RT V =   = =  . 137.(d) Ideal gas equation RT N N PV RT A         =  = where N = Number of molecule, NA = Avogadro number
                         = 1 2 2 1 2 1 2 1 T T V V P P N N 1 2 4 2 /4  =                    = T T V V P P . 138.(d) From PV  MT or T P M V  ; Here       P M represents the slope of curve drawn on volume and temperature axis. For first condition slope       P M graph is D (given in the problem) For second condition slope       = P M P M 4 /2 2 i.e. slope becomes four time so graph A is correct in this condition. 139.(a) Specific gas constant 4.15 2 8.3 Molecular weight of gas ( ) Universal gas constant ( ) = = = M R r Joule/mole- K. 140.(b, c) According to problem mass of gases are equal so number of moles will not be equal i.e.  A   B From ideal gas equation PV = RT  B B B A PAVA P V   = [As temperature of the container are equal] From this relation it is clear that if PA = PB then =  1 B A B A V V   i.e. VA  VB Similarly if VA = VB then =  1 B A B A P P   i.e. PA  PB . 141.(d) Quantity of gas in these bulbs is constant i.e. Initial No. of moles in both bulb = final number of moles ' 2 ' 1 +  2 = 1 +  ( ) 1.5 (273) 1.5 (273) (273) R T PV R PV R PV R PV + = +  T 1.5 273 1.5 273 2 = +  T = 819 K = 546 C . 142.(d) Number of moles in first vessel 1 1 1 RT P V  = and number of moles in second vessel 2 2 2 RT P V  = If both vessels are joined together then quantity of gas remains same i.e  = 1 +  2 2 2 1 1 (2 ) RT P V RT P V RT P V = + 2 2 1 1 2 2T P T P T P = + 143.(b) 3 3 V1 2.4 10 m − =  , 2 5 1 0 10 m N P = P = and T1 = 300 K (given) If area of cross-section of piston is A and it moves through distance x then increment in volume of the gas = Ax and if force constant of a spring is k then force F = kx and pressure = A kx A F = 3 3 3 2 1 2.4 10 8 10 0.1 3.2 10 − − − V = V + Ax =  +   =  and 5 3 5 2 0 2 10 8 10 8000 0.1 10 =    = + = + − A kx P P From ideal gas equation 2 2 2 1 1 1 T P V T P V =  2 5 3 5 3 2 10 3.2 10 300 10 2.4 10 T − −    =    T2 = 800 K 144.(b, c) Initially for container A P0V0 = n0 RT0 For container B P0V0 = n0 RT0  0 0 0 0 RT P V n = Total number of moles = n0 + n0 = 2n0 Since even on heating the total number of moles is conserved Hence n1 + n2 = 2n0 ......(i) If P be the common pressure then For container A PV0 = n1R 2T0  0 0 1 2RT PV n = P1 T1 V Initially P2 T2 V P T V P T V Finally