JEE Physics Kinetic Theory MCQs Set 03

Answer: (a) remains the same because the motion of the vessel as a whole does not affect the relative motion of the gas and the walls

Explanation:
1. Pressure is determined by the motion of the molecules relative to the walls of the vessel.
2. The uniform motion of the whole vessel adds the same velocity to the gas and to the walls, so nothing changes for the molecules hitting the walls. The pressure stays the same, irrespective of the magnitude of the speed.

Teacher's Note:
1. T
2. h
3. e
4.
5. s
6. t
7. a
8. t
9. e
10.
11. o
12. f
13.
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15.
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17. a
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19.
20. d
21. e
22. p
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25. d
26. s
27.
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31. m
32. o
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34. i
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37.
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43. i
44. v
45. e
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48. o
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53.
54. c
55. o
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58. a
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63. ,
64.
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68.
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71.
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75.
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79. i
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81. n
82.
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85.
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89.
90. c
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95. i
96. n
97. e
98. r
99.
100. i
101. t
102. s
103. e
104. l
105. f
106. .

Answer: (b) \( C_v\gt \frac52R \) for a diatomic gas

Explanation:
1. A monatomic gas has no rotational or vibrational energy, so \( C_v=\frac32R \) at all temperatures: the first, third... options for monatomic gases are wrong.
2. For a diatomic gas the vibrational mode adds two quadratic terms (kinetic and potential), so the heat capacity increases beyond the value \( \frac52R \) of translation plus rotation: \( C_v\gt \frac52R \).

Teacher's Note:
1. E
2. a
3. c
4. h
5.
6. e
7. x
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9. r
10. a
11.
12. d
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49.
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72.
73. c
74. a
75. p
76. a
77. c
78. i
79. t
80. y
81. .

Answer: (c) average translational kinetic energy

Explanation:
1. The average translational kinetic energy per molecule is \( \frac32k_BT \), independent of the gas.
2. Mass differs between gases, and so do the speeds \( \sqrt{3k_BT/m} \) and the momenta.

Teacher's Note:
1. T
2. e
3. m
4. p
5. e
6. r
7. a
8. t
9. u
10. r
11. e
12.
13. i
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15.
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17.
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28.
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35. e
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42. r
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62.
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64. n
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67. g
68. y
69.
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72.
73. t
74. h
75. e
76.
77. m
78. o
79. l
80. e
81. c
82. u
83. l
84. e
85. s
86. ,
87.
88. w
89. h
90. a
91. t
92. e
93. v
94. e
95. r
96.
97. t
98. h
99. e
100.
101. g
102. a
103. s
104. .

Answer: (d) low pressure and high temperature

Explanation:
1. Ideal behaviour requires that the molecules are far apart (so that their volume and mutual forces are negligible).
2. Low pressure gives a large separation, and high temperature gives a large kinetic energy compared with the intermolecular potential energy.

Teacher's Note:
1. R
2. e
3. a
4. l
5.
6. g
7. a
8. s
9. e
10. s
11.
12. a
13. p
14. p
15. r
16. o
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20.
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26.
27. b
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60. d
61.
62. h
63. o
64. t
65. .

Answer: (a) temperature

Explanation:
1. An ideal gas has no intermolecular forces, so its internal energy is the kinetic energy of its molecules.
2. This depends only on the temperature: \( U=\frac f2nRT \).

Teacher's Note:
1. T
2. h
3. i
4. s
5.
6. i
7. s
8.
9. w
10. h
11. y
12.
13. \
14. (
15.
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17. D
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21. a
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24. =
25. 0
26.
27. \
28. )
29.
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31. n
32.
33. a
34. n
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36.
37. i
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44. m
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55.
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58.
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60. n
61.
62. i
63. d
64. e
65. a
66. l
67.
68. g
69. a
70. s
71. .

Answer: (b) \( 5\times10^4\ \text{J} \)

Explanation:
1. Volume \( V=\frac{1}{4}=0.25 \) m³.
2. Thermal energy of a diatomic gas \( =\frac52nRT=\frac52PV=\frac52\times8\times10^4\times0.25=5\times10^4 \) J.

Teacher's Note:
1. F
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37.
38. \
39. (
40.
41. \
42. f
43. r
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45. c
46.
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48. 2
49. P
50. V
51.
52. \
53. )
54. ,
55.
56. w
57. i
58. t
59. h
60.
61. f
62.
63. =
64.
65. 5
66.
67. f
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72.
73. d
74. i
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77. o
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102. e
103. r
104. a
105. t
106. u
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110. .

Answer: (c) \( 4\times10^5\ \text{Pa} \)

Explanation:
1. Let \( V_A=V \), \( V_B=4V \). Moles are conserved: \( \frac{pV}{300}+\frac{p\,4V}{400}=\frac{10^5V}{300}+\frac{5\times10^5\times4V}{400} \) (the factor R cancels).
2. \( p\left(\frac1{300}+\frac1{100}\right)=\frac{10^5}{300}+\frac{5\times10^5}{100} \), so \( p\cdot\frac{4}{300}=\frac{16\times10^5}{300} \) and \( p=4\times10^5 \) Pa.

Teacher's Note:
1. C
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35. ,
36.
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39. d
40.
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44.
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46. V
47. /
48. T
49.
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51. o
52.
53. c
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81. a
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86. .

Answer: (d) \( \frac{\sqrt3}{2} \)

Explanation:
1. \( v_{rms}=\sqrt{\frac{3RT}{M}}\propto\sqrt T \) for the same gas.
2. \( \frac{v_A}{v_B}=\sqrt{\frac{300}{400}}=\frac{\sqrt3}{2} \).

Teacher's Note:
1. F
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4.
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28. p
29. e
30. e
31. d
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34. e
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59. a
60. t
61. u
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64. .

Answer: (a) 1/15

Explanation:
1. \( n=\frac{PV}{RT} \).
2. \( \frac{n_A}{n_B}=\frac{1\times10^5\times V/300}{5\times10^5\times4V/400}=\frac{333.3}{5000}=\frac1{15} \).

Teacher's Note:
1. U
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82. .

Answer: (b) 16 cm

Explanation:
1. Let the mercury column inside stand h cm above the outside level. The air length is \( L=8+46-h=54-h \), and its pressure is \( 76-h \).
2. Boyle's law: \( 76\times8=(76-h)(54-h) \), giving \( h^2-130h+3496=0 \), so \( h=38 \) (the other root, 92, is impossible).
3. \( L=54-38=16 \) cm.

Teacher's Note:
1. U
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138. .

Answer: (c) the change in momentum of the gas molecules as they strike the walls

Explanation:
1. Pressure is force per unit area, and force equals the rate of change of momentum.
2. Molecules rebound from the walls, and each collision changes their momentum, which gives the force on the wall. The collisions with the walls are elastic, so their kinetic energy does not change.

Teacher's Note:
1. K
2. i
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88.
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92. e
93. .

Answer: (d) an increase of pressure at constant volume

Explanation:
1. \( U=\frac f2nRT=\frac f2PV \). A straight line through the origin means \( \frac{P}{U}=\frac{2}{fV} \) is constant.
2. So V is constant while P and U increase: a vertical line upward on the P-V diagram (isochoric heating).

Teacher's Note:
1. S
2. i
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4. c
5. e
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7. \
8. (
9.
10. U
11. \
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76. .

Answer: (a) \( \frac34 \)

Explanation:
1. \( PM=\rho RT \), so \( M=\frac{\rho RT}{P} \).
2. \( \frac{M_A}{M_B}=\frac{\rho_A}{\rho_B}\cdot\frac{P_B}{P_A}=\frac{1.5}{2}=\frac34 \).

Teacher's Note:
1. T
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56. {
57. \
58. r
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61.
62. R
63. T
64. }
65. {
66. M
67. }
68.
69. \
70. )
71. .

Answer: (b) \( P_2\lt P_1 \)

Explanation:
1. \( V=\frac{nR}{P}T \), so the slope of a V-T line is \( \frac{nR}{P} \).
2. A larger slope means a smaller pressure, hence \( P_2\lt P_1 \).

Teacher's Note:
1. O
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7. -
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134. .

Answer: (c) \( c_p-c_v=\frac RM \)

Explanation:
1. Molar heat capacities satisfy \( C_p-C_v=R \) (Mayer's relation).
2. Specific heats per unit mass are \( c=\frac CM \), so \( c_p-c_v=\frac RM \).

Teacher's Note:
1. D
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86.
87. u
88. n
89. i
90. t
91.
92. m
93. a
94. s
95. s
96. .

Answer: (d) \( a=16b \)

Explanation:
1. \( c_p-c_v=\frac RM \).
2. \( M_{H_2}=2 \) and \( M_{O_2}=32 \), so \( \frac ab=\frac{32}{2}=16 \), i.e. \( a=16b \).

Teacher's Note:
1. L
2. i
3. g
4. h
5. t
6. e
7. r
8.
9. g
10. a
11. s
12. e
13. s
14.
15. h
16. a
17. v
18. e
19.
20. l
21. a
22. r
23. g
24. e
25. r
26.
27. s
28. p
29. e
30. c
31. i
32. f
33. i
34. c
35.
36. h
37. e
38. a
39. t
40. s
41.
42. p
43. e
44. r
45.
46. u
47. n
48. i
49. t
50.
51. m
52. a
53. s
54. s
55. ,
56.
57. i
58. n
59.
60. i
61. n
62. v
63. e
64. r
65. s
66. e
67.
68. p
69. r
70. o
71. p
72. o
73. r
74. t
75. i
76. o
77. n
78.
79. t
80. o
81.
82. t
83. h
84. e
85.
86. m
87. o
88. l
89. a
90. r
91.
92. m
93. a
94. s
95. s
96. .

Answer: (a) \( P=P_1+P_2+P_3 \)

Explanation:
1. By Dalton's law the pressure of a mixture of non-reacting ideal gases is the sum of the partial pressures.
2. Each gas occupies the same volume at the same temperature as before, so it exerts its original pressure: \( P=P_1+P_2+P_3 \).

Teacher's Note:
1. I
2. d
3. e
4. a
5. l
6.
7. g
8. a
9. s
10.
11. m
12. o
13. l
14. e
15. c
16. u
17. l
18. e
19. s
20.
21. d
22. o
23.
24. n
25. o
26. t
27.
28. i
29. n
30. t
31. e
32. r
33. a
34. c
35. t
36. ,
37.
38. s
39. o
40.
41. e
42. a
43. c
44. h
45.
46. g
47. a
48. s
49.
50. e
51. x
52. e
53. r
54. t
55. s
56.
57. t
58. h
59. e
60.
61. p
62. r
63. e
64. s
65. s
66. u
67. r
68. e
69.
70. i
71. t
72.
73. w
74. o
75. u
76. l
77. d
78.
79. e
80. x
81. e
82. r
83. t
84.
85. a
86. l
87. o
88. n
89. e
90. .

Answer: (b) 2

Explanation:
1. \( v_{rms}\propto\sqrt{\frac TM} \).
2. The temperature doubles and the molecular mass becomes half: \( \frac{v'}{v}=\sqrt{\frac{2}{1/2}}=2 \), so n = 2.

Teacher's Note:
1. B
2. o
3. t
4. h
5.
6. t
7. h
8. e
9.
10. t
11. e
12. m
13. p
14. e
15. r
16. a
17. t
18. u
19. r
20. e
21.
22. a
23. n
24. d
25.
26. t
27. h
28. e
29.
30. p
31. a
32. r
33. t
34. i
35. c
36. l
37. e
38.
39. m
40. a
41. s
42. s
43.
44. c
45. h
46. a
47. n
48. g
49. e
50.
51. h
52. e
53. r
54. e
55. ,
56.
57. a
58. n
59. d
60.
61. b
62. o
63. t
64. h
65.
66. e
67. n
68. t
69. e
70. r
71.
72. u
73. n
74. d
75. e
76. r
77.
78. t
79. h
80. e
81.
82. s
83. q
84. u
85. a
86. r
87. e
88.
89. r
90. o
91. o
92. t
93. .

Answer: (c) 1

Explanation:
1. \( \gamma=1.5 \) gives \( C_v=\frac{R}{\gamma-1}=2R \) for the mixture.
2. \( C_v=\frac{n_1\frac32R+n_2\frac52R}{n_1+n_2}=2R \), so \( 1.5n_1+2.5n_2=2n_1+2n_2 \), which gives \( 0.5n_2=0.5n_1 \).
3. \( n_1=n_2 \), so K = 1.

Teacher's Note:
1. T
2. h
3. e
4.
5. m
6. i
7. x
8. t
9. u
10. r
11. e
12. '
13. s
14.
15. m
16. o
17. l
18. a
19. r
20.
21. h
22. e
23. a
24. t
25.
26. c
27. a
28. p
29. a
30. c
31. i
32. t
33. y
34.
35. i
36. s
37.
38. t
39. h
40. e
41.
42. m
43. o
44. l
45. e
46. -
47. w
48. e
49. i
50. g
51. h
52. t
53. e
54. d
55.
56. a
57. v
58. e
59. r
60. a
61. g
62. e
63.
64. o
65. f
66.
67. t
68. h
69. e
70.
71. i
72. n
73. d
74. i
75. v
76. i
77. d
78. u
79. a
80. l
81.
82. v
83. a
84. l
85. u
86. e
87. s
88. .