JEE Physics Thermodynamics MCQs Set 12

Answer: (a) \( C_I\gt C_{III}\gt C_{II} \)

Explanation:
1. \( C=\frac{Q}{m\Delta T} \) and \( Q=\Delta U+W \), so \( \Delta T\propto Q-W \).
2. (I) the gas does positive work, so less heat goes into internal energy and \( \Delta T \) is smallest: C is largest. (II) work is done on the gas (W negative), so \( \Delta T \) is largest: C is smallest. (III) \( W=0 \) lies in between.

Teacher's Note:
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Answer: (b) against atmospheric pressure plus the spring

Explanation:
1. The expanding piston must push back the atmosphere and compress the spring.
2. So the work is done against both the atmospheric pressure and the spring. Part of the heat also raises the internal energy, so the work is less than the heat supplied.

Teacher's Note:
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Answer: (c) I and II only

Explanation:
1. At constant pressure \( dQ=nC_pdT=n\frac52R\,dT \) and \( dU=nC_vdT=n\frac32R\,dT \).
2. \( \frac{dU}{dQ}=\frac35 \) and \( \frac{dW}{dQ}=1-\frac35=\frac25 \): (I) and (II) hold. (III) is false. (IV) violates the first law, because \( dQ=dU+dW \) and not \( dW+dU+dQ=0 \).

Teacher's Note:
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Answer: (d) energy

Explanation:
1. The law \( \Delta Q=\Delta U+\Delta W \) relates the heat supplied, the change of internal energy and the work done.
2. It expresses conservation of energy, including heat as a form of energy.

Teacher's Note:
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83. .

Answer: (a) the gas does positive work

Explanation:
1. For an ideal gas \( U \) depends only on T, so \( \Delta U=0 \) in an isothermal process.
2. Then \( Q=W \): all the heat supplied appears as positive work done by the gas.

Teacher's Note:
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Answer: (b) \( \text{The change in internal energy in the process BC is } -500R \)

Explanation:
1. For a diatomic gas \( \Delta U=\frac52R\,\Delta T \) per mole.
2. BC: \( \frac52R(600-800)=-500R \). CA: \( \frac52R(400-600)=-500R \). AB: \( \frac52R(400)=+1000R \). The total over the cycle is zero.

Teacher's Note:
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123. .

Answer: (c) \( 4P_0V_0 \)

Explanation:
1. Work is the area under the line from \( V_0 \) to \( 3V_0 \): a trapezium with parallel sides \( P_0 \) and \( 3P_0 \).
2. \( W=\frac12(P_0+3P_0)(3V_0-V_0)=4P_0V_0 \).

Teacher's Note:
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Answer: (d) infinite

Explanation:
1. Heat capacity of a process is \( C=\frac{dQ}{dT} \).
2. In an isothermal process \( dT=0 \) while heat is exchanged, so C is infinite.

Teacher's Note:
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Answer: (a) \( \frac{13}{2}p_0V_0 \)

Explanation:
1. Heat is absorbed in the isochoric rise: \( \Delta Q=\frac32(2p_0V_0-p_0V_0)=\frac32p_0V_0 \).
2. Heat is absorbed in the isobaric expansion: \( \Delta Q=\frac52(2p_0)(V_0)=5p_0V_0 \).
3. Total absorbed \( =\frac32p_0V_0+5p_0V_0=\frac{13}{2}p_0V_0 \). The other two steps reject heat.

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103. .

Answer: (b) \( \frac1{2\pi}\sqrt{\frac{A^2\gamma p_0}{MV_0}} \)

Explanation:
1. Isolated means adiabatic: \( pV^\gamma=\text{const} \), so \( dp=-\gamma p_0\frac{dV}{V_0} \) with \( dV=A\,x \).
2. Restoring force \( F=A\,dp=-\frac{\gamma p_0A^2}{V_0}x \), so \( k=\frac{\gamma p_0A^2}{V_0} \).
3. \( f=\frac1{2\pi}\sqrt{\frac kM}=\frac1{2\pi}\sqrt{\frac{A^2\gamma p_0}{MV_0}} \).

Teacher's Note:
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Answer: (c) \( 2RT_0 \)

Explanation:
1. With \( T=\frac{PV}{R} \): \( P^2=\frac{cPV}{R} \), so \( P\propto V \), say \( P=kV \).
2. Work \( W=\int P\,dV=\frac k2(V_2^2-V_1^2)=\frac12(P_2V_2-P_1V_1)=\frac R2(2T_0-T_0)=\frac{RT_0}{2} \).
3. \( \Delta U=\frac32R(T_0)=\frac32RT_0 \), so \( Q=\Delta U+W=2RT_0 \).

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Answer: (d) adiabatic

Explanation:
1. Isochoric: no volume change, no work. Isobaric at the initial pressure: \( W=\frac{p_iV_i}{2} \). Isothermal: \( W=p_iV_i\ln2\approx0.69p_iV_i \).
2. Adiabatic: the pressure rises faster than in the other cases (the gas also heats up), so the area under the curve is the largest: for \( \gamma=1.4 \), \( W=\frac{p_iV_i(2^{0.4}-1)}{0.4}\approx0.80p_iV_i \).

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Answer: (a) 90 J

Explanation:
1. Efficiency \( \eta=1-\frac{T_2}{T_1}=\frac1{10} \), so \( \frac{T_2}{T_1}=\frac9{10} \).
2. Coefficient of performance \( =\frac{T_2}{T_1-T_2}=9 \), so \( Q_2=9\times10=90 \) J.

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Answer: (b) \( -20\ \text{kJ} \)

Explanation:
1. \( \Delta U=\frac52(p_BV_B-p_AV_A)=\frac52(2\times6-5\times4)\ \text{kPa m}^3 \).
2. \( =\frac52(12-20)=-20 \) kJ (since 1 kPa m³ = 1 kJ).

Teacher's Note:
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4.
5. a
6. n
7.
8. i
9. d
10. e
11. a
12. l
13.
14. g
15. a
16. s
17.
18. t
19. h
20. e
21.
22. c
23. h
24. a
25. n
26. g
27. e
28.
29. i
30. n
31.
32. i
33. n
34. t
35. e
36. r
37. n
38. a
39. l
40.
41. e
42. n
43. e
44. r
45. g
46. y
47.
48. d
49. e
50. p
51. e
52. n
53. d
54. s
55.
56. o
57. n
58. l
59. y
60.
61. o
62. n
63.
64. t
65. h
66. e
67.
68. e
69. n
70. d
71.
72. s
73. t
74. a
75. t
76. e
77. s
78. :
79.
80. \
81. (
82.
83. \
84. f
85. r
86. a
87. c
88.
89. f
90. 2
91. \
92. D
93. e
94. l
95. t
96. a
97. (
98. P
99. V
100. )
101.
102. \
103. )
104. .

Answer: (c) 460 J

Explanation:
1. Along ABC: \( W=6\times10^4\times2\times10^{-3}=120 \) J (only BC does work). \( \Delta U=Q-W=500-120=380 \) J.
2. Along AC: \( W=\frac12(2+6)\times10^4\times2\times10^{-3}=80 \) J.
3. \( Q_{AC}=\Delta U+W=380+80=460 \) J.

Teacher's Note:
1. I
2. n
3. t
4. e
5. r
6. n
7. a
8. l
9.
10. e
11. n
12. e
13. r
14. g
15. y
16.
17. c
18. h
19. a
20. n
21. g
22. e
23.
24. i
25. s
26.
27. t
28. h
29. e
30.
31. s
32. a
33. m
34. e
35.
36. a
37. l
38. o
39. n
40. g
41.
42. e
43. v
44. e
45. r
46. y
47.
48. p
49. a
50. t
51. h
52.
53. b
54. e
55. t
56. w
57. e
58. e
59. n
60.
61. t
62. w
63. o
64.
65. s
66. t
67. a
68. t
69. e
70. s
71. ,
72.
73. s
74. o
75.
76. f
77. i
78. n
79. d
80.
81. i
82. t
83.
84. f
85. r
86. o
87. m
88.
89. t
90. h
91. e
92.
93. p
94. a
95. t
96. h
97.
98. w
99. h
100. e
101. r
102. e
103.
104. t
105. h
106. e
107.
108. h
109. e
110. a
111. t
112.
113. i
114. s
115.
116. k
117. n
118. o
119. w
120. n
121. .

Answer: (d) zero

Explanation:
1. A to B: \( W=\frac12(P_0+3P_0)V_0=2P_0V_0 \). B to C (leftwards at \( 3P_0 \)): \( W=-3P_0V_0 \).
2. C to D: \( W=\frac12(3P_0+P_0)V_0=2P_0V_0 \). D to A (leftwards at \( P_0 \)): \( W=-P_0V_0 \).
3. Total \( =2-3+2-1=0 \). The figure of eight consists of two loops of equal area traversed in opposite senses.

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

Answer: (a) 1000 J

Explanation:
1. The cycle is a right triangle: base \( 5\times10^{-3} \) m³ and height \( 4\times10^5 \) Pa.
2. The cycle runs clockwise (the expansion AB is at higher pressure than the compression CA), so the work is positive: \( W=\frac12\times5\times10^{-3}\times4\times10^5=1000 \) J.

Teacher's Note:
1. N
2. e
3. t
4.
5. w
6. o
7. r
8. k
9.
10. o
11. f
12.
13. a
14.
15. c
16. y
17. c
18. l
19. e
20.
21. e
22. q
23. u
24. a
25. l
26. s
27.
28. i
29. t
30. s
31.
32. e
33. n
34. c
35. l
36. o
37. s
38. e
39. d
40.
41. a
42. r
43. e
44. a
45. ,
46.
47. p
48. o
49. s
50. i
51. t
52. i
53. v
54. e
55.
56. i
57. f
58.
59. t
60. h
61. e
62.
63. p
64. a
65. t
66. h
67.
68. i
69. s
70.
71. c
72. l
73. o
74. c
75. k
76. w
77. i
78. s
79. e
80.
81. o
82. n
83.
84. t
85. h
86. e
87.
88. P
89. -
90. V
91.
92. d
93. i
94. a
95. g
96. r
97. a
98. m
99. .

Answer: (b) \( \frac38N_ak_B(T_2-T_1) \)

Explanation:
1. Helium has molar mass 4 g, so 1 g is \( \frac14 \) mole.
2. Monatomic gas, heated at constant volume: \( Q=n\,\frac32R\,\Delta T=\frac14\cdot\frac32N_ak_B(T_2-T_1)=\frac38N_ak_B(T_2-T_1) \).

Teacher's Note:
1. R
2. e
3. m
4. e
5. m
6. b
7. e
8. r
9.
10. t
11. o
12.
13. c
14. o
15. n
16. v
17. e
18. r
19. t
20.
21. t
22. h
23. e
24.
25. m
26. a
27. s
28. s
29.
30. t
31. o
32.
33. m
34. o
35. l
36. e
37. s
38. ,
39.
40. a
41. n
42. d
43.
44. u
45. s
46. e
47.
48. \
49. (
50.
51. R
52. =
53. N
54. _
55. a
56. k
57. _
58. B
59.
60. \
61. )
62.
63. f
64. o
65. r
66.
67. t
68. h
69. e
70.
71. m
72. o
73. l
74. a
75. r
76.
77. f
78. o
79. r
80. m
81. .

Answer: (c) 6 J

Explanation:
1. Along adc: \( \Delta U=Q-W=50-20=30 \) J, so \( U_c=40 \) J.
2. Along abc: \( \Delta U \) is also 30 J, so \( W=Q-\Delta U=36-30=6 \) J.

Teacher's Note:
1. T
2. h
3. e
4.
5. c
6. h
7. a
8. n
9. g
10. e
11.
12. i
13. n
14.
15. i
16. n
17. t
18. e
19. r
20. n
21. a
22. l
23.
24. e
25. n
26. e
27. r
28. g
29. y
30.
31. i
32. s
33.
34. p
35. a
36. t
37. h
38.
39. i
40. n
41. d
42. e
43. p
44. e
45. n
46. d
47. e
48. n
49. t
50. .
51.
52. U
53. s
54. e
55.
56. t
57. h
58. e
59.
60. p
61. a
62. t
63. h
64.
65. w
66. h
67. e
68. r
69. e
70.
71. b
72. o
73. t
74. h
75.
76. Q
77.
78. a
79. n
80. d
81.
82. W
83.
84. a
85. r
86. e
87.
88. k
89. n
90. o
91. w
92. n
93.
94. t
95. o
96.
97. f
98. i
99. n
100. d
101.
102. i
103. t
104. .

Answer: (d) \( Q_1\gt Q_2\gt Q_3\ \text{and}\ \Delta U_1=\Delta U_2=\Delta U_3 \)

Explanation:
1. The end states are the same, so \( \Delta U \) is the same for all three processes.
2. The gas expands in all the cases, and the work equals the area under the curve, \( W_1\gt W_2\gt W_3 \). Since \( Q=\Delta U+W \), \( Q_1\gt Q_2\gt Q_3 \).

Teacher's Note:
1. W
2. i
3. t
4. h
5.
6. t
7. h
8. e
9.
10. s
11. a
12. m
13. e
14.
15. c
16. h
17. a
18. n
19. g
20. e
21.
22. o
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24.
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27. t
28. e
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30. n
31. a
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33.
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35. n
36. e
37. r
38. g
39. y
40. ,
41.
42. m
43. o
44. r
45. e
46.
47. w
48. o
49. r
50. k
51.
52. d
53. o
54. n
55. e
56.
57. b
58. y
59.
60. t
61. h
62. e
63.
64. g
65. a
66. s
67.
68. m
69. e
70. a
71. n
72. s
73.
74. m
75. o
76. r
77. e
78.
79. h
80. e
81. a
82. t
83.
84. m
85. u
86. s
87. t
88.
89. b
90. e
91.
92. s
93. u
94. p
95. p
96. l
97. i
98. e
99. d
100. .

Answer: (a) 150 J of heat has been added to the gas

Explanation:
1. \( \Delta U=0 \) for an isothermal process of an ideal gas.
2. So \( Q=W=150 \) J, and this much heat was added to the gas.

Teacher's Note:
1. I
2. s
3. o
4. t
5. h
6. e
7. r
8. m
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10. l
11.
12. d
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20.
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33.
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79. h
80. e
81.
82. w
83. o
84. r
85. k
86.
87. d
88. o
89. n
90. e
91. .

Answer: (b) \( \Delta U=-\Delta W\ \text{in an adiabatic process} \)

Explanation:
1. First law: \( Q=\Delta U+W \).
2. Adiabatic: \( Q=0 \), so \( \Delta U=-W \). Isothermal (ideal gas): \( \Delta U=0 \), so neither of the isothermal statements holds.

Teacher's Note:
1. I
2. n
3.
4. a
5. n
6.
7. a
8. d
9. i
10. a
11. b
12. a
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55.
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81. a
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83.
84. e
85. n
86. e
87. r
88. g
89. y
90. .

Answer: (c) 7900 J

Explanation:
1. \( Q=2000\times4.2=8400 \) J.
2. \( \Delta U=Q-W=8400-500=7900 \) J.

Teacher's Note:
1. C
2. o
3. n
4. v
5. e
6. r
7. t
8.
9. c
10. a
11. l
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45. f
46. i
47. r
48. s
49. t
50.
51. l
52. a
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54. .

Answer: (d) 3

Explanation:
1. Adiabatic: \( dU=-P\,dV \). With \( U=K+nPV \): \( n(P\,dV+V\,dP)=-P\,dV \), so \( nV\,dP=-(n+1)P\,dV \).
2. Integrating: \( PV^{(n+1)/n}=\text{const} \), so \( \gamma=\frac{n+1}{n}=\frac43 \) and n = 3.

Teacher's Note:
1. W
2. r
3. i
4. t
5. e
6.
7. t
8. h
9. e
10.
11. a
12. d
13. i
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16. a
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20.
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22. o
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24. d
25. i
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27. i
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30.
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73. e
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76. i
77. t
78. h
79.
80. \
81. (
82.
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84. V
85. ^
86. \
87. g
88. a
89. m
90. m
91. a
92. =
93. \
94. t
95. e
96. x
97. t
98. {
99. c
100. o
101. n
102. s
103. t
104. }
105.
106. \
107. )
108. .