Original practice sets for Kinetic Theory and Thermodynamics are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
1. The ideal gas equation is:
Explanation: For n moles of ideal gas at pressure P, volume V, temperature T: PV = nRT where R = 8.314 J/mol·K.
2. Gas pressure is caused by:
Explanation: Each molecular collision with a wall transfers momentum. Averaged over all collisions, this gives macroscopic pressure.
3. In an ideal gas, intermolecular forces are:
Explanation: Key assumption of KTG: molecules are point masses with no intermolecular forces except perfectly elastic collisions.
4. Pressure of an ideal gas in terms of mean square speed (v²̄) is:
Explanation: From kinetic theory: P = (1/3)ρ = (1/3)(m/V)N .
5. Average kinetic energy of a molecule of an ideal gas at temperature T is:
Explanation: For 3 translational degrees of freedom: KE_avg = (3/2)kT where k = Boltzmann constant = 1.38×10⁻²³ J/K.
6. RMS speed of gas molecules is:
Explanation: v_rms = √(3RT/M) where M is molar mass (kg/mol). Derived from (1/2)Mv_rms² = (3/2)RT.
7. The correct order of molecular speeds in a gas is:
Explanation: v_mp = √(2RT/M), v_avg = √(8RT/πM), v_rms = √(3RT/M). Numerically: 1 : 1.128 : 1.225. Always v_mp
8. Degrees of freedom for a diatomic molecule at moderate temperature:
Explanation: Diatomic molecule: 3 translational + 2 rotational = 5 degrees of freedom at moderate T (vibrational modes frozen).
9. Law of equipartition of energy states that each degree of freedom contributes to energy:
Explanation: Each quadratic energy term (each degree of freedom) has average energy = kT/2 = ½RT per mole.
10. At constant temperature, PV = constant. This is:
Explanation: Boyle's law: PV = constant at constant T and n.
11. Ratio of RMS speeds of oxygen (M = 32) and hydrogen (M = 2) at same temperature is:
Explanation: v_rms ∝ 1/√M. v_O₂/v_H₂ = √(M_H₂/M_O₂) = √(2/32) = 1/4.
12. Internal energy of n moles of monatomic ideal gas is:
Explanation: Monatomic gas: 3 degrees of freedom. U = n × (3/2)RT per degree of freedom rule.
13. Mean free path λ of gas molecules is:
Explanation: λ = kT/(√2 πd²P). At constant T, λ ∝ 1/P. Higher pressure means more collisions, shorter mean free path.
14. Cv for a diatomic ideal gas is:
Explanation: Each degree of freedom contributes R/2 to Cv. Diatomic (f=5): Cv = (5/2)R = 20.8 J/mol·K.
15. At STP (0°C, 1 atm), molar volume of ideal gas is:
Explanation: V = nRT/P = 1×8.314×273/(101325) ≈ 22.4×10⁻³ m³ = 22.4 L. This is Avogadro's molar volume.
16. At temperature T, average kinetic energy of nitrogen molecule (M = 28 g/mol, R = 8.314 J/mol·K) at 300 K is:
Explanation: KE = (3/2)kT = 1.5 × 1.38×10⁻²³ × 300 = 6.21×10⁻²¹ J.
17. Mean free path of air at room temperature is of the order:
Explanation: Mean free path of air molecules at atmospheric pressure is ~70 nm ≈ 100 nm. This is much smaller than everyday length scales but large compared to molecular size.
18. Van der Waals equation (P + a/V²)(V − b) = RT corrects for:
Explanation: a/V² accounts for intermolecular attraction (reducing pressure). b accounts for finite volume of molecules (reducing available volume).
19. Real gases behave like ideal gas at:
Explanation: At low P: molecules are far apart (intermolecular forces negligible). At high T: thermal KE >> intermolecular PE. Both conditions approximate ideal behaviour.
20. For a solid (Einstein model), energy per atom is 3kT because:
Explanation: Each atom in a solid can oscillate in 3 directions, each with KE and PE energy terms → 6 × (kT/2) = 3kT total. This gives Dulong-Petit molar heat capacity 3R ≈ 25 J/mol·K.
21. First law of thermodynamics is a statement of:
Explanation: First law: ΔQ = ΔU + W. Heat supplied = increase in internal energy + work done by system.
22. In first law ΔQ = ΔU + W, W represents:
Explanation: W = ∫PdV = work done BY the gas on surroundings. If gas expands, W > 0; if compressed, W
23. In an isothermal process, internal energy of an ideal gas:
Explanation: For ideal gas, U depends only on T. In isothermal process, T = constant → ΔU = 0 → ΔQ = W.
24. In an adiabatic process, heat exchange with surroundings is:
Explanation: Adiabatic means thermally insulated — Q = 0. All work comes from (or goes into) internal energy.
25. In an isochoric (constant volume) process, work done by gas is:
Explanation: W = ∫PdV = 0 when ΔV = 0. All heat supplied goes to increase internal energy: ΔQ = ΔU.
26. Work done by gas in isobaric expansion from V₁ to V₂ is:
Explanation: In isobaric process, P = constant. W = ∫PdV = P(V₂ − V₁) = PΔV.
27. Second law of thermodynamics states that heat cannot spontaneously flow from:
Explanation: Clausius statement: heat cannot spontaneously flow from a cold body to a hot body without work input.
28. Entropy of an isolated system in a spontaneous process:
Explanation: Second law: total entropy of isolated system always increases (or stays same in reversible process). ΔS ≥ 0.
29. Efficiency of Carnot engine working between T₁ (hot) and T₂ (cold) is:
Explanation: η_Carnot = 1 − T₂/T₁ (T in Kelvin). This is the maximum possible efficiency for any engine operating between these temperatures.
30. A Carnot engine absorbs 1000 J at T₁ = 500 K and rejects heat at T₂ = 300 K. Work done is:
Explanation: η = 1 − 300/500 = 0.4. W = η × Q₁ = 0.4 × 1000 = 400 J.
31. For adiabatic process in ideal gas, relation between P and V is:
Explanation: Adiabatic equation: PVγ = constant, where γ = Cp/Cv. Also TV^(γ−1) = constant and T^γP^(1−γ) = constant.
32. Efficiency of a Carnot engine can be 100% only if cold reservoir temperature T₂ is:
Explanation: η = 1 − T₂/T₁. η = 1 only when T₂ = 0 K. Absolute zero is unattainable by third law, so 100% efficiency is impossible.
33. For a reversible process, change in entropy of universe is:
Explanation: Reversible process: ΔS_universe = 0. Only irreversible processes increase total entropy.
34. Area enclosed by a cycle on a P-V diagram represents:
Explanation: Work W = ∮PdV = area enclosed by the cycle on P-V diagram. Sign depends on direction (clockwise = positive work = engine).
35. For a cyclic process, ΔU = 0. Therefore:
Explanation: First law: ΔQ = ΔU + W. For cycle, ΔU = 0, so ΔQ = W. Net heat absorbed = net work done.
36. For ideal gas, Cp − Cv = R. This is because:
Explanation: At constant P, gas must expand when heated. Extra work done = PΔV = nRΔT. Per mole, this extra heat per degree = R. So Cp = Cv + R.
37. A refrigerator operates between −10°C (cold) and 30°C (hot). Minimum work needed to remove 500 J from the cold reservoir is:
Explanation: For a reversible refrigerator, $COP=T_c/(T_h-T_c)=263/40=6.575$. Therefore $W=Q_c/COP=500/6.575≈ 76.0\text{ J} $.
38. Speed of sound in gas is given by √(γP/ρ) rather than √(P/ρ) because sound propagation is:
Explanation: Sound waves alternate compression/rarefaction so fast that no heat exchange occurs — they are adiabatic. Newton's isothermal formula underestimates by factor √γ.
39. Kelvin-Planck statement of second law says:
Explanation: Kelvin-Planck: It is impossible to construct a heat engine that, operating in a cycle, converts all absorbed heat into work.
40. When 100 g of ice melts at 0°C (latent heat = 336 J/g), entropy change of ice is:
Explanation: ΔS = Q/T = (100 × 336)/273 ≈ 123 J/K. Temperature must be in Kelvin.
41. Work done by an ideal gas during isothermal expansion from V₁ to V₂ at temperature T is:
Explanation: Isothermal: W = ∫(nRT/V)dV = nRT ln(V₂/V₁). Since T = const, PV = const and pressure varies.
42. Work done in an adiabatic process (Q = 0) equals:
Explanation: First law: Q = ΔU + W. With Q = 0: W = −ΔU. Work done equals decrease in internal energy.
43. On a P-V diagram, an isothermal curve is called:
Explanation: Isothermal process: PV = const at fixed T. This is a rectangular hyperbola (isotherm) on P-V graph.
44. For the same expansion from V₁ to V₂, an adiabat is steeper than isotherm on P-V diagram because:
Explanation: Adiabat: PVγ = const. Isotherm: PV = const. Since γ > 1, adiabatic pressure drop is steeper.
45. Work done against constant atmospheric pressure P₀ when liquid of volume V vaporises (vapour volume V_g >> V_l):
Explanation: W = P₀ΔV = P₀(V_gas − V_liquid). Since V_g >> V_l, W ≈ P₀V_g.
46. Carnot cycle consists of:
Explanation: Carnot cycle: isothermal expansion → adiabatic expansion → isothermal compression → adiabatic compression.
47. For a diatomic ideal gas, ratio γ = Cp/Cv is:
Explanation: Diatomic: Cv = (5/2)R, Cp = (7/2)R. γ = Cp/Cv = 7/5 = 1.4.
48. For an isochoric process on P-V diagram, the curve is:
Explanation: Isochoric: V = constant. This is a vertical line on the P-V diagram.
49. A heat engine absorbs Q₁ = 1000 J and rejects Q₂ = 600 J. Efficiency is:
Explanation: η = W/Q₁ = (Q₁−Q₂)/Q₁ = 400/1000 = 40%.
50. COP (coefficient of performance) of a refrigerator is defined as:
Explanation: COP_ref = Q_cold / W = Q₂/(Q₁−Q₂). Higher COP = more efficient refrigerator.
51. In an adiabatic compression, temperature of gas:
Explanation: Q = 0, W done ON gas > 0 → ΔU > 0 → T increases. Diesel engine compression is an example.
52. Throttling (Joule-Thomson effect) for an ideal gas: temperature change is:
Explanation: For an ideal gas, Joule-Thomson coefficient μ_JT = 0 — no temperature change during throttling. Real gases show cooling or warming depending on temperature relative to inversion temperature.
53. In a P-V diagram, work done is positive when the cycle is traversed:
Explanation: Clockwise on P-V diagram: expansion at higher P and compression at lower P → net positive work (engine cycle).
54. Two Carnot engines A (between T₁ and T) and B (between T and T₂) have same efficiency. Temperature T is:
Explanation: η_A = 1−T/T₁ = η_B = 1−T₂/T. So T/T₁ = T₂/T → T² = T₁T₂ → T = √(T₁T₂) (geometric mean).
55. 1 mole of gas at 0°C expands isothermally from 1 L to 10 L. Work done is (R = 8.314 J/mol·K):
Explanation: W = nRT ln(V₂/V₁) = 1×8.314×273×ln(10) = 8.314×273×2.303 ≈ 5229 J.
56. For mixing of two ideal gases at same T and P, entropy change is:
Explanation: Mixing increases disorder → entropy of mixing ΔS_mix = −nR(x₁lnx₁ + x₂lnx₂) > 0 where x₁, x₂ are mole fractions.
57. Why Cv of monatomic gas (3R/2) is less than diatomic (5R/2)?
Explanation: Monatomic: 3 translational DOF → Cv = 3R/2. Diatomic: 3 translational + 2 rotational DOF → Cv = 5R/2. More modes = larger Cv.
58. On T-S (temperature-entropy) diagram, area enclosed by a Carnot cycle equals:
Explanation: On T-S diagram, dQ = TdS. Area = ∮TdS = Q_absorbed − Q_rejected = net work done.
59. A steam engine has a source at 200°C and a sink at 20°C. Its actual efficiency is 25%. The ratio of actual efficiency to Carnot efficiency is approximately:
Explanation: $η_C=1-293/473≈0.3805$. Hence $η_{actual}/η_C=0.25/0.3805≈ 0.66 $.
60. Third law of thermodynamics states that at absolute zero:
Explanation: Nernst-Planck statement (third law): as T → 0 K, entropy of a perfect crystal approaches zero — only one microstate available.
61. A gas at 27°C (T₁) is heated to 127°C (T₂) at constant volume. Ratio P₂/P₁ is:
Explanation: At constant V: P ∝ T (Kelvin). T₁ = 300 K, T₂ = 400 K. P₂/P₁ = 400/300 = 4/3.
62. Heat supplied to system is 200 J and work done by system is 80 J. Change in internal energy is:
Explanation: ΔU = Q − W = 200 − 80 = 120 J.
63. Carnot engine has efficiency 40%. If hot reservoir temperature is 500 K, cold reservoir is:
Explanation: η = 1 − T₂/T₁ → 0.4 = 1 − T₂/500 → T₂ = 300 K.
64. At what temperature is RMS speed of nitrogen molecules equal to RMS speed of oxygen at 300 K? (M_N₂=28, M_O₂=32)
Explanation: v_rms ∝ √(T/M). √(T_N₂/28) = √(300/32). T_N₂ = 300×28/32 = 262.5 K.
65. Work done in a cyclic process on P-V graph where cycle is traversed anticlockwise is:
Explanation: Anticlockwise P-V cycle: expansion at lower pressure, compression at higher pressure → net negative work by gas = work done on gas (refrigerator cycle).
66. For a monatomic ideal gas, ratio Cp/Cv is:
Explanation: Monatomic: Cv = 3R/2, Cp = 5R/2. γ = 5/3 ≈ 1.67.
67. In isothermal expansion of ideal gas, temperature is constant, so ΔU = 0. Heat absorbed:
Explanation: ΔU = 0 → Q = W. All heat absorbed by the gas is converted into work done during isothermal expansion.
68. Viscosity of a gas is independent of pressure because as pressure increases:
Explanation: η ∝ (number density) × (mean free path). These have opposite dependencies on P and their product is constant.
69. Which statement is true about entropy?
Explanation: ΔS_universe ≥ 0 always. Equal in reversible processes, positive in irreversible. Never negative.
70. For adiabatic process, TV^(γ−1) = constant. If volume doubles adiabatically, temperature ratio T₂/T₁ for diatomic gas (γ = 1.4) is:
Explanation: T₁V₁^(γ−1) = T₂V₂^(γ−1). T₂/T₁ = (V₁/V₂)^(γ−1) = (1/2)^0.4 = 2^(−0.4).
71. Two identical bodies at temperatures T₁ and T₂ (T₁ > T₂) are used as source and sink. After Carnot engine runs until both reach same temperature T_f, maximum work output is:
Explanation: T_f = √(T₁T₂) for maximum work. W = nCv(T₁ + T₂ − 2T_f) = nCv(T₁ + T₂ − 2√(T₁T₂)).
72. Pressure exerted by gas of n molecules per unit volume, each of mass m, with mean square speed , is:
Explanation: P = (1/3)ρ = (1/3)(nm) where n = number density (molecules per m³) and m = mass per molecule.
73. Molar heat capacity at constant pressure for water (liquid) is about 75 J/mol·K. This is much larger than 3R ≈ 25 J/mol·K for solids because:
Explanation: Water molecules have multiple vibrational modes and strong hydrogen bonds that absorb energy, giving a large Cp relative to simple solids.
74. Helmholtz free energy F = U − TS. For a spontaneous isothermal process, ΔF:
Explanation: At constant T and V, equilibrium corresponds to minimum F. Spontaneous processes decrease F; equilibrium means ΔF = 0.
75. Dalton's law of partial pressures states that total pressure of gas mixture equals:
Explanation: For ideal gas mixture: P_total = P₁ + P₂ + ... Each gas acts independently (no intermolecular interactions between different species).
76. A gas undergoes cycle: A→B (isothermal expansion), B→C (isochoric cooling), C→A (isobaric compression). Net work done by gas is positive if A→B expansion provides more work than C→A compression requires. This cycle is analogous to:
Explanation: Stirling cycle: isothermal expansion + isochoric process + isothermal compression + isochoric process. The described cycle is a variant.
77. Effusion rate of gas through a small hole (Graham's law) is proportional to:
Explanation: Graham's law: rate of effusion ∝ 1/√M. Lighter molecules effuse faster. Used in uranium enrichment via UF₆ effusion.
78. When 1 kg water at 100°C converts to steam at 100°C (L = 2.26×10⁶ J/kg), entropy change is:
Explanation: ΔS = Q/T = mL/T = 1×2.26×10⁶/373 ≈ 6057 J/K ≈ 6045 J/K.
79. During adiabatic free expansion of an ideal gas into vacuum, temperature:
Explanation: Free expansion: no work done (vacuum), no heat exchange. For ideal gas, ΔU = 0 → ΔT = 0. (Real gas: Joule expansion causes cooling due to intermolecular forces.)
80. Gibbs free energy G = H − TS. For spontaneous process at constant T and P:
Explanation: At constant T and P, spontaneous processes decrease Gibbs free energy: ΔG 0.
81. Maxwell-Boltzmann speed distribution shows that as temperature increases:
Explanation: At higher T, the most probable speed v_mp = √(2RT/M) increases, and the distribution broadens because more molecules gain high speeds.
82. In a Carnot refrigerator operating in reverse, COP is related to Carnot engine efficiency by:
Explanation: η = 1 − T₂/T₁ and COP = T₂/(T₁−T₂). So COP = (T₂/T₁)/(1−T₂/T₁) = (1−η)/η.
83. The statistical definition of entropy by Boltzmann is:
Explanation: S = k ln Ω where Ω = number of microstates. More microstates = more disorder = higher entropy. This is engraved on Boltzmann's tombstone.
84. For an ideal gas undergoing polytropic process PVⁿ = constant, specific heat C is:
Explanation: For polytropic process, C = Cv(γ−n)/(1−n). When n=0 (isobaric): C=Cp. When n=∞ (isochoric): C=Cv. When n=1 (isothermal): C=∞. When n=γ (adiabatic): C=0.
85. Inversion temperature of a van der Waals gas (above which Joule-Thomson expansion causes warming) is:
Explanation: T_inv = 2a/(Rb). Below T_inv, JT expansion causes cooling (throttling refrigerators exploit this). Above T_inv, gas warms on expansion.
86. At very low temperatures, Cv of diatomic gas approaches 3R/2 (monatomic value) because:
Explanation: Quantum mechanics: rotational energy is quantised. At low T, thermal energy kT
87. Clausius inequality states that for any cyclic process:
Explanation: Clausius inequality: ∮dQ_rev/T = 0 (reversible) and ∮dQ_irrev/T
88. A hot (90°C) and cold (10°C) body exchange heat reversibly via Carnot engine until they reach equilibrium temperature T_eq. T_eq is:
Explanation: For equal heat capacities, entropy is maximised (and Carnot work is maximised) at T_eq = √(T₁T₂) = √(363×283) ≈ 320 K = 47°C.
89. Partial pressure of water vapour in a mixture is related to humidity. Relative humidity is:
Explanation: RH% = (P_vapour / P_sat) × 100. At saturation (100% RH), water starts condensing as dew.
90. Otto cycle (petrol engine) operates with:
Explanation: Otto cycle: adiabatic compression → isochoric heat addition (spark ignition) → adiabatic expansion → isochoric heat rejection. Efficiency = 1 − 1/r^(γ−1) where r is compression ratio.
91. Diesel cycle has higher efficiency than Otto at same compression ratio because:
Explanation: In the Diesel cycle, fuel ignites at constant pressure (isobaric), then the gas expands further adiabatically. The expansion ratio > compression ratio, capturing more work than Otto for same input.
92. Entropy change in reversible isothermal expansion of n moles ideal gas from V₁ to V₂ is:
Explanation: ΔS = Q_rev/T = W/T = nRT ln(V₂/V₁) / T = nR ln(V₂/V₁).
93. Critical constants of a van der Waals gas (P_c, V_c, T_c) are related to a and b by:
Explanation: Critical constants: T_c = 8a/(27Rb), V_c = 3b, P_c = a/(27b²). At critical point, gas-liquid distinction vanishes.
94. Zartman-Ko experiment (1930) verified:
Explanation: The Zartman-Ko rotating disk experiment directly measured the speed distribution of bismuth atoms, confirming the Maxwell-Boltzmann distribution.
95. The statement 'entropy of the universe tends to a maximum' is a paraphrase of:
Explanation: Clausius formulated: 'The entropy of the universe tends to a maximum.' This is a concise statement of the second law.
96. For an ideal gas, the slope of adiabat at a point (P,V) on P-V diagram is:
Explanation: PVγ = const. Differentiating: γPV^(γ−1)dV + V^γdP = 0 → dP/dV = −γP/V. Steeper than isothermal slope (−P/V) by factor γ.
97. Maxwell's demon thought experiment failed because:
Explanation: Landauer (1961): erasure of one bit of information requires minimum k ln2 of energy dissipated as heat, compensating the entropy decrease the demon creates.
98. Brownian motion was explained by Einstein (1905) as evidence for:
Explanation: Einstein showed that random jiggling of pollen grains in water (Brownian motion) is caused by random impacts of water molecules — direct evidence for the molecular nature of matter.
99. Brayton cycle (gas turbine) consists of:
Explanation: Brayton cycle: adiabatic compression → isobaric heat addition (combustion chamber) → adiabatic expansion (turbine) → isobaric heat rejection. Used in jet engines and gas turbines.