Class 11 Chemistry Chapter 5: Thermodynamics MCQs with answers
10 multiple-choice questions on Thermodynamics, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.
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Answer: (b) a closed system
Glass conducts heat, so energy passes between the bath and the mixture, but the seal stops any matter leaving or entering: a closed system.
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Answer: (d) molar volume, pressure, temperature
Pressure and temperature do not depend on the amount of matter, and a molar property such as molar volume is an extensive property divided by n, so it is independent of the amount. Enthalpy, heat capacity and internal energy are extensive.
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Answer: (a) w is positive and q is negative
Work done on the system is positive; heat leaving the system for the surroundings is negative.
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Answer: (c) ΔS of the gas
Expansion into vacuum is free expansion: pex = 0, so w = 0; Joule found q = 0, so ΔU = q + w = 0. The expansion is spontaneous and irreversible, so ΔStotal > 0; the surroundings receive no heat (ΔSsurr = 0), so the whole increase is in the gas, which has spread into a larger volume: ΔS of the gas > 0.
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Answer: (c) w
ΔU, ΔH and q + w depend only on the initial and final states. q and w separately depend on how the change is carried out.
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Answer: (d) freezing of water at 273 K
Melting, sublimation and vaporisation absorb heat (positive ΔH). Freezing is the reverse of melting and gives the same heat out to the surroundings, so ΔH is negative.
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Answer: (b) S(rhombic)
ΔfH⊖ is zero only for an element in its reference state, its most stable form at 25 °C and 1 bar. For sulphur that is rhombic sulphur; C(diamond) is +1.89 and Br₂(g) is +30.91 kJ mol⁻¹ (bromine's reference state is the liquid), and oxygen's reference state is O₂(g), not O₃.
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Answer: (a) a pure crystalline solid
The third law is confined to pure crystalline solids, which are perfectly ordered at 0 K; the entropy of solutions and supercooled liquids is not zero at absolute zero.
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Answer: (c) 0.90 J g⁻¹ K⁻¹
q = c × m × ΔT, so c = 3600 J / (200 g × 20 K) = 0.90 J g⁻¹ K⁻¹.
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Answer: (b) is endothermic, with ΔrH = +50 kJ mol⁻¹
Hp above Hr means ΔrH = Hp − Hr = +50 kJ mol⁻¹: heat is absorbed from the surroundings, an endothermic reaction.
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