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Class XI · Chemistry

Chemical Thermodynamics: NEET MCQ practice

249 practice questions on Chemical Thermodynamics, each with an explanation. Practise them chapter by chapter in the GyaanLoop app.

249

Practice questions

95

Easy

134

Medium

20

Hard

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What Chemical Thermodynamics covers

Topics in the NEET (UG) 2026–27 syllabus for this chapter.

  • System and surroundings, extensive and intensive properties, state functions, types of processes
  • First law of thermodynamics- work, heat, internal energy, enthalpy
  • Hess's law of constant heat summation
  • Enthalpies of bond dissociation, combustion, formation, atomization, sublimation, phase transition, hydration, ionization, solution
  • Second law of thermodynamics- spontaneity, entropy change, Gibbs energy change and equilibrium constant

Sample questions on Chemical Thermodynamics

Try each one, then open the answer and explanation.

Question 1 · Practice question

Volume of an ideal gas is to be decreased by 10% by increase of pressure by x% under isothermal condition. Thus, x is

  1. A.100/9
  2. B.9/100
  3. C.10
  4. D.1/10
Show answer

Answer: A. 100/9

Under isothermal conditions, Boyle's law applies: PV = constant. A 10% decrease in volume means the new volume is 90% of the original. To keep PV constant, pressure must increase to 10/9 of the original, so the increase is (1/9) of the original, which is (100/9)% increase. Option (c) 10 might tempt you if you confuse percentage for a factor.

Question 2 · Practice question

Specific heat of water is 4.184 JK -1 g -1, Rise in temperature when 1 kJ of heat is absorbed by 2 moles of H₂O is

  1. A.119.5 o
  2. B.0.37 o
  3. C.0.12 o
  4. D.6.64 o
Show answer

Answer: D. 6.64 o

Mass of 2 moles of water = 2 × 18 = 36 g. Using Q = m c ΔT, ΔT = 1000 J / (36 g × 4.184 J K⁻¹ g⁻¹) ≈ 6.64°C. Option (d) is correct. Option (a) is wrong if one forgets to multiply molar mass by moles; option (b) or (c) arise from mixing up units or incorrectly converting heat to calories.

Question 3 · Practice question

The amount of the heat released when 20 ml 0.5 M NaOH is mixed with 100 ml 0.1 M HCl is x kJ. The heat of neutralization is

  1. A.– 100 x kJ/mol
  2. B.– 50 x kJ/mol
  3. C.+ 100 x kJ/mol
  4. D.+50 x kJ/mol
Show answer

Answer: A. – 100 x kJ/mol

After stoichiometry, neutralization heat is heat per mole of water formed. Since 0.01 mole produces x kJ, 1 mole would produce 100x kJ, with negative sign because it's exothermic. Wrong options ignore stoichiometry or sign.

Question 4 · Practice question

The heat capacity ratio, r was determined for cyanogen as 1.177. Thus, C p for this gas is

  1. A.55.3 JK -1 mol -1
  2. B.33.26 JK -1 mol -1
  3. C.8.314 JK -1 mol -1
  4. D.24.94 JK -1 mol -1
Show answer

Answer: A. 55.3 JK -1 mol -1

The heat capacity ratio γ = Cp/Cv = 1.177, and for any ideal gas, Cp – Cv = R (8.314 J K⁻¹ mol⁻¹). Solving these two equations gives Cp ≈ 55.3 J K⁻¹ mol⁻¹. The wrong options come from misusing the formula or forgetting R, e.g., 33.26 might be Cv and 24.94 might be an incorrect Cp = γR.

Question 5 · Practice question

Assuming the composition of air to be X (N₂)= 0.80, X (O₂)= 0.18 and X (CO₂)= 0.02, molar heat capacity of air at constant pressure is

  1. A.3.50R
  2. B.4.00R
  3. C.2.50R
  4. D.3.51R
Show answer

Answer: D. 3.51R

The molar heat capacity at constant pressure for a gas mixture is the mole-fraction-weighted average of individual Cₚ values. For diatomic N₂ and O₂, Cₚ = 7/2 R = 3.5R, and for CO₂, Cₚ ≈ 4R at moderate temperatures. The calculation yields 0.80×3.5R + 0.18×3.5R + 0.02×4R = 3.51R. Option (a) 3.50R is incorrect because it ignores the slightly higher contribution from CO₂; option (b) 4.00R wrongly assumes all gases have Cₚ of CO₂; option (c) 2.50R mistakenly uses Cᵥ values for diatomic gases.

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