NEET Thermodynamics PYQs & Practice Questions

10 free NEET Chemistry questions on Thermodynamics, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Spontaneity · Basic Concepts and State Functions · Enthalpy and Hess's Law · First Law of Thermodynamics · Entropy and Spontaneity · Relationship between ΔH and ΔU · Energy Profile Diagrams - Exothermic Reaction · Scope of Thermodynamics · Types of Systems · State Variables · Specific Heat as Intensive Property · Enthalpy of Combustion

1. Q1NEET 2026Entropy Change at Equilibrium (Protein Denaturation)

   A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to N ⇌ D. At 60 °C, the concentrations of both N and D are equal at equilibrium, and the standard enthalpy change of denaturation is 666 kJ mol⁻¹. The standard entropy change (ΔS° in kJ K⁻¹mol⁻¹) of the protein upon denaturation at 60 °C is closest to:

   • A. 2000.0
   • B. 333.0
   • C. 11.1
   • D. 2.0
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   Answer: C — 11.1

   Since [N] = [D] at equilibrium, K = 1, so ΔG° = -RT ln(1) = 0. Using ΔG° = ΔH° - TΔS° = 0, we get ΔS° = ΔH°/T = 666 kJ mol⁻¹ / 333 K = 2.0 kJ K⁻¹ mol⁻¹ (T = 60°C = 333 K).

2. Q2NEET 2026Carnot Cycle - Work in Isothermal and Adiabatic Processes

   Consider the reversible processes for 1.0 mol of an ideal gas shown in a p-V diagram forming a closed cycle: (p1,V1,T1) --process 1 (isothermal, T1)--> (p2,V2,T1) --process 2 (adiabatic)--> (p3,V3,T2) --process 3 (isothermal, T2)--> (p4,V4,T2) --process 4 (adiabatic)--> back to (p1,V1,T1). w1, w2, w3 and w4 represent the work done (in calories) in processes 1, 2, 3 and 4 respectively (work done on the gas convention); ΔU2 and ΔU4 are the changes in internal energy for processes 2 and 4, respectively. [Use R = 2 cal K⁻¹ mol⁻¹] Which relation is correct?

   • A. w2 + w4 = ΔU2 − ΔU4
   • B. w1 + w2 = 2T1 ln(V2/V1)
   • C. w1 + w2 + w3 + w4 = 0
   • D. w1 + w3 = −2T1 ln(V2/V1) − 2T2 ln(V4/V3)
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   Answer: C — w1 + w2 + w3 + w4 = 0

   This is a Carnot-type cycle: processes 1 and 3 are isothermal (ΔU=0 for each, since T is constant for an ideal gas), while processes 2 and 4 are adiabatic. For an isothermal process, the work done by the gas is nRT ln(Vf/Vi); using the work-done-on-the-gas sign convention, w1 = -2T1 ln(V2/V1) (process 1, expansion at T1) and w3 = -2T2 ln(V4/V3) (process 3, compression at T2). Adding these isothermal work terms gives w1 + w3 = -2T1 ln(V2/V1) - 2T2 ln(V4/V3), consistent with option (4). (The adiabatic legs do not contribute logarithmic PV work terms of this form, since q=0 there and w2=-ΔU2, w4=-ΔU4.)

3. Q3NEET 2026Entropy Change in Free Expansion of an Ideal Gas

   Two moles of an ideal gas undergo free expansion from 10 L to 100 L at 300 K. The values of ΔS_system and ΔS_surroundings are: (R is universal gas constant)

   • A. ΔS_system = 4.606 R; ΔS_surroundings = −4.606 R
   • B. ΔS_system = 0; ΔS_surroundings = 4.606 R
   • C. ΔS_system = 4.606 R; ΔS_surroundings = 0
   • D. ΔS_system = 0; ΔS_surroundings = 0
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   Answer: B — ΔS_system = 0; ΔS_surroundings = 4.606 R

   Free expansion (expansion into vacuum) has q=0 and w=0, so for an ideal gas (isothermal), ΔU=0. The entropy change of the system depends only on initial and final states: ΔS_system = nR ln(V2/V1) = 2R ln(100/10) = 2R ln(10) = 2 × 2.303R = 4.606R. Since free expansion is irreversible with no actual heat exchanged with the surroundings (q_surr=0), ΔS_surroundings = 0.

4. Q4NEET 2026Spontaneity from ΔG°

   Consider the following reaction: 2A(g)+B(g)→2D(g), ΔU°=-10 kJ mol⁻¹ and ΔS°=-44 J K⁻¹ at 298 K. Identify the correct option with ΔG° for the reaction and spontaneity of the reaction at 298 K. (R=8.31 J mol⁻¹K⁻¹)

   • A. +0.63568 kJ mol⁻¹, non-spontaneous
   • B. -0.63568 kJ mol⁻¹, spontaneous
   • C. -1.635 kJ mol⁻¹, spontaneous
   • D. +1.635 kJ mol⁻¹, non-spontaneous
   Show answer

   Answer: —

   ΔH=ΔU+ΔnRT=-10+(-1)(8.31×298/1000)=-12.48 kJ/mol; ΔG=ΔH-TΔS=-12.48-298×(-0.044)=+0.636 kJ/mol, non-spontaneous.

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5. Q5NEET 2026Change in internal energy

   At a certain temperature, T(K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy is:

   • A. 700 J
   • B. 400 J
   • C. 300 J
   • D. 500 J
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   Answer: B — 400 J

   ΔU=q-w=500-200=300 J.

6. Q6NEET 2025Energy Profile Diagrams - Exothermic Reaction

   C(s)+2H2(g)→CH4(g); ΔH=−74.8 kJ mol⁻¹. Which of the following diagrams gives an accurate representation of the above reaction? [R → reactants; P → products]

   • A. R at lower energy than P, with 74.8 gap marked between the peak and the trough above P (R below P, hump between)
   • B. R at lower energy, P at higher energy, 74.8 marked near P at top
   • C. R at a level, P below R by 74.8, hump in between (R higher starting point, correct exothermic drop)
   • D. R at lower energy than P with hump, 74.8 near P
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   Answer: B — R at lower energy, P at higher energy, 74.8 marked near P at top

   Since the reaction is exothermic (ΔH negative), products (P) must be at lower energy than reactants (R), with an activation energy hump in between, and the 74.8 kJ/mol gap representing the energy difference between R and P (reactants higher, products 74.8 kJ/mol lower).

7. Q7NEET 2025Heat of Formation / Hess's Law

   The standard heat of formation, in kcal/mol of Ba2+ is : [Given: standard heat of formation SO4²⁻ ion (aq) = −216 kcal/mol, standard heat of crystallisation of BaSO4(s) = −4.5 kcal/mol, standard heat of formation of BaSO4(s) = −349 kcal/mol]

   • A. −128.5
   • B. −133.0
   • C. +133.0
   • D. +220.5
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   Answer: —

   Ba2+(aq)+SO4²⁻(aq)→BaSO4(s), ΔH(crystallisation) = −4.5 kcal. ΔH(cre.) = ΔHf(BaSO4) − ΔHf(Ba2+) − ΔHf(SO4²⁻). −4.5 = −349 − x − (−216) ⟹ −4.5 = −349 − x + 216 ⟹ x = −128.5 kcal/mol.

8. Q8NEET 2023Relation between Enthalpy and Internal Energy

   Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?

   • A. ΔH = ΔU + Δn(g)RT
   • B. ΔH - ΔU = -ΔnRT
   • C. ΔH + ΔU = ΔnR
   • D. ΔH = ΔU - Δn(g)RT
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   Answer: —

   The relation between enthalpy change and internal energy change for a reaction involving gases at constant temperature and pressure is ΔH = ΔU + Δn(g)RT, where Δn(g) is the change in the number of moles of gaseous species.

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9. Q9NEET 2022Work Done in Isothermal Process

   Which of the following p-V curve represents maximum work done? The options show different isothermal process curves: (1) A smooth isothermal expansion curve from high pressure/low volume to low pressure/high volume with area under curve shown (2) A vertical line (constant volume, no work) (3) A curve showing isothermal with a sharp initial drop then gradual curve (stepwise/irreversible-like with smaller area) (4) An elliptical/cyclic loop curve

   • A. Curve 1: smooth continuous isothermal expansion curve with maximum area under curve
   • B. Curve 2: vertical line representing constant volume process
   • C. Curve 3: curve with sharp initial pressure drop followed by gradual isothermal path, smaller area
   • D. Curve 4: elliptical/cyclic loop curve
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   Answer: —

   Work done in a p-V process equals the area under the curve. Curve (1), representing a smooth reversible isothermal expansion, encloses the maximum area under the curve among the given options, so it represents maximum work done.

10. Q10NEET 2022Work Done in Isothermal Process - PV Diagram

    Which of the following p-V curve represents maximum work done? (Four isothermal curve options: (1) an elliptical/cyclic loop (2) a decreasing curve with larger area under it (3) a vertical line (4) a decreasing curve with smaller area under it, all under an isothermal label)

    • A. Elliptical/cyclic loop curve
    • B. Decreasing curve with larger area under it (larger volume change)
    • C. Vertical line (no volume change)
    • D. Decreasing curve with smaller area under it
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    Answer: A — Elliptical/cyclic loop curve

    Work done in an expansion process equals the area under the P-V curve. The curve showing the largest area under it (largest extent of expansion along V) represents the maximum work done, which is option 2.