Thermodynamics NEET PYQs (2005–2026) with Solutions

10 solved Thermodynamics PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumSpontaneity & Gibbs Energy

   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. (Given: R = 8.31 J mol⁻¹ K⁻¹)

   1. AOption A: −1.635 kJ mol⁻¹, spontaneous
   2. BOption B: −0.63568 kJ mol⁻¹, spontaneous
   3. COption C: +0.63568 kJ mol⁻¹, non-spontaneous
   4. DOption D: +1.635 kJ mol⁻¹, non-spontaneous

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   Correct answer: (C) +0.63568 kJ mol⁻¹, non-spontaneous

   Explanation

   Δn(g)=2−3=−1. Therefore ΔH°=ΔU°+ΔnRT=−10−(1×8.31×298/1000)=−12.48 kJ mol⁻¹. Then ΔG°=ΔH°−TΔS°=−12.48−[298×(−44)/1000]=+0.632 kJ mol⁻¹≈+0.63568 kJ mol⁻¹. Since ΔG°>0, the reaction is non-spontaneous. Hence option (C) is correct.

2. Question 2 (NEET 2026)

   NEET 2026MediumEntropy change in free expansion

   Two moles of an ideal gas undergo free expansion from 10 L to 100 L at 300 K. The values of $\Delta S_{\mathrm{system}}$ and $\Delta S_{\mathrm{surroundings}}$ are ($R$ is universal gas constant)

   1. AOption A: $\Delta S_{\mathrm{system}} = 0$; $\Delta S_{\mathrm{surroundings}} = 0$
   2. BOption B: $\Delta S_{\mathrm{system}} = 4.606\ R$; $\Delta S_{\mathrm{surroundings}} = -4.606\ R$
   3. COption C: $\Delta S_{\mathrm{system}} = 0$; $\Delta S_{\mathrm{surroundings}} = 4.606\ R$
   4. DOption D: $\Delta S_{\mathrm{system}} = 4.606\ R$; $\Delta S_{\mathrm{surroundings}} = 0$

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   Correct answer: (D) $\Delta S_{\mathrm{system}} = 4.606\ R$; $\Delta S_{\mathrm{surroundings}} = 0$

   Explanation

   In free expansion the gas expands into vacuum ($p_{ex} = 0$), so $w = 0$. For an ideal gas at constant temperature $\Delta U = 0$, so $q = 0$. Entropy is a state function. Its change for the system is the same as for a reversible isothermal path between the same states: $$\Delta S_{\mathrm{system}} = nR\ln\frac{V_2}{V_1} = 2R \times 2.303\log\frac{100}{10} = 4.606\ R$$ No heat passes to or from the surroundings ($q = 0$), so$\Delta S_{\mathrm{surroundings}} = 0$.$\Delta S_{\mathrm{total}} = 4.606\ R > 0$, as expected for a spontaneous, irreversible process. So the answer is option (D).

3. Question 3 (NEET 2026)

   NEET 2026HardWork in reversible isothermal and adiabatic processes

   Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure. $w_1$, $w_2$, $w_3$ and $w_4$ represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; $\Delta U_2$ and $\Delta U_4$ are changes in the internal energy for the processes 2 and 4, respectively. [use $R = 2\ \mathrm{cal\ K^{-1}\ mol^{-1}}$] The correct option is

   1. AOption A: $w_1 + w_3 = -2T_1\ln\frac{V_2}{V_1} - 2T_2\ln\frac{V_4}{V_3}$
   2. BOption B: $w_2 + w_4 = \Delta U_2 - \Delta U_4$
   3. COption C: $w_1 + w_2 = 2T_1\ln\frac{V_2}{V_1}$
   4. DOption D: $w_1 + w_2 + w_3 + w_4 = 0$

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   Correct answer: (A) $w_1 + w_3 = -2T_1\ln\frac{V_2}{V_1} - 2T_2\ln\frac{V_4}{V_3}$

   Explanation

   This is a Carnot-type cycle. Processes 1 ($T_1$, $V_1 \rightarrow V_2$) and 3 ($T_2$, $V_3 \rightarrow V_4$) are reversible isothermal; 2 and 4 are adiabatic. Work done on the system is taken as positive (NCERT convention). Reversible isothermal work: $w = -nRT\ln\frac{V_f}{V_i}$, with $n = 1$ and $R = 2\ \mathrm{cal\ K^{-1}\ mol^{-1}}$: $w_1 = -2T_1\ln\frac{V_2}{V_1}$ and $w_3 = -2T_2\ln\frac{V_4}{V_3}$. Adding them gives option (A) exactly. Option (B) is wrong. For adiabatic steps $q = 0$, so $w = \Delta U$, giving $w_2 + w_4 = \Delta U_2 + \Delta U_4$. This sum is 0 because $\Delta U_2 = C_V(T_2 - T_1) = -\Delta U_4$, whereas $\Delta U_2 - \Delta U_4 = 2\Delta U_2 \neq 0$. Option (C) is wrong. $w_1 + w_2 = -2T_1\ln\frac{V_2}{V_1} + \Delta U_2$: the sign is wrong and the $\Delta U_2$ term is missing. Option (D) is wrong. For the cycle $\Delta U = 0$, so total $w = -q_{\mathrm{total}} = -[2T_1\ln\frac{V_2}{V_1} + 2T_2\ln\frac{V_4}{V_3}]$, which is not zero because $T_1 \neq T_2$. So the answer is option (A).

4. Question 4 (NEET 2026)

   NEET 2026MediumGibbs energy and equilibrium constant

   A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to $\mathrm{N \rightleftharpoons D}$. At $60^\circ\mathrm{C}$, the concentrations of both N and D are equal at equilibrium, and the standard enthalpy change of denaturation is $666\ \mathrm{kJ\ mol^{-1}}$. The standard entropy change ($\Delta S^\circ$ in $\mathrm{kJ\ K^{-1}\ mol^{-1}}$) of the protein upon denaturation at $60^\circ\mathrm{C}$ is closest to

   1. AOption A: $2.0$
   2. BOption B: $2000.0$
   3. COption C: $333.0$
   4. DOption D: $11.1$

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   Correct answer: (A) $2.0$

   Explanation

   At equilibrium $[\mathrm{N}] = [\mathrm{D}]$, so $K = \frac{[\mathrm{D}]}{[\mathrm{N}]} = 1$. $\Delta G^\circ = -RT\ln K = -RT\ln 1 = 0$. Since $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0$, we get $\Delta S^\circ = \frac{\Delta H^\circ}{T}$. $T = 60 + 273 = 333\ \mathrm{K}$, so $\Delta S^\circ = \frac{666\ \mathrm{kJ\ mol^{-1}}}{333\ \mathrm{K}} = 2.0\ \mathrm{kJ\ K^{-1}\ mol^{-1}}$, option (A).

5. Question 5 (NEET 2025)

   NEET 2025MediumCalorimetry & Enthalpy

   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]

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (A) Option (1)

   Explanation

   The reaction has ΔH = -74.8 kJ mol⁻¹, which means it is exothermic. Therefore, the products must lie at a lower energy level than the reactants by 74.8 kJ mol⁻¹. Option (A) correctly represents this energy profile.

6. Question 6 (NEET 2025)

   NEET 2025HardHess Law & Reaction Enthalpies

   The standard heat of formation, in kcal/mol, of $\mathrm{Ba^{2+}}$ is: Given: Standard heat of formation of $\mathrm{SO_4^{2-}(aq)} = -216\ \text{kcal/mol}$ Standard heat of crystallisation of $\mathrm{BaSO_4(s)} = -4.5\ \text{kcal/mol}$ Standard heat of formation of $\mathrm{BaSO_4(s)} = -349\ \text{kcal/mol}$

   1. AOption A: $-128.5$
   2. BOption B: $-133.0$
   3. COption C: $+133.0$
   4. DOption D: $+220.5$

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   Correct answer: (A) $-128.5$

   Explanation

   Using Hess’s law: For crystallisation: $$\mathrm{Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)}$$ $$\Delta H = -4.5\ \text{kcal/mol}$$ Applying enthalpy relation: $$\Delta H = \Delta H_f^\circ[\mathrm{BaSO_4(s)}] - \left(\Delta H_f^\circ[\mathrm{Ba^{2+}}] + \Delta H_f^\circ[\mathrm{SO_4^{2-}}]\right)$$ Substituting values: $$-4.5 = -349 - \left(x + (-216)\right)$$ $$-4.5 = -349 - x + 216$$ $$-4.5 = -133 - x$$ $$x = -128.5\ \text{kcal/mol}$$ Hence, the correct answer is option (A).

7. Question 7 (NEET 2024)

   NEET 2024EasyThermodynamic Terms & Processes

   Match List I with List II. Choose the correct answer from the options given below:

   1. AOption A: A-I, B-II, C-III, D-IV
   2. BOption B: A-II, B-III, C-IV, D-I
   3. COption C: A-IV, B-III, C-II, D-I
   4. DOption D: A-IV, B-II, C-III, D-I

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   Correct answer: (B) A-II, B-III, C-IV, D-I

   Explanation

   Isothermal process → constant temperature. Isochoric process → constant volume. Isobaric process → constant pressure. Adiabatic process → no heat exchange. Thus: A-II, B-III, C-IV, D-I Hence, option (B) is correct.

8. Question 8 (NEET 2024)

   NEET 2024MediumEntropy

   In which of the following processes entropy increases? A. A liquid evaporates to vapour. B. Temperature of a crystalline solid lowered from 130 K to 0 K. C. 2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g) D. Cl₂(g) → 2 Cl(g) Choose the correct answer from the options given below:

   1. AOption A: A, C and D
   2. BOption B: C and D
   3. COption C: A and C
   4. DOption D: A, B and D

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   Correct answer: (A) A, C and D

   Explanation

   Entropy increases when randomness increases. A. Liquid to vapour increases randomness → entropy increases. B. Lowering temperature decreases entropy. C. Formation of gaseous products increases entropy. D. Dissociation of Cl₂ into atoms increases number of particles and randomness. Thus entropy increases in A, C and D. Hence, option (A) is correct.

9. Question 9 (NEET 2024)

   NEET 2024MediumThermodynamic Terms & Processes

   The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is: (Given: R = 2 cal K⁻¹ mol⁻¹)

   1. AOption A: 413.14 calories
   2. BOption B: 100 calories
   3. COption C: 0 calorie
   4. DOption D: −413.14 calories

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   Correct answer: (D) −413.14 calories

   Explanation

   For reversible isothermal expansion: $$w = -nRT \ln\frac{V_2}{V_1}$$ Since: $$\frac{V_2}{V_1} = \frac{P_1}{P_2} = \frac{20}{10}=2$$ Substituting values: $$w = -(1)(2)(298)\ln 2$$ Using: $$\ln 2 = 0.693$$ $$w = -412.8\,\text{calories} \approx -413.14\,\text{calories}$$ Hence, option (D) is correct.

10. Question 10 (NEET 2021)

    NEET 2021MediumEntropy and Irreversible Processes

    For irreversible expansion of an ideal gas under isothermal condition, the correct option is:

    1. AOption A: $\Delta U \neq 0,\ \Delta S_{\text{total}} = 0$
    2. BOption B: $\Delta U = 0,\ \Delta S_{\text{total}} = 0$
    3. COption C: $\Delta U \neq 0,\ \Delta S_{\text{total}} \neq 0$
    4. DOption D: $\Delta U = 0,\ \Delta S_{\text{total}} \neq 0$

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    Correct answer: (D) $\Delta U = 0,\ \Delta S_{\text{total}} \neq 0$

    Explanation

    For an ideal gas, internal energy depends only on temperature. Under isothermal conditions: $T = \text{constant} \Rightarrow \Delta U = 0$ For an irreversible process, total entropy always increases: $\Delta S_{\text{total}} > 0$ Hence, $\Delta S_{\text{total}} \neq 0$ Therefore, the correct option is **(4)**.