Thermodynamics NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Thermodynamics question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Thermodynamics previous year questions
38 questions. Tap an option to check your answer.
Q1NEET 2021
Which one among the following is the correct option for right relationship between Cp and Cv for one mole of ideal gas ?
Not quite — the answer is B.
For one mole of an ideal gas, C_p exceeds C_v by the work done in expansion at constant pressure, which is R per kelvin. So Mayer's relation is C_p − C_v = R; a product relation like C_p = RC_v is dimensionally inconsistent.
Q2NEET 2021
For irreversible expansion of an ideal gas under isothermal condition, the correct option is :
Not quite — the answer is C.
For an ideal gas, internal energy depends only on temperature, so ΔU = 0 in any isothermal process. An irreversible process always increases the total entropy of system + surroundings, so ΔS(total) > 0. ΔS(total) = 0 holds only for a reversible process.
Q3NEET 2020 · Q94
Hydrolysis of sucrose is given by the following reaction. Sucrose + H₂O ⇌ Glucose + Fructose. If the equilibrium constant (K꜀) is 2×10¹³ at 300 K, the value of ΔrG° at the same temperature will be :
Not quite — the answer is A.
ΔrG° = −RT ln K, so with K = 2×10¹³ at 300 K, ΔrG° = −8.314 × 300 × ln(2×10¹³) J mol⁻¹, a large negative value as expected for K ≫ 1. The same expression without the minus sign would make a reaction with a huge K non-spontaneous, which is wrong.
Q4NEET 2020 · Q104
The correct option for free expansion of an ideal gas under adiabatic condition is :
Not quite — the answer is A.
In free expansion the gas expands into vacuum, so no work is done (w = 0), and the adiabatic condition gives q = 0. Then ΔU = 0, and since the internal energy of an ideal gas depends only on temperature, ΔT = 0. A temperature drop would need work done by the gas against an external pressure, which is absent here.
Q5NEET 2020 · Q127
For the reaction, 2Cl(g) → Cl₂(g), the correct option is :
Not quite — the answer is D.
Forming a Cl–Cl bond releases energy, so ΔrH < 0, and two gaseous particles become one, so disorder falls and ΔrS < 0. A positive enthalpy change would correspond to bond breaking, the reverse process.
Q6NEET 2019 · Q86
In which case change in entropy is negative?
Not quite — the answer is D.
Combining two gaseous atoms into one molecule reduces the number of particles and hence disorder, so ΔS < 0. Evaporation, sublimation and isothermal expansion all increase entropy.
Q7NEET 2018 · Q85
The bond dissociation energies of X₂, Y₂ and XY are in the ratio of 1 : 0.5 : 1. ΔH for the formation of XY is −200 kJ mol⁻¹. The bond dissociation energy of X₂ will be
Not quite — the answer is A.
Let BE(X₂) = a, BE(Y₂) = 0.5a, BE(XY) = a. For ½X₂ + ½Y₂ → XY: ΔH = ½a + ¼a − a = −a/4 = −200, so a = 800 kJ mol⁻¹. Forgetting the ½ coefficients (forming 2XY) gives 400 kJ mol⁻¹.
Q8NEET 2017 · Q24
For a given reaction, ΔH = 35.5 kJ mol⁻¹ and ΔS = 83.6 J K⁻¹ mol⁻¹. The reaction is spontaneous at: (Assume that ΔH and ΔS do not vary with temperature)
Not quite — the answer is B.
ΔG = ΔH − TΔS is negative when T > ΔH/ΔS = 35500 / 83.6 ≈ 425 K. Both ΔH and ΔS are positive, so the reaction is non-spontaneous at low temperature and becomes spontaneous only above 425 K; it is not spontaneous at all temperatures.
Q9NEET 2017 (Gujarati paper) · Q12
Of the following, the largest value of entropy at 25°C and 1 atm is that of:
Not quite — the answer is B.
Standard molar entropy rises with molar mass and the number of atoms (more vibrational and rotational freedom). C₂H₆ (about 229 J K⁻¹ mol⁻¹) exceeds C₂H₂ (about 201), CH₄ (about 186) and H₂ (about 131).
Q10NEET 2017 (Gujarati paper) · Q31
Under isothermal and reversible conditions, the term "free energy" in thermodynamics signifies:
Not quite — the answer is A.
At constant temperature and pressure for a reversible change, −ΔG equals the maximum useful (non-expansion) work the system can do, such as electrical work in a cell. Expansion (PV) work is already accounted for in the enthalpy term and is not what ΔG measures.
Q11NEET 2016 Phase 2 · Q52
For a sample of perfect gas when its pressure is changed isothermally from pᵢ to p_f, the entropy change is given by
Not quite — the answer is B.
For an isothermal reversible change ΔS = q_rev/T = nR ln(V_f/Vᵢ), and at constant T, V_f/Vᵢ = pᵢ/p_f, so ΔS = nR ln(pᵢ/p_f). Expressions containing T have units of energy, not entropy, and the ratio p_f/pᵢ has the wrong sign.
Q12NEET 2016 Phase 1 · Q64
The correct thermodynamic conditions for the spontaneous reaction at all temperatures is:
Not quite — the answer is C.
ΔG = ΔH − TΔS. With ΔH < 0 and ΔS > 0, both terms make ΔG negative at every temperature. If ΔS < 0, the −TΔS term becomes positive and the reaction stops being spontaneous at high temperature.
Q13AIPMT 2014
For the reaction: X₂O₄(l) → 2XO₂(g), ΔU = 2.1 kcal, ΔS = 20 cal K⁻¹ at 300 K. Hence, ΔG is:
Not quite — the answer is A.
Δn(g) = 2, so ΔH = ΔU + ΔnRT = 2.1 + 2 × 1.987 × 10⁻³ × 300 ≈ 3.3 kcal. ΔG = ΔH − TΔS = 3.3 − 300 × 0.020 = −2.7 kcal; using ΔU directly would give −3.9 kcal.
Q14AIPMT 2012 Mains
Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats (Cp/Cv) of the mixture will be
Not quite — the answer is D.
For any monoatomic ideal gas Cv = (3/2)R and Cp = (5/2)R, so a mixture of two monoatomic gases still has Cp/Cv = 5/3 ≈ 1.67. 1.50 is not the value for any simple ideal gas, and values below 1 are impossible.
Q15AIPMT 2012 Prelims
The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is
Not quite — the answer is D.
ΔS_fus = ΔH_fus/T = 1435 cal mol⁻¹/273 K = 5.26 cal mol⁻¹ K⁻¹. 0.526 results from a misplaced decimal and 10.52 from doubling.
Q16AIPMT 2012 Prelims
In which of the following reactions, standard reaction entropy change (ΔS°) is positive and standard Gibbs energy change (ΔG°) decreases sharply with increasing temperature?
Not quite — the answer is B.
In C(graphite) + ½O₂(g) → CO(g), gaseous moles increase from ½ to 1, so ΔS° is clearly positive and ΔG° = ΔH° − TΔS° falls steeply as T rises. Forming CO₂ from C keeps gas moles unchanged (ΔS° ≈ 0), while CO → CO₂ and Mg → MgO decrease gas moles (ΔS° < 0).
Q17AIPMT 2012 Prelims
Standard enthalpy of vapourisation ΔvapH° for water at 100°C is 40.66 kJ mol⁻¹. The internal energy of vapourisation of water at 100°C (in kJ mol⁻¹) is
Not quite — the answer is B.
For H₂O(l) → H₂O(g), Δn_g = 1. ΔU = ΔH − Δn_gRT = 40.66 − (8.314 × 10⁻³ × 373) = 40.66 − 3.10 = 37.56 kJ mol⁻¹. 43.76 comes from adding RT instead of subtracting it.
Q18AIPMT 2011 Mains
Consider the following processes: (1/2)A → B; ΔH = +150 kJ/mol 3B → 2C + D; ΔH = −125 kJ/mol E + A → 2D; ΔH = +350 kJ/mol For B + D → E + 2C, ΔH will be
Not quite — the answer is D.
Double the first equation (A → 2B, +300), add the second (3B → 2C + D, −125) and reverse the third (2D → E + A, −350). Adding gives B + D → E + 2C with ΔH = 300 − 125 − 350 = −175 kJ/mol. Forgetting to double the first equation leads to −325 kJ/mol.
Q19AIPMT 2011 Prelims
Which of the following is correct option for free expansion of an ideal gas under adiabatic condition?
Not quite — the answer is D.
In free expansion into vacuum, no work is done (w = 0) and in adiabatic conditions q = 0. Thus ΔU = 0, and for an ideal gas U depends only on T, so ΔT = 0 as well. A temperature drop would occur only for a real gas or reversible adiabatic expansion.
Q20AIPMT 2011 Prelims
If the enthalpy change for the transition of liquid water to steam is 30 kJ mol⁻¹ at 27°C, the entropy change for the process would be
Not quite — the answer is A.
At the transition temperature ΔS = ΔH/T = 30000 J mol⁻¹/300 K = 100 J mol⁻¹ K⁻¹. Forgetting to convert kJ to J gives 0.1, the trap.
Q21AIPMT 2011 Prelims
Enthalpy change for the reaction, 4H(g) → 2H₂(g) is −869.6 kJ. The dissociation energy of H–H bond is
Not quite — the answer is D.
4H(g) → 2H₂(g) forms two H–H bonds and releases 869.6 kJ. Breaking one mole of H–H bonds therefore needs 869.6/2 = +434.8 kJ. Dissociation energy is always positive, so the negative values are wrong.
Q22AIPMT 2010 Mains · Q98
Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be
Not quite — the answer is D.
Work in expansion is w = −p(ext)ΔV, and for free expansion into vacuum p(ext) = 0, so no work is done whatever the number of moles. Values like 3 J or 9 J wrongly link the work to the amount of gas.
Q23AIPMT 2010 Mains · Q103
For vaporization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63 kJ mol⁻¹ and 108.8 J K⁻¹ mol⁻¹, respectively. The temperature when Gibbs energy change (ΔG) for this transformation will be zero, is
Not quite — the answer is C.
ΔG = ΔH − TΔS = 0 gives T = ΔH/ΔS = 40630/108.8 = 373.4 K. Forgetting to convert kJ to J (or mixing units) gives nonsensical values; 373 K is also the normal boiling point of water, as expected.
Q24AIPMT 2010 Mains · Q117
The following two reactions are known: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g); ΔH = −26.8 kJ FeO(s) + CO(g) → Fe(s) + CO₂(g); ΔH = −16.5 kJ The value of ΔH for the following reaction Fe₂O₃(s) + CO(g) → 2FeO(s) + CO₂(g) is
Not quite — the answer is D.
Target = (first equation) − 2 × (second equation): Fe₂O₃ + 3CO − 2FeO − 2CO → 2Fe + 3CO₂ − 2Fe − 2CO₂, i.e. Fe₂O₃ + CO → 2FeO + CO₂. ΔH = −26.8 − 2(−16.5) = −26.8 + 33.0 = +6.2 kJ. −10.3 kJ comes from subtracting the second equation only once.
Q25AIPMT 2010 Prelims · Q59
Standard entropies of X₂, Y₂ and XY₃ are 60, 40 and 50 J K⁻¹ mol⁻¹ respectively. For the reaction (1/2)X₂ + (3/2)Y₂ ⇌ XY₃, ΔH = −30 kJ to be at equilibrium, the temperature should be
Not quite — the answer is B.
ΔS° = 50 − (½ × 60 + 3/2 × 40) = 50 − 90 = −40 J K⁻¹ mol⁻¹. At equilibrium ΔG = 0, so T = ΔH/ΔS = (−30000)/(−40) = 750 K. Forgetting the stoichiometric halves gives a different ΔS and a wrong T.
Q26AIPMT 2009 Prelims · Q176
The values of ΔH and ΔS for the reaction, C(graphite) + CO₂(g) → 2CO(g) are 170 kJ and 170 J K⁻¹, respectively. This reaction will be spontaneous at:
Not quite — the answer is B.
ΔG = ΔH − TΔS becomes negative when T > ΔH/ΔS = 170000 J/170 J K⁻¹ = 1000 K. Only 1110 K is above 1000 K. Remember to convert kJ to J before dividing.
Q27AIPMT 2009 Prelims · Q191
From the following bond energies: H–H bond energy: 431.37 kJ mol⁻¹ C=C bond energy: 606.10 kJ mol⁻¹ C–C bond energy: 336.49 kJ mol⁻¹ C–H bond energy: 410.50 kJ mol⁻¹ Enthalpy for the reaction H₂C=CH₂ + H–H → H₃C–CH₃ will be:
Not quite — the answer is B.
ΔH = bonds broken − bonds formed = [BE(C=C) + BE(H–H)] − [BE(C–C) + 2BE(C–H)] = (606.10 + 431.37) − (336.49 + 821.00) = −120.0 kJ mol⁻¹. The four C–H bonds common to both sides cancel out.
Q28AIPMT 2008 Prelims · Q57
Which of the following are not state functions? (I) q + w (II) q (III) w (IV) H − TS
Not quite — the answer is A.
Heat q and work w depend on the path, so they are not state functions. Their sum q + w equals ΔU, and H − TS is the Gibbs energy G; both depend only on the state. So only q and w are not state functions.
Q29AIPMT 2008 Prelims · Q60
Bond dissociation enthalpy of H₂, Cl₂ and HCl are 434, 242 and 431 kJ mol⁻¹ respectively. Enthalpy of formation of HCl is
Not quite — the answer is D.
For ½H₂ + ½Cl₂ → HCl, ΔHf = ½(434) + ½(242) − 431 = 338 − 431 = −93 kJ mol⁻¹. The ±245 values come from using the full bond enthalpies of H₂ and Cl₂ with only one H–Cl bond (434 + 242 − 431 = 245); that equation actually forms 2 mol HCl.
Q30AIPMT 2008 Prelims · Q65
For the gas phase reaction, PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), which of the following conditions is correct?
Not quite — the answer is C.
Dissociation of PCl₅ breaks P–Cl bonds, so it absorbs heat (∆H > 0). One mole of gas gives two moles of gas, so disorder increases (∆S > 0). A negative ∆S would require fewer gas molecules on the product side.
Q31AIPMT 2007 Prelims · Q57
Consider the following reactions: (i) H⁺(aq) + OH⁻(aq) = H₂O(l), ΔH = −X₁ kJ mol⁻¹ (ii) H₂(g) + ½O₂(g) = H₂O(l), ΔH = −X₂ kJ mol⁻¹ (iii) CO₂(g) + H₂(g) = CO(g) + H₂O, ΔH = −X₃ kJ mol⁻¹ (iv) C₂H₂(g) + (5/2)O₂(g) = 2CO₂(g) + H₂O(l), ΔH = −X₄ kJ mol⁻¹ Enthalpy of formation of H₂O(l) is
Not quite — the answer is D.
Enthalpy of formation is for forming 1 mol of the compound from elements in their standard states: H₂(g) + ½O₂(g) → H₂O(l), ΔH = −X₂. The neutralisation enthalpy −X₁ does not start from elements.
Q32AIPMT 2007 Prelims · Q67
Given that bond energies of H–H and Cl–Cl are 430 kJ mol⁻¹ and 240 kJ mol⁻¹ respectively and ΔHf for HCl is −90 kJ mol⁻¹, bond enthalpy of HCl is
Not quite — the answer is B.
½H₂ + ½Cl₂ → HCl: ΔHf = ½(430) + ½(240) − x = −90, so x = 335 + 90 = 425 kJ mol⁻¹. 245 kJ mol⁻¹ subtracts the 90 instead of adding it.
Q33AIPMT 2006 Prelims · Q51
Identify the correct statement for change of Gibbs energy for a system (ΔG_system) at constant temperature and pressure :
Not quite — the answer is B.
At constant temperature and pressure ΔG < 0 means spontaneous, ΔG > 0 means non-spontaneous, and ΔG = 0 means the system is at equilibrium. So ΔG = 0 marks equilibrium, with no net change in any direction. The other statements reverse these signs.
Q34AIPMT 2006 Prelims · Q54
Assume each reaction is carried out in an open container. For which reaction will ΔH = ΔE ?
Not quite — the answer is A.
ΔH = ΔE + Δn_gRT, so ΔH = ΔE only when the number of moles of gas is unchanged. For H₂ + Br₂ → 2HBr, Δn_g = 2 − 2 = 0. The water-gas reaction and the PCl₅ decomposition have Δn_g = +1, and the CO combustion has Δn_g = −1.
Q35AIPMT 2006 Prelims · Q64
The enthalpy and entropy change for the reaction : Br₂(l) + Cl₂(g) → 2BrCl(g) are 30 kJ mol⁻¹ and 105 J K⁻¹ mol⁻¹ respectively. The temperature at which the reaction will be in equilibrium is :
Not quite — the answer is A.
At equilibrium ΔG = ΔH − TΔS = 0, so T = ΔH/ΔS = 30 000 J mol⁻¹/105 J K⁻¹ mol⁻¹ = 285.7 K. Forgetting to convert kJ to J gives a meaningless 0.29 K.
Q36AIPMT 2006 Prelims · Q75
The enthalpy of combustion of H₂, cyclohexene (C₆H₁₀) and cyclohexane (C₆H₁₂) are −241, −3800 and −3920 kJ per mol respectively. Heat of hydrogenation of cyclohexene is :
Not quite — the answer is A.
For C₆H₁₀ + H₂ → C₆H₁₂, ΔH = Σ ΔcH(reactants) − Σ ΔcH(products) = (−3800 − 241) − (−3920) = −121 kJ mol⁻¹. Reversing the subtraction gives +121 kJ mol⁻¹, but hydrogenation of an alkene is exothermic.
Q37AIPMT 2005 Prelims · Q66
A reaction occurs spontaneously if :
Not quite — the answer is B.
ΔG = ΔH − TΔS. When ΔH and ΔS are both positive, ΔG < 0 only when TΔS > ΔH. With ΔH positive and ΔS negative, ΔG is always positive.
Q38AIPMT 2005 Prelims · Q70
Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction ?
Not quite — the answer is C.
With ΔH < 0 (exothermic) and ΔS > 0 (increasing disorder), ΔG = ΔH − TΔS is negative at every temperature. The other combinations are spontaneous only in some temperature range or never.
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