NEET Previous Year Questions (2014-2025): Thermodynamics | Chemistry 31 Years NEET Chapterwise Solved Papers PDF Download

2024

Q1: Match List I with List II.            (NEET 2024)

Choose the correct answer from the options given below:
(a) A-IV, B-II, C-III, D-I
(b) A-I, B-II, C-III, D-IV
(c) A-II, B-III, C-IV, D-I
(d) A-IV, B-III, C-II, D-I
Ans: (c)
- Isothermal process occurs at constant temperature.
- Isochoric process occurs at constant volume.
- Isobaric process occurs at constant pressure.
- Adiabatic process occurs with no heat exchange. 
Conclusion: The correct option is (c) A-II, B-III, C-IV, D-I

Q2: In which of the following processes entropy increases?        (NEET 2024)
(a) A, B and D
(b) A, C and D
(c) C and D
(d) A and C
Ans: (b)
Entropy (ΔS) is a measure of disorder or randomness in a system. Entropy increases when there is an increase in disorder.
A. A liquid evaporates to vapour.
When a liquid evaporates, it changes to a gaseous state, which has higher disorder than a liquid. Therefore, entropy increases. (ΔS > 0)
B. Temperature of a crystalline solid lowered from 130 K to 0 K.
As the temperature of a crystalline solid is lowered, the particles become more ordered, reducing entropy. (ΔS < 0)
C. 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)
In this reaction, a solid reactant decomposes into a solid and gaseous products. Gases have much higher disorder than solids, so entropy increases. (ΔS > 0)
D. Cl₂(g) → 2Cl(g)
A diatomic molecule dissociates into two individual gaseous atoms, which increases the number of particles and disorder. Therefore, entropy increases. (ΔS > 0)
Conclusion: The processes where entropy increases are A, C, and D

Q3: 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 = 2.0 cal K^–1 mol^–1) (NEET 2024)
(a) –413.14 calories
(b) 413.14 calories
(c) 100 calories
(d) 0 calorie
Ans: (a)
The work done in a reversible isothermal process is given by:
W = -nRT ln(V_f / V_i)
Where:

• W is the work done by the gas.
• n is the number of moles of the gas.
• R is the universal gas constant.
• T is the temperature in Kelvin.
• V_i and V_f are the initial and final volumes of the gas.

Then, the work done can be written as:

W = -nRT ln(P_i / P_f)

Where:
Pi and Pf are the initial and final pressures, respectively.
Given:

• n = 1 (one mole of hydrogen)
• T = 25°C = 298.15 K
• R = 2.0 cal K⁻¹ mol⁻¹
• P_i = 20 atm
• P_f = 10 atm

Substituting the values into the formula, we get:
W = -1 × 2.0 × 298.15 × ln(20/10)
W = -2.0 × 298.15 × ln(2)
W ≈ -2.0 × 298.15 × 0.693
W ≈ -413.14 calories
Thus, the work done during the process is -413.14 calories, indicating that energy is done by the system (expansion work done by the gas against external pressure).
The correct option is Option A: -413.14 calories.

Q4: Enthalpy of combustion of carbon to carbon dioxide is −390.0kJ mol⁻¹. The amount of heat released when 35.0 g of CO₂ is formed from the reaction of carbon and dioxygen gas, is: (NEET 2024)
(a) 310 kJ
(b) 490 kJ
(c) 245 kJ
(d) 700 kJ
Ans: (a)
The enthalpy of combustion of carbon to carbon dioxide is given as −390.0 kJ/mol. This means that for every mole of carbon burned, 390 kJ of energy is released.
Now, we are given that 35.0 g of CO₂ is formed. First, we need to calculate the number of moles of CO₂ produced.
The molar mass of CO₂ is:

• Carbon (C) = 12 g/mol
• Oxygen (O) = 16 g/mol × 2 = 32 g/mol
• Molar mass of CO₂ = 12 + 32 = 44 g/mol

Now, calculate the number of moles of CO₂ in 35.0 g:
Moles of CO₂ = 35.0 g ÷ 44 g/mol = 0.795 mol
Since the enthalpy of combustion is −390.0 kJ/mol, the heat released for 0.795 mol of CO₂ is:
Heat released = 0.795 mol × 390 kJ/mol = 310.05 kJ
So, the correct answer is: (a) 310 kJ

Q5: Choose the correct statement for the work done in the expansion and heat absorbed or released when 5 liters of an ideal gas at 10 atmospheric pressure isothermally expands into a vacuum until the volume is 15 liters:        (NEET 2024)
(a) Both the heat and work done will be greater than zero.
(b) Heat absorbed will be less than zero and work done will be positive.
(c) Work done will be zero and heat absorbed or evolved will also be zero.
(d) Work done will be greater than zero and heat absorbed will remain zero.
Ans: (c)
The scenario described involves an isothermal expansion of an ideal gas into a vacuum.
For an isothermal process, the temperature remains constant throughout the process.
Key points:

• The gas expands into a vacuum, meaning no external pressure is exerted on the gas (it's a free expansion).
• In an ideal gas expansion into a vacuum, there is no work done because there is no opposing force (no pressure to do work against).

Now, let’s analyze the options:

• Work done: In the case of an isothermal expansion into a vacuum, since the gas is expanding freely and no external pressure is applied, no work is done by the gas.
• Heat absorbed or released: Since no work is done, the energy is not used for work, and because the process is isothermal, there must be no temperature change. As a result, no heat is exchanged either.

The correct statement is: (c) Work done will be zero and heat absorbed or evolved will also be zero.
This is because, in a free expansion (like into a vacuum), the gas doesn't perform any work, and no heat exchange is required to maintain the constant temperature.

Q6: For an endothermic reaction:        (NEET 2024)
A. q_p (heat at constant pressure) is negative.
B. Δ_rH (enthalpy change of the reaction) is positive.
C. Δ_rH is negative.
D. q_p is positive.
Which of the following combinations is correct?
(a) B and D
(b) C and D
(c) A and B
(d) A and C
Ans: (a)
For an endothermic reaction, the system absorbs heat, so the heat at constant pressure (q_p) is positive.
The enthalpy change (Δ_rH) for an endothermic reaction is positive because the system gains energy from its surroundings.
OR
For endothermic reactions, Δ_rH = positive (Heat is absorbed)
△H = E_a(f) − E_a(b)

q_p = +ve (positive)

Q7: For the following reaction at 300K:        (NEET 2024)
A₂(g) + 3 B₂(g) → 2 AB₃(g)
The enthalpy change is +15 kJ, then the internal energy change is:
(a) 19.98 kJ
(b) 200 J
(c) 1999 J
(d) 1.9988 kJ
Ans: (a)
Reaction:A₂(g) + 3 B₂(g) → 2 AB₃(g)
Change in the number of moles of gas (Δn) : Δn_(g) = n_(P) − n_(R)
Δn = 2 − 3 − 1 = −2
Enthalpy-Internal Energy Relation: ΔH = ΔU + Δn_gRT
Given: ΔH = 15 kJ = 15 × 1000 J (since 1 kJ = 1000 J)
Δn = -2
R = 8.314 J/mol·K
T = 300 K
Substitute the values into the equation:
15 × 1000 = ΔU - 2 × 8.314 × 300
ΔU = 15,000 + 600 × 8.314
ΔU = 15,000 + 4984.4
ΔU = 19,988.4 J
So, the internal energy change (ΔU) is approximately 19.98 kJ.
Correct option is (a) 19.98 kJ.

2023

Q1: Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?    (NEET 2023)
A: △H = △U – △n_gRT
B: △H = △U + △n_gRT
C: △H – △U = –△nRT
D: △H + △U = △nR
Ans: B
Correct relation between change in enthalpy and change in internal energy is
△H = △U + △n_gRT 

Q2: Consider the following reaction:   (NEET 2023)
2H₂(g) + O₂(g) → 2H₂O(g)Δ_rH° = -483.64 kJ.
What is the enthalpy change for the decomposition of one mole of water?
(a) 120.9 kJ
(b) 241.82 kJ
(c) 18 kJ
(d) 100 kJ
Ans: (b)
The given reaction is:
2H₂(g) + O₂(g) → 2H₂O(g), Δ_rH° = -483.64 kJ
This means that the enthalpy change for the formation of 2 moles of water from hydrogen and oxygen is -483.64 kJ.
To find the enthalpy change for the decomposition of 1 mole of water, we need to divide the total enthalpy change by 2, as the given reaction involves the formation of 2 moles of water.
Thus, for the decomposition of 1 mole of water:

$halpy\; change$Therefore, the enthalpy change for the decomposition of one mole of water is 241.82 kJ.
Correct Answer: (b) 241.82 kJ.

2022

Q1: Which of the following P-V curve represents maximum work done?       (NEET 2022)
(a)
(b)
(c)
(d)

Ans: d
The area covered under P-V graph is more.

Q2: The work done when 1 mole of gas expands reversibly and isothermally from a pressure of 5 atm to 1 atm at 300 K is:
(Given log 5 = 0.6989 and R = 8.314 J K⁻¹ mol⁻¹)     (NEET 2022)
(a) zero J
(b) 150 J
(c) +4014.6 J
(d) -4014.6 J
Ans: (d)
Formula:W = - 2.303 nRT log(P₁ / P₂)
Where:

• n = number of moles = 1 mol
• R = 8.314 J/mol·K
• T = 300 K
• P₁ = 5 atm
• P₂ = 1 atm

Substituting values:
W = - 2.303 × 1 mol × 8.314 J/mol·K × 300 K × log(5/1)
W = - 2.303 × 8.314 × 300 × log(5)
W = - 2.303 × 8.314 × 300 × 0.6989
W = - 4014.58 J

Thus, the work done is -4014.58 J, and the negative sign indicates that the work is done by the gas (expansion work).
Correct Answer is (d) -4014.6 J

2019

Q1: Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25 L against a constant external pressure of 2 bar. The work done by the gas is (Given that 1 L bar = 100 J)          (NEET 2019)
(a) -30 J
(b) 5 kJ
(c) 25 J
(d) 30 J
Ans: a
∴ W_irr = - P_ext ΔV
= - 2 bar × (0.25 – 0.1) L
= - 2 × 0.15
= - 0.30 L bar (1 L bar = 100J)
= - 0.30 × 100 J
= - 30 J

Q2: In which case change in entropy is negative?    (2019)
(a) Evaporation of water
(b) Expansion of a gas at constant temperature
(c) Sublimation of solid to gas
(d) 2H(g) → H₂(g)
Ans: d
•
• Expansion of gas at constant temperature,  ΔS > 0
• Sublimation of solid to gas, ΔS > 0
• 2H(g) → H₂(g), ΔS < 0 (∵ Δn_g < 0)

2018

Q1: 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 X₂ will be (NEET 2018)
(a) 200 kJ mol^-1
(b) 100 kJ mol^-1
(c) 800 kJ mol^-1
(d) 400 kJ mol^-1
Ans: c

2017

Q1: For a given reaction, ΔH = 35.5 kJ mol^-1 and ΔS = 83.6 JK^-1mol^-1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with tempearature) (NEET 2017)
(a) T > 425 K
(b) T > 298 K
(c) T < 425 K
(d) All temperatures
Ans: a


Q2: A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be:-       (NEET 2017)
(a) -500J
(b) -505J
(c) +505J
(d) 1136.25J
Ans: b

2016

Q1: The correct thermodynamic conditions for the spontaneous reaction at all temperatures is:    (NEET 2016)
(a) ΔH < 0 and ΔS < 0
(b) ΔH < 0 and ΔS = 0
(c) ΔH > 0 and ΔS < 0
(d) ΔH < 0 and ΔS > 0
Ans: d
ΔG = ΔH - TΔS
For spontaneous process (ΔG = -ve) at all temperature, ΔH < 0 & ΔS > 0.

Q2: Consider the following liquid-vapour equilibrium.
Liquid ⇌ Vapour
Which of the following relations is correct ?     (NEET 2016)
(a)
(b)
(c)
(d)
Ans: a
The given phase equilibria is
Liquid ⇌ Vapour
This equilibrium states that, when liquid is heated, it converts into vapour but on cooling, it further converts into liquid, which is derived by Clausius clapeyron and the relationship is written as,

2016

Q1: Which of the following statements is correct for a reversible process in a state of equilibrium?    (NEET 2015)
(a) ΔG° = 2.30 RT log K
(b) ΔG = -2.30 RT log K
(c) ΔG = 2.30 RT log K
(d) ΔG° = -2.30 RT log K
Ans: d
ΔG=ΔG° + 2.303 RT log K
0 = ΔG° + 2.303 RT log K
ΔG° = -2.303 RT log K

2014

Q1: For the reaction: X₂O₄ (l) → 2XO₂ (g)

Hence, ΔG is :        (NEET 2014)
(a) 9.3 k cal
(b) -9.3 k cal
(c) 2.7 k cal
(d) -2.7 k cal
Ans: d