NEET Previous Year Questions (2014-2025): Thermodynamics

Table of Contents
1. 2025
2. 2024
3. 2023
4. 2022
5. 2021
6. 2020
7. 2019
8. 2018
9. 2017
10. 2016
11. 2015
12. 2014
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2025

Two gases A and B are filled at the same pressure in separate cylinders with movable pistons of radius , respectively. On supplying an equal amount of heat to both systems reversibly under constant pressure, the pistons of gas A and B are displaced by 16 cm and 9 cm, respectively. If the change in their internal energy is the same, then the ratio   is equal to:

(a)

(b)

(c) 4/3

(d) 3/4

Ans: (d)

Using the First Law of Thermodynamics:

ΔQ = ΔU + PΔV

ΔQ is the same

ΔU is also the same

⇒ W_A = W_B

⇒ (PΔV)_A = (PΔV)_B

P is also the same

⇒ A_Ad_A = A_Bd_B

πr_A²d_A = πr_B²d_B

r_A / r_B = √(d_B / d_A) = √(9 / 16)

= 3 / 4

2024

(a) Zero
(b) 30 J
(c) -90 J
(d) -60 J
Ans: (a)
Path bc is an isochoric process.
∴ Work done by gas along path bc is zero.

Q2: For an adiabatic process, the factor which remains constant is (all the notations have their usual meaning):       [NEET 2024]
(a) P^(γ-1) T^γ
(b) T V^(1-γ)
(c) V^(γ-1) T^γ
(d) P^(1-γ) T^γ
Ans: (d)
In an adiabatic process, the heat exchange Q is zero, meaning no heat is exchanged between the system and its surroundings. For an ideal gas undergoing an adiabatic process, the relationship between pressure P, volume V, and temperature T is governed by the following equations:

• The adiabatic condition is given by:
  P V^γ = constant (where γ = C_P / C_V, the ratio of specific heats).
• Additionally, for an ideal gas undergoing an adiabatic process, we also have the relation:
  T V^(γ-1) = constant or equivalently, P T^γ = constant.

Thus, the factor that remains constant in an adiabatic process is P^(1-γ) T^γ. This is because both pressure and temperature are involved in a relation that stays constant throughout the process for an ideal gas.
Therefore, the correct answer is (d) P^(1-γ) T^γ.

2023

Q2: Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?                [NEET 2023]
(a)

(b)

(c)  

(d)  
Ans: (a)

Q3: The temperature of a gas is -50^∘C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?
(a) 3295^∘C
(b) 3097 K
(c) 223 K
(d) 669^∘C [NEET 2023]
Ans: (a)

Q4: A Carnot engine has an efficiency of 50% when its source is at a temperature 327^∘C. The temperature of the sink is :- [NEET 2023]
(a) 15^∘C
(b) 100^∘C
(c) 200^∘C
(d) 27^∘C [NEET 2023]
Ans: (d)
Efficiency of carnot engine

So temperature of sink is = 300 - 273 = 27^∘C

Q5: For the given cycle, the work done during the isobaric process is:     [NEET 2023](a) 200 J
(b) Zero
(c) 400 J
(d) 600 J
Ans: (d)
To find the work done during the isobaric process, we use the formula:
Work = P × ΔV
Where:

• P is the constant pressure,
• ΔV is the change in volume.

From the graph, the pressure is constant during the isobaric process. The pressure value is 300 N/m²
The change in volume ΔV is still (V_B - V_A), which is:

• V_B = 3 m³
• V_A = 1 m³
• ΔV = (3 - 1) m³ = 2 m³

Now, let's calculate the work done:
Work = (300 N/m²) × (2 m³) = 600 J
So, the correct answer should indeed be (d) 600 J. 

2022

Q1: Which of the following p-V curve represents maximum work done?             [NEET 2022 Phase 1]
(a)

(b)

(c)

(d)

Ans: (b)
Work done under any thermodynamic process can be determined by area under the 'p-V' graph. As it can be observed maximum area is covered in option '2'.

Q2: One mole of an ideal gas at 300 K is expanded isothermally from 1 L to 10 L volume. ΔU for this process is :

(Use R = 8.314 J k^-1 mol^-1) [NEET 2022 Phase 2]
(a) 0 J
(b) 1260 J
(c) 2520 J
(d) 5040 J
Ans: (a)
ΔU = nC_vΔT
For isothermal condition; ΔT = 0
ΔU = 0

Q3: A vessel contains 3.2 g of dioxygen gas at STP (273.15 K and 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes one third of the original pressure. The volume of new vessel in L is : (Given : molar volume at STP is 22.4 L)      [NEET 2022 Phase 2]
(a) 67.2
(b) 6.72
(c) 2.24
(d) 22.4
Ans: (b)
At constant temperature and amount

2021

Q1: Which one among the following is the correct option for right relationship between C_p and C_V for one mole of ideal gas ? [NEET 2021]
(a) C_P = RC_V
(b) C_P  = RC_V
(c) C_P + C_V = R
(d) C_P - C_V = R
Ans: (d)

At constant volume, q_V = C_VΔT = ΔU
At constant pressure, q_P = C_P ΔT = ΔH
For a mole of an ideal gas,
ΔH = ΔU + Δ( PV )
=ΔU+Δ(RT)
=ΔU+RΔT
On putting the values of ΔH and ΔU, we have C_PΔT = C_VΔT+RΔT
C_P = C_V + R
C_P-C_V = R

Q2: For irreversible expansion of an ideal gas under isothermal condition, the correct option is :     [NEET 2021]
(a) ΔU = 0, Δ S_total ≠ 0
(b) ∆U ≠ 0, ∆ S_total = 0
(c) ∆U = 0, ∆ S_total = 0
(d) ∆U ≠ 0, ∆ S_total ≠ 0
Ans: (a)
Hint: ∆U = _nC_V∆T
In the case of isothermal process, ΔT is zero. The value of ΔU is also zero from the relation, ∆U=_nC_V∆T.
Thus, for reversible and irreversible expansion for an ideal gas, under isothermal conditions, ∆U = 0. But the value ∆S for irreversible expansion of an ideal gas under isothermal conditions is not equal to zero.
The total entropy change ( ∆S total) for the system and surroundings of a spontaneous process is given by

2020

Q1: For the reaction, 2Cl(g) → Cl₂(g), the correct option is: [NEET 2020]
(a) Δ_rH < 0 and Δ_rS > 0
(b) Δ_rH < 0 and Δ_rS < 0
(c) Δ_rH > 0 and Δ_rS > 0
(d) Δ_rH > 0 and Δ_rS < 0
Ans: (b)

2Cl(g) → Cl₂(g) + Heat
Due to bond formation stability increases which results in release of heat.
∴ ΔH_r = -ve or exothermic process
ΔH_r < O
& ΔS < O, because number of Cl atoms decreases in the formation of Cl₂(g)

Q2: The correct option for free expansion of an ideal gas under adiabatic condition is :     [NEET 2020]
(a) q = 0, Δ T < 0 and w > 0
(b) q < 0, Δ T = 0 and w = 0
(c) q > 0, Δ T > 0 and w > 0
(d) q = 0, Δ T = 0 and w = 0
Ans: (d)
Free expansion, so p_ex = 0
So, w = - p_exΔ V = 0
Since, adiabatic process, so q = 0
Since both q and w are equal to zero. Then according to first law of thermodynamics Δ E = q + w
Δ U = 0
Hence, Δ T = 0 

2019

Q1: In which case change in entropy is negative?   [NEET 2019]
(a) Sublimation of solid to gas
(b) 2H(g) $\to$ H₂(g)
(c) Evaporation of water
(d) Expansion of a gas at temperature
Ans: (b)
2H(g) → H₂(g)
Due to bond formation, entropy decreases.

Q2: 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) 25 J
(b) 30 J
(c) -30 J
(d) 5 KJ
Ans: (c)
W = -P_ext (V₂-V₁)
P_ext = 2 bar
V₁ = 0.1 L
V₂ = 0.25 L
W = -2 bar[0.25 - 0.1] L
W = -2 × 0.15 bar L
W = -0.30 bar L
W = (-0.30) × 100 = -30 J 

2018

Q1: The bond dissociation energies of X₂ , Y₂ and XY are in the ratio of 1 : 0.5 : 1. $\mathrm{\Delta <br><br>}$H for the formation of XY is -200 kJ mol^-1. The bond dissociation energy of X₂ will be [NEET 2018]
(a) 800 kJ mol^-1
(b) 200 kJ mol^-1
(c) 400 kJ mol^-1
(d) 100 kJ mol^-1
Ans: (a)
Let bond dissociation energies of X₂ , Y₂ and XY are x kJ mol^-1 , 0.5x kJ mol^-1 and x kJ mol^-1 respectively. 

2017

Q1: 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) -500 J
(b) -505 J
(c) +505 J
(d) 1136.25 J
Ans: (b)
w = - P_extΔV = -2.5(4.50 - 2.50)
$\Rightarrow$ - 5 L atm = - 5 $\times$ 1.01.325 J = - 506.625 J
ΔU = q + w
As, the container is insulted, thus q = 0
Hence, ΔU = w = -506.625 J = - 505.5 J ( nearly equals)

Q2: 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.)   [NEET 2017]
(a) T > 425 K
(b) Ball temperatures
(c) T > 298 K
(d) T < 425 K
Ans: (a)
For a spontaneous reaction,

2016

Q1: For a sample of perfect gas when its pressure is changed isothermally from pi to pf, the entropy change is given by    [NEET 2016]
(a)

(b)

(c)

(d)

Ans: (b)
For an ideal gas undergoing reversible expansion, when temperature changes from T_i to T_f and pressure changes from P_i to P_f. 

2015

2014