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 2026EasyWork done in adiabatic process

   In an adiabatic expansion, the temperature of one mole of an ideal monatomic gas ($\gamma = 5/3$) decreases from $60\ \mathrm{K}$ to $50\ \mathrm{K}$. The work done by the gas in the process is: (Take the universal gas constant as $R = 8.3\ \mathrm{J\ mol^{-1}\ K^{-1}}$)

   1. AOption A: $41.5\ \mathrm{J}$
   2. BOption B: $83\ \mathrm{J}$
   3. COption C: $124.5\ \mathrm{J}$
   4. DOption D: $166\ \mathrm{J}$

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   Correct answer: (C) $124.5\ \mathrm{J}$

   Explanation

   Work done by a gas in an adiabatic process: $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$. $W = \frac{1 \times 8.3 \times (60 - 50)}{\frac{5}{3} - 1} = \frac{83}{2/3} = 124.5\ \mathrm{J}$. (Equivalently, $W = -\Delta U = nC_V(T_1 - T_2) = \frac{3}{2}(8.3)(10) = 124.5\ \mathrm{J}$.) Option (C).

2. Question 2 (NEET 2026)

   NEET 2026EasyFirst law of thermodynamics

   An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, then the rate at which internal energy increases will be:

   1. AOption A: 75 W
   2. BOption B: 100 W
   3. COption C: 125 W
   4. DOption D: 25 W

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   Correct answer: (D) 25 W

   Explanation

   Using first law of thermodynamics: Q=ΔU+W ⇒ 100=ΔU+75 ⇒ ΔU=25 W. Hence option (D) is correct.

3. Question 3 (NEET 2026)

   NEET 2026MediumCyclic process and first law

   One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is:

   1. AOption A: $400\ \mathrm{J}$
   2. BOption B: $500\ \mathrm{J}$
   3. COption C: $600\ \mathrm{J}$
   4. DOption D: $800\ \mathrm{J}$

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   Correct answer: (C) $600\ \mathrm{J}$

   Explanation

   For a complete cycle the gas returns to its initial state, so $\Delta U = 0$ and, by the first law, the net heat supplied equals the net work done: $Q = W$. The cycle $a \to b \to c \to d \to a$ is clockwise on the $P$–$V$ diagram, so $W$ is positive and equals the enclosed area: $W = (300 - 100)\ \mathrm{N\,m^{-2}} \times (5 - 2)\ \mathrm{m^3} = 200 \times 3 = 600\ \mathrm{J}$. Total (net) heat supplied $= 600\ \mathrm{J}$, option (C).

4. Question 4 (NEET 2025)

   NEET 2025MediumIdeal Gas Equation

   A container has two chambers of volumes V₁ = 2 litres and V₂ = 3 litres separated by a partition made of a thermal insulator. The chambers contains n₁ = 5 and n₂ = 4 moles of ideal gas at pressures P₁ = 1 atm and P₂ = 2 atm, respectively. When the partition is removed, the equilibrium pressure of the system is:

   1. AOption A: 1.3 atm
   2. BOption B: 1.6 atm
   3. COption C: 1.4 atm
   4. DOption D: 1.8 atm

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   Correct answer: (B) 1.6 atm

   Explanation

   For ideal gases in insulated chambers, final equilibrium pressure is: $$P_f = \frac{P_1V_1 + P_2V_2}{V_1 + V_2}$$ Substituting values: $$P_f = \frac{(1)(2) + (2)(3)}{2 + 3} = \frac{8}{5} = 1.6 \text{ atm}$$

5. Question 5 (NEET 2025)

   NEET 2025MediumHeat, work, and internal energy

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

   1. AOption A: $\dfrac{4}{3}$
   2. BOption B: $\dfrac{3}{4}$
   3. COption C: $\dfrac{2}{\sqrt{3}}$
   4. DOption D: $\dfrac{\sqrt{3}}{2}$

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   Correct answer: (B) $\dfrac{3}{4}$

   Explanation

   Given that equal heat is supplied and the change in internal energy is the same for both gases: $$Q = \Delta U + P\Delta V$$ Since $Q$and$\Delta U$are equal for both systems,$$P\Delta V_A = P\Delta V_B$$As pressure is same:$$\Delta V_A = \Delta V_B$$For a cylinder:$$\Delta V = \pi r^2 h$$Hence,$$\pi r_A^2 (16) = \pi r_B^2 (9)$$$$\frac{r_A^2}{r_B^2} = \frac{9}{16}$$$$\frac{r_A}{r_B} = \frac{3}{4}$$ Therefore, the correct answer is option (B).

6. Question 6 (NEET 2025)

   NEET 2025MediumSecond Law & Entropy

   Given below are two statements : Statement I : In ecosystem, there is unidirectional flow of energy of sun from producers to consumers. Statement II : Ecosystems are exempted from 2nd law of thermodynamics. In the light of the above statements, choose the most appropriate answer from the options given below :

   1. AOption A: Both statement I and statement II are correct
   2. BOption B: Both statement I and statement II are incorrect
   3. COption C: Statement I is correct but statement II is incorrect
   4. DOption D: Statement I is incorrect but statement II is correct

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   Correct answer: (C) Statement I is correct but statement II is incorrect

   Explanation

   Energy flow in an ecosystem is unidirectional from producers to consumers. Ecosystems are not exempt from the second law of thermodynamics because energy transformations involve entropy increase.

7. Question 7 (NEET 2024)

   NEET 2024MediumThermodynamic Processes

   A thermodynamic system is taken through the cycle abcda. The work done by the gas along the path bc is:

   1. AOption A: −90 J
   2. BOption B: −60 J
   3. COption C: zero
   4. DOption D: 30 J

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   Correct answer: (C) zero

   Explanation

   Along path bc, volume remains constant (isochoric process). Work done is: $$W = \int P\,dV$$ Since: $$dV = 0$$ Therefore: $$W = 0$$ Hence, option (C) is correct.

8. Question 8 (NEET 2024)

   NEET 2024MediumCharles Law

   The following graph represents the T-V curves of an ideal gas (where T is the temperature and V the volume) at three pressures P₁, P₂ and P₃ compared with those of Charles’s law represented by dotted lines. Then the correct relation is:

   1. AOption A: P₂ > P₁ > P₃
   2. BOption B: P₁ > P₂ > P₃
   3. COption C: P₃ > P₂ > P₁
   4. DOption D: P₁ > P₃ > P₂

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   Correct answer: (B) P₁ > P₂ > P₃

   Explanation

   For an ideal gas: $$PV = nRT$$ At constant pressure: $$T \propto V$$ The slope of the T-V graph is: $$\frac{T}{V} = \frac{P}{nR}$$ Thus, greater slope corresponds to greater pressure. From the graph: $$\text{slope of } P_1 > \text{slope of } P_2 > \text{slope of } P_3$$ Therefore: $$P_1 > P_2 > P_3$$ Hence, option (B) is correct.

9. Question 9 (NEET 2023)

   NEET 2023MediumCarnot Engine

   A Carnot engine has an efficiency of 50% when the source is at a temperature 327°C. The temperature of the sink is :

   1. AOption A: 27°C
   2. BOption B: 15°C
   3. COption C: 100°C
   4. DOption D: 200°C

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   Correct answer: (A) 27°C

   Explanation

   Efficiency of Carnot engine: η = 1 - (T₂/T₁) Given: η = 0.5 T₁ = 327°C = 600 K 0.5 = 1 - T₂/600 T₂ = 300 K = 27°C Hence, the sink temperature is 27°C.

10. Question 10 (NEET 2021)

    NEET 2021EasySpecific Heat Capacity (Cp and Cv)

    Which one among the following is the correct option for right relationship between $C_p$ and $C_v$ for one mole of ideal gas?

    1. AOption A: $C_v = R C_p$
    2. BOption B: $C_p + C_v = R$
    3. COption C: $C_p - C_v = R$
    4. DOption D: $C_p = R C_v$

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    Correct answer: (C) $C_p - C_v = R$

    Explanation

    For one mole of an ideal gas, Mayer's relation gives: where: - $C_p$ = molar specific heat at constant pressure - $C_v$ = molar specific heat at constant volume - $R$ = universal gas constant Hence, the correct option is **(3)**.