Thermodynamics JEE Main previous year questions with solutions

5 solved JEE Main questions on Thermodynamics, free to read — no sign-in needed. The full chapter has 196 questions; sign in to attempt the remaining 191 in the exam simulator.

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1. Q1JEE Main 2026 (08 Apr, Shift 2)First Law of Thermodynamics
   Initial pressure and volume of a monoatomic ideal gas are $P$ and $V$. The change in internal energy of this gas in adiabatic expansion to volume $V_{final}=27V$ is ________ J.
   1. A.$-2PV(3\sqrt{3}-1)$
   2. B.$\dfrac{4}{3}PV$
   3. C.$-\dfrac{4}{3}PV$
   4. D.$\dfrac{3}{4}PV$
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   Answer: (C)

   For a monoatomic ideal gas, the ratio of specific heats is $\gamma = \dfrac{5}{3}$. In an adiabatic process, $PV^{\gamma} = \text{constant}$. $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$ $P V^{5/3} = P_2 (27V)^{5/3}$ $P_2 = P \left(\dfrac{1}{27}\right)^{5/3} = \dfrac{P}{243}$ The change in internal energy is given by: $\Delta U = \dfrac{P_2 V_2 - P_1 V_1}{\gamma - 1}$ Substituting the values: $\Delta U = \dfrac{\left(\dfrac{P}{243}\right)(27V) - PV}{\dfrac{5}{3} - 1}$ $\Delta U = \dfrac{\dfrac{PV}{9} - PV}{\dfrac{2}{3}}$ $\Delta U = \dfrac{-\dfrac{8}{9}PV}{\dfrac{2}{3}} = -\dfrac{4}{3}PV$ Answer: $-\dfrac{4}{3}PV$
2. Q2JEE Main 2025 (23 Jan, Shift 2)Heat Engine, Refrigerator and Second Law of Thermodynamics
   Water of mass $m$ gram is slowly heated to increase the temperature from $T_1$ to $T_z$ The change in entropy of the water, given specific heat of water is $1 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$, is :
   1. A.$\mathrm{m} \ln \left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)$
   2. B.zero
   3. C.$\mathrm{m} \ln \left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)$
   4. D.$\mathrm{m}\left(\mathrm{T}_2-\mathrm{T}_1\right)$
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   Answer: (A)

   $\begin{aligned} & d \mathrm{Q}=\mathrm{msdT} \\ & \mathrm{dS}=\frac{\mathrm{dQ}}{\mathrm{T}}=\frac{\mathrm{msdT}}{\mathrm{T}} \\ & \Delta \mathrm{S}=\int \frac{\mathrm{msdT}}{\mathrm{T}}=\mathrm{ms} \ln \frac{\mathrm{T}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{i}}} \\ & \Delta \mathrm{S}=\mathrm{m} \ln \frac{\mathrm{T}_2}{\mathrm{~T}_1}\end{aligned}$
3. Q3JEE Main 2024 (09 Apr, Shift 1)Thermodynamic Processes
   The volume of an ideal gas $(\gamma=1.5)$ is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is:
   1. A.$\frac{16}{25}$
   2. B.$\frac{4}{5}$
   3. C.$\frac{8}{5 \sqrt{5}}$
   4. D.$\frac{2}{\sqrt{5}}$
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   Answer: (C)

   For Adiabatic process $\begin{aligned} & P_i V_i=P_f V_f^\gamma \\ & P_i(5)^{1.5}=P_f(4)^{1.5} \\ & \frac{P_i}{P_f}=\left(\frac{4}{5}\right)^{\frac{3}{2}}=\frac{4}{5} \cdot\left(\frac{4}{5}\right)^{\frac{1}{2}} \Rightarrow \frac{8}{5 \sqrt{5}}\end{aligned}$
4. Q4JEE Main 2023 (11 Apr, Shift 1)Calorimetry
   $1kg$ of water at $100^{\circ}C$ is converted into steam at $100^{\circ}C$ by boiling at atmospheric pressure. The volume of water changes from $1.00\times {10}^{-3}{m}^{3}$ as a liquid to $1.671{m}^{3}$ as steam. The change in internal energy of the system during the process will be (Given latent heat of vaporisation $=2257kJ/kg$, Atmospheric pressure $\left.=1\times {10}^{5}Pa\right)$
   1. A.$-2426kJ$
   2. B.$+2090kJ$
   3. C.$-2090kJ$
   4. D.$+2476kJ$
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   Answer: (B)

   The work to be done in the process is given by $\begin{matrix}dW & = & PdV \\ & = & 1\times {10}^{5}Pa\times (1.671-0.001){m}^{3} \\ & = & 1.670\times {10}^{5}J\end{matrix}$ The change in heat energy during the vaporisation process can be calculated as follows- $\begin{matrix}\Delta {Q}_{\text{supplied}} & = & 2257\times 1\times {10}^{3}J \\ & = & 22.57\times {10}^{5}J\end{matrix}$ Hence, the change in internal energy in the process is given by $\begin{matrix}\Delta U & = & \Delta {Q}_{supplied}-\Delta W \\ & = & (22.57-1.67)\times {10}^{5}J \\ & = & 20.9\times {10}^{5}J \\ & = & 2090kJ\end{matrix}$
5. Q5JEE Main 2022 (24 Jun, Shift 2)Degree of Freedom and Specific Heat
   A monoatomic gas performs a work of $\dfrac{Q}{4}$ where $Q$ is the heat supplied to it. The molar heat capacity of the gas will be _____ $R$ during this transformation. Where $R$ is the gas constant.
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   Answer: 2

   Using first law of thermodynamics, $Q=∆U+W$ $\Rightarrow Q=\Delta U+\dfrac{Q}{4}$ $\Rightarrow \Delta U=\dfrac{3Q}{4}$ $\Rightarrow ∆U=\dfrac{3}{2}nR\Delta T=\dfrac{3Q}{4}$ Therefore, molar heat capacity, $C=\dfrac{Q}{n∆T}$ $\Rightarrow C=2R$