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 187 questions; sign in to attempt the remaining 182 in the exam simulator.

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 (24 Jan, Shift 2)Thermodynamic Systems
   Let ${\gamma }_{1}$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and ${\gamma }_{2}$ be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio $\dfrac{{\gamma }_{1}}{{\gamma }_{2}}$ is:
   1. A.$\dfrac{27}{35}$
   2. B.$\dfrac{35}{27}$
   3. C.$\dfrac{25}{21}$
   4. D.$\dfrac{21}{25}$
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   Answer: (C)

   As we know, ${C}_{v}=f\dfrac{R}{2}$ and ${C}_{p}={C}_{v}+R=f\dfrac{R}{2}+R$ Therefore, $\gamma =\dfrac{{C}_{p}}{{C}_{v}}=1+\dfrac{2}{f}$ For monatomic gas, $f=3$ and hence ${\gamma }_{1}=\dfrac{5}{3}$ For diatomic gas, $f=5$ and hence ${\gamma }_{2}=\dfrac{7}{5}$ Required ratio, $\dfrac{{\gamma }_{1}}{{\gamma }_{2}}=\dfrac{25}{21}$
5. Q5JEE Main 2026 (08 Apr, Shift 2)First Law of Thermodynamics
   One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is $4$ cm$^2$. The gas is heated slowly to raise the temperature by $1.2\,^\circ$C during which the piston moves by $25$ mm. The amount of heat supplied to the gas is ________ J. (Atmospheric pressure $=100$ kPa, $R=8.3$ J/mol·K) (Neglect mass of the piston)
   1. A.$24.8$
   2. B.$25$
   3. C.$15.04$
   4. D.$29.98$
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   Answer: (B)

   Using the first law of thermodynamics: $Q = \Delta U + W$ Since the gas is diatomic with rotational modes only, the degrees of freedom are $f = 5$, so: $C_V = \dfrac{5}{2}R$ For $n = 1\text{ mole}$ and $\Delta T = 1.2\text{ K}$: $\Delta U = nC_V\Delta T = 1 \times \dfrac{5}{2} \times 8.3 \times 1.2 = 24.9\text{ J}$ Work done by the gas: $W = P\,\Delta V$ Cross-sectional area of the piston: $A = 4\text{ cm}^2 = 4 \times 10^{-4}\text{ m}^2$ Piston displacement: $x = 25\text{ mm} = 2.5 \times 10^{-2}\text{ m}$ Change in volume: $\Delta V = A \cdot x = (4 \times 10^{-4})(2.5 \times 10^{-2}) = 1 \times 10^{-5}\text{ m}^3$ With $P = 100\text{ kPa} = 10^5\text{ Pa}$: $W = (10^5)(1 \times 10^{-5}) = 1\text{ J}$ Therefore, the heat supplied is: $Q = \Delta U + W = 24.9 + 1 = 25.9\text{ J}$ Hence, the heat supplied to the gas is $25.9\text{ J}$.