Thermodynamics (C) JEE Main previous year questions with solutions

5 solved JEE Main questions on Thermodynamics (C), free to read — no sign-in needed. The full chapter has 264 questions; sign in to attempt the remaining 259 in the exam simulator.

1. Q1JEE Main 2026 (28 Jan, Shift 2)Entropy and Second law of thermodynamics
   The plot of $\log _{10} \mathrm{~K}$ vs $\frac{1}{\mathrm{~T}}$ gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).
   1. A.$\frac{\Delta \mathrm{S}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}},-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}$
   2. B.$-\frac{\Delta \mathrm{S}^{\circ} \mathrm{R}}{2 \cdot 303}, \frac{\Delta \mathrm{H}^{\circ} \mathrm{R}}{2 \cdot 303}$
   3. C.$\frac{2 \cdot 303 \mathrm{R}}{\Delta \mathrm{H}^{\mathrm{o}}}, \frac{2 \cdot 303 \mathrm{R}}{\Delta \mathrm{~S}^{\mathrm{o}}}$
   4. D.$-\frac{\Delta \mathrm{H}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}, \frac{\Delta \mathrm{~S}^{\mathrm{o}}}{2 \cdot 303 \mathrm{R}}$
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   Answer: (A)

   The relationship between the standard Gibbs free energy change and the equilibrium constant is given by $\Delta G^o = -RT \ln K$. We also know that $\Delta G^o = \Delta H^o - T\Delta S^o$. Equating the two expressions: $-RT \ln K = \Delta H^o - T\Delta S^o$. Dividing by $-RT$, we get: $\ln K = -\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}$. To convert natural log to base 10, use $\ln K = 2.303 \log_{10} K$: $2.303 \log_{10} K = -\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}$. Dividing by $2.303$: $\log_{10} K = -\frac{\Delta H^o}{2.303RT} + \frac{\Delta S^o}{2.303R}$. Rearranging in the form of a straight line equation $y = mx + c$ where $y = \log_{10} K$ and $x = \frac{1}{T}$: $\log_{10} K = \left( -\frac{\Delta H^o}{2.303R} \right) \frac{1}{T} + \frac{\Delta S^o}{2.303R}$. Comparing the terms: Intercept ($c$) = $\frac{\Delta S^o}{2.303R}$ Slope ($m$) = $-\frac{\Delta H^o}{2.303R}$.
2. Q2JEE Main 2025 (29 Jan, Shift 1)First Law and Basic Fundamentals of Thermodynamics
   500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
   1. A.378 K and 500 J
   2. B.368 K and 500 J
   3. C.348 K and 300 J
   4. D.378 K and 300 J
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   Answer: (C)

   $\begin{aligned} & \mathrm{q}_{\mathrm{p}}=\mathrm{n} \times \mathrm{c}_{\mathrm{p}} \times \Delta \mathrm{T} \\ & \Rightarrow 500=0.5 \times \frac{5}{2} \times 8.3\left(\mathrm{~T}_{\mathrm{f}}-298\right) \\ & \Rightarrow \mathrm{T}_{\mathrm{f}} \simeq 346.2 \mathrm{~K} \\ & \frac{\Delta \mathrm{H}}{\Delta \mathrm{U}}=\frac{\mathrm{C}_{\mathrm{p}}}{\mathrm{C}_{\mathrm{v}}}=\left(\frac{5}{3}\right) \\ & \Rightarrow \Delta \mathrm{U}=\frac{3}{5} \times 500=300 \mathrm{~J}\end{aligned}$
3. Q3JEE Main 2024 (09 Apr, Shift 1)Laws of Thermochemistry and Enthalpy Change
   When equal volume of $1 \mathrm{M} \mathrm{HCl}$ and $1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ are separately neutralised by excess volume of $1 \mathrm{M}$ $\mathrm{NaOH}$ solution. $x$ and $y \mathrm{~kJ}$ of heat is liberated respectively. The value of $y / x$ is _______
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   Answer: 2

   $\begin{aligned} & \mathrm{H}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_2 \mathrm{O} \Rightarrow \mathrm{x} \\ & 2 \mathrm{H}^{+}+2 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} \Rightarrow 2 \mathrm{x}=\mathrm{y} \\ & \mathrm{y} / \mathrm{x}=2\end{aligned}$
4. Q4JEE Main 2023 (11 Apr, Shift 2)System and surrounding
   The total number of intensive properties from the following is........... Volume, Molar heat capacity, molarity, ${E}_{cell}^{o}$,Gibbs free energy change, Molar mass, Mole
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   Answer: 4

   An intensive property is a physical quantity whose value does not depend on the amount of substance which was measured. Molar heat capacity, molarity, ${E}_{cell}^{o}$ and molar mass are intensive properties. An extensive property of a system depends on the system size or the amount of matter in the system. Volume, Gibbs free energy change, Mole are extensive properties.
5. Q5JEE Main 2022 (26 Jun, Shift 2)Third law of thermodynamics
   $40$ of $HI$ undergoes decomposition to ${H}_{2}$ and ${I}_{2}$ at $300K.{\Delta G}^{\Theta }$ for this decompostion reaction at one atmopsphere pressure is____${Jmol}^{-1}$-[nearest integer] $\left(UseR=8.31{JK}^{-1}{mol}^{-1};\log 2=0.3010,\ln 10=2.3,\log 3=0.477\right)$
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   Answer: 2735

   $HI\left(g\right)\rightleftharpoons \dfrac{1}{2}{H}_{2}\left(g\right)+\dfrac{1}{2}{I}_{2}\left(g\right)$ Initial moles $1$ $0$ $0$ moles at equilibrium $\left(1-0.4\right)$ $0.2$ $0.2$ ${K}_{P}=\dfrac{{\left(0.2\right)}^{\dfrac{1}{2}}\times {\left(0.2\right)}^{\dfrac{1}{2}}}{\left(0.6\right)}$ ${K}_{P}=\dfrac{0.2}{0.6}=\dfrac{1}{3}$ $\Delta G^{\circ}=-2.3{RTlogK}_{P}$ $=-2.3\times 8.31\times 300\log \dfrac{1}{3}$ =$2375{Jmol}^{-1}$