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 240 questions; sign in to attempt the remaining 235 in the exam simulator.

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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 (25 Jan, Shift 1)Preparation and Reactions of Glucose
   <p>An athlete is given $100g of glucose \left({C}_{6}{H}_{12}{O}_{6}\right) for energy. This is equivalent to 1800kJ of energy. The 50$.</p>
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   Answer: 360

   <p>Overall reaction of metabolism of glucose in the body: ${C}_{6}{H}_{12}{O}_{6}(s)+6{O}_{2}\rightarrow 6{CO}_{2}(g)+6{H}_{2}O(l) Extra energy used to convert {H}_{2}O(l) into {H}_{2}O(g) , energy left =\dfrac{1800}{2}=900kJ Moles of water vaporised from enthalpy of evaporation: \Rightarrow 900={n}_{{H}_{2}O}\times 45 {n}_{{H}_{2}O}=\dfrac{900}{45}=20\text{mole} {W}_{{H}_{2}O}=20\times 18=360g$</p>
5. Q5JEE Main 2022 (29 Jul, Shift 2)Mole concept
   $C\left(s\right)+{O}_{2}\left(g\right)\rightarrow {CO}_{2}\left(g\right)+400kJ$ $C\left(s\right)+\dfrac{1}{2}{O}_{2}\left(g\right)\rightarrow CO\left(g\right)+100kJ$ When coal of purity $60$ is allowed to burn in presence of insufficient oxygen, $60$ of carbon is converted into ' $CO$' and the remaining is converted into '${CO}_{2}$'. The heat generated when $0.6kg$ of coal is burnt is
   1. A.$1600kJ$
   2. B.$3200kJ$
   3. C.$4400kJ$
   4. D.$6600kJ$
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   Answer: (D)

   $C\left(S\right)+{O}_{2}\left(g\right)\rightarrow {CO}_{2}\left(g\right)+400kJ$ $1$ mole $C\left(s\right)+\dfrac{1}{2}{O}_{2}\left(g\right)\rightarrow CO\left(g\right)+100kJ...\left(2\right)$ $0.6Kg\times 1000$ $=600gm$ Mass of $60$ pure coal =$600\times \dfrac{60}{100}$ (Pure Carbon) $=360gm Numberofmoles=\dfrac{360}{12}=30molesofpurecarbon$ Moles of 60% pure Carbon converted into ${CO}_{2}=\left(30-30\times \dfrac{60}{100}\right)$ $=12$ mole and Moles of 60% pure carbon converted in to $CO=30\times \dfrac{60}{100}=18$ mole Energy generated during $2$ equation $=18\times 100$ $=1800kJ$ Energy generated during ${1}^{st}$ reaction. $=12\times 400$ $=4800$ Total $=1800+4800=6600kJ$