Thermodynamics (C) JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Thermodynamics (C), free to read — no sign-in needed. The full chapter has 46 questions; sign in to attempt the remaining 41 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 1)
   List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy ($\Delta H$) and entropy ($\Delta S$). Match each entry in List-I to the appropriate entry in List-II, and choose the correct option. List-IList-II(P) Physisorption(1) $\Delta H \gt 0$ and $\Delta S \gt 0$(Q) Diamond $\longrightarrow$ Graphite(2) $\Delta H \lt 0$ and $\Delta S \lt 0$(R) Denaturation of protein(3) $\Delta H \lt 0$ and $\Delta S = 0$(S) Propene $\longrightarrow$ Cyclopropane(4) $\Delta H \gt 0$ and $\Delta S \lt 0$(5) $\Delta H \lt 0$ and $\Delta S \gt 0$
   1. A.P $\rightarrow$ 2; Q $\rightarrow$ 3; R $\rightarrow$ 5; S $\rightarrow$ 4
   2. B.P $\rightarrow$ 4; Q $\rightarrow$ 3; R $\rightarrow$ 5; S $\rightarrow$ 1
   3. C.P $\rightarrow$ 2; Q $\rightarrow$ 5; R $\rightarrow$ 1; S $\rightarrow$ 4
   4. D.P $\rightarrow$ 2; Q $\rightarrow$ 5; R $\rightarrow$ 1; S $\rightarrow$ 3
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   Answer: (C)

   (P) Physisorption is an exothermic process because attractive forces are formed between the adsorbate and the adsorbent, so $\Delta H \lt 0$. The gas molecules are restricted on the solid surface, decreasing randomness, so $\Delta S \lt 0$. This matches (2). (Q) Diamond $\longrightarrow$ Graphite is an exothermic process because graphite is the thermodynamically more stable allotrope of carbon at standard conditions ($\Delta H \lt 0$). Graphite has a layered structure and is less dense, so it has higher entropy than the rigid 3D network of diamond ($\Delta S \gt 0$). This matches (5). (R) Denaturation of protein involves the breaking of hydrogen bonds and the unfolding of its specific 3D structure into a more random coil. Breaking these interactions requires energy ($\Delta H \gt 0$) and the unfolding increases randomness ($\Delta S \gt 0$). This matches (1). (S) Propene $\longrightarrow$ Cyclopropane is an endothermic process because cyclopropane has significant ring strain compared to the open-chain propene, making it less stable ($\Delta H \gt 0$). The formation of a rigid ring restricts the degrees of freedom (such as rotation around single bonds), decreasing entropy ($\Delta S \lt 0$). This matches (4). Thus, P $\rightarrow$ 2, Q $\rightarrow$ 5, R $\rightarrow$ 1, S $\rightarrow$ 4. Answer: P $\rightarrow$ 2; Q $\rightarrow$ 5; R $\rightarrow$ 1; S $\rightarrow$ 4
2. Q2JEE Advanced Adv 2026 (Paper 1)
   An ideal gas ($0.5$ mol), initially at $2$ bar pressure, is compressed at a constant temperature of $600$ K in two steps: first, against a constant external pressure of $P$ bar ($2 \lt P \lt 8$), and then against constant external pressure of $8$ bar. At each step, the compression is stopped only when the pressure of the gas becomes equal to the external pressure. The total work done on the gas in these steps is $W$. Considering all possible values of $P$ ($2 \lt P \lt 8$) and taking the gas constant as $R$ (in J K$^{-1}$ mol$^{-1}$), the minimum value of $|W|$ (in J) is
   1. A.$207R$
   2. B.$600R$
   3. C.$630R$
   4. D.$900R$
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   Answer: (B)

   Let the initial, intermediate, and final volumes of the gas be $V_1$, $V_2$, and $V_3$ respectively. Using the ideal gas equation $V = \dfrac{nRT}{P}$, we have: $V_1 = \dfrac{nRT}{2}$ $V_2 = \dfrac{nRT}{P}$ $V_3 = \dfrac{nRT}{8}$ The work done on the gas during the first step against a constant external pressure $P$ is: $W_1 = -P_{ext,1}(V_2 - V_1) = -P \left( \dfrac{nRT}{P} - \dfrac{nRT}{2} \right) = nRT \left( \dfrac{P}{2} - 1 \right)$ The work done on the gas during the second step against a constant external pressure of $8$ bar is: $W_2 = -P_{ext,2}(V_3 - V_2) = -8 \left( \dfrac{nRT}{8} - \dfrac{nRT}{P} \right) = nRT \left( \dfrac{8}{P} - 1 \right)$ The total work done on the gas is: $W = W_1 + W_2 = nRT \left( \dfrac{P}{2} - 1 \right) + nRT \left( \dfrac{8}{P} - 1 \right) = nRT \left( \dfrac{P}{2} + \dfrac{8}{P} - 2 \right)$ To find the minimum value of $W$, we need to minimize the expression $\left( \dfrac{P}{2} + \dfrac{8}{P} \right)$. Using the AM-GM inequality: $\dfrac{P}{2} + \dfrac{8}{P} \ge 2 \sqrt{\dfrac{P}{2} \times \dfrac{8}{P}} = 2 \sqrt{4} = 4$ The minimum value is $4$, which occurs when $\dfrac{P}{2} = \dfrac{8}{P} \Rightarrow P^2 = 16 \Rightarrow P = 4$ bar. This value of $P$ lies within the given range $(2 \lt P \lt 8)$. Substituting this minimum value back into the work equation: $W_{min} = nRT (4 - 2) = 2nRT$ Given $n = 0.5$ mol and $T = 600$ K: $W_{min} = 2 \times 0.5 \times R \times 600 = 600R$ Answer: $600R$
3. Q3JEE Advanced Adv 2025 (Paper 1)
   Considering ideal gas behavior, the expansion work done (in kJ) when 144 g of water is electrolyzed completely under constant pressure at 300 K is _____ Use: Universal gas constant $(\mathrm{R})=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$; Atomic mass (in amu): $\mathrm{H}=1, \mathrm{O}=16$ (mark absolute value as answer)
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   Answer: 29.88

   in official answer key answer was [-29.95 to -29.8] OR [29.8 to 29.95], we have updated the question to make it valid. $\begin{array}{lcc} \mathrm{H}_2 \mathrm{O}(l) \rightarrow & \mathrm{H}_2(\mathrm{~g})+ & \frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \\ 144 \mathrm{~g} & - & - \\ 8 \mathrm{~mol} & 8 \mathrm{~mol} & 4 \mathrm{~mol} \\ \mathrm{~W}=-\mathrm{P} \Delta \mathrm{~V}= & -(\Delta \mathrm{n}) \mathrm{RT} \end{array}$ $\begin{aligned} &\text { Change in gaseous moles }=12\\ &\Rightarrow \mathrm{W}=-\frac{12 \times 8.3 \times 300}{1000} \mathrm{~kJ}=-29.88 \mathrm{~kJ} . \end{aligned}$
4. Q4JEE Advanced Adv 2022 (Paper 2)
   The correct option(s) about entropy (S) is(are) [$R=$ gas constant, $F=$ Faraday constant, $T=$ Temperature]
   1. A.For the reaction, $M\left(s\right)+2{H}^{+}\left(aq\right)\rightarrow {H}_{2}\left(g\right)+{M}^{2+}\left(aq\right)$, if $\dfrac{{dE}_{cell}}{dT}=\dfrac{R}{F}$ then the entropy change of the reaction is $R$ (assume that entropy and internal energy changes are temperature independent).
   2. B.The cell reaction, $Pt\left(s\right)∣{H}_{2}\left(g,1bar\right)\left|{H}^{+}\left(aq,0.01M\right)‖{H}^{+}\left(aq,0.1M\right)\right|{H}_{2}\left(g,1bar\right)∣Pt\left(s\right)$ , is an entropy driven process.
   3. C.For racemization of an optically active compound, $\Delta S\gt 0$
   4. D.$\Delta \mathrm{S}>0$, for $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right) 6\right]^{2+}+3 \mathrm{en} \rightarrow[\mathrm{Ni}(\mathrm{en})_{3}]^{2+}+6 \mathrm{H}_2 \mathrm{O}$ (where en $=$ ethylenediamine)
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   Answer: B,C,D

   (A) $M\left(s\right)+2{H}^{+}\left(aq\right)\rightarrow {H}_{2}\left(g\right)+{M}^{2+}\left(aq\right)$ if $\dfrac{{dE}_{cell}}{dT}=\dfrac{R}{F}$ $\Delta S=nF\dfrac{dE}{dT}=2F\left(\dfrac{R}{F}\right)=2R$ (B) ${E}_{cell}=\dfrac{-2.303RT}{F}\log \dfrac{0.01}{0.1}=\dfrac{2.303RT}{F}$ $\dfrac{{dE}_{cell}}{dT}=\dfrac{2.303R}{F}$ $∴\Delta S=nF\dfrac{dE}{dT}\gt 0$ It is an entropy driven process. (C) It is correct Racemisation is a thermodynamically favourable method, and it proceeds spontaneously if a suitable pathway is accessible for the interconversion of the enantiomers. During racemisation of optically active compound, disorderness increases and hence, entropy increases. (D) For ${\left[Ni{\left({H}_{2}O\right)}_{6}\right]}^{2+}+3en\rightarrow {\left[Ni{\left(en\right)}_{3}\right]}^{+3}+6{H}_{2}O$ Entropy increases when bidentate ligands replace monodentate ligands due to increase in the number of molecules on the product side. Hence, (B, C, D) are correct.
5. Q5JEE Advanced Adv 2022 (Paper 1)
   $2mol$ of $Hg\left(g\right)$ is combusted in a fixed volume bomb calorimeter with excess of ${O}_{2}$ at $298K$ and $1atm$ into $HgO\left(s\right)$. During the reaction, temperature increases from $298.0K$ to $312.8K$. If heat capacity of the bomb calorimeter and enthalpy of formation of $Hg\left(g\right)$ are $20.00kJ{K}^{-1}$ and $61.32kJ{mol}^{-1}$ at $298K$, respectively, the calculated standard molar enthalpy of formation of $HgO\left(s\right)$ at $298K$ is $X{kJmol}^{-1}$. The value of $\left|X\right|$ is___[Given: Gas constant $R=8.3J{K}^{-1}{mol}^{-1}$]
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   Answer: 90.39

   $2Hg\left(g\right)+{O}_{2}\left(g\right)⟶2HgO\left(s\right)$ Heat capacity of calorimeter $=20kJ{K}^{-1}$ Rise in temperature $=14.8K$ Heat evolved $=20\times 14.8=296kJ$ $\Delta H^{\circ}=\Delta U^{\circ}+{\Delta n}_{g}RT$ $=-296-3\times 8.3\times 298\times {10}^{-3}$ $=-303.42kJ$ $\Delta H^{\circ}=2{\Delta H}_{f}^{\circ}\left(HgO\left(s\right)\right)-2{\Delta H}_{f}^{\circ}\left(Hg\left(g\right)\right)$ $-303.42=2{\Delta H}_{f}^{\circ}\left(HgO\left(s\right)\right)-2\times 61.32$ $2{\Delta H}_{f}^{\circ}\left(HgO\left(s\right)\right)=-180.78kJ$ $\left|{\Delta H}_{f}^{\circ}\left(HgO\left(s\right)\right)\right|=90.39kJ{mol}^{-1}$