Thermodynamics JEE Main PYQs (2020–2026)

Solved Thermodynamics PYQs

3 recent questions with answers. The full set of 159 is free to practise.

1. JEE Main 2026EasyEnthalpy (H), Heat Capacity (Cp, Cv) and Reaction Enthalpies
   Q1

   Consider the reaction
   
   $$2H_{2}S(g)+3O_2(g)\to 2H_{2}O(l)+2{SO}_2(g)$$
   
   The magnitude of enthalpy change for the reaction in $kJ{mol}^{-1}$ is $\_\_\_\_$ . (Nearest integer)
   
   $$\begin{array}{rl}Given:& {\Delta }_f{H}^{\ominus }\left(H_{2}S\right)=-20.1kJ{mol}^{-1}\\ & {\Delta }_f{H}^{\ominus }\left(H_{2}O\right)=-286.0kJ{mol}^{-1}\\ & {\Delta }_f{H}^{\ominus }\left({SO}_2\right)=-297.0kJ{mol}^{-1}\end{array}$$

   Numerical answer type

   Show answer & solution

   Correct answer: 1126

   We are given the reaction:
   
   $$2H_{2}S(g)+3O_2(g)\to 2H_{2}O(l)+2{SO}_2(g)$$
   
   We are also given the standard enthalpies of formation:
   
   $$\begin{array}{rl}{\Delta }_f{H}^{\circ }(H_{2}S)& =-20.1kJmo{l}^{-1}\\ {\Delta }_f{H}^{\circ }(H_{2}O)& =-286.0kJmo{l}^{-1}\\ {\Delta }_f{H}^{\circ }(S{O}_2)& =-297.0kJmo{l}^{-1}\end{array}$$
   
   According to NCERT convention,
   
   $${\Delta }_r{H}^{\circ }=\sum {\Delta }_f{H}^{\circ }(\text{products})-\sum {\Delta }_f{H}^{\circ }(\text{reactants})$$
   
   Products:
   
   $$2H_{2}O(l)+2S{O}_2(g)$$
   
   $$\sum {\Delta }_f{H}^{\circ }(\text{products})=[2(-286.0)+2(-297.0)]kJ$$
   
   $$=(-572.0-594.0)=-1166.0kJ$$
   
   Reactants:
   
   $$2H_{2}S(g)+3O_2(g)$$
   
   $$\sum {\Delta }_f{H}^{\circ }(\text{reactants})=[2(-20.1)+3(0)]kJ=-40.2kJ$$
   
   $${\Delta }_r{H}^{\circ }=(-1166.0)-(-40.2)=-1125.8kJ$$
   
   The magnitude of enthalpy change is
   
   $$|{\Delta }_r{H}^{\circ }|=1126kJmo{l}^{-1}(\text{nearest integer})$$
   
   Final Answer
   
   $$\boxed{1126kJmo{l}^{-1}}$$
2. JEE Main 2026EasyGibbs Free Energy (G), Spontaneity Criteria & Calculations
   Q2

   At the transition temperature $T$ , $A\rightleftharpoons B$ and $\Delta G^0=105-35\log T$ where A and B are two states of substance X. The transition temperature in ${}^{\circ }C$ when pressure is 1 atm is ________. (Nearest integer)

   Numerical answer type

   Show answer & solution

   Correct answer: 727

   At the transition temperature, the two states $A$ and $B$ are in equilibrium.
   
   So, for the equilibrium condition,
   
   $$\Delta G^0=0$$
   
   Given:
   
   $$\Delta G^0=105-35\log T$$
   
   At transition temperature,
   
   $$105-35\log T=0$$
   
   Now solve:
   
   $$35\log T=105$$
   
   $$\log T=\frac{105}{35}=3$$
   
   So,
   
   $$T=10^3=1000\text{ K}$$
   
   Now convert into degree Celsius:
   
   $$T{(}^{\circ }C)=1000-273=727^{\circ }C$$
   
   Nearest integer:
   
   $$\boxed{727}$$
3. JEE Main 2026MediumFirst Law of Thermodynamics: Internal Energy, Heat (q) and Work (w)
   Q3

   Gas ‘A’ undergoes change from state ‘X’ to state ‘Y’. In this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state ‘X’ by another process during which 6 J of heat is evolved. In the reverse process of ‘Y’ to ‘X’,

   • A18 J of the work is done by the gas ‘A’.
   • B2 J of the work is done by the gas ‘A’.
   • C12 J of the work is done on the gas ‘A’ by the surrounding.
   • D14 J of the work is done on the gas ‘A’ by the surrounding.

   Show answer & solution

   Correct answer: (D)

   For a thermodynamic process, using the NCERT sign convention:
   
   Heat absorbed by the gas: $Q>0$
   
   Heat evolved by the gas: $Q<0$
   
   Work done by the gas: $W>0$
   
   Also, from the first law of thermodynamics:
   
   $$\Delta U=Q-W$$
   
   Step 1: Process from $X$ to $Y$
   
   Given:
   
   Heat absorbed, $Q_{XY}=+10\text{J}$
   
   Work done by gas, $W_{XY}=+18\text{J}$
   
   So,
   
   $$\Delta U_{XY}=Q_{XY}-W_{XY}$$
   
   $$\Delta U_{XY}=10-18=-8\text{J}$$
   
   Thus,
   
   $$U_Y-U_X=-8\text{J}$$
   
   Step 2: Reverse process from $Y$ to $X$
   
   For the reverse process,
   
   $$\Delta U_{YX}=-\Delta U_{XY}=+8\text{J}$$
   
   Also, during this process, $6\text{J}$ of heat is evolved, so
   
   $$Q_{YX}=-6\text{J}$$
   
   Using first law again:
   
   $$\Delta U_{YX}=Q_{YX}-W_{YX}$$
   
   Substitute the values:
   
   $$8=-6-W_{YX}$$
   
   $$-W_{YX}=14$$
   
   $$W_{YX}=-14\text{J}$$
   
   Negative sign means work is done on the gas by the surroundings.
   
   So, in the process from $Y$ to $X$,
   
   $$14\text{J}$$
   
   of work is done on the gas by the surroundings.
   
   Correct option: D
   
   $$\boxed{\text{14 J of work is done on the gas by the surrounding}}$$

Topic-wise question count

Where JEE Main has asked from Thermodynamics, most frequent first.

• Enthalpy (H), Heat Capacity (Cp, Cv) and Reaction Enthalpies45
• First Law of Thermodynamics: Internal Energy, Heat (q) and Work (w)40
• Gibbs Free Energy (G), Spontaneity Criteria & Calculations33
• Hess's Law of Constant Heat Summation & Bond Enthalpies28
• Thermodynamic Terms, State Functions & Extensive/Intensive Properties6
• Second Law of Thermodynamics, Entropy (S) & Entropy Changes5
• Third Law of Thermodynamics & Standard States2

Year-wise Thermodynamics questions in JEE Main

Number of questions from this chapter in each year's paper(s).