Thermodynamics Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Chemical Thermodynamics with concise step-by-step solutions covering first law, Hess's law, Gibbs free energy, entropy, heat capacities and isothermal work.
Practice the actual JEE Main 2026 Thermodynamics questions below, each worked out step by step so you can check both the final answer and the method.
Solutions are AI-generated and pending review.
Solution
For an ideal gas the Mayer relation holds:
$$C_p = C_V + R$$Since $R > 0$, we always have $C_p > C_V$. So Statement I (claiming $C_V > C_p$) is false.
At constant volume, $w = -p_{ext}\,\Delta V = 0$, so the first law gives:
$$q_V = \Delta U = nC_V\,\Delta T$$All the heat added raises the internal (chaotic) molecular energy and hence the temperature. Statement II is true.
Answer: D
Solution
For an isothermal gas moving between the same two states, the work magnitude depends on the external pressure profile.
Expansion (1 → 2, gas pushes out):
- Single stage against the low final pressure $p_2$: $|w_A| = p_2(V_2 - V_1)$ — smallest expansion work.
- Multi stage / reversible (external pressure tracks the gas): $|w_B| = nRT\ln\dfrac{V_2}{V_1}$ — larger.
Compression (2 → 1, surroundings push in):
- Multi stage / reversible: $|w_D| = nRT\ln\dfrac{V_2}{V_1} = |w_B|$ — minimum compression work.
- Single stage against the high pressure $p_1$: $|w_C| = p_1(V_2 - V_1)$ — maximum.
So the ranking is:
$$|w_C| > |w_B| = |w_D| > |w_A|$$The single-stage compression is largest and the single-stage expansion is smallest. Among the choices, this matches:
$$|w_C| > |w_D| > |w_B| > |w_A|$$Answer: C
Solution
First find $\Delta G^\theta$ from the equilibrium constant:
$$\Delta G^\theta = -RT\ln K$$Evaluate $\ln K = 2.303\log(1.8\times10^{-7})$ using the given logs:
$$\log 1.8 = \log 2 + 2\log 3 - 1 = 0.30 + 0.96 - 1 = 0.26$$$$\log K = 0.26 - 7 = -6.74 \implies \ln K = 2.3 \times (-6.74) = -15.5$$$$\Delta G^\theta = -(8.3)(300)(-15.5) = 38{,}600 \text{ J mol}^{-1}$$Now use $\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta$:
$$\Delta S^\theta = \frac{\Delta H^\theta - \Delta G^\theta}{T} = \frac{28{,}400 - 38{,}600}{300} = -34 \text{ J K}^{-1}\text{mol}^{-1}$$Magnitude $= 34$.
Answer: 34
Solution
Target reaction:
$$2\mathrm{Al}(s) + 3\mathrm{Cl}_2(g) \rightarrow \mathrm{Al}_2\mathrm{Cl}_6(s)$$Assign $\Delta H$ values (heat released ⇒ negative):
$$\Delta H_1=-1200,\quad \Delta H_2=-164,\quad \Delta H_3=-83,\quad \Delta H_4=-663 \text{ kJ}$$Combine: take (i), add $3\times$(ii) to make 6 HCl(g), add $6\times$(iii) to convert HCl(g) → HCl(aq), and reverse (iv) to precipitate the solid. The aqueous $\mathrm{Al}_2\mathrm{Cl}_6(aq)$, $3\mathrm{H}_2$, HCl(g) and HCl(aq) species all cancel, leaving the target:
$$\Delta H_f = \Delta H_1 + 3\Delta H_2 + 6\Delta H_3 - \Delta H_4$$$$\Delta H_f = -1200 + 3(-164) + 6(-83) - (-663)$$$$= -1200 - 492 - 498 + 663 = -1527 \text{ kJ mol}^{-1}$$Answer: $-1527$ kJ mol$^{-1}$ (D)
Solution
At the transition temperature the two states are in equilibrium, so $\Delta G^0 = 0$:
$$105 - 35\log T = 0 \implies \log T = 3 \implies T = 10^{3} = 1000 \text{ K}$$Convert to Celsius:
$$t = 1000 - 273 = 727\ ^\circ\text{C}$$Answer: 727
Solution
Molar mass of C$_2$H$_5$OH $= 46$ g mol$^{-1}$, so moles burnt:
$$n = \frac{3.365}{46} = 0.07315 \text{ mol}$$A bomb calorimeter is a constant-volume system, so the measured heat is $\Delta U_c$:
$$\Delta U_c = \frac{-99.472}{0.07315} = -1359.8 \text{ kJ mol}^{-1}$$Convert to $\Delta H_c$ for $\mathrm{C_2H_5OH}(l) + 3\mathrm{O_2}(g) \rightarrow 2\mathrm{CO_2}(g) + 3\mathrm{H_2O}(l)$, with $\Delta n_g = 2-3 = -1$:
$$\Delta H_c = \Delta U_c + \Delta n_g RT = -1359.8 + (-1)(8.314\times10^{-3})(298.15) \approx -1362.3 \text{ kJ mol}^{-1}$$Apply Hess’s law, $\Delta H_c = 2\Delta H_f(\mathrm{CO_2}) + 3\Delta H_f(\mathrm{H_2O}) - \Delta H_f(\text{ethanol})$:
$$\Delta H_f(\text{ethanol}) = 2(-393.5) + 3(-285.8) - (-1362.3) = -282.1 \text{ kJ mol}^{-1}$$$$|\Delta H_f^\ominus| = 282.1 \text{ kJ mol}^{-1} = 2.82\times10^2 \approx 3\times10^2$$Answer: 3
Solution
Internal energy is a state function. For X → Y (using $\Delta U = q - w_{by}$):
$$\Delta U_{X\to Y} = q - w_{by} = 10 - 18 = -8 \text{ J}$$For the return Y → X, the change reverses sign:
$$\Delta U_{Y\to X} = +8 \text{ J}, \qquad q = -6 \text{ J (evolved)}$$Apply the first law again:
$$\Delta U_{Y\to X} = q - w_{by} \implies 8 = -6 - w_{by} \implies w_{by} = -14 \text{ J}$$A negative $w_{by}$ means 14 J of work is done on the gas by the surroundings.
Answer: D
Solution
- A. Reversible isothermal expansion: $\Delta U = 0$, so $q = -w = nRT\ln\dfrac{V_2}{V_1}$ → II.
- B. Free expansion: $p_{ext}=0 \Rightarrow w=0$; isothermal ideal gas has $\Delta U=0$, so $q=0$ → I.
- C. Irreversible compression: work against constant external pressure, $w=-p_{ext}(V_2-V_1)=-p_{ext}(V_1-V_2)$ as written → III.
- D. Cyclic reversible: entropy is a state function, so over a cycle $\displaystyle\oint \dfrac{q_{rev}}{T}=0$ → IV.
So A-II, B-I, C-III, D-IV.
Answer: C
Solution
Enthalpy of reaction:
$$\Delta_r H^\circ = 2(42) - (8 + 80) = 84 - 88 = -4 \text{ kJ mol}^{-1}$$Entropy of reaction:
$$\Delta_r S^\circ = 2(200) - (140 + 250) = 400 - 390 = 10 \text{ J mol}^{-1}\text{K}^{-1}$$Gibbs free energy at 600 K:
$$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ = -4 - 600\times\frac{10}{1000} = -4 - 6 = -10 \text{ kJ mol}^{-1}$$Answer: $-10$ (B)
Solution
Compare typical molar heat capacities at 298 K:
- Helium(g) — monatomic ideal gas, $C_{p,m} = \tfrac{5}{2}R \approx 20.8$ J mol$^{-1}$K$^{-1}$ (only translational modes).
- Copper(s) — solid, $C_{p,m} \approx 24.4$ J mol$^{-1}$K$^{-1}$ (Dulong–Petit, $\approx 3R$).
- Bromine(l) — a polyatomic liquid with many active vibrational/rotational modes, $C_{p,m} \approx 75.7$ J mol$^{-1}$K$^{-1}$, the highest.
So the order is:
$$\text{Bromine(l)} > \text{Copper(s)} > \text{Helium(g)}$$Answer: B
Solution
Using $\Delta_r H = \sum \Delta_f H_{\text{products}} - \sum \Delta_f H_{\text{reactants}}$, with $\Delta_f H(\mathrm{O_2}) = 0$:
$$\Delta_r H = \big[2(-286.0) + 2(-297.0)\big] - \big[2(-20.1) + 3(0)\big]$$$$= (-572.0 - 594.0) - (-40.2) = -1166.0 + 40.2 = -1125.8 \text{ kJ mol}^{-1}$$Magnitude $= 1125.8 \approx 1126$.
Answer: 1126