Thermodynamics Previous Year Questions — JEE Main Chemistry

44 past-year Thermodynamics questions from JEE Main (Chemistry) — page 9.

Q63 (2026)

Match the List-I with List-II <div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">List-I (Isothermal process for ideal gas system)</th><th style="border: 1px solid #888; padding: 8px;">List-II Work done ($V_f > V_i$)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. Reversible expansion</td><td style="border: 1px solid #888; padding: 8px;">I. $w = 0$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Free expansion</td><td style="border: 1px solid #888; padding: 8px;">II. $w = -nRT \ln \frac{V_f}{V_i}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Irreversible expansion</td><td style="border: 1px solid #888; padding: 8px;">III. $w = -p_{ex}(V_f - V_i)$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Irreversible compression</td><td style="border: 1px solid #888; padding: 8px;">IV. $w = -p_{ex}(V_i - V_f)$</td></tr></tbody></table></div> Choose the correct answer from the options given below :
  1. $\text{A-IV, B-I, C-III, D-II}$
  2. $\text{A-IV, B-II, C-III, D-I}$
  3. $\text{A-I, B-III, C-II, D-IV}$
  4. $\text{A-II, B-I, C-III, D-IV}$
### Related Formula $$W_{\text{Rev}} = -\int P_{\text{gas}} dV$$ $$W_{\text{Irrev}} = -P_{\text{ext}} \Delta V$$ ### Core Logic (A) Reversible isothermal expansion: $$W_{\text{Rev}} = -nRT \ln \left[ \frac{V_f}{V_i} \right]$$ Matches with II. (B) Free expansion (expansion into vacuum): $$P_{\text{ext}} = 0$$ $W = 0$ Matches with I. (C) Irreversible expansion against constant external pressure: $$W_{\text{irrev}} = -P_{\text{ext}} \Delta V$$ $$W_{\text{irrev}} = -P_{\text{ext}} (V_f - V_i)$$ Matches with III. (D) Irreversible compression: Same core formula but since it is a compression from $V_f$ back to $V_i$, the volume change term flips to $\Delta V = (V_i - V_f)$, leading to: $$W_{\text{irrev}} = -P_{\text{ext}} (V_i - V_f)$$ Matches with IV. {{SOL_IMG1}} ### Step 1: Final Conclusion The correct matches are A-II, B-I, C-III, D-IV. ### Pattern Recognition Free expansion always implies zero work because $P_{\text{ext}} = 0$. Reversible implies a continuous integration over volume resulting in the natural log formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

Q67 (2026)

The heat of atomisation of methane and ethane are 'x' kJ mol $^{-1}$ and 'y' kJ mol $^{-1}$ respectively. The longest wavelength ( $\lambda$ ) of light capable of breaking the C-C bond can be expressed in SI unit as:
  1. $\frac{\mathrm{hc}}{1000}\left(\frac{\mathrm{y}-6\mathrm{x}}{4}\right)^{-1}$
  2. $\frac{N_{A}hc}{250(4y-6x)}$
  3. $\frac{N_{A}hc}{250(y-6x)}$
  4. $N_{A}hc\left(y-\frac{6x}{4}\right)^{-1}$
### Related Formula $$E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}$$ Where E is the energy required to break one bond. ### Core Logic For Methane ($\mathrm{CH}_4$): $\mathrm{CH}_4(\mathrm{g}) \rightarrow \mathrm{C(g)} + 4\mathrm{H(g)} \quad \Delta_{\mathrm{r}}\mathrm{H} = x \text{ kJ / mole}$ The energy is used to break 4 C-H bonds: $4 \times \varepsilon_{\mathrm{C-H}} = 1000x \text{ J / mole}$ For Ethane ($\mathrm{C}_2\mathrm{H}_6$): $\mathrm{C}_2\mathrm{H}_6(\mathrm{g}) \rightarrow 2\mathrm{C(g)} + 6\mathrm{H(g)} \quad \Delta_{\mathrm{r}}\mathrm{H} = y \text{ kJ / mole}$ The energy is used to break 1 C-C bond and 6 C-H bonds: $\varepsilon_{\mathrm{C-C}} + 6 \times \varepsilon_{\mathrm{C-H}} = 1000y \text{ J / mole}$ ### Step 1: Extract C-C Bond Energy From methane: $\varepsilon_{\mathrm{C-H}} = \frac{1000x}{4} = 250x \text{ J / mole}$ Substitute into ethane equation: $\varepsilon_{\mathrm{C-C}} + 6 \times (250x) = 1000y$ $\varepsilon_{\mathrm{C-C}} = 1000y - 1500x = \left[y - \frac{3x}{2}\right] \times 1000 \text{ J / mole}$ ### Step 2: Wavelength Calculation The energy calculated above is per mole. Energy required to break *one* C-C bond is: $E = \frac{\varepsilon_{\mathrm{C-C}}}{N_{\mathrm{A}}}$ Now, equating this to photon energy: $\frac{\varepsilon_{\mathrm{C-C}}}{N_{\mathrm{A}}} = \frac{hc}{\lambda}$ $\lambda = \frac{hc \cdot N_{\mathrm{A}}}{\varepsilon_{\mathrm{C-C}}}$ Substitute $\varepsilon_{\mathrm{C-C}}$: $\lambda = \frac{hc \cdot N_{\mathrm{A}}}{\left[y - \frac{3x}{2}\right] \times 1000}$ $\lambda = \frac{hc \cdot N_{\mathrm{A}}}{\frac{2y - 3x}{2} \times 1000}$ $\lambda = \frac{2 \cdot hc \cdot N_{\mathrm{A}}}{1000(2y - 3x)} = \frac{hc \cdot N_{\mathrm{A}}}{500(2y - 3x)} = \frac{hc \cdot N_{\mathrm{A}}}{250(4y - 6x)}$ ### Pattern Recognition Always convert molar quantities to per-bond (atomic scale) quantities by dividing by Avogadro's number ($N_A$) when equating macroscopic enthalpy data to single photon quantum limits ($hc/\lambda$). Don't forget to multiply $kJ$ to $J$ by 1000 for standard SI unit consistency. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Structure of Atom

Q51 (2026)

$20.0~\mathrm{dm}^{3}$ of an ideal gas 'X' at $600\mathrm{~K}$ and $0.5\mathrm{~MPa}$ undergoes isothermal reversible expansion until pressure of the gas is $0.2\mathrm{~MPa}$. Which of the following option is correct? (Given: $\log~2=0.3010$ and $\log~5=0.6989$)
  1. $w=-9.1\mathrm{~kJ}$, $\Delta U=0$, $\Delta H=0$, $q=9.1\mathrm{~kJ}$
  2. $w=9.1\mathrm{~J}$, $\Delta U=9.1\mathrm{~J}$, $\Delta H=0$, $q=0$
  3. $w=+4.1\mathrm{~kJ}$, $\Delta U=0$, $\Delta H=0$, $q=-4.1\mathrm{~kJ}$
  4. $w=-3.9\mathrm{~kJ}$, $\Delta U=0$, $\Delta H=0$, $q=3.9\mathrm{~kJ}$
### Related Formula $$w_{\text{iso}} = -nRT \ln\left(\frac{P_1}{P_2}\right) = -P_1 V_1 \ln\left(\frac{P_1}{P_2}\right)$$ \n $$\Delta U = 0 \quad \text{and} \quad \Delta H = 0 \quad \text{(For isothermal process)}$$ ### Core Logic For an isothermal reversible process involving an ideal gas, the change in internal energy ($\Delta U$) and change in enthalpy ($\Delta H$) are zero. By First Law of Thermodynamics, $q = -w$. ### Step 1: Calculate Work Done $$w_{\text{iso}} = -0.5 \times 10^6 \times 20 \times 10^{-3} \ln\left(\frac{0.5}{0.2}\right)$$\n$$w_{\text{iso}} = -10^4 \times 2.303 \times (\log 5 - \log 2)$$\n$$w_{\text{iso}} = -10^4 \times 2.303 \times (0.6989 - 0.3010)$$\n$$w_{\text{iso}} = -10^4 \times 2.303 \times 0.3979$$\n$$w \approx -9163 \mathrm{~J} = -9.1 \mathrm{~kJ}$$ ### Step 2: Calculate Heat $q = -w = -(-9.1 \mathrm{~kJ}) = 9.1 \mathrm{~kJ}$ ### Pattern Recognition Isothermal expansion of an ideal gas always yields $w < 0$, $q > 0$, and $\Delta U = \Delta H = 0$. Matching signs instantly eliminates non-conforming options. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics

Q55 (2026)

The plot of $\log_{10}K$ vs $\frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).
  1. $(1)\text{ } \frac{2.303R}{\Delta H^{\circ}} , \frac{2.303R}{\Delta S^{\circ}}$
  2. $(2)\text{ } \frac{\Delta S^{\circ}}{2.303R} , -\frac{\Delta H^{\circ}}{2.303R}$
  3. $(3)\text{ } -\frac{\Delta S^{\circ}R}{2.303} , \frac{\Delta H^{\circ}R}{2.303}$
  4. $(4)\text{ } -\frac{\Delta H^{\circ}}{2.303R} , \frac{\Delta S^{\circ}}{2.303R}$
### Related Formula $$\Delta G^{\circ} = -2.303 RT \log_{10}K$$ $$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$$ ### Core Logic Equating both forms for $\Delta G^{\circ}$: $$-2.303 RT \log_{10}K = \Delta H^{\circ} - T\Delta S^{\circ}$$ Divide entirely by $-2.303 RT$: $$\log_{10}K = -\frac{\Delta H^{\circ}}{2.303RT} + \frac{\Delta S^{\circ}}{2.303R}$$ Comparing this with equation of straight line $y = mx + c$ where $y = \log_{10}K$ and $x = \frac{1}{T}$: Slope ($m$) $= -\frac{\Delta H^{\circ}}{2.303R}$ Intercept ($c$) $= \frac{\Delta S^{\circ}}{2.303R}$ ### Step 1: Final Conclusion The intercept is $\frac{\Delta S^{\circ}}{2.303R}$ and the slope is $-\frac{\Delta H^{\circ}}{2.303R}$. ### Pattern Recognition Standard Van't Hoff thermodynamic plot derivation. Always remember to carry the negative sign from $\Delta G$ to the Enthalpy term. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium
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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

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