Thermodynamics Previous Year Questions — JEE Main Chemistry
44 past-year Thermodynamics questions from JEE Main (Chemistry) — page 5.
Q47 (2025)
Standard entropies of $\mathrm{X}_2, \mathrm{Y}_2$ and $\mathrm{XY}_5$ are 70, 50 and $110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The temperature in Kelvin at which the reaction
$$\frac {1}{2} \mathrm {X} _ {2} + \frac {5}{2} \mathrm {Y} _ {2} \rightarrow \mathrm {X Y} _ {5} \quad \Delta \mathrm {H} ^ {\circ} = - 3 5 \mathrm {k J} \mathrm {m o l} ^ {- 1}$$
will be at equilibrium is (Nearest integer)
### Related Formula
$$\Delta S_{\text{rxn}}^{0} = \sum S_{\text{products}}^{0} - \sum S_{\text{reactants}}^{0} \quad \text{and} \quad T = \frac{\Delta H^0}{\Delta S^0} \quad \text{at equilibrium (}\Delta G^0 = 0\text{)}$$
### Core Logic
First, calculate the standard entropy change for the reaction system ($$\Delta S_{\text{rxn}}^{0}$$):
$$\Delta S_{\text{rxn}}^{0} = S^{0}(XY_5) - \left[ \frac{1}{2}S^{0}(X_2) + \frac{5}{2}S^{0}(Y_2) \right]$$
$$\Delta S_{\text{rxn}}^{0} = 110 - \left[ \left(\frac{1}{2} \times 70\right) + \left(\frac{5}{2} \times 50\right) \right] = 110 - [35 + 125]$$
$$\Delta S_{\text{rxn}}^{0} = 110 - 160 = -50\text{ J K}^{-1}\text{ mol}^{-1}$$
At thermodynamic equilibrium, the change in Gibbs free energy drops to zero ($\Delta G^0 = 0$):
$$0 = \Delta H^0 - T\Delta S^0 \implies T = \frac{\Delta H^0}{\Delta S^0}$$
Convert the enthalpy value into Joules ($$\Delta H^0 = -35 \times 10^3\text{ J/mol}$$) and substitute the parameters:
$$T = \frac{-35000\text{ J mol}^{-1}}{-50\text{ J K}^{-1}\text{ mol}^{-1}} = 700\text{ Kelvin}$$
### Pattern Recognition
Ensure all variables use matching energy units (Joules vs. Kilojoules) before setting up your final division step.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q1233 (2025)
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path $\mathrm{A}\to \mathrm{B}\to \mathrm{C} \rightarrow \mathrm{D}\rightarrow \mathrm{A}$ as shown in the three cases below.
Choose the correct option regarding $\Delta U$:
{{IMG1}}
- $\Delta U\text{ (Case-III)} > \Delta U\text{ (Case-II)} > \Delta U\text{ (Case-I)}$
- $\Delta U\text{ (Case-I)} > \Delta U\text{ (Case-II)} > \Delta U\text{ (Case-III)}$
- $\Delta U\text{ (Case-I)} > \Delta U\text{ (Case-III)} > \Delta U\text{ (Case-II)}$
- $\Delta U\text{ (Case-I)} = \Delta U\text{ (Case-II)} = \Delta U\text{ (Case-III)}$
### Related Formula
For any state function like Internal Energy ($U$), the cyclic integral over a complete closed loop is identically zero:
$$\oint dU = 0 \implies \Delta U_{\text{cyclic}} = 0$$
### Core Logic
Internal energy ($U$) depends only on the initial and final states of the thermodynamic system, not on the path followed.
In all three listed cases, the ideal gas undergoes a complete cyclic path that returns to its original configuration state $A$.
### Step 1: Final Evaluation
Since every transformation begins and ends at point $A$:
$$\Delta U_{\text{Case-I}} = 0$$
$$\Delta U_{\text{Case-II}} = 0$$
$$\Delta U_{\text{Case-III}} = 0$$
Therefore, $\Delta U\text{ (Case-I)} = \Delta U\text{ (Case-II)} = \Delta U\text{ (Case-III)}$.
### Pattern Recognition
Do not waste time calculating path areas or values if the question asks for a state function change ($\Delta U$, $\Delta H$, $\Delta S$, $\Delta G$) over a cyclic loop. The answer is instantly zero for all cases!
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q47 (2025)
Consider the following data:
Heat of formation of $CO_{2}(g) = -393.5\mathrm{~kJ~mol^{-1}}$
Heat of formation of $H_{2}O(l) = -286.0\mathrm{~kJ~mol^{-1}}$
Heat of combustion of benzene $= -3267.0\mathrm{~kJ~mol^{-1}}$
The heat of formation of benzene is ______ $\mathrm{kJ~mol^{-1}}$ (Nearest integer).
### Related Formula
Enthalpy of reaction from enthalpy of formation data:
$$\Delta H_{\text{reaction}} = \sum \Delta H_f(\text{Products}) - \sum \Delta H_f(\text{Reactants})$$
### Core Logic
Write out the balanced thermochemical equation for the combustion of benzene ($C_6H_6$):
$$C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$$
Given parameters:
- $\Delta H_c = -3267.0\mathrm{\ kJ/mol}$
- $\Delta H_f[CO_2] = -393.5\mathrm{\ kJ/mol}$
- $\Delta H_f[H_2O] = -286.0\mathrm{\ kJ/mol}$
- $\Delta H_f[O_2] = 0\mathrm{\ kJ/mol}$
### Step 1: Applying Hess's Law
Substitute these values into the reaction expression:
$$-3267 = [6(-393.5) + 3(-286.0)] - \Delta H_f[C_6H_6]$$
$$-3267 = [-2361.0 - 858.0] - \Delta H_f[C_6H_6]$$
$$-3267 = -3219.0 - \Delta H_f[C_6H_6]$$
$$\Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48\mathrm{\ kJ/mol}$$
### Pattern Recognition
Always set up products minus reactants when using heat of formation data. Pay close attention to stoichiometric coefficients (multiply $CO_2$ by $6$ and $H_2O$ by $3$) to ensure accurate bookkeeping.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q1313 (2025)
$500 \mathrm{~J}$ of energy is transferred as heat to $0.5 \mathrm{~mol}$ of Argon gas at $298 \mathrm{~K}$ and $1.00 \mathrm{atm}$ . The final temperature and the change in internal energy respectively are :
Given: $\mathrm{R} = 8.3 \, \mathrm{JK}^{-1} \mathrm{mol}^{-1}$
- $348\mathrm{K}$ and $300\mathrm{J}$
- $378\mathrm{K}$ and $300\mathrm{J}$
- $368\mathrm{K}$ and $500\mathrm{J}$
- $378\mathrm{K}$ and $500\mathrm{J}$
### Formulas Used
For an ideal gas undergoing a constant pressure process ($1.00\text{ atm}$):
$$q_p = n \cdot C_p \cdot \Delta T$$
Change in internal energy ($\Delta U$):
$$\Delta U = n \cdot C_v \cdot \Delta T$$
For a monoatomic gas like Argon:
* $C_v = \frac{3}{2} R$
* $C_p = \frac{5}{2} R$
### Core Logic
**Step 1: Calculate the final temperature ($T_f$)**
Heat transferred at constant pressure ($q_p$) = $500\text{ J}$
$$500 = 0.5 \times \left(\frac{5}{2} \times 8.3\right) \times (T_f - 298)$$
$$500 = 0.5 \times 20.75 \times (T_f - 298)$$
$$500 = 10.375 \times (T_f - 298)$$
$$T_f - 298 = \frac{500}{10.375} \approx 48.2\text{ K}$$
$$T_f = 298 + 48.2 = 346.2\text{ K} \approx 348\text{ K}$$
---
**Step 2: Calculate the change in internal energy ($\Delta U$)**
$$\Delta U = n \cdot C_v \cdot \Delta T$$
Alternatively, using the ratio of heat capacities:
$$\Delta U = \left(\frac{C_v}{C_p}\right) \times q_p = \frac{3}{5} \times 500\text{ J} = 300\text{ J}$$
Thus, the final temperature is **$348\text{ K}$** and the change in internal energy is **$300\text{ J}$**.
### Pattern Recognition
For a monoatomic ideal gas under constant pressure, exactly $60\%$ of the heat added ($\frac{C_v}{C_p} = \frac{3}{5}$) goes into increasing the internal energy ($\Delta U$), while $40\%$ is lost to expansion work ($W$).
**Correct Option:** **(A)**
Q79 (2024)
Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following:
- $q = 0, \Delta T \neq 0, w = 0$
- $q = 0, \Delta T < 0, w \neq 0$
- $q \neq 0, \Delta T = 0, w = 0$
- $q = 0, \Delta T = 0, w = 0$
### Core Logic
Free expansion means expansion against a vacuum ($P_{ext} = 0$).
Work done: $w = -P_{ext} \Delta V$. Since $P_{ext} = 0$, $w = 0$.
Adiabatic condition means there is no heat exchange with the surroundings.
Heat transfer: $q = 0$.
According to the First Law of Thermodynamics, $\Delta U = q + w$.
Since $q = 0$ and $w = 0$, the change in internal energy $\Delta U = 0$.
For an ideal gas, internal energy is a function of temperature only ($\Delta U = nC_v\Delta T$).
If $\Delta U = 0$, then $\Delta T = 0$.
### Step 1: Final Parameter Check
Evaluating all parameters simultaneously:
$q = 0$
$w = 0$
$\Delta T = 0$
### Pattern Recognition
Adiabatic + Free Expansion of IDEAL gas $\rightarrow$ Nothing changes thermodynamically except volume and pressure. $q = 0, w = 0, \Delta U = 0, \Delta T = 0, \Delta H = 0$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
