Thermodynamics Previous Year Questions — JEE Main Chemistry
44 past-year Thermodynamics questions from JEE Main (Chemistry).
Q38 (2025)
Arrange the following in order of magnitude of work done by the system / on the system at constant temperature :
(a) $|\mathrm{w}_{\mathrm{reversible}}|$ for expansion in infinite stage.
(b) $|\mathrm{w}_{\mathrm{irreversible}}|$ for expansion in single stage.
(c) $|\mathrm{w}_{\mathrm{reversible}}|$ for compression in infinite stage.
(d) $|\mathrm{w}_{\mathrm{irreversible}}|$ for compression in single stage.
Choose the correct answer from the options given below:
- $a > b > c > d$
- $\mathrm{d} > \mathrm{c} = \mathrm{a} > \mathrm{b}$
- $c = a > d > b$
- $a > c > b > d$
### Related Formula
$$w_{\text{rev}} = -nRT \ln\left(\frac{V_{\mathrm{f}}}{V_{\mathrm{i}}}\right)$$
$$w_{\text{irrev}} = -P_{\text{ext}} \left(V_{\mathrm{f}} - V_{\mathrm{i}}\right)$$
### Core Logic
For isothermal reversible and irreversible steps:
1. **Reversible Path**: Since a reversible compression path retraces the exact coordinates of the reversible expansion path, the magnitudes of work are equal:
$$|w_{\text{rev, expansion}}| = |w_{\text{rev, compression}}| \implies a = c$$
{{SOL_IMG1}}
2. **Isothermal Expansion**: Reversible work magnitude is the maximum possible work. Hence, for expansion:
$$|w_{\text{rev, expansion}}| > |w_{\text{irrev, expansion}}| \implies a > b$$
{{SOL_IMG2}}
3. **Isothermal Compression**: Irreversible compression requires more work than reversible compression because of sudden pressure adjustments against the surroundings:
$$|w_{\text{irrev, compression}}| > |w_{\text{rev, compression}}| \implies d > c$$
{{SOL_IMG3}}
{{SOL_IMG4}}
### Step 1: Combine the inequalities
Combining the results:
- We have $a = c$
- We have $d > c$
- We have $a > b$
This leads to the strict inequality sequence:
$d > c = a > b$
### Pattern Recognition
Thermodynamics Principle: Reversible expansion is the most efficient (gives maximum work magnitude), whereas reversible compression is the most efficient (requires minimum work magnitude). Single-stage irreversible compression is always the least efficient, demanding the absolute highest work input.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q456 (2025)
Two vessels A and B are connected via stopcock. The vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and is allowed to come to thermal equilibrium with water. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed in the thermometer. Which of the following statement is true?
{{IMG1}}
- $(1)\ \mathrm{dw} \neq 0$
- $(2)\ \mathrm{dq} \neq 0$
- $(3)\ \mathrm{dU} \neq 0$
- $(4)\ \text{The pressure in the vessel B before opening the stopcock is zero.}$
### Related Formula
First Law of Thermodynamics expression:
$$\mathrm{dU = dq + dw}$$
### Core Logic
The system parameters show an isothermal transformation layout with zero overall heat transfer step:
* No change in temperature signifies $\mathrm{dT} = 0$, hence internal energy change for an ideal gas satisfies:
$$\mathrm{dU} = nC_v\mathrm{dT} = 0$$
* Since it expands freely into an empty chamber (vessel B), external pressure $P_{\text{ext}} = 0$, meaning work done is:
$$\mathrm{dw} = -P_{\text{ext}}\mathrm{dV} = 0$$
* Combining these parameters in the First Law gives $\mathrm{dq} = 0$.
* This classic situation of "free expansion" implies vessel B was completely evacuated initially.
### Step 1: Statement Verification
Therefore, the pressure inside vessel B before opening the stopcock was precisely zero.
### Pattern Recognition
Isothermal + expansion against no opposing force = Free Expansion. For free expansion of an ideal gas, always remember: $w = 0$, $q = 0$, and $\Delta U = 0$ simultaneously.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Chemical Thermodynamics
Q41 (2025)
Given below are two statements:
Statement I: When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the system until whole ice gets melted.
Statement II: At melting point of ice, there is absorption of heat in order to overcome intermolecular forces of attraction within the molecules of water in ice and kinetic energy of molecules is not increased at melting point.
In the light of the above statements, choose the correct answer from the options given below:
- Statement I is true but Statement II is false
- Both Statement I and Statement II are false
- Both Statement I and Statement II are true
- Statement I is false but Statement II is true
### Related Formula
During phase transition, latent heat of fusion ($L_f$) is absorbed:
$Q = m L_f$
Temperature remains constant because the average kinetic energy of the molecules does not change; instead, potential energy changes during phase transition:
$$\text{Temperature } T \propto \text{Average Kinetic Energy of molecules}$$
### Core Logic
Statement I Analysis:
- During a phase transition (such as ice melting at $0^{\circ}\mathrm{C}$), any added heat is utilized as latent heat of fusion. The system remains at a constant temperature of $0^{\circ}\mathrm{C}$ as long as both solid and liquid phases coexist in equilibrium. Thus, Statement I is True.
### Step 1: Analyze Statement II
- Statement II explains *why* this happens: The thermal energy is spent entirely to break down the highly ordered crystalline hydrogen-bonded lattice of ice into liquid water. It does not increase the translational kinetic energy of the molecules. Since kinetic energy is constant, temperature remains constant. Thus, Statement II is True.
### Step 2: Conclusion
Therefore, both Statement I and Statement II are True, matching Option (3).
### Pattern Recognition
For any phase transition (melting, boiling, sublimation): Temperature stays flat. The added energy goes entirely into latent heat (potential energy change to overcome intermolecular forces), meaning average kinetic energy is constant.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Class 11 Physics: Thermal Properties of Matter
Q47 (2025)
A perfect gas ($0.1\mathrm{~mol}$) having $\bar{C}_{v}=1.50\mathrm{~R}$ (independent of temperature) undergoes the transformation from point 1 to point 4 as shown in the pressure-volume diagram below. If each step is reversible, the total work done (w) while going from point 1 to point 4 is ($-$) ________ J. (nearest integer) {{IMG1}}
[Given: $R=0.082\mathrm{~L~atm~K^{-1}~mol^{-1}} = 8.314\mathrm{~J~K^{-1}~mol^{-1}}$]
### Related Formula
Thermodynamic work done ($w$) for each step:
- Isochoric step ($V = \text{constant}$):
$w = 0$
- Isobaric step ($P = \text{constant}$):
$w = -P \Delta V$
### Core Logic
The process from point 1 to point 4 consists of three distinct segments:
1. Step $1 \rightarrow 2$: Isochoric cooling at constant volume $V_1 = 1000\mathrm{~cm}^3$. Work done $w_{1\rightarrow 2} = 0$.
2. Step $2 \rightarrow 3$: Isobaric compression at constant pressure $P = 3.00\mathrm{~atm}$ from volume $2000\mathrm{~cm}^3$ to $1000\mathrm{~cm}^3$.
3. Step $3 \rightarrow 4$: Isochoric step at constant volume. Work done $w_{3\rightarrow 4} = 0$.
### Step 1: Calculate work done in the isobaric step ($2 \rightarrow 3$)
The volume changes from $V_i = 2000\mathrm{~cm}^3 = 2.0\mathrm{~L}$ to $V_f = 1000\mathrm{~cm}^3 = 1.0\mathrm{~L}$:
$$w_{2\rightarrow 3} = -P \Delta V = -3.00\mathrm{~atm} \times (1.0\mathrm{~L} - 2.0\mathrm{~L}) = +3.00\mathrm{~L\cdot atm}$$$
### Step 2: Convert work to Joules and analyze direction
Convert $\mathrm{L\cdot atm}$ to Joules:
$$w_{2\rightarrow 3} = 3.00 \times 101.325\mathrm{~J} = 303.975\mathrm{~J} \approx 304\mathrm{~J}$$
The question asks for the total work done as ($-$) ________ J, meaning work done *by* the system (expansion) is negative and work done *on* the system (compression) is positive. Since this is compression, work done on the gas is $+304\mathrm{~J}$, which is represented as $-(-304)\mathrm{~J}$ in typical IUPAC convention where work of expansion is examined. The absolute magnitude of the work is $304\mathrm{~J}$.
### Pattern Recognition
During any cyclic or multi-step path on a P-V graph, work is done *only* when there is a change in volume ($W = -\int P dV$). Any vertical line (constant volume) represents an isochoric step where work is exactly zero. The horizontal segment directly represents rectangular area under the path.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Class 11 Physics: Thermodynamics
Q48 (2025)
A sample of n-octane ($1.14\mathrm{~g}$) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is $5\mathrm{~kJ~K^{-1}}$. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K. The magnitude of the heat of combustion of octane at constant volume is ________ $\mathrm{kJ~mol^{-1}}$. (nearest integer)
### Related Formula
Heat released at constant volume ($q_v$) in a bomb calorimeter is:
$$q_v = C_{\text{cal}} \cdot \Delta T$$
Molar heat of combustion at constant volume ($\Delta U_{\text{comb}}$):
$$\Delta U_{\text{comb}} = \frac{q_v}{n_{\text{fuel}}}$$
### Core Logic
Given parameters:
- Mass of n-octane $m = 1.14\mathrm{~g}$
- Heat capacity of calorimeter $C_{\text{cal}} = 5\mathrm{~kJ/K}$
- Temperature rise $\Delta T = 5\mathrm{~K}$
- Formula of octane: $\mathrm{C_8H_{18}}$ $\Rightarrow$ Molar mass $= 8(12) + 18(1) = 114\mathrm{~g/mol}$
### Step 1: Calculate heat absorbed by the calorimeter ($q_v$)
$$q_v = 5\mathrm{~kJ/K} \times 5\mathrm{~K} = 25\mathrm{~kJ}$$
### Step 2: Calculate moles of octane
$$n = \frac{1.14\mathrm{~g}}{114\mathrm{~g/mol}} = 0.01\mathrm{~mol}$$$
### Step 3: Calculate molar heat of combustion
$$\Delta U_{\text{comb}} = \frac{25\mathrm{~kJ}}{0.01\mathrm{~mol}} = 2500\mathrm{~kJ/mol}$$
The magnitude of the heat of combustion is $2500\mathrm{~kJ~mol^{-1}}$.
### Pattern Recognition
A bomb calorimeter operates at rigid constant volume, meaning boundary work $w = 0$. Thus, by the first law of thermodynamics, the measured heat flow represents the internal energy change ($\Delta U$), not the enthalpy change ($\Delta H$, which occurs at constant pressure).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
