Thermodynamics Previous Year Questions — JEE Main Physics
36 past-year Thermodynamics questions from JEE Main (Physics) — page 3.
Q1 (2025)
During the melting of a slab of ice at $273\mathrm{~K}$ at atmospheric pressure:
- Internal energy of ice-water system remains unchanged.
- Positive work is done by the ice-water system on the atmosphere.
- Internal energy of the ice-water system decreases.
- Positive work is done on the ice-water system by the atmosphere.
### Related Formula
$$\Delta U = \Delta Q + \Delta W_{\text{on system}}$$
where,
$\Delta U$ = change in internal energy,
$\Delta Q$ = heat exchange,
$\Delta W_{\text{on system}}$ = work done on the system.
### Core Logic
During the melting of ice at $273\mathrm{~K}$, the density of water is higher than the density of ice. This means the volume of the ice-water system decreases during melting:
$$V_f < V_i \implies \Delta V < 0$$
Since the system contracts, the atmosphere performs positive work on it:
$$W_{\text{on system}} = -P \Delta V > 0$$
Additionally, heat is absorbed by the system to melt the ice, so $\Delta Q > 0$. By the first law of thermodynamics, since both $\Delta Q$ and $\Delta W_{\text{on system}}$ are positive, the internal energy of the system increases:
$$\Delta U = \Delta Q + W_{\text{on system}} > 0$$
### Step 1: Evaluation of Options
Let's check the given options:
1. Internal energy remains unchanged $\rightarrow$ False (it increases).
2. Positive work is done by the system $\rightarrow$ False (work done by the system is negative since it contracts).
3. Internal energy decreases $\rightarrow$ False.
4. Positive work is done on the ice-water system by the atmosphere $\rightarrow$ True (since volume decreases under atmospheric pressure).
### Pattern Recognition
Remember: Ice contracts upon melting (unlike most solids). Shrinking volume ($V \downarrow$) under positive pressure ($P > 0$) means the surroundings (atmosphere) compress it, performing positive work on the system.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Class 11 Physics: Thermodynamics
Q12 (2025)
A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of $800~\mathrm{cm}^3$ and temperature $27^{\circ}\mathrm{C}$ . The change in temperature when the gas is adiabatically compressed to $200~\mathrm{cm}^3$ is: (Take $\gamma = 1.5$)
- $327\mathrm{~K}$
- $600\mathrm{~K}$
- $522\mathrm{~K}$
- $300\mathrm{~K}$
### Related Formula
For an adiabatic process:
$$T V^{\gamma - 1} = \text{constant}$$
where,
$T$ = absolute temperature in Kelvin,
$V$ = volume of the gas,
$\gamma$ = adiabatic exponent.
### Core Logic
Given values:
- Initial volume, $V_1 = 800\mathrm{~cm}^3$
- Final volume, $V_2 = 200\mathrm{~cm}^3$
- Initial temperature, $T_1 = 27^{\circ}\mathrm{C} = 27 + 273 = 300\mathrm{~K}$
- Adiabatic exponent, $\gamma = 1.5 \implies \gamma - 1 = 0.5$
### Step 1: Calculating Final Temperature
Apply the adiabatic relation:
$$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$$
$$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1}$$
Substitute the values:
$$T_2 = 300 \left(\frac{800}{200}\right)^{0.5} = 300 \times (4)^{0.5}$$
$$T_2 = 300 \times 2 = 600\mathrm{~K}$$
### Step 2: Calculating Change in Temperature
Now compute the change in temperature ($\Delta T$):
$$\Delta T = T_2 - T_1 = 600\mathrm{~K} - 300\mathrm{~K} = 300\mathrm{~K}$$
### Pattern Recognition
Always read carefully to see if the question asks for the **final temperature** or the **change in temperature**. Many students lose marks by choosing $600\mathrm{~K}$ (the final temperature) instead of the difference $300\mathrm{~K}$! Stay sharp.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q17 (2025)
Match 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</th><th style="border: 1px solid #888; padding: 8px;">List-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">(A) Isobaric</td><td style="border: 1px solid #888; padding: 8px;">(I) $\Delta Q=\Delta W$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(B) Isochoric</td><td style="border: 1px solid #888; padding: 8px;">(II) $\Delta Q=\Delta U$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(C) Adiabatic</td><td style="border: 1px solid #888; padding: 8px;">(III) $\Delta Q=\text{zero}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(D) Isothermal</td><td style="border: 1px solid #888; padding: 8px;">(IV) $\Delta Q=\Delta U+P\Delta V$</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- \text{(A)-(IV), (B)-(III), (C)-(II), (D)-(I)}
- \text{(A)-(IV), (B)-(I), (C)-(III), (D)-(II)}
- \text{(A)-(IV), (B)-(II), (C)-(III), (D)-(I)}
- \text{(A)-(II), (B)-(IV), (C)-(III), (D)-(I)}
### Related Formula
First Law of Thermodynamics:
$$\Delta Q = \Delta U + \Delta W$$
### Core Logic
- **Isobaric**: Pressure is constant, work done $\Delta W = P\Delta V$. Thus, $\Delta Q = \Delta U + P\Delta V$ (Matches IV).
- **Isochoric**: Volume is constant, $\Delta V = 0 \implies \Delta W = 0$. Thus, $\Delta Q = \Delta U$ (Matches II).
- **Adiabatic**: No heat transfer, $\Delta Q = 0$ (Matches III).
- **Isothermal**: Temperature is constant, internal energy change $\Delta U = 0$ for an ideal gas. Thus, $\Delta Q = \Delta W$ (Matches I).
### Step 1: Match Compilation
Combining everything gives: (A)-(IV), (B)-(II), (C)-(III), (D)-(I).
### Pattern Recognition
Adiabatic definition is always heat-isolated ($Q=0$), isochoric implies rigid boundaries ($W=0$). Identifying these two immediately isolates the correct match permutation in seconds.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q17 (2025)
Match 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</th><th style="border: 1px solid #888; padding: 8px;">List-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">(A) Isothermal</td><td style="border: 1px solid #888; padding: 8px;">(I) $\Delta W$ (work done) $=0$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(B) Adiabatic</td><td style="border: 1px solid #888; padding: 8px;">(II) $\Delta Q$ (supplied heat) $=0$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(C) Isobaric</td><td style="border: 1px solid #888; padding: 8px;">(III) $\Delta U$ (change in internal energy) $\ne0$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">(D) Isochoric</td><td style="border: 1px solid #888; padding: 8px;">(IV) $\Delta U=0$</td></tr></tbody></table></div>
Choose the correct answer from the options given below: [cite: 151, 152]
- (A)-(III), (B)-(II), (C)-(I), (D)-(IV) [cite: 153]
- (A)-(IV), (B)-(I), (C)-(III), (D)-(II) [cite: 154]
- (A)-(IV), (B)-(II), (C)-(III), (D)-(I) [cite: 155]
- (A)-(II), (B)-(IV), (C)-(I), (D)-(III) [cite: 156]
### Core Logic
Let's evaluate each process condition based on the first law of thermodynamics:
* **(A) Isothermal:** Continuous constant temperature ($\Delta T = 0$) implies that the internal energy change of an ideal gas is zero, so $\Delta U = 0 \implies \text{(IV)}$ [cite: 730].
* **(B) Adiabatic:** No thermal energy transfer occurs between the system and surroundings, meaning $\Delta Q = 0 \implies \text{(II)}$ [cite: 731].
* **(C) Isobaric:** Constant pressure process where both volume and temperature typically vary, so internal energy changes continuously, $\Delta U \neq 0 \implies \text{(III)}$ [cite: 732].
* **(D) Isochoric:** Rigid boundary condition at constant volume ($\Delta V = 0$) ensures work done $\Delta W = P\Delta V = 0 \implies \text{(I)}$ [cite: 733].
Putting these together yields: (A)-(IV), (B)-(II), (C)-(III), (D)-(I)[cite: 155, 729].
### Pattern Recognition
Matching 'Isochoric' with zero work done ($\Delta W=0$) or 'Adiabatic' with zero heat exchange ($\Delta Q=0$) are fundamental definitions that let you rapidly break down multi-choice grids[cite: 731, 733].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q6 (2025)
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases.
Reason (R) : Free expansion of an ideal gas is an irreversible and an adiabatic process.
In the light of the above statement, choose the correct answer from the options given below :
- Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$
- $(A)$ is true but $(R)$ is false
- $(A)$ is false but $(R)$ is true
- Both $(A)$ and $(R)$ are true but $(R)$ is NOT the correct explanation of $(A)$
### Related Formula
$$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$$
### Core Logic
Assertion (A) review:
An insulated container means the process is adiabatic ($Q=0$). When the gas is shrunk (compressed) to half its volume ($V_2 = V_1 / 2$),
$$T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} = T_1 (2)^{\gamma-1}$$
Since $\gamma > 1$, $T_2 > T_1$. Therefore, temperature increases during adiabatic compression, which makes Assertion (A) false.
Reason (R) review:
Free expansion occurs when a gas expands into a vacuum inside an insulated container. No work is done ($W=0$) and no heat is exchanged ($Q=0$), hence it is adiabatic. It cannot spontaneously reverse, so it is irreversible. Thus, Reason (R) is true.
### Pattern Recognition
Adiabatic compression always raises temperature due to work being done on the gas, while free expansion keeps the temperature of an ideal gas constant ($dI=0$ as $W=0, Q=0$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
