Thermodynamics Previous Year Questions — JEE Main Physics
36 past-year Thermodynamics questions from JEE Main (Physics) — page 6.
Q49 (2026)
10 mole of oxygen is heated at constant volume from $30^{\circ}\text{C}$ to $40^{\circ}\text{C}$. The change in the internal energy of the gas is ____ cal. (The molecular specific heat of oxygen at constant pressure, $C_{p} = 7\text{ cal./mol }^{\circ}\text{C}$ and $R = 2\text{ cal./mol }^{\circ}\text{C}$.)
### Related Formula
$$\Delta U = n C_v \Delta T$$
$C_v = C_p - R$
### Core Logic
Given values:
$n = 10\text{ moles}$
$\Delta T = 40^{\circ}\text{C} - 30^{\circ}\text{C} = 10^{\circ}\text{C}$
$C_p = 7\text{ cal/mol}\cdot^{\circ}\text{C}$
$R = 2\text{ cal/mol}\cdot^{\circ}\text{C}$
First, find $C_v$ using Mayer's relation:
$$C_v = C_p - R = 7 - 2 = 5\text{ cal/mol}\cdot^{\circ}\text{C}$$
### Step 1: Calculate Internal Energy Change
$$\Delta U = n C_v \Delta T$$
$$\Delta U = 10 \times (7 - 2) \times (40 - 30)$$
$$\Delta U = 10 \times 5 \times 10 = 500\text{ cal}$$
### Pattern Recognition
Change in internal energy of an ideal gas is ALWAYS $\Delta U = n C_v \Delta T$, regardless of the process (constant volume or not). Use $C_v = C_p - R$ when $C_p$ is given.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Class 11 Physics: Kinetic Theory of Gases
Q50 (2026)
A diatomic gas ($\gamma = 1.4$) does $100 \text{ J}$ of work when it is expanded isobarically. Then the heat given to the gas ________ J.
### Related Formula
$$W = P\Delta V = nR\Delta T$$
$$Q = nC_p\Delta T$$
$$C_p = \left(\frac{f}{2} + 1\right)R$$
### Core Logic
For an isobaric (constant pressure) process, work done is given by:
$$W = nR\Delta T = 100 \text{ J}$$
For a diatomic gas, the degrees of freedom $f = 5$. Thus, the molar heat capacity at constant pressure is:
$$C_p = \left(\frac{5}{2} + 1\right)R = \frac{7}{2}R$$
### Step 1: Calculating Heat Transfer
The heat supplied to the gas is:
$$Q = nC_p\Delta T = n \left(\frac{7}{2}R\right)\Delta T = \frac{7}{2} (nR\Delta T)$$
### Step 2: Final Conclusion
Substitute the value of work done:
$$Q = \frac{7}{2} \times (100) = 350 \text{ J}$$
### Pattern Recognition
In an isobaric process, the ratio of Work : Internal Energy Change : Heat Added ($W : \Delta U : Q$) is always $2 : f : (f+2)$. For diatomic gases, $f=5$, so the ratio is $2:5:7$. Thus $Q = \frac{7}{2}W$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Class 11 Physics: Kinetic Theory of Gases
Q41 (2026)
The volume of an ideal gas increases 8 times and temperature becomes $(1/4)^{\text{th}}$ of initial temperature during a reversible change. If there is no exchange of heat in this process ($\Delta Q = 0$) then identify the gas from the following options (Assuming the gases given in the options are ideal gases):
- $\mathrm{CO}_{2}$
- $\mathrm{O}_{2}$
- $\mathrm{NH}_{3}$
- He
### Related Formula
$$TV^{\gamma-1} = \text{constant}$$
### Core Logic
{{SOL_IMG1}}
Using adiabatic relation $TV^{\gamma-1} = \text{constant}$:
$$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$$
$$T(V)^{\gamma-1} = \left(\frac{T}{4}\right)(8V)^{\gamma-1}$$
$$4 = 8^{\gamma-1} \implies 2^2 = 2^{3(\gamma-1)} \implies \gamma = \frac{5}{3}$$
Since $\gamma = 5/3$, the gas is monoatomic (Helium / He).
### Pattern Recognition
Sees: Adiabatic expansion with volume and temperature changes.
Shortcut: Apply $TV^{\gamma-1}$ relation to determine adiabatic exponent $\gamma$.
Check: Matches option (4). ✓
### Chapter Mix
Class 11 Physics: Thermodynamics
Q50 (2026)
An insulated cylinder of volume $60 \mathrm{~cm}^3$ is filled with a gas at $27^\circ\mathrm{C}$ and 2 atmospheric pressure. Then the gas is compressed making the final volume as $20 \mathrm{~cm}^3$ while allowing the temperature to rise to $77^\circ\mathrm{C}$. The final pressure is ____ atmospheric pressure.
### Related Formula
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
$$T(\mathrm{K}) = T(^\circ\mathrm{C}) + 273$$
### Core Logic
Converting initial and final temperatures to Kelvin:
$$T_1 = 27^\circ\mathrm{C} + 273 = 300 \mathrm{~K}$$
$$T_2 = 77^\circ\mathrm{C} + 273 = 350 \mathrm{~K}$$
Applying ideal gas law relation:
$$\frac{2 \times 60}{300} = \frac{P_2 \times 20}{350}$$
$$\frac{120}{300} = \frac{20 P_2}{350}$$
$$\frac{2}{5} = \frac{20 P_2}{350} \implies 20 P_2 = 140 \implies P_2 = 7 \mathrm{~atm}$$
### Step 1: Final Conclusion
The final pressure is $7$ atmospheric pressure.
### Pattern Recognition
Combined Gas Law: $P_2 = P_1 \left(\frac{V_1}{V_2}\right) \left(\frac{T_2}{T_1}\right)$.
$P_2 = 2 \times \left(\frac{60}{20}\right) \times \left(\frac{350}{300}\right) = 2 \times 3 \times \frac{7}{6} = 7\text{ atm}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q26 (2026)
The internal energy of a monoatomic gas is $3nRT$. One mole of helium is kept in a cylinder having internal cross section area of $17 \, \mathrm{cm}^2$ and fitted with a light movable frictionless piston. The gas is heated slowly by suppling $126 \, \mathrm{J}$ heat. If the temperature rises by $4^{\circ}\mathrm{C}$ , then the piston will move ____ cm.
(atmospheric pressure = $10^{5} \, \mathrm{Pa}$)
- $14.5$
- $1.55$
- $15.5$
- $1.45$
### Related Formula
$$\Delta Q = \Delta U + W$$
$$\Delta U = n C_v \Delta T$$
$$W = P \Delta V = P (A \cdot \Delta x)$$
### Core Logic
For a monoatomic gas like Helium, $C_v = \frac{3}{2}R$. Given internal energy is $3nRT$, but standard change is $\Delta U = n C_v \Delta T$. Wait, the problem states internal energy $U = 3nRT$ (which implies $C_v = 3R$ for this specific state/setup, or it's a typo in the paper and they mean $U = \frac{3}{2}nRT$). Let's follow the solution's lead: $\Delta U = 3nR\Delta T$ is directly given.
### Step 1: Calculate Change in Internal Energy
Using the given relation:
$$\Delta U = 3nR \Delta T$$
Substitute $n=1$, $R = \frac{25}{3} \, \mathrm{J/mol \cdot K}$, and $\Delta T = 4^{\circ}\mathrm{C}$:
$$\Delta U = 3 \times 1 \times \frac{25}{3} \times 4 = 100 \, \mathrm{J}$$
### Step 2: Apply First Law of Thermodynamics
We know $\Delta Q = 126 \, \mathrm{J}$.
$$W = \Delta Q - \Delta U$$
$$W = 126 - 100 = 26 \, \mathrm{J}$$
### Step 3: Calculate Piston Displacement
Work done is isobaric since the piston is freely moving:
$$W = P \Delta V = P \cdot A \cdot \Delta x$$
$$26 = 10^{5} \times 17 \times 10^{-4} \times \Delta x$$
$$26 = 170 \cdot \Delta x$$
$$\Delta x = \frac{26}{170} \, \mathrm{m} = \frac{26}{170} \times 100 \, \mathrm{cm} = 15.3 \, \mathrm{cm}$$
Closest given option is $15.5 \, \mathrm{cm}$.
### Pattern Recognition
When dealing with a free movable piston, pressure is constant. Apply First Law directly. Watch out for modified $U$ definitions given in the prompt, replacing standard $C_v$ logic.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
