Kinetic Theory of Gases Previous Year Questions — JEE Main Physics
26 past-year Kinetic Theory of Gases questions from JEE Main (Physics).
Q22 (2025)
The internal energy of air in $4\mathrm{m}\times 4\mathrm{m}\times 3\mathrm{m}$ sized room at $1$ atmospheric pressure will be $\_ \times 10^{6}\mathrm{J}.$
(Consider air as diatomic molecule)
### Related Formula
1. Ideal Gas Law:
$P V = n R T$
2. Internal Energy ($U$) of a diatomic gas ($f = 5$ degrees of freedom):
$$U = n C_v T = n \left(\frac{5}{2} R\right) T = \frac{5}{2} P V$$
### Core Logic
Given parameters:
- Dimensions of the room = $4 \ \mathrm{m} \times 4 \ \mathrm{m} \times 3 \ \mathrm{m}$
- Volume of air in the room $V = 4 \times 4 \times 3 = 48 \ \mathrm{m}^3$
- Room pressure $P = 1 \ \mathrm{atm} = 10^5 \ \mathrm{N/m^2}$
### Step 1: Calculate internal energy
Using the thermodynamic relationship:
$$U = \frac{5}{2} P V$$
Substitute the volume and pressure parameters:
$$U = \frac{5}{2} \times 10^5 \ \mathrm{N/m^2} \times 48 \ \mathrm{m}^3$$
$$U = 5 \times 10^5 \times 24 = 120 \times 10^5 \ \mathrm{J} = 12 \times 10^6 \ \mathrm{J}$$
Thus, the internal energy is $12 \times 10^6 \mathrm{~J}$.
### Pattern Recognition
Sees: Internal energy of diatomic gas occupying a macroscopic room volume.
Trap: Attempting to calculate thermodynamic variables like temperature or density explicitly. Internal energy is completely determined by pressure and volume via degrees of freedom ($U = \frac{f}{2}PV$).
Shortcut: A diatomic gas has $f=5$, so $U = 2.5 PV$. Putting in numbers, $U = 2.5 \times 10^5 \times 48 = 120 \times 10^5 = 12 \times 10^6 \mathrm{~J}$, leaving a coefficient of 12.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q24 (2025)
$\gamma_A$ is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. $\gamma_B$ is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If
$$\frac{\gamma_A}{\gamma_B} = \left(1 + \frac{1}{n}\right)$$
then the value of $n$ is
### Related Formula
$$\gamma = 1 + \frac{2}{f}$$
### Core Logic
Let's find the specific heat ratio for each gas based on degrees of freedom:
1. **Monoatomic gas A:**
- Degrees of freedom, $f_A = 3$ (translational only)
$$\gamma_A = 1 + \frac{2}{3} = \frac{5}{3}$$
2. **Polyatomic gas B:**
- Translational degrees of freedom = 3
- Rotational degrees of freedom = 3
- Vibrational modes = 1.
*Note: Each active vibrational mode has 2 degrees of freedom (kinetic + potential energy terms).* This contributes $2 \times 1 = 2$ degrees of freedom.
- Therefore, the total active degrees of freedom is:
$$f_B = 3 + 3 + 2 = 8$$
The specific heat ratio of B is:
$$\gamma_B = 1 + \frac{2}{f_B} = 1 + \frac{2}{8} = 1 + \frac{1}{4} = \frac{5}{4}$$
Now, find the ratio of specific heat capacities:
$$\frac{\gamma_A}{\gamma_B} = \frac{5/3}{5/4} = \frac{4}{3}$$
We are given:
$$\frac{\gamma_A}{\gamma_B} = 1 + \frac{1}{n} \implies \frac{4}{3} = 1 + \frac{1}{n} \implies \frac{1}{n} = \frac{1}{3} \implies n = 3$$
### Step 1: Final Conclusion
The value of $n$ is 3.
### Pattern Recognition
Always remember that each vibrational mode contributes exactly 2 degrees of freedom because it holds both kinetic and potential energy components ($f_{\text{vib}} = 2 \times \text{modes}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Class 11 Physics: Thermodynamics
Q17 (2025)
Pressure of an ideal gas, contained in a closed vessel, is increased by 0.4% when heated by $1^{\circ}\mathrm{C}$. Its initial temperature must be :
- $25^{\circ}\mathrm{C}$
- 2500 K
- 250 K
- $250^{\circ}\mathrm{C}$
### Related Formula
For an ideal gas in a closed container, the volume $V$ remains constant (isochoric process). By Gay-Lussac's Law:
$$P \propto T \Rightarrow \frac{\Delta P}{P} = \frac{\Delta T}{T}$$
where $T$ must be in Kelvin.
### Core Logic
Given parameters:
- Percent increase in pressure: $\frac{\Delta P}{P} \times 100 = 0.4\% \Rightarrow \frac{\Delta P}{P} = 0.004$
- Increase in temperature $\Delta T = 1^{\circ}\mathrm{C} = 1\mathrm{~K}$
### Step 1: Calculate initial temperature ($T$)
Substitute the values into Gay-Lussac's fractional variance formula:
$$0.004 = \frac{1}{T}$$
$$T = \frac{1}{0.004} = 250\mathrm{~K}$$
### Pattern Recognition
A standard percentage-increase layout. A change of $0.4\%$ means $\frac{1}{250}$ of the original quantity. Hence, a $1\mathrm{~K}$ rise corresponds to an initial temperature of $250\mathrm{~K}$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q238 (2025)
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</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. Triatomic rigid gas</td><td style="border: 1px solid #888; padding: 8px;">I. $\frac{C_{P}}{C_{V}}=\frac{5}{3}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. Diatomic non-rigid gas</td><td style="border: 1px solid #888; padding: 8px;">II. $\frac{C_{P}}{C_{V}}=\frac{7}{5}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Monoatomic gas</td><td style="border: 1px solid #888; padding: 8px;">III. $\frac{C_{P}}{C_{V}}=\frac{4}{3}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Diatomic rigid gas</td><td style="border: 1px solid #888; padding: 8px;">IV. $\frac{C_{P}}{C_{V}}=\frac{9}{7}$</td></tr></tbody></table></div>
Choose the correct answer from the options given below:
- A-III, B-IV, C-I, D-II
- A-III, B-II, C-IV, D-I
- A-II, B-IV, C-I, D-III
- A-IV, B-II, C-III, D-I
### Related Formula
The ratio of specific heats $\gamma$ is related to degrees of freedom $f$ by:
$$\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$$$
### Core Logic
Determine the degrees of freedom $f$ for each type of gas:
- **Monoatomic gas**: Translational only $\implies f = 3$
$$\gamma = 1 + \frac{2}{3} = \frac{5}{3} \quad \text{(Matches C-I)}$$
- **Diatomic rigid gas**: Translational (3) + Rotational (2) $\implies f = 5$
$$\gamma = 1 + \frac{2}{5} = \frac{7}{5} \quad \text{(Matches D-II)}$$
### Step 1: Check Remaining Categories
- **Diatomic non-rigid gas**: Translational (3) + Rotational (2) + Vibrational (2) $\implies f = 7$
$$\gamma = 1 + \frac{2}{7} = \frac{9}{7} \quad \text{(Matches B-IV)}$$
- **Triatomic rigid gas**: Translational (3) + Rotational (3) $\implies f = 6$
$$\gamma = 1 + \frac{2}{6} = 1 + \frac{1}{3} = \frac{4}{3} \quad \text{(Matches A-III)}$$
This yields the matching order: A-III, B-IV, C-I, D-II.
### Pattern Recognition
Sees: Degrees of freedom and $\gamma$ values.
Shortcut: Lower degrees of freedom result in higher $\gamma$ values. Order of degrees of freedom: Monoatomic (3) < Diatomic rigid (5) < Triatomic rigid (6) < Diatomic non-rigid (7). Corresponding $\gamma$: $\frac{5}{3} > \frac{7}{5} > \frac{4}{3} > \frac{9}{7}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory
Q684 (2025)
For a particular ideal gas which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?
- $\text{Graph (3)}$
- $\text{Graph (4)}$
### Related Formula
$$\mathrm{V}_{\mathrm{rms}} = \sqrt{\frac{3\mathrm{R}\mathrm{T}}{\mathrm{M}}} \implies \mathrm{V}_{\mathrm{rms}}^2 = \frac{3\mathrm{R}}{\mathrm{M}}\mathrm{T}$$
### Core Logic
The parameter asked is the mean squared velocity, which corresponds directly to $\mathrm{V}_{\mathrm{rms}}^2$.
From the ideal gas kinematics relation, we observe:
$$\mathrm{V}_{\mathrm{rms}}^2 \propto \mathrm{T}$$
Comparing this format against standard geometric linear templates ($y = mx$), the curve must map as a clean straight line originating from absolute zero zero coordinates.
### Step 1: Final Conclusion
This linear profile matches Graph (1), selecting option (1).
### Pattern Recognition
Watch the vertical ordinate labels carefully: Root-mean-square velocity scales as a sub-linear curve ($\sqrt{\mathrm{T}}$), while mean squared metric trends linearly directly ($y \propto x$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory
