Kinetic Theory of Gases Previous Year Questions — JEE Main Physics
26 past-year Kinetic Theory of Gases questions from JEE Main (Physics) — page 2.
Q8 (2025)
There are two vessels filled with an ideal gas where volume of one is double the volume of other. The large vessel contains the gas at 8 kPa at 1000 K while the smaller vessel contains the gas at 7 kPa at 500 K. If the vessels are connected to each other by a thin tube allowing the gas to flow and the temperature of both vessels is maintained at 600 K, at steady state the pressure in the vessels will be (in kPa).
- 4.4
- 6
- 24
- 18
### Related Formula
Ideal Gas Law:
$$n = \frac{PV}{RT}$$
Conservation of moles:
$n_1 + n_2 = n_f$
### Core Logic
Let the volume of the smaller vessel be $V_1 = V$, then the volume of the larger vessel is $V_2 = 2V$.
Initial moles in large vessel:
$$n_2 = \frac{8 \times 2V}{R \times 1000} = \frac{16V}{1000R}$$
Initial moles in small vessel:
$$n_1 = \frac{7 \times V}{R \times 500} = \frac{14V}{1000R}$$
Total total initial moles:
$$n_{\text{total}} = n_1 + n_2 = \frac{30V}{1000R}$$
### Step 1: Connect Vessels to Dynamic Equilibrium
When connected, the total final volume is $V_f = V + 2V = 3V$.
The final temperature is $T_f = 600\text{ K}$.
Using mole conservation:
$$\frac{30V}{1000R} = \frac{P_f (3V)}{R \times 600}$$
$$\frac{30}{1000} = \frac{3P_f}{600} \implies \frac{30}{1000} = \frac{P_f}{200}$$
$$P_f = \frac{30 \times 200}{1000} = 6\text{ kPa}$$ {{SOL_IMG1}}
### Pattern Recognition
Connecting chambers preserves the net mass/moles ($\sum n_i = \text{constant}$). Keep everything relative to a common volume multiplier $V$ to easily cancel terms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory
Q1 (2025)
The mean free path and the average speed of oxygen molecules at $300\mathrm{~K}$ and $1\mathrm{~atm}$ are $3 \times 10^{-7}\mathrm{~m}$ and $600\mathrm{~m/s}$, respectively. Find the frequency of its collisions.
- $2 \times 10^{10}/\mathrm{s}$
- $9 \times 10^{5}/\mathrm{s}$
- $2 \times 10^{9}/\mathrm{s}$
- $5 \times 10^{8}/\mathrm{s}$
### Related Formula
$$f = \frac{1}{T} = \frac{v_{\text{avg}}}{\lambda}$$
where:
* $f$ = frequency of collisions
* $v_{\text{avg}}$ = average speed of the molecules
* $\lambda$ = mean free path
### Core Logic
Given parameters:
* Average speed, $v_{\text{avg}} = 600\mathrm{~m/s}$
* Mean free path, $\lambda = 3 \times 10^{-7}\mathrm{~m}$
### Step 1: Calculate Frequency
Substitute the values into the formula:
$$f = \frac{600}{3 \times 10^{-7}} = 2 \times 10^{9}\mathrm{~s}^{-1}$$
Hence, the collision frequency is $2 \times 10^{9}/\mathrm{s}$.
### Pattern Recognition
Collision frequency is simply distance covered per unit time (average velocity) divided by the average distance between consecutive collisions (mean free path).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory
Q7 (2025)
The helium and argon are put in the flask at the same room temperature (300 K).
The ratio of average kinetic energies (per molecule) of helium and argon is :
(Give: Molar mass of helium = 4 g/mol, Molar mass of argon $=40~g/mol)$ [cite: 74, 75, 76, 77]
- 1:10 [cite: 79]
- 10:1 [cite: 81]
- 1: $\sqrt{10}$ [cite: 80]
- 1:1 [cite: 82]
### Related Formula
$$\text{K.E.} = \frac{f}{2} k_B T$$ [cite: 688]
### Core Logic
The average kinetic energy per molecule depends only on the temperature $T$ and the degrees of freedom $f$ of the gas[cite: 75, 688]. Both Helium ($\mathrm{He}$) and Argon ($\mathrm{Ar}$) are monoatomic noble gases, meaning both share the same degrees of freedom ($f = 3$)[cite: 689]. Since they sit in the same flask at identical room temperature ($T = 300\ \text{K}$), their translational kinetic energies per molecule are exactly equal [cite: 74, 688]:
$$\frac{\text{K.E.}_{\mathrm{He}}}{\text{K.E.}_{\mathrm{Ar}}} = \frac{1}{1}$$ [cite: 688]
### Pattern Recognition
Do not get distracted by the molar masses given in the question stem[cite: 76, 77]. Kinetic energy per molecule is purely temperature-dependent for an ideal gas, unlike the root-mean-square velocity ($v_{\text{rms}}$) which explicitly includes molecular weight.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q4 (2025)
The ratio of vapour densities of two gases at the same temperature is $\frac{4}{25}$ , then the ratio of r.m.s. velocities will be: [cite: 59-61]
- $\frac{25}{4}$
- $\frac{2}{5}$
- $\frac{5}{2}$
- $\frac{4}{25}$
### Related Formula
The root-mean-square (r.m.s.) velocity of gas molecules is given by:
$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$
Since molecular weight $M$ is directly proportional to the vapour density ($\rho$), the r.m.s. velocity is inversely proportional to the square root of its vapour density:
$$\frac{v_{\text{rms1}}}{v_{\text{rms2}}} = \sqrt{\frac{\rho_2}{\rho_1}}$$
### Core Logic
Given the ratio of vapour densities :
$$\frac{\rho_1}{\rho_2} = \frac{4}{25}$$
Therefore, the ratio of their r.m.s. velocities is:
$$\frac{v_{\text{rms1}}}{v_{\text{rms2}}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$
### Pattern Recognition
R.M.S. velocity changes inversely with the square root of mass or density. Whenever a density ratio is given, simply invert the fraction and take the square root to immediately find the velocity ratio.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q5 (2025)
The kinetic energy of translation of the molecules in $50 \mathrm{~g}$ of $\mathrm{CO}_{2}$ gas at $17^{\circ} \mathrm{C}$ is :
- $3986.3 \mathrm{J}$
- $4102.8 \mathrm{J}$
- $4205.5 \mathrm{~J}$
- $3582.7 \mathrm{J}$
### Related Formula
The total translational kinetic energy of a gas sample depends only on the number of moles and absolute temperature, regardless of whether the molecule is monoatomic or polyatomic:
$$K.E._{\text{translational}} = \frac{3}{2} n R T$$
### Core Logic
Given data:
* Mass of $\mathrm{CO}_2$ gas, $m = 50 \text{ g}$
* Molar mass of $\mathrm{CO}_2$, $M = 44 \text{ g/mol}$
* Absolute Temperature, $T = 17 + 273.15 = 290.15 \text{ K}$
* Universal gas constant, $R \approx 8.314 \text{ J/(mol}\cdot\text{K)}$
Calculate total moles $n$:
$$n = \frac{50}{44} \approx 1.1364 \text{ moles}$$
Substitute values into the expression :
$$K.E._{\text{translational}} = \frac{3}{2} \times \frac{50}{44} \times 8.314 \times 290.15$$
$$K.E._{\text{translational}} = 1.5 \times 1.1364 \times 8.314 \times 290.15 \approx 4108.6 \text{ J}$$
The closest matching value specified in the test alternatives is $4102.8 \mathrm{J}$.
### Pattern Recognition
A common mistake is using $\frac{5}{2}nRT$ or $\frac{7}{2}nRT$ because $\mathrm{CO}_2$ is a triatomic linear molecule. Remember that **translational** kinetic energy is always $\frac{3}{2}nRT$ for any gas sample, as translation has exactly 3 degrees of freedom regardless of the molecular layout.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
