Kinetic Theory of Gases Previous Year Questions — JEE Main Physics
26 past-year Kinetic Theory of Gases questions from JEE Main (Physics) — page 5.
Q30 (2026)
The r.m.s speed of oxygen molecules at $47^{\circ}\text{C}$ is equal to that of the hydrogen molecules kept at ________ $^{\circ}\text{C}$. (Mass of oxygen molecule / mass of hydrogen molecule = 32 / 2)
- $-235$
- $-100$
- $-253$
- $-20$
### Related Formula
$$V_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$
### Core Logic
Given that the RMS speed of oxygen molecules equals the RMS speed of hydrogen molecules:
$$V_{\text{rms}(O_2)} = V_{\text{rms}(H_2)}$$
$$\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}$$
Squaring both sides:
$$\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$$
### Step 1: Absolute Temperature Conversion
Temperature of oxygen in Kelvin:
$$T_{O_2} = 273 + 47 = 320 \text{ K}$$
### Step 2: Evaluating for Hydrogen
Substitute the given mass ratio and absolute temperature:
$$\frac{320}{32} = \frac{T_{H_2}}{2}$$
$$10 = \frac{T_{H_2}}{2}$$
$$T_{H_2} = 20 \text{ K}$$
### Step 3: Final Conclusion
Convert back to Celsius:
$$T_{H_2} \text{ in }^{\circ}\text{C} = 20 - 273 = -253^{\circ}\text{C}$$
### Pattern Recognition
Direct proportionality between Temperature and Molecular Mass for equal RMS speeds. $T_1/M_1 = T_2/M_2$. Always convert to Kelvin before evaluating.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q37 (2026)
A cylindrical tube AB of length l, closed at both ends contains an ideal gas of 1 mol having molecular weight M. The tube is rotated in a horizontal plane with constant angular velocity $\omega$ about an axis perpendicular to AB and passing through the edge at end A, as shown in the figure. If $P_{A}$ and $P_{B}$ are the pressures at A and B respectively, then (Consider the temperature is same at all points in the tube)
{{IMG1}}
- $P_{B} = P_{A}\exp (M\omega^{2}l^{2} / 2RT)$
- $P_{B} = P_{A}$
- $P_{B} = P_{A}\exp (M\omega^{2}l^{2} / 3RT)$
- $P_{B} = P_{A}\exp (M\omega^{2}l^{2} / RT)$
### Related Formula
$$dP = \rho \omega^2 x dx, \quad PM = \rho RT$$
### Core Logic
{{SOL_IMG1}}
Setting up differential pressure equation:
$$A[(P+dP) - P] = (dm)(\omega^2 x) \implies dP = \frac{dm}{A} \omega^2 x$$
Substituting $\rho = \frac{PM}{RT}$:
$$\int_{P_A}^{P_B} \frac{dP}{P} = \frac{\omega^2 M}{RT} \int_0^l x dx$$
$$\ln\left(\frac{P_B}{P_A}\right) = \frac{\omega^2 l^2 M}{2RT} \implies P_B = P_A e^{\frac{M\omega^2 l^2}{2RT}}$$
### Pattern Recognition
Sees: Rotating gas column with centrifugal pressure gradient.
Shortcut: Integrate hydrostatic equation with centripetal acceleration term.
Check: Matches option (1). ✓
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q28 (2026)
Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half of that of B and mass of molecule A is four times that of B. If the collision frequency in gas B is $32 \times 10^{18}/\mathrm{s}$ then collision frequency in gas A is ____/s.
- $32 \times 10^{8}$
- $4 \times 10^{8}$
- $2 \times 10^{8}$
- $8 \times 10^{8}$
### Related Formula
$$Z = \sqrt{2}\pi d^{2} N \sqrt{\frac{8RT}{\pi M}}$$
where $d$ is molecular diameter, $N$ is number density, $T$ is temperature, and $M$ is molar mass.
### Core Logic
Given that $T$ and $N$ are identical for both gases:
$$Z \propto \frac{d^2}{\sqrt{M}}$$
From the problem statement:
$$d_A = \frac{d_B}{2}, \quad M_A = 4M_B$$
Calculating the ratio of collision frequencies:
$$\frac{Z_A}{Z_B} = \left(\frac{d_A}{d_B}\right)^2 \times \sqrt{\frac{M_B}{M_A}} = \left(\frac{1}{2}\right)^2 \times \sqrt{\frac{1}{4}} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$$
Substituting $Z_B = 32 \times 10^8 /\mathrm{s}$:
$$Z_A = \frac{32 \times 10^8}{8} = 4 \times 10^8 /\mathrm{s}$$
### Step 1: Final Conclusion
The collision frequency in gas A is $4 \times 10^8 /\mathrm{s}$.
### Pattern Recognition
Proportionality check: $Z \propto d^2 / \sqrt{M}$.
Diameter halved $\implies 1/4$ factor. Mass quadrupled $\implies 1/\sqrt{4} = 1/2$ factor.
Combined factor $= 1/4 \times 1/2 = 1/8$. $32 / 8 = 4$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Q36 (2026)
An air bubble of volume $2.9 \, \mathrm{cm}^{3}$ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is $17 \, ^{\circ}\mathrm{C}$ . The volume of the bubble when it reaches the surface, where the water temperature is $27 \, ^{\circ}\mathrm{C}$ , is ____ $\mathrm{cm}^{3}$ .
($g = 10 \, \mathrm{m/s}^{2}$ , density of water = $10^{3} \, \mathrm{kg/m}^{3}$ , and 1 atm pressure is $10^{5} \, \mathrm{Pa}$ )
- $4.2$
- $2.0$
- $3.0$
- $4.5$
### Related Formula
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
$$P = P_{\text{atm}} + \rho g h$$
### Core Logic
{{SOL_IMG1}}
For an air bubble rising in water, the number of moles of gas remains constant.
$P_1$ = Pressure at bottom = $P_{\text{atm}} + \rho gh$
$T_1$ = Temperature at bottom = $17^{\circ}\mathrm{C} = 290 \, \mathrm{K}$
$V_1$ = $2.9 \, \mathrm{cm}^3$
$P_2$ = Pressure at surface = $P_{\text{atm}}$
$T_2$ = Temperature at surface = $27^{\circ}\mathrm{C} = 300 \, \mathrm{K}$
### Step 1: Calculate Pressures
$$P_1 = 10^5 + (10^3 \times 10 \times 5) = 10^5 + 50,000 = 1.5 \times 10^5 \, \mathrm{Pa}$$
$$P_2 = 10^5 \, \mathrm{Pa}$$
### Step 2: Apply Ideal Gas Law
$$\frac{(1.5 \times 10^5)(2.9)}{290} = \frac{(10^5)(V_2)}{300}$$
$$\frac{1.5 \times 2.9}{290} = \frac{V_2}{300}$$
$$V_2 = \frac{1.5 \times 2.9 \times 300}{290} = 1.5 \times 3 = 4.5 \, \mathrm{cm}^3$$
### Pattern Recognition
Standard combined gas law application. Always remember to convert Celsius to Kelvin and carefully compute gauge pressure plus atmospheric pressure for the submerged state.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory of Gases
Class 11 Physics: Mechanical Properties of Fluids
Q47 (2026)
A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature $50^{\circ}\text{C}$. The gas is now heated to double its temperature. The modified pressure is ____ Pa.
### Related Formula
$$P \propto T \quad (\text{Gay-Lussac's Law for constant V})$$
### Core Logic
Since the gas is filled in a closed cylinder, the volume $V$ remains constant.
Therefore, $P \propto T$ (where T must be in Kelvin).
Initial temperature $T_i = 50^{\circ}\text{C} = 273 + 50 = 323 \text{ K}$.
Final temperature $T_f = 100^{\circ}\text{C} = 273 + 100 = 373 \text{ K}$.
Note: The phrasing "double its temperature" is slightly ambiguous (Celsius vs Kelvin). As per standard interpretations in such exams, "double its temperature" given in Celsius means $100^{\circ}\text{C}$ (which is 373 K). If it meant doubling absolute temperature, final would be 646 K. The official interpretation used here doubles the Celsius scale value.
### Step 1: Calculate Final Pressure
Using the relation:
$$\frac{P_f}{P_i} = \frac{T_f}{T_i}$$
$$\frac{P_f}{3.23 \times 10^3} = \frac{373}{323}$$
$$P_f = 3.23 \times 10^3 \times \frac{373}{323} = 3.73 \times 10^3 \text{ Pa}$$
$$P_f = 3730 \text{ Pa}$$
### Pattern Recognition
Be highly alert to 'double temperature' traps when given in Celsius. The question specifically intends $50^\circ \rightarrow 100^\circ$, not $323\text{K} \rightarrow 646\text{K}$. Calculating with 373K nicely cancels with the $3.23\text{kPa}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Kinetic Theory
Class 11 Physics: Thermodynamics
