Kinetic Theory of Gases JEE Main previous year questions with solutions

5 solved JEE Main questions on Kinetic Theory of Gases, free to read — no sign-in needed. The full chapter has 156 questions; sign in to attempt the remaining 151 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Main 2026 (05 Apr, Shift 1)Degree of Freedom and Specific Heat
   Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Statement I: Change in internal energy of a system containing $n$ mole of ideal gas can be written as $\Delta U = n C_v (T_f - T_i) = \dfrac{nR}{\gamma - 1}(T_f - T_i)$, where $\gamma = \dfrac{C_p}{C_v}$, $T_i =$ initial temperature, $T_f =$ final temperature. Statement II: Relation between degree of freedom $f$ and $\gamma (= C_p/C_v)$ is $\left(\gamma = 1 + \dfrac{2}{f}\right)$ Choose the correct answer from the options given below
   1. A.Both A and R are true and R is the correct explanation of A
   2. B.Both A and R are true but R is NOT the correct explanation of A
   3. C.A is true but R is false
   4. D.A is false but R is true
   Show answer & solution

   Answer: (B)

   Statement I is true because for an ideal gas, the change in internal energy is given by $\Delta U = n C_v \Delta T$. Using Mayer's relation $C_p - C_v = R$ and the ratio of specific heats $\gamma = \dfrac{C_p}{C_v}$, we can write $C_v = \dfrac{R}{\gamma - 1}$. Substituting this, we get $\Delta U = \dfrac{nR}{\gamma - 1}(T_f - T_i)$. Statement II is true because according to the law of equipartition of energy, the molar heat capacity at constant volume is $C_v = \dfrac{f}{2}R$ and at constant pressure is $C_p = \left(\dfrac{f}{2} + 1\right)R$. Therefore, $\gamma = \dfrac{C_p}{C_v} = 1 + \dfrac{2}{f}$. Statement II does not explain Statement I, as Statement I is derived purely from macroscopic thermodynamic relations (Mayer's relation) and is independent of the microscopic concept of degrees of freedom. Hence, both statements are true but Statement II is NOT the correct explanation of Statement I. Answer: Both A and R are true but R is NOT the correct explanation of A
2. Q2JEE Main 2025 (07 Apr, Shift 2)Pressure and Energy of gas
   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 $=4 \mathrm{~g} / \mathrm{mol}$, Molar mass of argon $=40 \mathrm{~g} / \mathrm{mol}$)
   1. A.$1: 10$
   2. B.$10: 1$
   3. C.$1: \sqrt{10}$
   4. D.$1: 1$
   Show answer & solution

   Answer: (D)

   $K. E=\frac{f}{2} K T$ For He and $\mathrm{Ar} \mathrm{f}=3$ $\frac{\mathrm{K} \cdot \mathrm{E}_{\mathrm{He}}}{\mathrm{~K} \cdot \mathrm{E}_{\mathrm{Ar}}}=\frac{1}{1}$
3. Q3JEE Main 2024 (05 Apr, Shift 1)Gas Laws
   If the collision frequency of hydrogen molecules in a closed chamber at $27^{\circ} \mathrm{C}$ is $\mathrm{Z}$, then the collision frequency of the same system at $127^{\circ} \mathrm{C}$ is :
   1. A.$\frac{\sqrt{3}}{2} \mathrm{z}$
   2. B.$\frac{2}{\sqrt{3}} \mathrm{Z}$
   3. C.$\frac{3}{4} \mathrm{Z}$
   4. D.$\frac{4}{3} \mathrm{Z}$
   Show answer & solution

   Answer: (B)

   Assuming mean free path constant. $\begin{aligned} & \mathrm{f} \propto \mathrm{V} \propto \sqrt{\mathrm{T}} \\ & \frac{\mathrm{f}_1}{\mathrm{f}_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}}=\sqrt{\frac{300}{400}} \\ & \mathrm{f}_2=\sqrt{\frac{4}{3}}=\mathrm{f}_1=\frac{2}{\sqrt{3}} \mathrm{Z}\end{aligned}$
4. Q4JEE Main 2023 (25 Jan, Shift 2)Calorimetry
   According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-
   1. A.$\dfrac{9}{2}R$
   2. B.$\dfrac{5}{2}R$
   3. C.$\dfrac{3}{2}R$
   4. D.$\dfrac{7}{2}R$
   Show answer & solution

   Answer: (D)

   A diatomic gas has 5 degrees of freedom apart from vibrational degree of freedom. For each additional vibrational mode, degree of freedom is increased by $2$. So new degree of freedom, $f=5+2=7$ Therefore, we require molar specific heat capacity at constant volume, which is given by ${C}_{v}=\dfrac{f}{2}R=\dfrac{7}{2}R$
5. Q5JEE Main 2022 (29 Jul, Shift 2)Speed of Gas
   The root mean square speed of smoke particles of mass $5\times {10}^{-17}kg$ in their Brownian motion in air at NTP is approximately. [Given $k=1.38\times {10}^{-23}J{K}^{-1}$]
   1. A.$60mm{s}^{-1}$
   2. B.$12mm{s}^{-1}$
   3. C.$15mm{s}^{-1}$
   4. D.$36mm{s}^{-1}$
   Show answer & solution

   Answer: (C)

   According to Kinetic theory, Average kinetic energy of a gas molecule $=\dfrac{1}{2}m{v}_{rms}^{2}=\dfrac{3}{2}kT$ Thus, ${v}_{rms}=\sqrt{\dfrac{3kT}{m}}$ At NTP, $T=293K$ Putting the values, we have $=\sqrt{\dfrac{3\times 1.38\times {10}^{-23}\times 293}{5\times {10}^{-17}}}$ $\approx 15mm{s}^{-1}$