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 166 questions; sign in to attempt the remaining 161 in the exam simulator.

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
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   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.$1: 1$
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   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.$\frac{4}{3} \mathrm{Z}$
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   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 (13 Apr, Shift 2)Speed of Gas
   The mean free path of molecules of a certain gas at STP is $1500d$, where $d$is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at $373K$ is approximately:
   1. A.$750d$
   2. B.$1098d$
   3. C.$2049d$
   4. D.$1500d$
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   Answer: (C)

   The mean free path is given by the relation, $\lambda =\dfrac{RT}{\sqrt{2}\pi {d}^{2}{N}_{A}P}$ So, $\lambda \propto T$ Let $\lambda '$ be the new mean free path. Hence, $\dfrac{1500d}{\lambda '}=\dfrac{273}{373}$ $\Rightarrow {\lambda }^{'}=2049.45d\approx 2049d$
5. Q5JEE Main 2026 (22 Jan, Shift 1)Degree of Freedom and Specific Heat
   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 $(\Delta \mathrm{Q}=0)$ then identify the gas from the following options (Assuming the gases given in the options are ideal gases):
   1. A.$\mathrm{O}_{2}$
   2. B.$\mathrm{NH}_{3}$
   3. C.$\mathrm{CO}_{2}$
   4. D.He
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   Answer: (D)

   For an adiabatic process ($\Delta Q = 0$) with an ideal gas: $TV^{\gamma-1} = \text{constant}$ Given: $V_f = 8V_i$ and $T_f = \frac{1}{4}T_i$ Applying the adiabatic relation: $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$ $T_i V_i^{\gamma-1} = \frac{T_i}{4}(8V_i)^{\gamma-1}$ $1 = \frac{1}{4} \cdot 8^{\gamma-1}$ $4 = 8^{\gamma-1}$ $2^2 = 2^{3(\gamma-1)}$ $2 = 3(\gamma-1)$ $\gamma = \frac{5}{3} = 1.67$ This value corresponds to a monatomic gas. Among the options, only He is monatomic ($\gamma = 5/3$ for monatomic ideal gases). O₂ and N₂ are diatomic ($\gamma = 7/5$), CO₂ is polyatomic ($\gamma \approx 1.30$), and NH₃ is polyatomic.