Kinetic Theory of Gases Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Kinetic Theory of Gases with step-by-step solutions covering gas mixtures, ideal gas law, rms speed, kinetic energy, and coefficient of volume expansion.
Practice the most recent JEE Main 2026 Kinetic Theory of Gases questions with clean, step-by-step worked solutions.
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Solution
The total number of moles is conserved when the two gases are mixed:
$$n = n_1 + n_2$$Using the ideal gas law $PV = nRT$ for each component and for the mixture:
$$n_1 = \frac{P_1 V_1}{R T_1}, \qquad n_2 = \frac{P_2 V_2}{R T_2}, \qquad n = \frac{P V}{R T}$$Substituting into $n = n_1 + n_2$:
$$\frac{P V}{R T} = \frac{P_1 V_1}{R T_1} + \frac{P_2 V_2}{R T_2}$$Cancelling $R$ and combining the right side over a common denominator:
$$\frac{P V}{T} = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{T_1 T_2}$$Solving for the mixture temperature $T$:
$$T = \frac{T_1 T_2 \, P V}{T_2 P_1 V_1 + T_1 P_2 V_2}$$Answer: B ($T_1 T_2 PV/(T_2 P_1 V_1 + T_1 P_2 V_2)$)
Solution
Let each vessel have volume $V$. Initially both are at $P_0 = 90\,\text{kPa}$ and $T = 400\,\text{K}$, so the total number of moles is:
$$n = \frac{P_0 V}{R (400)} + \frac{P_0 V}{R (400)} = \frac{2 P_0 V}{400\,R}$$The vessels stay connected, so after heating they share a common final pressure $P$. Vessel 1 remains at 400 K and vessel 2 is raised to 500 K:
$$n = \frac{P V}{400\,R} + \frac{P V}{500\,R}$$Since the total moles are conserved, equate the two expressions and cancel $V/R$:
$$\frac{2 P_0}{400} = \frac{P}{400} + \frac{P}{500}$$$$\frac{2(90)}{400} = P\left(\frac{1}{400} + \frac{1}{500}\right) = P\left(\frac{9}{2000}\right)$$$$0.45 = P \times \frac{9}{2000} \quad\Rightarrow\quad P = 0.45 \times \frac{2000}{9} = 100\,\text{kPa}$$Answer: A (100)
Solution
The average translational kinetic energy of a gas molecule depends only on temperature:
$$\langle KE \rangle = \frac{3}{2} k_B T$$It is independent of the molecular mass and of the container size. So if H$_2$ and O$_2$ have the same average kinetic energy, they must be at the same temperature. Assertion A is true.
The root-mean-square speed does depend on the molecular mass:
$$v_{rms} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}}$$At the same temperature, since $M_{\text{H}_2} = 2\,\text{g/mol} \ne M_{\text{O}_2} = 32\,\text{g/mol}$, the lighter H$_2$ molecules move much faster than O$_2$. Their rms speeds are not the same. Reason R is false.
Therefore A is true but R is false.
Answer: C (A is true but R is false)
Solution
First find the total number of moles from the ideal gas law $PV = nRT$, with $P = 100\times10^3\,\text{Pa}$, $V = 8310\,\text{cm}^3 = 8310\times10^{-6}\,\text{m}^3$:
$$n = \frac{PV}{RT} = \frac{(100\times10^3)(8310\times10^{-6})}{(8.31)(300)} = \frac{831}{2493} = \frac{1}{3}\,\text{mol}$$Let $a$ = moles of CO$_2$ (molar mass $44\,\text{g/mol}$) and $b$ = moles of O$_2$ (molar mass $32\,\text{g/mol}$). Then:
$$a + b = \frac{1}{3}, \qquad 44a + 32b = 13.2$$Substitute $b = \tfrac{1}{3} - a$ into the mass equation:
$$44a + 32\left(\frac{1}{3} - a\right) = 13.2 \quad\Rightarrow\quad 12a + \frac{32}{3} = 13.2$$$$12a = 13.2 - 10.667 = 2.533 \quad\Rightarrow\quad a = 0.211 \approx 0.21$$$$b = \frac{1}{3} - 0.211 = 0.122 \approx 0.12$$So there are about $0.21$ mol of CO$_2$ and $0.12$ mol of O$_2$.
Answer: C ($0.21$ and $0.12$)
Solution
When the two gases are mixed without any external work or heat exchange, the total internal energy is conserved. For an ideal monoatomic gas $U = \dfrac{3}{2} n R T$, so:
$$U_1 + U_2 = U_{\text{mix}}$$$$\frac{3}{2} n_1 R T_1 + \frac{3}{2} n_2 R T_2 = \frac{3}{2}(n_1 + n_2) R T_{\text{mix}}$$Cancelling $\dfrac{3}{2}R$ throughout:
$$n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_{\text{mix}}$$Substitute $n_1 = 2,\ T_1 = T,\ n_2 = 6,\ T_2 = 2T$:
$$(2)(T) + (6)(2T) = (2 + 6)\, T_{\text{mix}}$$$$2T + 12T = 8\, T_{\text{mix}} \quad\Rightarrow\quad 14T = 8\, T_{\text{mix}}$$$$T_{\text{mix}} = \frac{14T}{8} = \frac{7}{4}T$$Answer: D ($\dfrac{7}{4}T$)
Solution
For an ideal gas, $PV = nRT$, so the pressure can be written as:
$$P = \frac{nRT}{V}$$Substitute this into the given condition $PT^3 = \text{constant}$:
$$\frac{nRT}{V}\, T^3 = \text{constant} \quad\Rightarrow\quad \frac{T^4}{V} = \text{constant}$$This means $V \propto T^4$, so we can write $V = c\,T^4$ for some constant $c$. The coefficient of volume expansion is defined as:
$$\gamma = \frac{1}{V}\frac{dV}{dT}$$Differentiating $V = c\,T^4$:
$$\frac{dV}{dT} = 4 c\, T^3 = \frac{4V}{T}$$Therefore:
$$\gamma = \frac{1}{V}\cdot\frac{4V}{T} = \frac{4}{T}$$Answer: C ($\dfrac{4}{T}$)