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Kinetic Theory of Gases - JEE Advanced Previous Year Questions with Solutions

Kinetic Theory of Gases JEE Advanced previous year questions test concepts like rms speed, gas mixture properties, ideal gas conditions, and energy per mole. JEE Advanced has consistently asked 1–2 questions from this topic, often multi-correct type. Mastering the core formulae and their applications is enough to score full marks here.
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Kinetic Theory of Gases - JEE Advanced Previous Year Questions with Solutions

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JEE Advanced Previous Year Questions of Physics with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Physics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. Get detailed Class 11th & 12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET etc. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects.     Click Here for JEE main Previous Year Topic Wise Questions of Physics with Solutions    Download eSaral app  for free study material and video tutorials.   Simulator   Previous Years JEE Advanced Questions

Q. A real gas behaves like an ideal gas if its ? (A) pressure and temperature are both high (B) pressure and temperature are both low (C) pressure is high and temperature is low (D) pressure is low and temperature is high
Ans. (D) $\operatorname{High} \vec{P},$ Low $P$
Q. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds $\left(\frac{v_{r m s}(\text {helium})}{v_{r m s}(\text {arg on})}\right)$ is ? (A) 0.32         (B) 0.45            (C) 2.24              (D) 3.16 [JEE-2012]
Ans. (D) $v_{m s}=\sqrt{\frac{3 R T}{M}} ; \frac{\left(v_{m s}\right)_{H e}}{\left(v_{m s}\right)_{A r}}=\sqrt{\frac{M_{A r}}{M_{H e}}}=\sqrt{\frac{40}{4}}=\sqrt{10}=3.16$
Q. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is :- (A) 1 : 4          (B) 1 : 2           (C) 6 : 9            (D) 8 : 9 [JEE-2013]
Ans. (D) $\mathrm{P}=\frac{\rho}{\mathrm{M}} \mathrm{RT}$ $\Rightarrow \frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}=\frac{\rho_{1} \mathrm{M}_{2}}{\rho_{2} \mathrm{M}_{1}}=\frac{4}{3} \Rightarrow \frac{\rho_{1}}{\rho_{2}}=\frac{4}{3} \times \frac{2}{3}=\frac{8}{9}$
Q. A container of fixed volume has a mixutre of one mole of hydrogen and one mole of helium in equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s) is (are) :- (A) The average energy per mole of the gas mixture is 2RT. (B) The ratio of speed of sound in the gas mixture to that in helium gas is $\sqrt{6 / 5}$. (C) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2. (D) The ratio of the rms speed of helium atoms to that of hydrogen molecules is $1 / \sqrt{2}$. [JEE-Advance-2015]
Ans. (A,B,D) $C_{V(\operatorname{mix})}=\frac{(1)\left(\frac{3}{2} R\right)+(1)\left(\frac{5}{2} R\right)}{2}=2 R$ $\mathrm{C}_{\mathrm{P}(\mathrm{mix})}=3 \mathrm{R} \quad \gamma_{\operatorname{mix}}=\frac{3}{2} \Rightarrow \mathrm{f}=4$ Average energy/mole $=\mathrm{f} \frac{1}{2} \mathrm{RT}=2 \mathrm{RT}$ $\frac{\left(\mathrm{V}_{\mathrm{sound}}\right)_{\mathrm{mixture}}}{\left(\mathrm{V}_{\mathrm{sound}}\right)_{\mathrm{He}}}=\frac{\sqrt{\frac{\mathrm{RT}}{2}}}{\sqrt{\frac{5 \mathrm{RT}}{12}}}=\sqrt{\frac{6}{5}}$ $\frac{\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{He}}}{\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{H}_{2}}}=\frac{\sqrt{\frac{3 \mathrm{RT}}{4}}}{\sqrt{\frac{3 \mathrm{RT}}{2}}}=\frac{1}{\sqrt{2}}$ $\therefore(\mathrm{A}, \mathrm{B}, \mathrm{D})$

Frequently Asked Questions

Find answers to common questions.

Does the number of moles affect the rms speed of a gas?

No. The rms speed formula is $v_{rms} = \sqrt{3RT/M}$, which depends only on temperature T and molar mass M. The number of moles is irrelevant when comparing rms speeds of two gases at the same temperature, as seen in the 2012 JEE Advanced helium-argon question.

What is the condition for a real gas to behave like an ideal gas?

A real gas behaves like an ideal gas at low pressure and high temperature. Low pressure increases the intermolecular distance (reducing attraction forces), and high temperature ensures kinetic energy dominates over intermolecular potential energy. This is the direct answer to the 2012 JEE Advanced question on this topic.

How do you calculate the speed of sound in a gas mixture?

First, find the effective $\gamma_{mix}$ and $M_{mix}$ of the mixture using weighted averages. Then apply $v_{sound} = \sqrt{\gamma_{mix} RT / M_{mix}}$. For a 1:1 molar mixture of He and H₂, $\gamma_{mix} = 3/2$ and $M_{mix} = 3$ g/mol, giving a sound speed ratio of $\sqrt{6/5}$ compared to pure He.

How many questions come from Kinetic Theory in JEE Advanced?

Typically 1–2 questions per year appear from Kinetic Theory in JEE Advanced. From 2012–2015, every paper had at least one question from this topic. Since 2016, it sometimes appears as part of a multi-topic thermodynamics question. Consistent preparation ensures you never drop marks here.

Where can I find more JEE Advanced Physics previous year questions with solutions?

All JEE Advanced Physics PYQs with chapter-wise solutions are available on eSaral, taught by IIT Bombay faculty with verified All India Ranks. You can also supplement your preparation using NCERT Solutions for Class 12 Physics for thermodynamics topics that connect with Kinetic Theory at the advanced level.

What are the degrees of freedom for monoatomic, diatomic, and triatomic gases?

Monoatomic gases (He, Ar) have 3 degrees of freedom (translational only). Diatomic gases (H₂, N₂, O₂) have 5 degrees of freedom at room temperature (3 translational + 2 rotational). Non-linear triatomic gases have 6. This directly determines $C_V = (f/2)R$ and $\gamma = 1 + 2/f$.

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Rajat
Sept. 26, 2020, 8:35 a.m.
Thanks
Saranya
Sept. 25, 2020, 3:25 p.m.
Thanq so much sir
Maddu.Akhila
Sept. 20, 2020, 6:19 a.m.
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wes
Sept. 17, 2020, 5:07 p.m.
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Akshay Gopal raj
Sept. 16, 2020, 9:22 p.m.
THANKYOU so much. It is really helpful for jee ADVANCED...
Saranya
Sept. 6, 2020, 3:56 p.m.
Tqq Soo much sir these are very helpful
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