States of Matter JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on States of Matter, free to read — no sign-in needed. The full chapter has 21 questions; sign in to attempt the remaining 16 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 1)
   Two cylinders, both fitted with frictionless pistons, are filled with mixtures of He and Ar gases. In the first cylinder, the masses of He and Ar are $m_1$ and $m_2$, respectively. In the second cylinder, the masses of He and Ar are $m_2$ and $m_1$, respectively. The molar mass of Ar is $10$ times the molar mass of He. The external pressure applied by the piston on the first cylinder needs to be $5$ times that on the second cylinder so that the volume of the gas mixtures in both the cylinders are equal at the same temperature. Assuming He and Ar behave like ideal gases, the value of $(m_1/m_2)$ is ____.
   Show answer & solution

   Answer: 9.8

   Let the molar mass of He be $M$. Then the molar mass of Ar is $10M$. In the first cylinder, the total number of moles is given by: $n_1 = \dfrac{m_1}{M} + \dfrac{m_2}{10M} = \dfrac{10m_1 + m_2}{10M}$ In the second cylinder, the total number of moles is given by: $n_2 = \dfrac{m_2}{M} + \dfrac{m_1}{10M} = \dfrac{10m_2 + m_1}{10M}$ Using the ideal gas equation $PV = nRT$, since the volume and temperature are the same for both cylinders, the pressure is directly proportional to the number of moles. Thus, $\dfrac{P_1}{P_2} = \dfrac{n_1}{n_2}$. Given that $P_1 = 5P_2$, we have: $\dfrac{n_1}{n_2} = 5$ Substituting the expressions for $n_1$ and $n_2$: $\dfrac{10m_1 + m_2}{10m_2 + m_1} = 5$ $10m_1 + m_2 = 50m_2 + 5m_1$ $5m_1 = 49m_2$ $\dfrac{m_1}{m_2} = \dfrac{49}{5} = 9.8$ Answer: $9.8$
2. Q2JEE Advanced Adv 2025 (Paper 1)
   Molar volume $\left(V_m\right)$ of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with $V_m$ as the variable. The ratio (in $\mathrm{mol} \mathrm{dm}^{-3}$ ) of the coefficient of $V_m^2$ to the coefficient of $V_m$ for a gas having van der Waals constants $a=6.0 \mathrm{dm}^6 \mathrm{~atm} \mathrm{~mol}^{-2}$ and $b=0.060 \mathrm{dm}^3 \mathrm{~mol}^{-1}$ at 300 K and 300 atm is _____ . Use: Universal gas constant $(R)=0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~mol}{ }^{-1} \mathrm{~K}^{-1}$
   Show answer & solution

   Answer: -7.1

   $\begin{aligned} & \left(P+\frac{a}{V_m^2}\right)\left(V_m-b\right)=R T \\ & P V_m-b P+\frac{a}{V_m}-\frac{a b}{V_m^2}-R T=0 \\ & \Rightarrow \quad P V_m^2-(b P+R T) V_m^2+a V_m-a b=0 \end{aligned}$ Coefficient of $\mathrm{V}_{\mathrm{m}}^2=-(\mathrm{bP}+\mathrm{RT})$ Coefficient of $\mathrm{V}_{\mathrm{m}}=\mathrm{a}$ $\text { Ratio }=-\frac{(\mathrm{bP}+\mathrm{RT})}{\mathrm{a}}=-\left[\frac{0.06 \times 300+24.6}{6}\right]=-7.1 .$
3. Q3JEE Advanced Adv 2024 (Paper 1)
   A closed vessel contains $10 \mathrm{~g}$ of an ideal gas $\mathbf{X}$ at $300 \mathrm{~K}$, which exerts 2 atm pressure. At the same temperature, $80 \mathrm{~g}$ of another ideal gas $\mathbf{Y}$ is added to it and the pressure becomes $6 \mathrm{~atm}$. The ratio of root mean square velocities of $\mathbf{X}$ and $\mathbf{Y}$ at $300 \mathrm{~K}$ is
   1. A.$2 \sqrt{2}: \sqrt{3}$
   2. B.$2 \sqrt{2}: 1$
   3. C.$1: 2$
   4. D.$2: 1$
   Show answer & solution

   Answer: (D)

   For Ideal Gas $\mathrm{PV}=\mathrm{nRT}$ $\therefore \mathrm{n} \propto \mathrm{P}$ at constant $\mathrm{T} \& \mathrm{~V}$. $\because \quad$ mole $=\frac{\text { Mass }}{\text { Molar mass }}$ For gas $\mathrm{X}: \frac{10}{\mathrm{M}_{\mathrm{X}}} \propto 2 \mathrm{~atm}$ $\quad$ $\ldots$(1) For gas $\mathrm{X} \& \mathrm{Y}: \frac{10}{\mathrm{M}_{\mathrm{X}}}+\frac{80}{\mathrm{M}_{\mathrm{Y}}} \propto 6 \mathrm{~atm}$ $\ldots$(2) From $(2)-(1)$ $\frac{80}{\mathrm{M}_{\mathrm{y}}} \propto 4$ $\ldots$(3) On dividing (1) by (3) $\frac{\mathrm{M}_{\mathrm{Y}}}{8 \mathrm{M}_{\mathrm{X}}}=\frac{1}{2}$ $\therefore \quad \frac{\mathrm{M}_{\mathrm{Y}}}{\mathrm{M}_{\mathrm{X}}}=4$ $\ldots$(4) $\because \mathrm{V}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}} \Rightarrow \mathrm{v}_{\mathrm{rms}} \propto \frac{1}{\sqrt{\mathrm{M}}}$ $\therefore \quad \frac{\left(\mathrm{V}_{\mathrm{ms}}\right)_{\mathrm{X}}}{\left(\mathrm{V}_{\mathrm{ms}}\right)_{\mathrm{Y}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{Y}}}{\mathrm{M}_{\mathrm{x}}}}=\sqrt{\frac{4}{1}}=\frac{2}{1}$
4. Q4JEE Advanced Adv 2023 (Paper 1)
   A gas has a compressibility factor of $0.5$ and a molar volume of $0.4{dm}^{3}{mol}^{-1}$ at a temperature of $800K$ and pressure $xatm$. If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be $y{dm}^{3}{mol}^{-1}$ . The value of$x/y$ is _________ [Use: Gas constant, $R=8\times {10}^{-2}Latm{K}^{-1}{mol}^{-1}$ ]
   Show answer & solution

   Answer: 100

   Compressibility factor $(Z)=\dfrac{{V}_{real}}{{V}_{ideal}}=0.5$ ${V}_{real}=0.4{dm}^{3}{mol}^{-1}=0.4L/mol$ $∴{V}_{ideal}=\dfrac{0.4}{0.5}=0.8L/mol$ $∴y=0.8L/mol$ Using ideal gas equation : $PV=nRT$ $P=\dfrac{1\times 8\times {10}^{-2}\times 800}{0.8}$ $x=80atm$ $∴\dfrac{x}{y}=\dfrac{80}{0.8}=100$
5. Q5JEE Advanced Adv 2019 (Paper 1)
   Which of the following statement(s) (are) correct regarding the root mean square speed $\left({U}_{rms}\right)$ and average translational kinetic energy $\left({\epsilon }_{av}\right)$ of a molecule in a gas at equilibrium?
   1. A.${U}_{rms}$ is doubled when its temperature is increased four times
   2. B.${\epsilon }_{av}$ at a given temperature does not depend on its molecular mass
   3. C.${U}_{rms}$ is inversely proportional to the square root of its molecular mass
   4. D.${\epsilon }_{av}$ is doubled when its temperature is increased four times
   Show answer & solution

   Answer: A,B,C

   ${\text{U}}_{\text{rms}}=\sqrt{\dfrac{\text{3RT}}{\text{M}}},$ ${\text{E}}_{\text{av}}=\dfrac{3}{2}\text{RT}$ If temperature is made 4 times, ${\text{U}}_{\text{rms}}$ between twice. Avg. kinetic energy $({\text{F}}_{\text{av}})$ is independent of molecular man at a given temperature, ${\text{U}}_{\text{rms}}$ inversely proportional to the square not of molecular mass.