States of Matter JEE Main previous year questions with solutions

5 solved JEE Main questions on States of Matter, free to read — no sign-in needed. The full chapter has 83 questions; sign in to attempt the remaining 78 in the exam simulator.

1. Q1JEE Main 2023 (12 Apr, Shift 1)
   At $600K$, the root mean square $(rms)$ speed of gas $X$ (molar mass $=40$) is equal to the most probable speed of gas $Y$ at $90K$. The molar mass of the gas $Y$ is _____ $g{mol}^{–1}$. (Nearest integer)
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   Answer: 4

   The root mean square velocity ${V}_{rms}=\sqrt{\dfrac{3RT}{M}}$ The most probable speed ${V}_{mp}=\sqrt{\dfrac{2RT}{M}}$ $T=$Temperature in Kelvin $M=$Molar mass $\dfrac{{V}_{rms}ofX}{{V}_{mp}ofY}=\sqrt{\dfrac{3{T}_{X}{M}_{Y}}{2{T}_{Y}{M}_{X}}}$ $1=\sqrt{\dfrac{3\times 600\times {M}_{Y}}{2\times 90\times 40}}=\sqrt{\dfrac{{M}_{Y}}{4}}$ ${M}_{Y}=4gm{mol}^{-1}$
2. Q2JEE Main 2023 (10 Apr, Shift 1)
   At constant temperature, a gas is at a pressure of $940.3mmHg$. The pressure at which its volume decreases by $40$ is ______ $mmHg$. (Nearest integer)
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   Answer: 1567

   From Boyle's Law, ${P}_{2}{V}_{2}={P}_{1}{V}_{1}$ ${P}_{1}=940.3$ mm of $Hg$ ${P}_{2}=$ Pressure at which volume is reduced by $40$ ${V}_{2}=$ Volume decreased by $40$ $\left({V}_{2}={V}_{1}-0.4{V}_{1}=0.6{V}_{1}\right)$ Then,${P}_{2}=\dfrac{940.3\times {V}_{1}}{0.6{V}_{1}}$ $=1567.16$ $\approx 1567mmHg$
3. Q3JEE Main 2023 (08 Apr, Shift 2)
   Arrange the following gases in increasing order of van der Waals constant $'a'$ (A) $Ar$ (B) ${CH}_{4}$ (C) ${H}_{2}O$ (D) ${C}_{6}{H}_{6}$ Choose the correct option from the following.
   1. A.D, C, B and A
   2. B.B, C, D and A
   3. C.C, D, B and A
   4. D.A, B, C and D
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   Answer: (D)

   The van der Waals' constant $'a'$is directly proportional to the intermolecular force of attraction; hence, the gas with greater value of $'a'$ can be most easily liquefied.The values of Vander Waal's constant $'a'$ for the given gases are: Gases Vander Waal's Constant $'a'$ A. $Ar$ $1.35{L}^{2}$ atm ${mol}^{2}$ B. ${CH}_{4}$ $2.25{L}^{2}$ atm ${mol}^{2}$ C. ${H}_{2}O$ $5.536{L}^{2}$ atm ${mol}^{2}$ D. ${C}_{6}{H}_{6}$ $18.57{L}^{2}$ atm ${mol}^{2}$ $∴$Increasing order of $'a'$ for the given gases $A\lt B\lt C\lt D$
4. Q4JEE Main 2022 (27 Jul, Shift 2)
   for a real gas at $25^{\circ}C$ temperature and high pressure ($99$ bar) the value of compressibility factor is $2$, so the value of Van 1der Waal's constant '$b$' should be_____$\times {10}^{-2}L{mol}^{-1}$ (Given $R=0.083L$ bar ${K}^{-1}{mol}^{-1}$)
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   Answer: 25

   For real gas under high pressure we can neglect pressure corrections due to intermolecular forces. But since volume will be low, we can not neglect volume corrections due to molecular size. $Z=1+\dfrac{Pb}{RT}=2 \Rightarrow b=\dfrac{RT}{P}$ $=\dfrac{0.083\times 298}{99}$ $=0.25\times {10}^{-2}L{mol}^{-1}$
5. Q5JEE Main 2022 (27 Jul, Shift 1)
   Given below are two statements. One is labelled as Assertion $A$ and the other is labelled as Reason $R$. Assertion $A$ : Activated charcoal adsorbs ${SO}_{2}$ more efficiently than ${CH}_{4}$. Reason $R$ : Gases with lower critical temperatures are readily adsorbed by activated charcoal. In the light of the above statements, choose the correct answer from the options given below.
   1. A.Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$.
   2. B.Both $A$ and $R$ are correct but $R$ is NOT the correct explanation of $A$.
   3. C.$A$ is correct but $R$ is not correct.
   4. D.$A$ is not correct but $R$ is correct.
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   Answer: (C)

   Lesser the value of critical temperature of gases lesser will be the extent of adsorption. Gas ${SO}_{2}$ ${CH}_{4}$ Critical temperature/K $630$ $190$ The critical temperature depends on the strength of the intermolecular interactions that hold a substance together as a liquid. ${SO}_{2}$ has stronger intermolecular force of attractions than ${CH}_{4}$, due to the more molar mass and non-zero dipole moment. So, Activated charcoal adsorbs ${SO}_{2}$ more efficiently than ${CH}_{4}$. is correct statement but Reason statement is not correct.