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 79 questions; sign in to attempt the remaining 74 in the exam simulator.

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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 (26 Jul, Shift 2)
   A $10g$ mixture of hydrogen and helium is contained in a vessel of capacity $0.0125{m}^{3}$ at $6$ bar and $27^{\circ}C$. The mass of helium in the mixture is ..... $g$. (Given : $R=8.3{JK}^{-1}{mol}^{-1}$ (Atomic masses of $H$ and $He$ are $1u$ and $4u$, respectively)
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   Answer: 8

   $PV={n}_{mix}RT$ ${n}_{mix}=\dfrac{6\times 12.5}{0.083\times 300}\approx 3$ Out of total 3 moles Let mole of $He=x$ Mole of ${H}_{2}=3-x$ Then total mass of the mixture will be = $4x+2\left(3-x\right)=10$ Number of moles of He gas $x=2mol$ Mass of $He=numberofmoles\times molarmass=8g$