States of Matter – JEE Main Previous Year Questions with Solutions
JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral.Simulator Previous Years AIEEE/JEE Mains Questions
Q. The molecular velocity of any gas is :-(1) inversely proportional to the square root of temperature(2) inversely proportional to absolute temperature(3) directly proportional to square of temperature(4) directly proportional to square root of temperature [AIEEE-2011]

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Sol. (4)

Q. , v and u represent most probable velocity, average velocity and root meansquarevelocity respectively of a gas at a particular temperature. The correct order among thefollowing is –(1)  > u > v(2) v > u > (3) u > v > (4) u >  > v [JEE(Main)-2012]

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Sol. (3)

Q. An open vessel at 300 K is heated till $\frac{2}{5}$th of the air in it is expelled.Assuming that thevolume of the vessel remains constant, the temperature to which the vessel is heated is :-(1) 750 K(2) 400 K(3) 500 K(4) 1500K [JEE(Main-online)-2012]

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Sol. (3) Q. For 1 mol of an ideal gas at constant temperature T, the plot of (log P)against(logV) is a (P : Pressure, V : Volume) :-(1) Straight line parallel to x-axis(2) Curve starting at origin](3) Straight line with a negative slope(4) Straight line passing through origin [JEE(Main-online)-2012]

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Sol. (3)PV = nRTlog (PV) = log (nRT)log P + log V = log (nRT)log P = –log V + log (nRT)y = – mx + Cslope $=\tan \theta=-1$

Q. The relationship among most probable velocity, average velocity and root meansquarevelocity is respectively :-(1) $\sqrt{2}: \sqrt{8 / \pi}: \sqrt{3}$(2) $\sqrt{2}: \sqrt{3}: \sqrt{8 / \pi}$(3) $\sqrt{3}: \sqrt{8 / \pi}: \sqrt{2}$(4) $\sqrt{8 / \pi}: \sqrt{3}: \sqrt{2}$ [JEE(Main-online)-2012]

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Sol. (1)

Q. Which one of the following is the wrong assumption of kinetic theory of gases ?(1) All the molecules move in straight line between collision and with same velocity.(2) Molecules are separated by great distances compared to their sizes.(3) Pressure is the result of elastic collision of molecules with the container’s wall.(4) Momentum and energy always remain conserved. [JEE(Main-online)-2013]

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Sol. (1)

Q. By how many folds the temperature of a gas would increase when the root meansquarevelocity of the gas molecules in a container of fixed volume is increased from $5 \times 10^{4} \mathrm{cm} /$s to $10 \times 10^{4} \mathrm{cm} / \mathrm{s} ?$(1) Four(2) three(3) Two(4) Six [JEE(Main-online)-2013]

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Sol. (1)$\mathrm{u}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}}$$\left(\frac{5 \times 10^{4}}{10 \times 10^{4}}\right)=\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}}$$\frac{1}{4}=\frac{T_{1}}{T_{2}}$$\mathrm{T}_{2}=4 \mathrm{T}_{1} Q. For gaseous state, if most probable speed is denoted by C, average speed by \overline{\mathrm{c}} andmeansquare speed by C, then for a large number of molecules the ratios of these speeds are :-(1) \mathrm{C}: \overline{\mathrm{C}}: \mathrm{C}=1.225: 1.128: 1(2) \mathrm{C}: \overline{\mathrm{C}}: \mathrm{C}=1.128: 1.225: 1(3) \mathrm{C}: \overline{\mathrm{C}}: \mathrm{C}=1: 1.128: 1.225(4) \mathrm{C}: \overline{\mathrm{C}}: \mathrm{C}=1: 1.225: 1.128 [JEE(Main-offline)-2013] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)\mathrm{u}_{\mathrm{avg}}=\sqrt{\frac{8 \mathrm{RT}}{\pi \mathrm{M}}}$$\mathrm{u}_{\mathrm{rms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$$\mathrm{u}_{\mathrm{mp}}=\sqrt{\frac{2 \mathrm{RT}}{\mathrm{M}}} Q. A gaseous compound of nitrogen and hydrogen contains 12.5%(by mass) ofhydrogen.The density of the compound relative to hydrogen is 16. The molecular formula of thecompound is :(1) \mathrm{NH}_{2}(2) \mathrm{NH}_{3}(3) \mathrm{N}_{3} \mathrm{H}(4) \mathrm{N}_{2} \mathrm{H}_{4} [JEE(Main-online)-2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)  Q. The initial volume of a gas cylinder is 750.0 mL. If the pressure of gas inside thecylinder changes from 840.0 mm Hg to 360.0 mm Hg, the final volume the gas will be(1)1.750 L(2) 7.50 L(3) 3.60 L(4) 4.032 L [JEE(Main-online)-2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}$$840 \times 750=360 \times \mathrm{V}_{2}$$\mathrm{V}_{2}=\frac{840 \times 750}{360} Q. The temperature at which oxygen molecules have the same root mean squarespeed ashelium atoms have at 300 K is :(Atomic masses : He = 4 u, O = 16 u)(1) 1200 K(2) 600 K(3) 300 K(4) 2400 K [JEE(Main-online)-2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4) Q. Two closed bulbs of equal volume(V) containing an ideal gas initially at pressurepi and temperature \mathrm{T}_{1} are connected through a narrow tube of negligiblevolume as shown in the figure below. The temperature of one of the bulbs is then raised to\mathrm{T}_{2}. The final pressure \mathrm{P}_{\mathrm{f}} is :-  [JEE(Main)-2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)Initial moles and final moles are equal\left(\mathrm{n}_{\mathrm{T}}\right)_{\mathrm{i}}=\left(\mathrm{n}_{\mathrm{T}}\right)_{\mathrm{f}}$$\frac{\mathrm{P}_{\mathrm{i}} \mathrm{V}}{\mathrm{RT}_{1}}+\frac{\mathrm{P}_{\mathrm{i}} \mathrm{V}}{\mathrm{RT}_{1}}=\frac{\mathrm{P}_{\mathrm{f}} \mathrm{V}}{\mathrm{RT}_{1}}+\frac{\mathrm{P}_{\mathrm{f}} \mathrm{V}}{\mathrm{RT}_{2}}$$2 \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{T}_{\mathrm{i}}}=\frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{T}_{\mathrm{i}}}+\frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{T}_{2}}$$\mathrm{P}_{\mathrm{f}}=\frac{2 \mathrm{P}_{\mathrm{i}} \mathrm{T}_{2}}{\mathrm{T}_{1}+\mathrm{T}_{2}}$

Q. ‘a’ and ‘b’ are Vander Waal’s constants for gases. Chlorine is more easily liquefied than ethanebecause :- [Aieee-2012]

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Sol. (2)Higher the ‘a’ value, more easily the gas is liquified, lower the ‘b’ value, more easily the gas is liquified

Q. The compressibility factor for a real gas at high pressure is :-(1) $1-\frac{\text { Pb }}{\mathrm{RT}}$(2) $1+\frac{\mathrm{RT}}{\mathrm{Pb}}$(3) 1(4) $1+\frac{\mathrm{Pb}}{\mathrm{RT}}$ [Aieee-2012]

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Sol. (4)At high pressure, repulsion dominate.$\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{Vm}^{2}}\right)(\mathrm{Vm}-\mathrm{b})=\mathrm{RT}$P(Vm – b) = RT$\mathrm{So}, \mathrm{Z}=1+\frac{\mathrm{Pb}}{\mathrm{RT}}$

Q. If Z is the compressibility factor, van der Waals’ equation at low pressure can be written as :$(1) Z=1-\frac{\mathrm{Pb}}{\mathrm{RT}}$$(2) Z=1+\frac{\mathrm{Pb}}{\mathrm{RT}}(3) \mathrm{Z}=1+\frac{\mathrm{RT}}{\mathrm{Pb}}(4) \mathrm{Z}=1-\frac{\mathrm{a}}{\mathrm{V}_{\mathrm{m}} \mathrm{RT}} [JEE-MAINS-2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4)At low pressure, attractions dominate.\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{Vm}^{2}}\right)(\mathrm{Vm}-\mathrm{b})=\mathrm{RT}$$\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{Vm}^{2}}\right)(\mathrm{Vm})=\mathrm{RT}$$\mathrm{So}, \mathrm{Z}=1-\frac{\mathrm{a}}{\mathrm{VmRT}}$

Q. When does a gas deviate the most from it’s ideal behaviour ?(1) At high pressure and low temperature(2) At high pressure and high temperature(3) At low pressure and low temperature(4) At low pressure and high temperature [JEEMAINS(online)-2015]

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Sol. (1)Gas behaves most ideally at high T and low P while deviates most from ideal behaviour at high P and low T.

• August 25, 2021 at 11:27 pm

Okay, I am preeettttyyyy suree there must be at least 10 times more questions here

• August 16, 2021 at 8:58 pm

Some answers of question are incorrect. I will not recommend the students who are preapring for jee to study from this site.

• June 8, 2021 at 10:07 pm

Excellent questions. Highly useful for examination

• August 16, 2021 at 9:00 pm

Also in one of the questions the volume of air expelled our is 2/5 but they have taken 3/5 as expelled air in their answer.

• May 11, 2021 at 10:11 pm

2nd answer is 750k u taken 3/5 but given is 2/5

• May 27, 2021 at 7:11 am

You have to 3/5 because according To question 2/5 is reduced so subtract it by 1 and u get 3/5 means 3/5 is left so you have to take 3/5 nad your answer will com 500 k

• August 25, 2021 at 11:26 pm

Come on, if it was so easy why would anyone practice anyway…You need to find how much remains in there , if 2/5 goes out, 3/5 of the moles of gas that remain will occupy V , so n2 will be 3/5

• March 27, 2021 at 7:59 pm

It is very super question and it has been helped for my exams

• October 6, 2020 at 7:14 pm

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thanks

• September 21, 2020 at 7:36 am

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Thanks 🙂

• September 12, 2020 at 11:26 am

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