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States of Matter: Gases and Liquids — JEE Previous Year Questions

Every States of Matter: Gases and Liquids question asked in JEE Main and JEE Advanced across the last 186 papers — 56 questions, each with its correct answer. Free to read, no account needed.

Questions

56

Papers it appeared in

52/186

Appearance rate

28%

All 56 States of Matter: Gases and Liquids questions

Most recent papers first.

Q1·ChemistryNumericalJEE Advanced 2026
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 m1m_1m1​ and m2m_2m2​, respectively. In the second cylinder, the masses of He and Ar are m2m_2m2​ and m1m_1m1​, 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 (m1/m2)(m_1/m_2)(m1​/m2​) is ____.

Correct answer: 9.8

Step-by-step solution →
Q2·ChemistryNumericalJEE Advanced 2025
Molar volume (Vm)(V_m)(Vm​) of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with VmV_mVm​ as the variable. The ratio (in mol dm−3dm^{-3}dm−3) of the coefficient of Vm2V_m^2Vm2​ to the coefficient of VmV_mVm​ for a gas having van der Waals constant a=6.0a = 6.0a=6.0 dm6dm^6dm6 atm mol−2mol^{-2}mol−2 and b=0.060b = 0.060b=0.060 dm3dm^3dm3 mol−1mol^{-1}mol−1 at 300 K and 300 atm is ______. Use: Universal gas constant (R) = 0.082 dm3dm^3dm3 atm mol−1mol^{-1}mol−1 K−1K^{-1}K−1.

Correct answer: -7.1

Step-by-step solution →
Q3·ChemistrySingle correctJEE Advanced 2024
A closed vessel contains 10 g of an ideal gas X\mathbf{X}X at 300 K, which exerts 2 atm pressure. At the same temperature, 80 g of another ideal gas Y\mathbf{Y}Y is added to it and the pressure becomes 6 atm. The ratio of root mean square velocities of X\mathbf{X}X and Y\mathbf{Y}Y at 300 K is
  1. (A)22:32\sqrt{2} : \sqrt{3}22​:3​
  2. (B)22:12\sqrt{2} : 122​:1
  3. (C)1:21 : 21:2
  4. (D)2:12 : 12:1

Correct answer: (D)

Step-by-step solution →
Q4·ChemistryNumericalJEE Advanced 2023
A gas has a compressibility factor of 0.5 and a molar volume of 0.4 dm3 mol−1\mathrm{dm^3\ mol^{-1}}dm3 mol−1 at a temperature of 800 K and pressure x atm. If it shows ideal gas behaviour at the same temperature and pressure, the molar volume will be y dm3 mol−1\mathrm{dm^3\ mol^{-1}}dm3 mol−1. The value of x/y is ___. [Use: Gas constant, R = 8×10−2 L atm K−1 mol−18 \times 10^{-2}\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}8×10−2 L atm K−1 mol−1]

Correct answer: 100

Step-by-step solution →
Q5·ChemistryNumericalJEE Main 2023
A certain quantity of real gas occupies a volume of 0.15 dm30.15\,dm^{3}0.15dm3 at 100 atm100\,atm100atm and 500 K500\,K500K when its compressibility factor is 1.871.871.87. Its volume at 300 atm300\,atm300atm and 300 K300\,K300K (When its compressibility factor is 1.41.41.4) is _____ ×10−4 dm3\times10^{-4}\,dm^{3}×10−4dm3 (Nearest integer).

Correct answer: 392

Step-by-step solution →
Q6·ChemistryNumericalJEE Main 2023
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^{-1}−1. (Nearest integer)

Correct answer: 4

Step-by-step solution →
Q7·ChemistryNumericalJEE Main 2023
At constant temperature a gas is at a pressure of 940.3 mm Hg. The pressure at which its volume decreases by 40% is _________ mm Hg. (Nearest Integer)

Correct answer: 1567

Step-by-step solution →
Q8·ChemistrySingle correctJEE Main 2023
Arrange the following gases in increasing order of van der Waals constant 'a': A. Ar B. CH4CH_4CH4​ C. H2OH_2OH2​O D. C6H6C_6H_6C6​H6​. Choose the correct option from the following:
  1. (A)B, C, D and A
  2. (B)C, D, B and A
  3. (C)A, B, C and D
  4. (D)D, C, B and A

Correct answer: (C)

Step-by-step solution →
Q9·ChemistryNumericalJEE Main 2023
Three bulbs are filled with CH4CH_4CH4​, CO2CO_2CO2​ and Ne as shown in the picture. The bulbs are connected through pipes of zero volume. When the stopcocks are opened and the temperature is kept constant throughout, the pressure of the system is found to be ____ atm. (Nearest integer)

Correct answer: 3

Step-by-step solution →
Q10·ChemistryNumericalJEE Main 2023
The total pressure of a mixture of non-reacting gases XXX (0.60.60.6 g) and YYY (0.450.450.45 g) in a vessel is 740740740 mm of Hg. The partial pressure of the gas XXX is ___ mm of Hg. (Nearest Integer) (Given : molar mass X=20X = 20X=20 and Y=45Y = 45Y=45 g mol−1\,mol^{-1}mol−1)

Correct answer: 555

Step-by-step solution →
Q11·ChemistrySingle correctJEE Main 2023
For 1 mol of gas, the plot of pVpVpV vs ppp is shown in the figure. ppp is the pressure and VVV is the volume of the gas. What is the value of compressibility factor at point A?
  1. (A)1+aRTV1 + \dfrac{a}{RTV}1+RTVa​
  2. (B)1−aRTV1 - \dfrac{a}{RTV}1−RTVa​
  3. (C)1+bV1 + \dfrac{b}{V}1+Vb​
  4. (D)1−bV1 - \dfrac{b}{V}1−Vb​

Correct answer: (B)

Step-by-step solution →
Q12·ChemistryNumericalJEE Main 2023
Based on the given figure, the number of correct statement/s is/are _____ A. Surface tension is the outcome of equal attractive and repulsive forces acting on the liquid molecule in bulk. B. Surface tension is due to uneven forces acting on the molecules present on the surface. C. The molecule in the bulk can never come to the liquid surface. D. The molecules on the surface are responsible for vapours pressure if system is a closed system.

Correct answer: 2

Step-by-step solution →
Q13·ChemistryNumericalJEE Main 2023
The number of statement/s, which are correct with respect to the compression of carbon dioxide from point (a) in the Andrews isotherm from the following is _________. A. Carbon dioxide remains as a gas upto point (b). B. Liquid carbon dioxide appears at point (c). C. Liquid and gaseous carbon dioxide coexist between points (b) and (c). D. As the volume decreases from (b) to (c), the amount of liquid decreases

Correct answer: 4

Step-by-step solution →
Q14·ChemistryNumericalJEE Main 2022
'x' g of molecular oxygen (O2_{2}2​) is mixed with 200 g of neon (Ne). The total pressure of the non-reactive mixture of O2_{2}2​ and Ne in the cylinder is 25 bar. The partial pressure of Ne is 20 bar at the same temperature and volume. The value of 'x' is_____. [Given: Molar mass of O2_{2}2​ = 32 g mol−1^{-1}−1. Molar mass of Ne = 20 g mol−1^{-1}−1]

Correct answer: 80

Step-by-step solution →
Q15·ChemistryNumericalJEE Main 2022
A 10 g mixture of hydrogen and helium is contained in a vessel of capacity 0.0125 m3m^{3}m3 at 6 bar and 27°C. The mass of helium in the mixture is ________ g. (nearest integer) Given : R = 8.3 JK−1mol−1JK^{-1}mol^{-1}JK−1mol−1 (Atomic masses of H and He are 1u and 4u, respectively)

Correct answer: 8

Step-by-step solution →
Q16·ChemistryNumericalJEE Main 2022
A mixture of hydrogen and oxygen contains 40% hydrogen by mass when the pressure is 2.2 bar. The partial pressure of hydrogen is bar. (Nearest Integer)

Correct answer: 2

Step-by-step solution →
Q17·ChemistryNumericalJEE Main 2022
The pressure of a moist gas at 27°C is 4 atm. The volume of the container is doubled at the same temperature. The new pressure of the moist gas is …. ×10−1\times 10^{-1}×10−1 atm. (Nearest integer) (Given : The vapour pressure of water at 27°C is 0.4 atm)

Correct answer: 22

Step-by-step solution →
Q18·ChemistryNumericalJEE Main 2022
Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 gL−1gL^{-1}gL−1 at 257°C and 100 mm Hg. The molar mass of geraniol is _______ (Nearest Integer) [Given R = 0.082 L atm K−1K^{-1}K−1 mol−1mol^{-1}mol−1]

Correct answer: 152

Step-by-step solution →
Q19·ChemistryNumericalJEE Main 2022
A box contains 0.90 g of liquid water in equilibrium with water vapour at 27°C. The equilibrium vapour pressure of water at 27°C 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be______ litre. [nearest integer] (Given: R = 0.082 L atm K−1^{-1}−1 mol−1^{-1}−1) (Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)

Correct answer: 29

Step-by-step solution →
Q20·ChemistryNumericalJEE Main 2022
100 g of an ideal gas is kept in a cylinder of 416 L volume at 27°C under 1.5 bar pressure. The molar mass of the gas is ______ g mol−1^{-1}−1. (Nearest integer) (Given : R = 0.083 L bar K−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 4

Step-by-step solution →
Q21·ChemistrySingle correctJEE Main 2022
Which amongst the given plots is the correct plot for pressure (p) vs density (d) for an ideal gas ?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (B)

Step-by-step solution →
Q22·ChemistryNumericalJEE Main 2021
An empty LPG cylinder weighs 14.8 kg. When full, it weighs 29.0 kg and shows a pressure of 3.47 atm. In the course of use at ambient temperature, the mass of the cylinder is reduced to 23.0 kg. The final pressure inside of the cylinder is ______ atm. (Nearest integer) (Assume LPG of be an ideal gas)

Correct answer: 2

Step-by-step solution →
Q23·ChemistrySingle correctJEE Main 2021
Which one of the following is the correct PV vs P plot at constant temperature for an ideal gas ? (P and V stand for pressure and volume of the gas respectively)
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q24·ChemistrySingle correctJEE Main 2021
The unit of the van der Waals gas equation parameter 'a' in (P+an2V2)(V−nb)=nRT\left(\mathrm{P}+\dfrac{\mathrm{an^2}}{\mathrm{V^2}}\right)(\mathrm{V-nb}) = \mathrm{nRT}(P+V2an2​)(V−nb)=nRT is :
  1. (A)kg m s−2\mathrm{kg\ m\ s^{-2}}kg m s−2
  2. (B)dm3 mol−1\mathrm{dm^3\ mol^{-1}}dm3 mol−1
  3. (C)kg m s−1\mathrm{kg\ m\ s^{-1}}kg m s−1
  4. (D)atm dm6 mol−2\mathrm{atm\ dm^6\ mol^{-2}}atm dm6 mol−2

Correct answer: (D)

Step-by-step solution →
Q25·ChemistryNumericalJEE Main 2021
Two flasks I and II shown below are connected by a valve of negligible volume. When the valve is opened, the final pressure of the system in bar is x × 10−2^{-2}−2. The value of x is _____ . (Integer answer) [Assume–Ideal gas; 1 bar = 105^{5}5 Pa; Molar mass of N2_{2}2​ = 28.0 g mol−1^{-1}−1 ; R = 8.31 J mol−1^{-1}−1K−1^{-1}−1]

Correct answer: 84

Step-by-step solution →
Q26·ChemistrySingle correctJEE Main 2021
The interaction energy of London forces between two particles is proportional to rx\mathrm{r^x}rx, where r is the distance between the particles. The value of x is :
  1. (A)3
  2. (B)−3-3−3
  3. (C)−6-6−6
  4. (D)6

Correct answer: (C)

Step-by-step solution →
Q27·ChemistryNumericalJEE Main 2021
2SO2_22​(g)+O2_22​(g) → 2SO3_33​(g) The above reaction is carried out in a vessel starting with partial pressure PSO2_{SO_2}SO2​​ = 250 m bar, PO2_{O_2}O2​​ = 750 m bar and PSO3_{SO_3}SO3​​ = 0 bar. When the reaction is complete, the total pressure in the reaction vessel is ____________ m bar. (Round off to the Nearest Integer).

Correct answer: 875

Step-by-step solution →
Q28·ChemistryNumericalJEE Main 2021
A home owner uses 4.00×103\times 10^{3}×103 m3^{3}3 of methane (CH4_44​) gas, (assume CH4_44​ is an ideal gas) in a year to heat his home. Under the pressure of 1.0 atm and 300 K, mass of gas used is x×105\times 10^{5}×105 g. The value of x is ____________. (Nearest integer) (Given R = 0.083 L atm K−1^{-1}−1mol−1^{-1}−1)

Correct answer: 26

Step-by-step solution →
Q29·ChemistryNumericalJEE Main 2021
An LPG cylinder contains gas at a pressure of 300 kPa at 27∘C27^{\circ}C27∘C. The cylinder can withstand the pressure of 1.2×1061.2\times10^{6}1.2×106 Pa. The room in which the cylinder is kept catches fire. The minimum temperature at which the bursting of cylinder will take place is______ ∘C^{\circ}C∘C. (Nearest integer)

Correct answer: 927

Step-by-step solution →
Q30·ChemistryNumericalJEE Main 2021
The pressure exerted by a non-reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27∘27^\circ27∘C is ________ kPa. (Round off to the Nearest Integer). [Assume gases are ideal, R = 8.314 J mol−1 K−1\mathrm{mol^{-1}\ K^{-1}}mol−1 K−1 Atomic masses : C : 12.0 u, H : 1.0 u, O : 16.0 u]

Correct answer: 150

Step-by-step solution →
Q31·ChemistryNumericalJEE Main 2021
A certain gas obeys P(Vm_mm​ − b) = RT. The value of (∂Z∂P)T\left(\frac{\partial Z}{\partial P}\right)_{T}(∂P∂Z​)T​ is xbRT\frac{xb}{RT}RTxb​ . The value of x is _______ .

Correct answer: 1

Step-by-step solution →
Q32·ChemistryNumericalJEE Main 2021
A car tyre is filled with nitrogen gas at 35 psi at 27°C. It will burst if pressure exceeds 40 psi. The temperature in °C at which the car tyre will burst is _______. (Rounded-off to the nearest integer)

Correct answer: 70

Step-by-step solution →
Q33·ChemistryNumericalJEE Main 2021
The volume occupied by 4.75 g of acetylene gas at 50ºC and 740 mmHg pressure is ________L. (Rounded off to the nearest integer) (Given R = 0.0826 L atm K−1K^{-1}K−1 mol−1mol^{-1}mol−1)

Correct answer: 5

Step-by-step solution →
Q34·ChemistrySingle correctJEE Advanced 2020
If the distribution of molecular speeds of a gas is as per the figure shown below, then the ratio of the most probable, the average and the roots mean square speeds, respectively, is
  1. (A)1 : 1 : 1
  2. (B)1 : 1 : 1.224
  3. (C)1 : 1.128 : 1.224
  4. (D)1 : 1.128 : 1

Correct answer: (B)

Step-by-step solution →
Q35·ChemistryNumericalJEE Main 2020
A spherical balloon of radius 3 cm containing helium gas has a pressure of 48×10−348\times10^{-3}48×10−3 bar. At the same temperature, the pressure, of a spherical balloon of radius 12 cm containing the same amount of gas will be ______________×10−6\times10^{-6}×10−6 bar.

Correct answer: 750

Step-by-step solution →
Q36·ChemistrySingle correctJEE Main 2020
A mixture of one mole each of H2_{2}2​, He and O2_{2}2​ each are enclosed in a cylinder of volume V at temperature T. If the partial pressure of H2_{2}2​ is 2 atm, the total pressure of the gases in the cylinder is:
  1. (A)14 atm
  2. (B)6 atm
  3. (C)22 atm
  4. (D)38 atm

Correct answer: (B)

Step-by-step solution →
Q37·ChemistrySingle correctJEE Main 2020
The predominant intermolecular forces present in ethyl acetate, a liquid, are:
  1. (A)hydrogen bonding and London dispersion
  2. (B)Dipole-dipole and hydrogen bonding
  3. (C)London dispersion, dipole-dipole and hydrogen bonding
  4. (D)London dispersion and dipole-dipole

Correct answer: (D)

Step-by-step solution →
Q38·ChemistrySingle correctJEE Main 2020
A graph of vapour pressure and temperature for three different liquids X, Y, and Z is shown below: The following inferences are made : (a) X has higher intermolecular interactions compared to Y. (b) X has lower intermolecular interactions compared to Y. (c) Z has lower intermolecular interactions compared to Y. The correct inference (s) is / are :
  1. (A)(c)
  2. (B)(a) and (c)
  3. (C)(b)
  4. (D)(a)

Correct answer: (C)

Step-by-step solution →
Q39·ChemistrySingle correctJEE Main 2020
The relative strength of interionic/intermolecular forces in decreasing order is
  1. (A)ion-dipole > dipole-dipole > ion-ion
  2. (B)dipole-dipole > ion-dipole > ion- ion
  3. (C)ion-ion > ion-dipole > dipole-dipole
  4. (D)ion-dipole > ion-ion > dipole-dipole

Correct answer: (C)

Step-by-step solution →
Q40·ChemistrySingle correctJEE Main 2020
Identify the correct labels of A, B and C in the following graph from the options given below Root mean square speed(VrmsV_{rms}Vrms​); most probable speed(VmpV_{\mathrm{m}\mathrm{p}}Vmp​); Average speed(VavV_{av}Vav​)
  1. (A)A-VavV_{av}Vav​, B-VrmsV_{rms}Vrms​, C-VmpV_{\mathrm{m}\mathrm{p}}Vmp​
  2. (B)A-VrmsV_{rms}Vrms​, B-VmpV_{\mathrm{m}\mathrm{p}}Vmp​, C-VavV_{av}Vav​
  3. (C)A-VmpV_{\mathrm{m}\mathrm{p}}Vmp​, B-VrmsV_{rms}Vrms​, C-VavV_{av}Vav​
  4. (D)A-VmpV_{\mathrm{m}\mathrm{p}}Vmp​, B-VavV_{av}Vav​, C-VrmsV_{rms}Vrms​

Correct answer: (D)

Step-by-step solution →
Q41·ChemistryMultiple correctJEE Advanced 2019
Which of the following statement(s) is(are) correct regarding the root mean square speed (Urms_{rms}rms​) and average translational kinetic energy (εav\varepsilon_{av}εav​) of a molecule in a gas at equilibrium ?
  1. (A)εav\varepsilon_{av}εav​ at a given temperature does not depend on its molecular mass
  2. (B)Urms_{rms}rms​ is doubled when its temperature is increased four times
  3. (C)εav\varepsilon_{av}εav​ is doubled when its temperature is increased four times
  4. (D)Urms_{rms}rms​ is inversely proportional to the square root of its molecular mass

Correct answer: (A), (B), (D)

Step-by-step solution →
Q42·ChemistrySingle correctJEE Main 2019
Consider the following table: a and b are van der Waals constants. The correct statement about the gases is:
Gasa / (k Pa dm6^66 mol−1^{-1}−1)b / (dm3^33 mol−1^{-1}−1)
A642.320.05196
B155.210.04136
C431.910.05196
D155.210.4382
  1. (A)Gas C will occupy more volume than gas A ; gas B will be lesser compressible than gas D
  2. (B)Gas C will occupy lesser volume than gas A ; gas B will be more compressible than gas D.
  3. (C)Gas C will occupy lesser volume than gas A ; gas B will be lesser compressible than gas D
  4. (D)Gas C will occupy more volume than gas A; gas B will be more compressible than gas D.

Correct answer: (D)

Step-by-step solution →
Q43·ChemistrySingle correctJEE Main 2019
Points I, II and III in the following plot respectively correspond to (Vmp_{\mathrm{m}\mathrm{p}}mp​ : most probable velocity)
  1. (A)Vmp_{\mathrm{m}\mathrm{p}}mp​ of N2_22​ (300K); Vmp_{\mathrm{m}\mathrm{p}}mp​ of H2_22​(300K); Vmp_{\mathrm{m}\mathrm{p}}mp​ of O2_22​(400K)
  2. (B)Vmp_{\mathrm{m}\mathrm{p}}mp​ of H2_22​ (300K); Vmp_{\mathrm{m}\mathrm{p}}mp​ of N2_22​(300K); Vmp_{\mathrm{m}\mathrm{p}}mp​ of O2_22​(400K)
  3. (C)Vmp_{\mathrm{m}\mathrm{p}}mp​ of O2_22​ (400K); Vmp_{\mathrm{m}\mathrm{p}}mp​ of N2_22​(300K); Vmp_{\mathrm{m}\mathrm{p}}mp​ of H2_22​(300K)
  4. (D)Vmp_{\mathrm{m}\mathrm{p}}mp​ of N2_22​ (300K); Vmp_{\mathrm{m}\mathrm{p}}mp​ of O2_22​(400K); Vmp_{\mathrm{m}\mathrm{p}}mp​ of H2_22​(300K)

Correct answer: (D)

Step-by-step solution →
Q44·ChemistrySingle correctJEE Main 2019
Consider the van der Waals constants, a and b, for the following gases. Which gas is expected to have the highest critical temperature?
GasArNeKrXe
a/(atm dm6^66mol−2^{-2}−2)1.30.25.14.1
b/(10−2^{-2}−2dm3^33mol−1^{-1}−1)3.21.71.05.0
  1. (A)Ar
  2. (B)Xe
  3. (C)Kr
  4. (D)Ne

Correct answer: (C)

Step-by-step solution →
Q45·ChemistrySingle correctJEE Main 2019
At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Their equation of state is given as P=RTV−bP=\dfrac{RT}{V-b}P=V−bRT​ at T. Here, b is the van der Waals constant. Which gas will exhibit steeper increase in the plot of Z (compression factor) versus P?
  1. (A)Ne
  2. (B)Ar
  3. (C)Xe
  4. (D)Kr

Correct answer: (C)

Step-by-step solution →
Q46·ChemistrySingle correctJEE Main 2019
An open vessel at 27°C is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is :
  1. (A)500°C
  2. (B)500 K
  3. (C)750°C
  4. (D)750 K

Correct answer: (B)

Step-by-step solution →
Q47·ChemistrySingle correctJEE Main 2019
The volume of gas A has is twice than that of gas B. The compressibility factor of gas A is thrice that that of gas B at same temperature. The pressure of gases for equal per moles are
  1. (A)3PA=2PB3P_A = 2P_B3PA​=2PB​
  2. (B)2PA=3PB2P_A = 3P_B2PA​=3PB​
  3. (C)PA=3PBP_A = 3P_BPA​=3PB​
  4. (D)PA=2PBP_A = 2P_BPA​=2PB​

Correct answer: (B)

Step-by-step solution →
Q48·ChemistrySingle correctJEE Main 2019
0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3^{3}3 at 1000 K. Given R is the gas constant in JK−1^{-1}−1mol−1^{-1}−1, x is
  1. (A)2R4+R\dfrac{2R}{4+R}4+R2R​
  2. (B)2R4−R\dfrac{2R}{4-R}4−R2R​
  3. (C)4+R2R\dfrac{4+R}{2R}2R4+R​
  4. (D)4−R2R\dfrac{4-R}{2R}2R4−R​

Correct answer: (D)

Step-by-step solution →
Q49·ChemistryNumericalJEE Advanced 2018
A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure 2), the volume (in m3m^{3}m3) of the compartment A after the system attains equilibrium is ___.

Correct answer: 2.22

Step-by-step solution →
Q50·ChemistryIntegerJEE Advanced 2016
The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases xxx times. The value of xxx is

Correct answer: 4

Step-by-step solution →
Q51·ChemistryMultiple correctJEE Advanced 2015
One mole of a monoatomic real gas satisfies the equation p(V−b)=RTp(V - b) = RTp(V−b)=RT where b is a constant. The relationship of interatomic potential V(r)V(r)V(r) and interatomic distance r for the gas is given by
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C)

Step-by-step solution →
Q52·ChemistryIntegerJEE Advanced 2015
A closed vessel with rigid walls contains 1 mol of 92238U^{238}_{92}\mathrm{U}92238​U and 1 mol of air at 298 K. Considering complete decay of 92238U^{238}_{92}\mathrm{U}92238​U to 82206Pb^{206}_{82}\mathrm{Pb}82206​Pb, the ratio of the final pressure to the initial pressure of the system at 298 K is

Correct answer: 9

Step-by-step solution →
Q53·ChemistrySingle correctJEE Advanced 2014
X and Y are two volatile liquids with molar weights of 10 g mol−110\ \mathrm{g\ mol^{-1}}10 g mol−1 and 40 g mol−140\ \mathrm{g\ mol^{-1}}40 g mol−1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L=24L = 24L=24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This is due to
  1. (A)larger mean free path for X as compared to that of Y.
  2. (B)larger mean free path for Y as compared to that of X.
  3. (C)increased collision frequency of Y with the inert gas as compared to that of X with the inert gas.
  4. (D)increased collision frequency of X with the inert gas as compared to that of Y with the inert gas.

Correct answer: (D)

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Q54·ChemistrySingle correctJEE Advanced 2014
X and Y are two volatile liquids with molar weights of 10 g mol−110\ \mathrm{g\ mol^{-1}}10 g mol−1 and 40 g mol−140\ \mathrm{g\ mol^{-1}}40 g mol−1 respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L=24L = 24L=24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. The value of d in cm (shown in the figure), as estimated from Graham's law, is
  1. (A)8
  2. (B)12
  3. (C)16
  4. (D)20

Correct answer: (C)

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Q55·ChemistrySingle correctJEE Advanced 2013
A fixed mass ‘m’ of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure The succeeding operations that enable this transformation of states are
  1. (A)Heating, cooling, heating, cooling
  2. (B)Cooling, heating, cooling, heating
  3. (C)Heating, cooling, cooling, heating
  4. (D)Cooling, heating, heating, cooling

Correct answer: (C)

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Q56·ChemistrySingle correctJEE Advanced 2013
A fixed mass ‘m’ of a gas is subjected to transformation of states from K to L to M to N and back to K as shown in the figure The pair of isochoric processes among the transformation of states is
  1. (A)K to L and L to M
  2. (B)L to M and N to K
  3. (C)L to M and M to N
  4. (D)M to N and N to K

Correct answer: (B)

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States of Matter: Gases and Liquids — frequently asked

How many questions from States of Matter: Gases and Liquids appear in JEE?

States of Matter: Gases and Liquids has appeared in 52 of the last 186 JEE Main and JEE Advanced papers — about 28% of them — contributing 56 questions in total across those papers.

Is States of Matter: Gases and Liquids an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 28% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these States of Matter: Gases and Liquids questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Chemistry chapters

  • Coordination Compounds 318
  • p-Block Elements 276
  • Redox Reactions and Electrochemistry 265
  • Chemical Bonding and Molecular Structure 207
  • d- and f-Block Elements 202
  • Aldehydes and Ketones 199
  • Equilibrium 199
  • Solutions 198

All 36 Chemistry chapters →

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