JEE Main States of Matter Previous Year Questions with Solutions

States of Matter Previous Year Questions with Solutions are given here. BYJU’S provides accurate solutions prepared by our subject matter experts. The three basic states of matter are solid, liquid and gas. The matter exists in nature in different forms. Some substances are rigid and have a fixed shape like rock and wood. Some other substances can flow and take the shape of their container like water, while some do not have definite shape or size, such as air.

Important topics in states of matter include, different states of matter- liquid, gases, solids, Van Der Waals forces, Properties of gases and gas laws, Kinetic Theory of Gases, Assumptions made for Kinetic Theory of Gases, Ideal Gas Law and Dalton’s Law of Partial Pressure, Graham’s law of diffusion and Gas Eudiometry, Deviation from ideal gas behaviour and Van Der Waal’s Equation. The questions in this article help students to understand the difficulty level of questions in the upcoming JEE Main and JEE Advanced exams. Students can easily download the solutions in PDF format.

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JEE Main Previous Year Solved Questions on States Of Matter

1. If Z is the compressibility factor, van der Waals’ equation at low pressure can be written as :

(1) Z=1-Pb/RT

(2) Z=1+Pb/RT

(3) Z = 1+RT/Pb

(4) Z = 1-a/VmRT

Solution:

(P+a/Vm2 )(Vm-b) = RT [Van der waals equation of state]

Vm-b ≈ Vm

So the equation becomes (P+a/Vm2 )Vm= RT

⇒PVm+a/Vm = RT

Divide all terms by RT

PVm/RT+a/VmRT = RT/RT

PVm/RT = 1- a/VmRT

Z = 1- a/VmRT [Z = PVm/RT]

Hence option (4) is the answer.

2. Which intermolecular force is most responsible in allowing xenon gas to liquefy?

(1) Instantaneous dipole induced dipole

(2) Ion dipole

(3) Ionic

(4) Dipole-dipole

Solution:

For the liquefaction of xenon, instantaneous dipole induced dipole forces are responsible.

Hence option (1) is the answer.

3. The temperature at which oxygen molecules have the same root mean square

speed as helium 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

Solution:

Given Atomic masses : He = 4 u, O = 16 u

(Vrms) O2 = (Vrms) He

√(3RT1/M1) = √(3RT2/M2)

T1 /M1 = T2/M2

T1/32 = 300/4

T1 = 300×32/4

= 2400 K

Hence option (4) is the answer.

4. The compressibility factor for a real gas at high pressure is :

(1) 1-Pb/RT

(2) 1+ RT/Pb

(3) 1

(4) 1+Pb/RT

Solution:

(P+a/V2)(V-b) = RT [Real gas equation]

a/V2 can be neglected at high pressure.

PV-Pb = RT

PV/RT = (RT/RT) + (Pb/RT)

PV/RT = 1 + (Pb/RT) …(1)

Z = PV/RT …(2)

Equating (1) and (2)

Z = 1 + (Pb/RT)

Hence option (4) is the answer.

5. The relationship among most probable velocity, average velocity and root mean

Square velocity is respectively :

(1) √2 : √(8/π) : √3

(2) √2 :√3 : √(8/π )

(3) √3 : √(8/π) : √2

(4) √(8/π) : √3 : √2

Solution:

Vmpv = √(2RT/M)

Vav = √(8RT/πM)

Vrms = √(3RT/M)

Vmpv : Vav : Vrms . = √(2RT/M) : √(8RT/πM) : √(3RT/M)

= .√2 : √(8/π) : √3

Hence option (1) is the answer.

6. Value of gas constant R is

(1) 0.082 L atm

(2) 0.987 cal mol-1 K-1

(3) 8.3 J mol-1 K-1

(4) 83 erg mol-1K-1

Solution:

R = 8.3 J mol-1 K-1

Hence option (3) is the answer.

7. By how many folds the temperature of a gas would increase when the root mean

Square velocity of the gas molecules in a container of fixed volume is increased from 5×104 cm/s to 10×104 cm/s?

(1) Four

(2) three

(3) Two

(4) Six

Solution:

Vrms ∝ √T

V1/V2 = √(T1/T2) = 5×104/10×104

squaring, we get

T1/T2 = 25/100 = ¼

T2 = 4T1

Hence option (1) is the answer.

8. Kinetic theory of gases proves

(1) Only boyle’s law

(2) Only Charle’s law

(3) Only Avogadro’s law

(4) all of these

Solution:

One of the postulates of kinetic theory of gases is average kinetic energy proportional to T.

This theory proves all the above given laws.

Hence option (4) is the answer.

9. Which one of the following is the wrong assumption of kinetic theory of gases?

(1) All the molecules move in a straight line between collision and with the 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.

Solution:

The molecules are always in random motion and obey Newton’s law of motion. They have velocities in all directions ranging from zero to infinity.

Hence option (1) is the answer.

10. ‘a’ and ‘b’ are Vander Waal’s constants for gases. Chlorine is more easily liquified than ethane because:

(1) a for Cl2 < a for C2H6 but b for Cl2 > b for C2H6

(2) a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6

(3) a and b for Cl2 > a and b for C2H6

(4) a and b for Cl2 < a and b for C2H6

Solution:

Greater the ‘a’ value, more easily the gas is liquified, lower the ‘b’ value, more easily the gas is liquified.

Hence option (2) is the answer.

11. A gaseous compound of nitrogen and hydrogen contains 12.5% (by mass) of

Hydrogen. The density of the compound relative to hydrogen is 16. The molecular formula of the compound is :

(1) NH2

(2) NH3

(3) N3H

(4) N2H4

Solution:

Given that gaseous compound of nitrogen and hydrogen contains 12.5% (by mass) of

Hydrogen.

Element Percentage Atomic ratio Simple ratio
H 12.5% 12.5/1 = 12.5 12.5/6.25 = 2
N 87.5% 87.5/14 = 6.25 6.25 / 6.25 = 1

Empirical formula = NH2

Empirical mass = 16

Molecular weight = 2× Vapour density = 2×16 = 32

So n = molecular mass / empirical mass = 32/16 = 2

Molecular formula = Empirical formula × n

= (NH2)×2

= N2H4

Hence option (4) is the answer.

12. The one that is extensively used as a piezo electric material is

(1) quartz

(2) amorphous silica

(3) trydymite

(4) mica

Solution:

Quartz is used as a piezo electric material.

Hence option (1) is the answer.

13. Which primitive unit cell has unequal edge lengths and all axial lengths different from 900.

(1) Monoclinic

(2) Triclinic

(3) Tetragonal

(4) Hexagonal

Solution:

Triclinic primitive unit cell has unequal edge lengths and all axial lengths different from 900.

Hence option (2) is the answer.

14. In Van der waals equation of state of the gas law, the constant b is a measure of

(1) Intermolecular repulsions

(2) Intermolecular attraction

(3) Volume occupied by molecules

(4) Intermolecular collisions per unit volume.

Solution:

Van der waals constant b is the measure of effective volume occupied by the gas molecules.

Hence option (3) is the answer.

15. A pressure cooker reduces cooking time for food because

(1) Heat is more evenly distributed in the cooking space.

(2) B.P of water involved in cooking is increased

(3) The higher pressure inside the cooker crushes the food.

(4) Cooking involves chemical changes helped by a rise in temperature.

Solution:

By Gay Lussac’s law, at constant pressure of a given mass of a gas is directly proportional to the absolute temperature of the gas. So on increasing pressure, temperature also increases. So the boiling point of water is also increased.

Hence option (2) is the answer.

16. According to kinetic theory of gases, in an ideal gas, between two successive collisions a gas molecule travels

(1) In a circular path

(2) In wavy path

(3) In a straight line path

(4) with an accelerated velocity

Solution:

According to kinetic theory of gases, in an ideal gas, between two successive collisions a gas molecule travels in a straight line path.

Hence option (3) is the answer.

Also Read:-

Important States of Matter Formulas for JEE Main and Advanced

FAQ States of Matter JEE