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JEE Main Test Series States of Matter: Previous Year Questions and Formulas for Revision
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Sonal Vaid

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States of Matter is one of the core topics of JEE Main chemistry syllabus. Usually, the topic has 1-3 questions covering a weightage of 4-12% in JEE Main question paper. Having thoroughly practised the subtopics will help students crack the section well. 

Some of the important subtopics from States of Matter include Properties of Gases, Gas Laws, Dalton’s Law for partial pressure, Ideal gas equation, Kinetic theory of gas, Concept of root mean square, Properties of liquid, Classification of solids, Unit cell and lattices, Voids, Electrical and magnetic properties, etc. Check JEE Main Chemistry Syllabus

Questions asked from States of Matter topic are usually direct in nature The difficulty level of States of Matter in JEE Main exam is moderate. The table below shows the weightage of the topic in the last five years:9

Year of an Exam Weightage (%)
2020 6.6 % (2 Questions)
2019 6.6 % (2 Questions)
2018 3 to 5 % (1-2 Questions)
2017 3% (1 Question)
2016 9.9 %(3 Questions)

Check JEE Main Chemistry Preparation

JEE Main Test Series States of Matter: Previous Year Questions 

Ques. The molecular velocity of any gas is ….. 

  1. Inversely proportional to the square root of the temperature
  2. Inversely proportional to the absolute temperature
  3. directly proportional to the square of the temperature
  4. directly proportional to the square root of the temperature

Solution : (D)

According to the law of gas “The Molecular velocity of any gas is directly proportional to the square root of the temperature ”.

Ques. 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: (D)

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

(Vrms) O2 = (Vrms) He

sqrt(3RT1/M1) = sqrt(3RT2/M2)

T1 /M1 = T2/M2

T1/32 = 300/4

T1 = 300×32/4

= 2400 K

Ques. 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: (C)

R = 8.3 J mol-1 K-1

Ques Which one of the following is the wrong assumption of the 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: (A)

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.

Ques. 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: (B)

By Gay Lussac’s law, at a 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.

Ques. ‘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: (B)

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

Ques. In Van der Waals equation of state of the gas law, the constant ‘b’ is a measure of 

  1. intermolecular repulsions
  2. intermolecular collisions per unit volume
  3. volume occupied by the molecules
  4. intermolecular attraction

Solution: (C)

 In Van der Waals equation, ‘b’ is the excluded volume. Excluded volume is the volume occupied by the molecules of gas

Ques. 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: (A)

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

Ques. 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: (D)

(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)

JEE Main States of Matter: Quick Formulas

While preparing for any examinations “Quick Formulas” are very important for last-minute exam preparation. Here some formulas for JEE Main States of Matter.

  • Boyle’s Law:

P1V1 = P2V2 (at constant T and n)

  • Charles's Law :

V1/T1=V2/T2(at Constant P and n)

  • Avogadro’s Law:

V = kn (at constant P and T)

  • Ideal Gas Equation:

PV=nRT

  • Combined Gas Equation:

P1V1/T1=P2V2/T2

  •  Dalton’s Law of Partial Pressures:

PTotal = P1 + P2 + P3 + …Pn

  • Compressibility Factor (Z):

Z=PV/RT

  • Most probable Speed:

Umps=sqrt(2KT/M)

 Difference between the different States of Matter:

Characteristics Solid State Liquid State Gas State
Shape Fixed Indefinite Indefinite
Volume Fixed Fixed Indefinite
Forces Applied Strong weak Almost Zero
Motion of Particles Restricted Very slow random
Packing of Particles Closely packed Less closely packed Very loosely packed
Compressibility Incompressible Compressible Highly Compressible
Density High Low Very Low

Check JEE Main Study Notes on States of Matter 

!! Best of Luck !! 

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