JEE Chemistry States Of Matter MCQs Set 08

Download JEE MCQs for JEE Chemistry: States Of Matter

Access targeted multiple-choice questions for States Of Matter designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Chapter-wise Objective Questions: States Of Matter

Access the complete set of multiple-choice questions for States Of Matter below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: Which of the following is not true for an ideal gas?
(a) It cannot be converted into a liquid
(b) There is no interaction between the molecules
(c) All molecules of the gas move with the same speed
(d) At a given temperature, PV is proportional to the amount of gas
Show Answer & Explanation

Answer: (c) All molecules of the gas move with the same speed

Explanation:
1. Molecules of an ideal gas move with a distribution of speeds (Maxwell distribution), not a single speed.
2. The other statements follow from the ideal gas model.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 2: The ratio among the most probable velocity, mean velocity and root mean square velocity is
(a) \( 1:2:3 \)
(b) \( 1:\sqrt2:\sqrt3 \)
(c) \( \sqrt2:\sqrt3:\sqrt{8/\pi} \)
(d) \( \sqrt2:\sqrt{8/\pi}:\sqrt3 \)
Show Answer & Explanation

Answer: (d) \( \sqrt2:\sqrt{8/\pi}:\sqrt3 \)

Explanation:
1. Most probable \( =\sqrt{2RT/M} \), mean \( =\sqrt{8RT/\pi M} \), rms \( =\sqrt{3RT/M} \).
2. Taking the common factor out gives \( \sqrt2:\sqrt{8/\pi}:\sqrt3 \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 3: Which of the following has the maximum root mean square velocity at the same temperature?
(a) \( \ce{SO2} \)
(b) \( \ce{CO2} \)
(c) \( \ce{O2} \)
(d) \( \ce{H2} \)
Show Answer & Explanation

Answer: (d) \( \ce{H2} \)

Explanation:
1. At fixed T, \( u_{rms}\propto 1/\sqrt{M} \).
2. Hydrogen has the smallest molar mass, so it has the greatest rms velocity.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 4: The temperature at which the rms velocity of \( \ce{SO2} \) molecules is half that of \( \ce{He} \) molecules at 300 K is
(a) 150 K
(b) 600 K
(c) 900 K
(d) 1200 K
Show Answer & Explanation

Answer: (d) 1200 K

Explanation:
1. Set \( \sqrt{T/64}=\tfrac12\sqrt{300/4} \).
2. Squaring: \( T/64=\tfrac14\times75=18.75 \), so \( T=1200\ K \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 5: At \( 27^\circ C \), the ratio of the rms velocities of ozone to oxygen is
(a) \( \sqrt{3/5} \)
(b) \( \sqrt{4/3} \)
(c) \( \sqrt{2/3} \)
(d) 0.25
Show Answer & Explanation

Answer: (c) \( \sqrt{2/3} \)

Explanation:
1. Same temperature, so \( u_{O_3}/u_{O_2}=\sqrt{M_{O_2}/M_{O_3}}=\sqrt{32/48} \).
2. This equals \( \sqrt{2/3} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 6: The average kinetic energy per molecule of an ideal gas at \( 25^\circ C \) is
(a) \( 6.17\times10^{-21}\ kJ \)
(b) \( 6.17\times10^{-21}\ J \)
(c) \( 6.17\times10^{-20}\ J \)
(d) \( 7.16\times10^{-20}\ J \)
Show Answer & Explanation

Answer: (b) \( 6.17\times10^{-21}\ J \)

Explanation:
1. Average KE per molecule = \( \tfrac32 kT \).
2. = 1.5 \( \times \) \( 1.38\times10^{-23} \) \( \times \) 298 = \( 6.17\times10^{-21}\ J \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 7: At what temperature will the rms velocity of \( \ce{SO2} \) be the same as that of \( \ce{O2} \) at 303 K?
(a) 273 K
(b) 606 K
(c) 303 K
(d) 403 K
Show Answer & Explanation

Answer: (b) 606 K

Explanation:
1. Equal rms velocities need equal \( T/M \).
2. T = 303 \( \times \) 64/32 = 606 K.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 8: Among the following gases, which has the lowest root mean square velocity at \( 25^\circ C \)?
(a) \( \ce{SO2} \)
(b) \( \ce{N2} \)
(c) \( \ce{O2} \)
(d) \( \ce{Cl2} \)
Show Answer & Explanation

Answer: (d) \( \ce{Cl2} \)

Explanation:
1. The lowest velocity belongs to the gas with the highest molar mass.
2. Molar masses: \( \ce{SO2} \) 64, \( \ce{N2} \) 28, \( \ce{O2} \) 32, \( \ce{Cl2} \) 71, so \( \ce{Cl2} \) is slowest.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 9: The root mean square velocity of a gas molecule is proportional to
(a) \( m^{1/2} \)
(b) \( m^{0} \)
(c) \( m^{-1/2} \)
(d) \( m \)
Show Answer & Explanation

Answer: (c) \( m^{-1/2} \)

Explanation:
1. \( u_{rms}=\sqrt{3kT/m} \), so it is proportional to \( m^{-1/2} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 10: Two gases at the same temperature have rms velocities \( u_1 \) and \( u_2 \) and masses \( m_1 \) and \( m_2 \). Which relation is correct?
(a) \( \frac{m_1}{u_1^2}=\frac{m_2}{u_2^2} \)
(b) \( m_1u_1=m_2u_2 \)
(c) \( \frac{m_1}{u_1}=\frac{m_2}{u_2} \)
(d) \( m_1u_1^2=m_2u_2^2 \)
Show Answer & Explanation

Answer: (d) \( m_1u_1^2=m_2u_2^2 \)

Explanation:
1. At the same temperature the average kinetic energies are equal.
2. So \( \tfrac12m_1u_1^2=\tfrac12m_2u_2^2 \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 11: The temperature of a gas is raised from \( 27^\circ C \) to \( 927^\circ C \). The root mean square velocity
(a) \( \sqrt{927/27} \) times the earlier value
(b) remains the same
(c) is halved
(d) is doubled
Show Answer & Explanation

Answer: (d) is doubled

Explanation:
1. Convert to kelvin: 300 K and 1200 K.
2. The ratio of rms velocities is \( \sqrt{1200/300}=2 \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 12: The ratio between the root mean square velocity of hydrogen (\( \ce{H2} \)) at 50 K and that of oxygen (\( \ce{O2} \)) at 800 K is
(a) 4
(b) 2
(c) 1
(d) \( 1/4 \)
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. Hydrogen: \( \sqrt{50/2}=5 \) (in units of the common factor).
2. Oxygen: \( \sqrt{800/32}=5 \).
3. The ratio is 1.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 13: At constant pressure, the root mean square velocity of an ideal gas varies with density d as
(a) \( d^2 \)
(b) \( d \)
(c) \( \sqrt d \)
(d) \( 1/\sqrt d \)
Show Answer & Explanation

Answer: (d) \( 1/\sqrt d \)

Explanation:
1. \( u_{rms}=\sqrt{3P/d} \).
2. At constant P, it is proportional to \( 1/\sqrt d \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 14: A mixture of \( \ce{SO2} \) and \( \ce{O2} \) is kept at room temperature. Compared with an oxygen molecule, an \( \ce{SO2} \) molecule hits the wall with
(a) smaller average speed
(b) greater average speed
(c) greater kinetic energy
(d) greater mass
Show Answer & Explanation

Answer: (a) smaller average speed

Explanation:
1. At the same temperature both have the same average kinetic energy.
2. SO2 is heavier, so its average speed is smaller.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 15: The rms speed of \( \ce{N2} \) molecules in a gas is u. If the temperature is doubled and the nitrogen molecules dissociate into atoms, the rms speed becomes
(a) \( u/2 \)
(b) \( 2u \)
(c) \( 4u \)
(d) \( 14u \)
Show Answer & Explanation

Answer: (b) \( 2u \)

Explanation:
1. \( u\propto\sqrt{T/M} \).
2. T doubles and M halves, so the factor is \( \sqrt{2\times2}=2 \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 16: The correct arrangement of the average speed (\( \bar u \)), most probable speed (\( u_p \)) and rms speed (\( u_{rms} \)) is
(a) \( \bar u>u_p>u_{rms} \)
(b) \( u_{rms}>\bar u>u_p \)
(c) \( u_p>\bar u>u_{rms} \)
(d) \( u_p>u_{rms}>\bar u \)
Show Answer & Explanation

Answer: (b) \( u_{rms}>\bar u>u_p \)

Explanation:
1. The ratio is \( u_p:\bar u:u_{rms}=1:1.128:1.224 \).
2. So \( u_{rms}>\bar u>u_p \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 17: The ratio of the most probable velocity to the average velocity of gas molecules is
(a) \( \pi/2 \)
(b) \( 2/\pi \)
(c) \( \sqrt\pi/2 \)
(d) \( 2/\sqrt\pi \)
Show Answer & Explanation

Answer: (c) \( \sqrt\pi/2 \)

Explanation:
1. \( \frac{\sqrt{2RT/M}}{\sqrt{8RT/\pi M}}=\sqrt{\frac{2\pi}{8}}=\frac{\sqrt\pi}{2} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 18: The rms velocity of a certain gas is v at 300 K. The temperature at which the rms velocity becomes double is
(a) 1200 K
(b) 900 K
(c) 600 K
(d) 150 K
Show Answer & Explanation

Answer: (a) 1200 K

Explanation:
1. Velocity doubles when T becomes four times: 4 \( \times \) 300 = 1200 K.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 19: The r.m.s. velocity of a gas depends upon
(a) temperature only
(b) molecular mass only
(c) temperature and molecular mass of the gas
(d) none of these
Show Answer & Explanation

Answer: (c) temperature and molecular mass of the gas

Explanation:
1. \( u_{rms}=\sqrt{3RT/M} \) depends on both T and M.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 20: The pressure of 2 mol of \( \ce{NH3} \) at \( 27^\circ C \) in a vessel of volume 5 L, using van der Waals equation (\( a=4.17\ atm\,L^2mol^{-2} \), \( b=0.0371\ L\,mol^{-1} \)), is
(a) 10.33 atm
(b) 9.33 atm
(c) 9.74 atm
(d) 9.2 atm
Show Answer & Explanation

Answer: (b) 9.33 atm

Explanation:
1. \( P=\frac{nRT}{V-nb}-\frac{an^2}{V^2} \).
2. nRT = 2 \( \times \) 0.0821 \( \times \) 300 = 49.26 and V - nb = 4.926, giving 10.00 atm.
3. Correction: 4.17 \( \times \) 4/25 = 0.667 atm, so P = 9.33 atm.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 21: The rms velocity of one mole of a monatomic gas of molar mass M is \( U_{rms} \). The relation between its average kinetic energy E and \( U_{rms} \) is
(a) \( U_{rms}=\sqrt{3E/2M} \)
(b) \( U_{rms}=\sqrt{2E/3M} \)
(c) \( U_{rms}=\sqrt{2E/M} \)
(d) \( U_{rms}=\sqrt{E/3M} \)
Show Answer & Explanation

Answer: (c) \( U_{rms}=\sqrt{2E/M} \)

Explanation:
1. For one mole, \( E=\tfrac12MU_{rms}^2 \).
2. Hence \( U_{rms}=\sqrt{2E/M} \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 22: The ratio of the average velocity to the most probable velocity is
(a) 1.128
(b) 1.0
(c) 1.224
(d) 1.112
Show Answer & Explanation

Answer: (a) 1.128

Explanation:
1. \( \frac{\sqrt{8RT/\pi M}}{\sqrt{2RT/M}}=\sqrt{4/\pi}=1.128 \).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 23: If \( v_{rms} \) is \( 30R^{1/2} \) at \( 27^\circ C \), the molar mass of the gas in kilograms is
(a) 1
(b) 2
(c) 4
(d) 0.001
Show Answer & Explanation

Answer: (d) 0.001

Explanation:
1. \( v_{rms}^2=3RT/M\Rightarrow 900R=3R(300)/M \).
2. M = 1 g/mol = 0.001 kg/mol.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 24: 0.50 mol of \( \ce{H2} \) gas and 1.0 mol of \( \ce{He} \) gas at STP
(a) have equal average kinetic energies
(b) have equal molecular speeds
(c) occupy equal volumes
(d) have equal effusion rates
Show Answer & Explanation

Answer: (a) have equal average kinetic energies

Explanation:
1. Average KE per molecule depends only on temperature, so it is equal at the same T.
2. Speeds, volumes and effusion rates differ (different mass and different moles).

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Question 25: With increase of pressure (at constant temperature), the mean free path of the molecules
(a) decreases
(b) increases
(c) does not change
(d) becomes zero
Show Answer & Explanation

Answer: (a) decreases

Explanation:
1. Mean free path is inversely proportional to the number density, so it decreases as pressure increases.

Teacher's Note:
1. Identify the key concept in the question first, then apply the standard relation step by step.
2. Check units and the number of significant figures before choosing the option.

Free study material for Chemistry

Multiple Choice Questions (MCQs) for JEE Chemistry States Of Matter

Chapter MCQs with Answers for JEE Chemistry

Review structured objective questions for JEE Chemistry States Of Matter. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

Expert Practice Material for JEE Chemistry

Each question includes structured solution keys mapped directly to standard JEE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Complete Your Chapter Revision

Follow up your worksheet practice by attempting the interactive online Chemistry MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Chemistry States Of Matter MCQs Set 08?

You can get most exhaustive JEE Chemistry States Of Matter MCQs Set 08 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry States Of Matter MCQs Set 08 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry States Of Matter MCQs Set 08, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry States Of Matter MCQs Set 08?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry States Of Matter MCQs Set 08 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.