NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 13 - Kinetic Theory

Welcome to the Chapter 13 - Kinetic Theory, Class 11 Physics NCERT Solutions page. Here, we provide detailed question answers for Chapter 13 - Kinetic Theory. The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.

Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Kinetic Theory and excel in their exams. By going through these Kinetic Theory question answers, you can strengthen your foundation and improve your performance in Class 11 Physics. Whether you’re revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.

Exercise 1

Q 1:

Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.

Diameter of an oxygen molecule, d= 3Å

Radius, r = d /2 = 3/2 = 1.5 Å = 1.5 × 10^-8 cm

Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm³

Molecular volume of oxygen gas, V = 4/3 π r³. N

Where, N is Avogadro's number = 6.023 × 10²³ molecules/mole

∴ V = 4/3 x 3.14 x (1.5 × 10^-8 )³x 6.023 × 10²³

= 8.51 cm³

Ratio of the molecular volume to the actual volume of oxygen = 8.51 / 22400 = 3.8 × 10^-4

Q 11:

A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Length of the narrow bore, L= 1 m = 100 cm

Length of the mercury thread, l= 76 cm

Length of the air column between mercury and the closed end, l_a= 15 cm

Since the boreis held vertically in air with the open end at the bottom, the mercury length that occupies the air space is: 100 - (76 + 15) = 9 cm

Hence, the total length of the air column = 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure.

∴Length of the air column in the bore= 24 + hcm

And, length of the mercury column = 76 - hcm

Initial pressure, P₁= 76 cm of mercury

Initial volume, V₁= 15 cm³

Final pressure, P₂= 76 - (76 - h) = h cm of mercury

Final volume, V₂= (24 + h) cm³

Temperature remains constant throughout the process.

∴P₁V₁= P₂V₂

= 76 × 15 = h (24 + h)

h²+ 24h - 1140 = 0

∴ h = - 24 +- underroot [(24)²+ 4 x 1 x 1140] / 2 x 1

= 23.8 cm or -47.8 cm

Height cannot be negative.

Hence, 23.8 cm of mercurywill flow out from the boreand 52.2 cm of mercury will remain in it.

The length of the air column will be 24 + 23.8 = 47.8 cm.

Q 12:

From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³s^-1. Identify the gas.

[Hint:Use Graham's law of diffusion: R₁/R₂= (M₂/M₁)^1/2, where R₁, R₂ are diffusion rates of gases 1 and 2, and M₁ and M₂their respective molecular masses. The law is a simple consequence of kinetic theory.]

Rate of diffusion of hydrogen, R₁ = 28.7 cm³s^-1

Rate of diffusion of another gas, R₂= 7.2 cm³s^-1

According to Graham's Law of diffusion, we have:

R₁/ R₂ = Underroot M₂/M₁

Where, M₁ is the molecular mass of hydrogen = 2.020 g

M₂ is the molecular mass of the unknown gas

∴ M₂ =M₁(R₁/ R₂)²

=2.02 (28.7 / 7.2)²

= 32.09 g

32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

Q 2:

Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.

The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as:

PV= nRT

Where, R is the universal gas constant = 8.314 J mol^-1K^-1

n= Number of moles = 1

T= Standard temperature = 273 K

P= Standard pressure = 1 atm = 1.013 × 10⁵Nm^-2

∴ V = nRT / P

= 1 x 8.314 x 273 / 1.013 x 10⁵

= 0.0224 m³ = 22.4 litres

Hence, the molar volume of a gas at STP is 22.4 litres.

Q 5:

An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

Volume of the air bubble, V₁ = 1.0 cm³ = 1.0 × 10^-6 m³

Bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T₁ = 12°C = 285 K

Temperature at the surface of the lake, T₂ = 35°C = 308 K

The pressure on the surface of the lake:

P₂ = 1 atm = 1 ×1.013 × 10⁵ Pa

The pressure at the depth of 40 m:

P1 = 1 atm + dpg

Where, p is the density of water = 10³ kg/m³

g is the acceleration due to gravity = 9.8 m/s²

∴P₁ = 1.013 × 10⁵ + 40 × 10³ × 9.8 = 493300 Pa

We have: P₁V₁/ T₁ = P₂V₂/ T₂

Where, V₂ is the volume of the air bubble when it reaches the surface

V₂ = P₁V₁T₂/ T₁P₂

= 493300 x (1.0 x 10^-6) 308 / 285 x 1.013 x 10⁵

= 5.263 × 10^-6 m3 or 5.263 cm³

Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm³.

Q 6:

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.

Volume of the room, V= 25.0 m³

Temperature of the room, T= 27°C = 300 K

Pressure in the room, P= 1 atm = 1 × 1.013 × 10⁵ Pa

The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can be written as:

PV = k_BNT

Where, K_B is Boltzmann constant = 1.38 × 10^-23 m² kg s^-2K^-1

N is the number of air molecules in the room

∴ N = PV / K_B T

= 1.013 x 10⁵ x 25 / 1.38 x 10^-23 x 300

= 6.11 × 10²⁶ molecules

Therefore, the total number of air molecules in the given room is 6.11 × 10²⁶.

Q 7:

Estimate the average thermal energy of a helium atom at

(i) room temperature (27 °C),

(ii) the temperature on the surface of the Sun (6000 K),

(iii) the temperature of 10 million Kelvin (the typical core temperature in the case of a star).

(i) At room temperature, T= 27°C = 300 K

Average thermal energy = 3/2 kT

Where k is Boltzmann constant = 1.38 × 10^-23m² kg s^-2K^-1

∴ 3/2 kT = 3/2 x 1.38 x 10^-38 x 300

= 6.21 × 10^-21J

Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 × 10^-21J.

(ii) On the surface of the sun, T= 6000 K

Average thermal energy = 3/2 kT

= 3/2 x 1.38 x 10^-38 x 6000

= 1.241 × 10^-19J

Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 × 10^-19J .

(iii) At temperature, T= 10⁷K

Average thermal energy = 3/2 kT

= 3/2 x 1.38 x 10^-38 x 10⁷

= 2.07 × 10^-16J

Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10^-16J.

Q 8:

Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v_rmsthe largest?

Yes.All contain the same number of the respective molecules.

No. The root mean square speed of neon is the largest.

Since the three vessels have the same capacity, they have the same volume.

Hence, each gas has the same pressure, volume, and temperature.

According to Avogadro's law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro's number, N= 6.023 × 10²³.

The root mean square speed (v_rms) of a gas of mass m, and temperature T, is given by the relation:

v_rms= underroot 3kT / m

Where, k is Boltzmann constant

For the given gases, k and T are constants.

Hence v_rmsdepends only on the mass of the atoms, i.e.,

v_rms∝ underroot 1/m

Therefore, the root mean square speed of the molecules in the three cases is not the same. Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest. Hence, neon has the largest root mean square speed among the given gases.

Frequently Asked Questions about Kinetic Theory - Class 11 Physics

- 1. How many questions are covered in Kinetic Theory solutions?
- All questions from Kinetic Theory are covered with detailed step-by-step solutions including exercise questions, additional questions, and examples.
- 2. Are the solutions for Kinetic Theory helpful for exam preparation?
- Yes, the solutions provide comprehensive explanations that help students understand concepts clearly and prepare effectively for both board and competitive exams.
- 3. Can I find solutions to all exercises in Kinetic Theory?
- Yes, we provide solutions to all exercises, examples, and additional questions from Kinetic Theory with detailed explanations.
- 4. How do these solutions help in understanding Kinetic Theory concepts?
- Our solutions break down complex problems into simple steps, provide clear explanations, and include relevant examples to help students grasp the concepts easily.
- 5. Are there any tips for studying Kinetic Theory effectively?
- Yes, practice regularly, understand the concepts before memorizing, solve additional problems, and refer to our step-by-step solutions for better understanding.

Exam Preparation Tips for Kinetic Theory

The Kinetic Theory is an important chapter of 11 Physics. This chapter’s important topics like Kinetic Theory are often featured in board exams. Practicing the question answers from this chapter will help you rank high in your board exams.

Latest Blog Posts

Stay updated with our latest educational content and study tips

Smart Questions to Ask in a Parent-Teacher Meeting | PTM Made Easy

Parent-Teacher Meetings (PTMs) are more than quick updates on marks — they’re a chance to build a real partnership between home and school. A good Parent-Teacher Meeting conversation helps parents see beyond grades. It opens up insights about a child’s strengths, struggles, emotions and even hidden talents. When parents participate actively, they don’t just track […]

The Secret to Smarter Learning — Building Strong Critical Thinking Skills

In today’s world of endless information , knowing how to think is more important than knowing what to think . From school projects to real – life decisions , critical thinking helps students question ideas , analyze facts and form logical conclusions . But what exactly does critical thinking mean ? Simply put , it’s […]

Study Smarter, Not Harder: Build Productive Habits That Stick

Every student dreams of better grades , stronger focus and more study time – but the real challenge isn’t starting, it’s staying consistent . Building productive study habits is not about studying all day , it’s about studying smart . In today’s fast – paced digital world, distractions are everywhere – from endless phone notifications […]

The Hidden Risks of Online Gaming for Children — Is your child safe while gaming online?

Online gaming has rapidly become one of the most popular pastimes among children. Whether it’s multiplayer mobile games , PC adventures or console challenges , kids are spending more time than ever in the virtual world . On the surface, gaming seems entertaining and even educational – improving hand- eye coordination , teamwork and problem […]

View All Blog Posts

Benefits of Using Our NCERT Solutions for Class

When it comes to excelling in your studies, having a well-structured study guide can make a huge difference. Our NCERT Solutions for Class provide you with a comprehensive, easy-to-understand, and exam-focused resource that is specifically tailored to help you maximize your potential. Here are some of the key benefits of using our NCERT solutions for effective learning and high scores:

NCERT Solutions for Effective Exam Preparation

Preparing for exams requires more than just reading through textbooks. It demands a structured approach to understanding concepts, solving problems, and revising thoroughly. Here’s how our NCERT solutions can enhance your exam preparation:

Clear Understanding of Concepts: Our NCERT solutions are designed to break down complex topics into simple, understandable language, making it easier for students to grasp essential concepts in . This helps in building a solid foundation for each chapter, which is crucial for scoring high marks.
Step-by-Step Solutions: Each solution is presented in a detailed, step-by-step manner. This approach not only helps you understand how to reach the answer but also equips you with the right techniques to tackle similar questions in exams.
Access to Important Questions: We provide a curated list of important questions and commonly asked questions in exams. By practicing these questions, you can familiarize yourself with the types of problems that are likely to appear in the exams and gain confidence in answering them.
Quick Revision Tool: Our NCERT solutions serve as an excellent tool for last-minute revision. The solutions cover all key points, definitions, and explanations, ensuring that you have everything you need to quickly review before exams.

Importance of Structured Answers for Scoring Higher Marks

In exams, it's not just about getting the right answer—it's also about presenting it in a well-structured and logical way. Our NCERT solutions for Class are designed to guide you on how to write answers that are organized and effective for scoring high marks.

Precise and Concise Answers: Our solutions are crafted to provide answers that are to the point, without unnecessary elaboration. This ensures that you don't waste time during exams and focus on delivering accurate answers that examiners appreciate.
Step-Wise Marks Distribution: We understand that exams often allot marks based on specific steps or points. Our NCERT solutions break down each answer into structured steps to ensure you cover all essential points required for full marks.
Improved Presentation Skills: By following the format of our NCERT solutions, you learn how to present your answers in a systematic and logical manner. This helps in making your answers easy to read and allows the examiner to quickly identify key points, resulting in better scores.
Alignment with NCERT Guidelines: Since exams are often set in alignment with NCERT guidelines, our solutions are tailored to follow the exact format and language that is expected in exams. This can improve your chances of scoring higher by meeting the examiner's expectations.