Have you ever questioned yourself as to why gases expand or put pressure on the sides of a container? The scientific explanation is given by the Kinetic Theory of Gases, which assumes that gases are made of small particles in everlasting, rapid movement. This theory is the work of great scientists such as Boyle, Newton, Maxwell, and Boltzmann, and it is used to explain how gases act in response to changes in pressure and temperature and how they diffuse or mix.
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The NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory are very valuable not only in the preparation for the CBSE board but also for competitive exams such as JEE and NEET. The NCERT solutions are the work of physics experts who make complex concepts simple and offer a step-by-step approach to resolving numerical and theoretical problems. These NCERT Solutions for Class 11 Physics Chapter 12 - Kinetic Theory are prepared as per the recent CBSE syllabus, making them conceptually clear and helping the students to develop their ability to solve problems. Through such NCERT solutions, students will be able to have confidence in the chapter as well as score highly in exams.
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The Class 11 Physics Chapter 12 - Kinetic Theory question answers include step-by-step solutions of the textbook questions, which help the students to have a clear picture of the behaviour of gases at the molecular scale. One can access these solutions online and download them as a PDF solution without any payment, making practice and revision convenient.
Kinetic Theory class 11 question answers provide all the detailed solutions to the exercises, which allow students to understand the behavior of gases easily at the microscopic level. The solutions make things easy to understand, such as explain concepts such as pressure, mean free path, gas laws, and the kinetic interpretation of temperature, developing competitive fundamentals to prepare students for exams and other competitive tests like the JEE or NEET.
Answer:
The diameter of an oxygen molecule, d = 3 Å.
The actual volume of a mole of oxygen molecules V actual is
The volume occupied by a mole of oxygen gas at STP is V molar = 22.4 litres
Answer:
As per the ideal gas equation
For one mole of a gas at STP we have
Q12.3 Figure 13.8 shows plot of
Answer:
(a) The dotted plot corresponds to the ideal gas behaviour.
(b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T 1 is closer to the horizontal line that the one corresponding to T 2 we conclude T 1 is greater than T 2.
(c) As per the ideal gas equation
The molar mass of oxygen = 32 g
R = 8.314
(d) If we obtained similar plots for
Molar Mass of Hydrogen M = 2 g
mass of hydrogen
Answer:
Initial volume, V 1 = Volume of Cylinder = 30 l
Initial Pressure P 1 = 15 atm
Initial Temperature T 1 = 27 o C = 300 K
The initial number of moles n 1 inside the cylinder is
Final volume, V 2 = Volume of Cylinder = 30 l
Final Pressure P 2 = 11 atm
Final Temperature T 2 = 17 o C = 290 K
The final number of moles n 2 inside the cylinder is
Moles of oxygen taken out of the cylinder = n 2 -n 1 = 18.28 - 13.86 = 4.42
Mass of oxygen taken out of the cylinder, m is
Answer:
Initial Volume of the bubble, V 1 = 1.0 cm 3
Initial temperature, T 1 = 12 o C = 273 + 12 = 285 K
The density of water is
Initial Pressure is P 1
Depth of the bottom of the lake = 40 m
Final Temperature, T 2 = 35 o C = 35 + 273 = 308 K
Final Pressure = Atmospheric Pressure
Let the final volume be V 2
As the number of moles inside the bubble remains constant, we have
Answer:
The volume of the room, V = 25.0 m 3
Temperature of the room, T = 27 o C = 300 K
The pressure inside the room, P = 1 atm
Let the number of moles of air molecules inside the room be n
Avogadro's Number,
The number of molecules inside the room is N
Answer:
The average energy of a Helium atom is given as
(i)
(ii)
(iii)
Answer:
As per Avogadro's Hypothesis, under similar conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. Since the volume of the vessels is the same and all vessels are kept at the same conditions of pressure and temperature, they would contain an equal number of molecules.
Root mean square velocity is given as
As we can see, vrms is inversely proportional to the square root of the molar mass; the root mean square velocity will be maximum in the case of Neon, as its molar mass is the least.
Answer:
As we know root mean square velocity is given as
Let at temperature T, the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at
Answer:
Pressure, P = 2atm
Temperature, T = 17 o C
The radius of the Nitrogen molecule, r=1 Å.
The molecular mass of N 2 = 28 u
The molar mass of N 2 = 28 g
From the ideal gas equation
The above tells us about the number of moles per unit volume; the number of molecules per unit volume would be given as
The mean free path
The root mean square velocity v rms is given as
The time between collisions T, is given as
Collision time T' is equal to the average time taken by a molecule to travel a distance equal to its diameter
The ratio of the average time between collisions to the collision time is
Thus, we can see that the time between collisions is much larger than the collision time.
The Class 11 Physics Chapter 12 - Kinetic Theory Additional Questions have been provided to boost the problem solving skills by discussing extra problems besides the NCERT textbook. The questions enhance an appreciation of such concepts as molecular motion, pressure of gases, mean free path, equipartition of energy, and are relevant in both board examinations and competitive entrance examinations such as JEE and NEET.
Answer:
Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P 1 = 1 atm = 76 cm of Mercury
Let the crossectional area of the tube be x cm 2
The initial volume of the air column, V 1 = 15x cm 3
Let's assume that once the tube is held vertical, y cm of Mercury flows out of it.
The pressure of the air column after y cm of Mercury has flown out of the column P 2 = 76 - (76 - y) cm of Mercury = y cm of mercury
Final volume of air column V 2 = (24 + y)x cm 3
Since the temperature of the air column does not change
Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm
Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.
Length of the air column = y + 24 = 47.8 cm.
Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it, and the length of the air column becomes 47.8 cm
Answer:
As per Graham's Law of diffusion, if two gases of Molar Mass M 1 and M 2 diffuse with rates R 1 and R 2 respectively, their diffusion rates are related by the following equation
In the given question
R 1 = 28.7 cm 3 s -1
R 2 = 7.2 cm 3 s -1
M 1 = 2 g
The above Molar Mass is close to 32, therefore, the gas is Oxygen.
Q 3) A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
Answer:
Let the suspended particles be spherical and have a radius r
The gravitational force acting on the suspended particles would be
The buoyant force acting on them would be
The net force acting on the particles become
Replacing mg in equation (i) with the above equation, we get
The above is the equation to be derived
Substance
|
Atomic Mass (u)
|
Density (10 3 Kg m-3 )
|
Carbon (diamond)
|
12.01
|
2.22
|
Gold
|
197
|
19.32
|
Nitrogen (liquid)
|
14.01
|
1
|
Lithium
|
6.94
|
0.53
|
Fluorine
|
19
|
1.14
|
Answer:
Let one mole of a substance of atomic radius r and density
Let us assume the atoms to be spherical
Avogadro's number is
For Carbon
For gold
For Nitrogen
For Lithium
For Fluorine
The HOTS questions of Class 11 Physics Chapter 12 - Kinetic Theory aim at engaging students in the real life application of the concept and solve the problems that challenge them beyond the core of the topic. These high order questions develop thinking, analytical and good conceptual clarity which are the requirements of board examination as well as competitive examinations, such as JEE and NEET.
Q1: Two moles of an ideal gas with
Answer:
For ideal gas:-
For the first case:-
For the second case:-
Q2: If temperature of the atmosphere varies with height as
Answer:
Q3: In two jars A and B, the pressure, volume and temperature in jar A are respectively P, V, and T, and those of B are 2P, V/4 and 2T. Then the ratio of the number of molecules in jars A and B will be,
Answer:
Q4: The temperature of an open room of volume 30 m3 increases from 17oC to 27oC due to the sunshine. The atmospheric pressure in the room remains 1×105 Pa. If
Answer:
PV = nRT
Q5: A gas mixture contains 3 moles of oxygen and x moles of monoatomic gas at temperature T. Considering only translational and rotational but not vibrational modes, the total energy of the system is 15 RT then the value of x is.
Answer:
According to the question,
The Class 11 Physics Chapter 12 - Kinetic Theory examines how the behaviour of gases can be considered at the molecular level, in which the behaviour at the microscopic level is related to the macroscopic behaviour in the form of pressure, volume and temperature. The chapter addresses all the basic assumptions of kinetic theory, derivations and applications that describe real-life phenomena like diffusion and pressure of gases. Knowledge of these topics does not only enhance clarity in concepts but also equips students to board examinations and other competitive exams such as JEE and NEET.
12.1 Introduction
12.2 Molecular Nature Of Matter
12.3 Behaviour Of Gases
12.4 Kinetic Theory Of An Ideal Gas
12.4.1 Pressure Of An Ideal Gas
12.4.2 Kinetic Interpretation Of Temperature
12.5 Law Of Equipartition Of Energy
12.6 Specific Heat Capacity
12.6.1 Monatomic Gases
12.6.2 Diatomic Gases
12.6.3 Polyatomic Gases
12.6.4 Specific Heat Capacity Of Solids
12.7 Mean Free Path
Kinetic Theory NCERT Solutions- Important Formulae contain important equations that describe the behavior of the molecular motion in relation to the macroscopic properties of gas such as pressure, temperature and volume. With the help of these formulas, students can solve numerical problems and see a correlation between the behaviour of microscopic particles and the observed physical phenomena.
Where:
Where
Where
Where:
Where:
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-
-
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The Kinetic Theory of Gases has several assumptions, equations and derivations that must be understood clearly to solve some of its questions. Students are expected to learn formulas not by memorization, but to work out the logical application of formulas and interpret physical meaning underlying a formula. Organized planning assists in dealing with numerical and conceptual issues with ease.
Get a clear understanding of the Assumptions: Start with the revision of the fundamental postulates of the kinetic theory, including the following: gases consist of small particles in a state of perpetual random motion, the volume is insignificant, and collisions are perfectly elastic. These assumptions assist you in relating concepts to formulas.
Determine the Type of Question: Determine whether the problem is based on the derivation of pressure, mean free path, root mean square (rms) velocity or laws, such as Boyle and Charles. Saving time is achieved through categorizing and guides to the correct formula.
Link Equations with Physical Meaning: Don’t just apply formulas like $P=\frac{1}{3} \rho c^2$ or $v_{r m s}=\sqrt{\frac{3 k T}{m}} ;$ consider what they are, relation of pressure with molecular movement or temperature with speed of particles.
Divide Numerical Problems into Steps: Begin with familiar values and change all the units into SI and then carefully replace them in the equation. Always verify dimensional consistency- it means that it is accurate.
Apply Graphical Interpretation to Theory based questions: Reasoning is involved in many questions such as pressure-temperature or volume-temperature graphs. Explanations are better understood and remembered through graphs.
Practice Derivations Comprehensively: Derivations such as the pressure exerted by the gas molecules, mean free path and degrees of freedom are common questions. The conceptual power is developed through writing them step by step.
NCERT textbooks are almost a must-have in studying and preparing competitive exams such as JEE and NEET, but are definitely lacking in preparing the numericals, tricky applications and deeper conceptual advancement. This is a list of what NCERT does not cover and what additional students are expected to learn in the chapter Kinetic Theory of Gases in preparation for JEE/NEET:
Chapter-wise links of NCERT Solutions Class 11 provides an easy way to access all chapters at one stop. These solutions are aimed at allowing students to easily get explanations, solved exercises, vital formulas, and additional questions to each chapter and therefore making exam preparation and concept reviewing easier and organized.
Frequently Asked Questions (FAQs)
Yes, NCERT chapter-wise solutions can be found in PDF format and can be easily studied offline anytime, anywhere.
They contain all chapters of the Mechanics to the Kinetic Theory with solved exercises, HOTS questions and additional questions and formulas.
These are step-by-step solutions accompanying every exercise, HOTS, and other questions in the NCERT textbooks that allow students to comprehend new information in a proper way.
They mainly consist of step-wise descriptions, crucial formulas and illustrated solutions that make studying easier, which makes it easy to revise and obtain a good score both in the board as
They do reinforce the essential ideas in Physics, upon which the questions are set in JEE, NEET and other such competitive tests.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
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