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NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory are essential resources for achieving high marks in school exams and competitive exams like JEE and NEET. On this Careers360 page, you'll find the comprehensive kinetic theory of gases class 11 solutions covering questions from 13.1 to 13.10 (exercise questions) and 13.11 to 13.14 (additional exercise questions).
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The solution of chapter 13 Physics Class 11 is prepared by our highly experienced Subject Matter which gives all the answers to NCERT Solution. By using of NCERT solutions for Class 11, specifically, the kinetic theory of gases class 11 NCERT solutions, can aid students in practising questions and grasping important syllabus concepts before their exams. These solutions are available for free and can be accessed below.
Maxwell, Boltzmann, and others developed the kinetic theory of gas in the 19th century. Before going to solutions of NCERT for Class 11 Physics Chapter 13 Kinetic Theory one must be familiar with ideal gas equations and gas laws. The Kinetic Theory interprets the pressure and temperature at the molecular level. Kinetic Theory is consistent with gas laws and it correctly explains specific heat and relates measurable properties of gases such as viscosity, diffusion, etc. The CBSE NCERT solutions for Class 11 Physics chapter 13 KTG gives an idea about how to apply the formulas studied in the chapter in numerical problems. class 11 physics ch 13 ncert solutions explain all the exercises and additional exercise questions.
** This chapter has been renumbered as Chapter 12 in accordance with the CBSE Syllabus 2023–24.
Free download kinetic theory of gases class 11 solutions PDF for CBSE exam.
NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory - Exercise Questions
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
Q13.3 Figure 13.8 shows plot of versus P for kg of oxygen gas at two different temperatures.
(a) What does the dotted plot signify?
(b) Which is true: ?
(c) What is the value of where the curves meet on the y-axis?
(d) If we obtained similar plots for kg of hydrogen, would we get the same
value of at the point where the curves meet on the y-axis? If not, what mass
of hydrogen yields the same value of (for low pressure high temperature
region of the plot) ? (Molecular mass of , 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 kg of hydrogen we would not get the same
value of at the point where the curves meet on the y-axis as 1 g of Hydrogen would contain more moles than 1 g of Oxygen because of having smaller molar mass.
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
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,
Number of molecules inside the room is N
Answer:
The average energy of a Helium atom is given as since it is monoatomic
(i)
(ii)
(iii)
Answer:
As per Avogadro's Hypothesis under similar conditions of temperature and pressure equal volumes of gases contain equal number of molecules. Since the volume of the vessels are the same and all vessels are kept at the same conditions of pressure and temperature they would contain equal number of molecules.
Root mean square velocity is given as
As we can see v _{ rms } is inversely proportional to the square root of the molar mass the root mean square velocity will be maximum in 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 ,
The molecular mass of N _{ 2 } = 28 u
The molar mass of N _{ 2 } = 28 g
From 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 is given as
The root mean square velocity v _{ rms } is given as
The time between collisions T is given as
Collision time T' is equal 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 time between collisions is much larger than the collision time.
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 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.
Q13.13 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
where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
where is the density of the suspended particle, and , that of the surrounding medium. [ s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]
Answer:
Let the suspended particles be spherical and have 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 have molar mass M
Let us assume the atoms to be spherical
Avogadro's number is
For Carbon
For gold
For Nitrogen
For Lithium
For Fluorine
Class 11 physics ch 13 ncert solutions are essential for comprehending how tiny particles in matter behave. This chapter helps students understand how gases work and their thermal properties. It also connects to other chapters in thermodynamics and has practical applications. By studying this chapter, students can do well in their exams and gain a solid understanding of how matter works on a fundamental level.
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | Kinetic Theory |
Chapter 14 | |
Chapter 15 |
where is the number of moles
N is the number of molecules
R and k _{ B } are universal constants.
where n is the number density, m is the mass of the molecule and is the mean of the squared speed.
Where T is the temperature.
where n is the number density and d is the diameter of the molecule
Comprehensive Coverage: These kinetic theory of gases exercise solutions cover all the important topics and concepts presented in the chapter, ensuring a thorough understanding of kinetic theory.
Exercise and Additional Exercise Solutions: Detailed kinetic theory of gases class 11 physics ncert solutions for exercise questions (13.1 to 13.10) and additional exercise questions (13.11 to 13.14) are provided, facilitating practice and self-assessment.
Clarity and Simplicity: The kinetic theory of gases class 11 questions and answers are explained in clear and simple language, making complex concepts more accessible to students.
Problem-Solving Skills: By practising with these kinetic theory class 11 solutions, students develop strong problem-solving skills, which are essential in physics and other subjects.
Exam Preparation: These kinetic theory of gases exercise solutions are designed to help students prepare effectively for their exams, including board exams and competitive exams like JEE and NEET.
Dropped Topics –
13.6.5 - Specific Heat Capacity of Water
Exercises - (13.11–13.14)
We trust that the kinetic theory of gases class 11 questions and answers prove to be beneficial for your studies. If you have any questions or require further clarification regarding these solutions, please feel free to leave a comment below, and we will promptly address your inquiries.
For both JEE Main and NEET one question can be expected from the chapter Kinetic Theory. To practice more problems on Kinetic Theory refer to NCERT book questions, NCERT exemplar questions and previous year papers of NEET and JEE Main.
The following are the topics covered in Class 11 chapter Kinetic Theory
The gas consists of rapidly moving atoms or molecules. Kinetic theory explains the behaviour of gas based on these rapidly moving molecules.
Pressure exists everywhere in a fluid, not only the walls of the container. Any layer of gas inside the volume of a container is in equilibrium because the pressure is the same on both sides of the layer.
To download chapter-wise solutions for class 11th Physics click on the given link.
https://school.careers360.com/ncert/ncert-solutions-class-11-physics
The behaviour of gases in Class 11 Physics involves studying how gases respond to changes in factors like pressure, temperature, and volume, as described by gas laws such as Boyle's, Charles's, Gay-Lussac's, and Avogadro's laws. This understanding helps predict and explain the behaviour of gases in various situations.
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