NCERT Solutions for class 11 Physics Chapter 13: Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 13: Kinetic Theory deals with the behaviour of gas molecules and ideal gas. An ideal gas is a gas that follows Boyle's law, Charles' law, Gay Lussac's law, and Avogadro’s law.

Class 11 Physics Chapter 13 Kinetic Theory belongs to Unit 9 Behaviour of Perfect Gases and Kinetic Theory of Gases. Along with Unit 7 and Unit 8, Unit 9 has a weightage of 20 marks. The NCERT Solutions for Chapter 13 deals with the molecular nature of matter and specific heat capacity.

Download PDF: NCERT Solutions for Class 11 Physics Chapter 13

NCERT Solutions for Class 11 Physics Chapter 13

Class 11 Physics Chapter 13 – Concepts Covered

Boyle’s Law: According to this law, the volume (V) of a fixed mass of a gas is inversely proportional to the pressure (P) of that gas, given that the temperature of the gas is kept constant.

V ∝ 1/P or PV = constant

Charle’s Law: The volume (V) of a given mass of a gas is directly proportional to the temperature of the gas, given that the pressure of the gas remains constant.

V ∝ T or VT = constant
V₁T₂= V₂T₂

Gay Lussac’s Law: The pressure P of a given mass of gas is directly proportional to its absolute temperature T, given that the volume V of the gas remains constant.

P ∝ T or PT = constant
P₁T₁= P₂T₂

Equation of State of An Ideal Gas: The relation between pressure, volume, and absolute temperature of a gas is known as its equation of state.

PV = nRT

n is the number of moles of the gas and R is the molar gas constant which is equal to 8.315 JK^-1mol^-1

Dalton’s Law of Partial Pressures states that the net pressure applied by a mixture of non-interacting gases is equivalent to the sum of their pressures.

P = P₁ + P₂ + ------- + P_n

Class 11 Physics Chapter 13 Related Guides –

Pressure of an Ideal Gas	Law of equipartition of energy	Monatomic Gases
Value of Boltzmann Constant	Properties of gases	Derivation of Van Der Waals Equation
Stefan Boltzmann Constant	Mean Free Path	Inelastic collision

Class 11 Physics Study Guides:

Difference between in Physics	SI Units in Physics	Relation between articles in Physics
NCERT Solutions for Class 11 Physics	Class 11 Physics Notes	Physics Constants
Derivations in Physics	Physics Formula	Physics Study Notes

CBSE CLASS XII Related Questions

1.
A 500 nm photon is incident normally on a perfectly reflecting surface and is reflected. The value of momentum transferred to the surface is:
- \( 3.87 \times 10^{-43} \, \text{kg} \, \text{ms}^{-1} \)
- \( 2.5 \times 10^{-30} \, \text{kg} \, \text{ms}^{-1} \)
- \( 2.65 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1} \)
- \( 1.33 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1} \)
View Solution
2.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.
View Solution
3.
The magnetic field in a plane electromagnetic wave travelling in glass (\( n = 1.5 \)) is given by \[ B_y = (2 \times 10^{-7} \text{ T}) \sin(\alpha x + 1.5 \times 10^{11} t) \] where \( x \) is in metres and \( t \) is in seconds. The value of \( \alpha \) is:
- \( 0.5 \times 10^3 \, \text{m}^{-1} \)
- \( 6.0 \times 10^2 \, \text{m}^{-1} \)
- \( 7.5 \times 10^2 \, \text{m}^{-1} \)
- \( 1.5 \times 10^3 \, \text{m}^{-1} \)
View Solution
4.
A circular coil of 100 turns and radius \( \left(\frac{10}{\sqrt{\pi}}\right) \, \text{cm}\) carrying current of \( 5.0 \, \text{A} \) is suspended vertically in a uniform horizontal magnetic field of \( 2.0 \, \text{T} \). The field makes an angle \( 30^\circ \) with the normal to the coil. Calculate:
the magnetic dipole moment of the coil, and
the magnitude of the counter torque that must be applied to prevent the coil from turning.
View Solution
5.
Write any two features of nuclear forces.
View Solution
6.
Determine the current in the \( 3 \, \Omega \) branch of a Wheatstone Bridge in the circuit shown in the figure.
View Solution