Course Content
Class 11 Physics Chapter 1 Physical World
Section Name Topic Name 1 Physical World 1.1 What is physics? 1.2 Scope and excitement of physics 1.3 Physics, technology and society 1.4 Fundamental forces in nature 1.5 Nature of physical laws
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Class 11 Physics Chapter 4 Motion In A Plane
4 Motion in a plane 4.1 Introduction 4.2 Scalars and vectors 4.3 Multiplication of vectors by real numbers 4.4 Addition and subtraction of vectors – graphical method 4.5 Resolution of vectors 4.6 Vector addition – analytical method 4.7 Motion in a plane 4.8 Motion in a plane with constant acceleration 4.9 Relative velocity in two dimensions 4.10 Projectile motion 4.11 Uniform circular motion
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Class 11 Physics Chapter 5 Laws of motion
Section Name Topic Name 5 Laws of motion 5.1 Introduction 5.2 Aristotle’s fallacy 5.3 The law of inertia 5.4 Newton’s first law of motion 5.5 Newton’s second law of motion 5.6 Newton’s third law of motion 5.7 Conservation of momentum 5.8 Equilibrium of a particle 5.9 Common forces in mechanics 5.10 Circular motion 5.11 Solving problems in mechanics
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Class 11 Physics Chapter 6 Work Energy and Power
Section Name Topic Name 6 Work Energy and power 6.1 Introduction 6.2 Notions of work and kinetic energy : The work-energy theorem 6.3 Work 6.4 Kinetic energy 6.5 Work done by a variable force 6.6 The work-energy theorem for a variable force 6.7 The concept of potential energy 6.8 The conservation of mechanical energy 6.9 The potential energy of a spring 6.10 Various forms of energy : the law of conservation of energy 6.11 Power 6.12 Collisions
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Class 11 Physics Chapter 7 Rotation motion
Topics Introduction Centre of mass Motion of COM Linear Momentum of System of Particles Vector Product Angular velocity Torque & Angular Momentum Conservation of Angular Momentum Equilibrium of Rigid Body Centre of Gravity Moment of Inertia Theorem of perpendicular axis Theorem of parallel axis Moment of Inertia of Objects Kinematics of Rotational Motion about a Fixed Axis Dynamics of Rotational Motion about a Fixed Axis Angular Momentum In Case of Rotation about a Fixed Axis Rolling motion
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Class 11 Physics Chapter 9 mechanics properties of solid
Section Name Topic Name 9 Mechanical Properties Of Solids 9.1 Introduction 9.2 Elastic behaviour of solids 9.3 Stress and strain 9.4 Hooke’s law 9.5 Stress-strain curve 9.6 Elastic moduli 9.7 Applications of elastic behaviour of materials
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Class 11 Physics Chapter 11 Thermal Properties of matter
Section Name Topic Name 11 Thermal Properties of matter 11.1 Introduction 11.2 Temperature and heat 11.3 Measurement of temperature 11.4 Ideal-gas equation and absolute temperature 11.5 Thermal expansion 11.6 Specific heat capacity 11.7 Calorimetry 11.8 Change of state 11.9 Heat transfer 11.10 Newton’s law of cooling
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Class 11 Physics Chapter 14 Oscillations
Section Name Topic Name 14 Oscillations 14.1 Introduction 14.2 Periodic and oscilatory motions 14.3 Simple harmonic motion 14.4 Simple harmonic motion and uniform circular motion 14.5 Velocity and acceleration in simple harmonic motion 14.6 Force law for simple harmonic motion 14.7 Energy in simple harmonic motion 14.8 Some systems executing Simple Harmonic Motion 14.9 Damped simple harmonic motion 14.10 Forced oscillations and resonance
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Class 11th Physics Online Class For 100% Result
About Lesson

Law of Equipartition of energy

According to this law, in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to (1/2)kBT.

  1. Each translational degree of freedom contributes (1/2) kB
  2. Each rotational degree of freedom contributes (1/2) kB
  3. Each vibrational degree of freedom contributes 2x (1/2)kB

Specific Heat Capacity for monoatomic gases

  • Monoatomic gases will only have translational degree of freedom.
  • Maximum they can have is three translational degrees of freedom.
  • Each degree of freedom will contribute (1/2) kB
  • Therefore 3 degrees of freedom will contribute (3/2) kB
  • By using law of equipartition of energy, the total internal energy of 1 mole of gas U=(3/2) kBTxNA=(3/2) RT
  • Specific heat capacity at constant volume CV= dU/dT=(3/2) R(i)
  • For an ideal gas CP-CV=R, By using equation(i)CP=(5/2)R
  • Ratio of specific heatsγ=CP/CV=(5/3)

Specific Heat of Diatomic gases (rigid)

  • A rigid diatomic gas means they will have translational as well as rotational degree of freedom but not vibrational.
  • They are rigid oscillator.
  • A rigid diatomic molecule will have 3 translational degrees of freedom and 2 rotational degrees of freedom. Total 5 degrees of freedom.
  • By law of equipartition of energy, each degree of freedom will contribute (1/2) kB
  • Therefore 5 degree of freedom will contribute (5/2) kB
  • Therefore the total internal energy of 1 mole of gas, U=(5/2) kBTxNA=(5/2)RT
  • Specific heat capacity at constant volume(CV) =dU/dT=(5/2)R
  • Specific heat capacity at constant pressure of a rigid diatomic is given as CP=(7/2)R
  • Ratio of specific heats γ=CP/CV=(7/5)

Specific Heat of Diatomic gases (non-rigid)

  • A no-rigid diatomic gas has translational, rotational as well as vibrational degrees of freedom.
  • There will be 3 translational degrees of freedom and 2 rotational degrees of freedom and 1 vibrational degree of freedom.
  • Total contribution by translational= (1/2) kBT, rotational=2x (1/2)kBT and vibrational =kB
  • Total Internal energy for 1 mole =(5/2)kBT+kBT = (7/2)kBT= (7/2)RT.
  • CV=dU/dT = (7/2) R.
  • CP=CV+R= (9/2) R.
  • γ= CP/CV =(9/7)
Class 11 Physics Chapter 13 Kinetic Theory Notes & NCERT Solution

Specific Heat Capacity for polyatomic gases

  • Polyatomic gases will have 3 translational degree of freedom, 3 rotational degrees of freedom and ‘f’ number of vibrational modes.
  • Total internal energy of 1 mole of gas =(3x(1/2)kBT + 3x(1/2)kBT+ fkBT)x NA

((3/2) + (3/2) +f)RT = (3+f) RT.

  • CV=dU/dT = (3+f)R
  • CP=CV+R=(4+f)R
  • γ= CP/CV = (4+f)/(3+f)

Specific Heat Capacity for solids

  • Consider there are N atoms in a solid. Each atom can oscillate about its mean position.
  • Therefore vibrational degree of freedom = kBT
  • In one-dimensional average energy=kBT, in three-dimensional average energy =3KBT
  • Therefore total internal energy (U) of 1 mole of solid = 3KBTxNA= 3RT
  • At constant pressure, ΔQ = ΔU + PΔV change in volume is very less in solids .Therefore ΔV = 0.
  • =>ΔQ = ΔU
  • CV=(dU/dT)v
  • CP=(dQ/dT)Vas ΔQ = ΔU, Therefore CV=dU/dT=3R
  • Therefore CP=CV=3R

Specific Heat Capacity of water

  • Consider water as solid,so it will have ‘N’ number of atoms.
  • Therefore for each atom average energy =3kBT
  • No of molecules in H2O= 3 atoms.
  • Total internal energy U=3kBTx3xNA =9RT.
  • CV=CP=9R.

Conclusion on Specific heat

  • According to classical mechanics, the specific heat which is calculated based on degree of freedom should be independent of temperature.
  • However T->0,degree of freedom becomes inefficient.
  • This shows classical mechanics is not enough; as a result quantum mechanics came into play.
  • According to quantum mechanics minimum non-zero energy is required for degree of freedom to come into play.
  • Specific heats of all substances approach zero as T->0.

 

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