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

First Law of Thermodynamics

  • The First law of thermodynamics is same as law of conservation of energy.
  • According to law of energy conservation: – Energy can neither be created nor be destroyed, only transformed to other forms.
  • According to first law of thermodynamics:- The change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work done by the system on its surroundings.
  • Examples:- Consider a ball falling from the roof of the building when at top of the building the ball has only potential energy and when it starts falling potential energy decreases and kinetic energy starts increasing. At the ground it has only kinetic energy.
Class 11 Physics Chapter 12 Thermodynamics Notes and NCERT Solution

Where:

  • ΔQ is the heat supplied to the system by the surroundings
  • ΔW is the work done by the system by the surroundings
  • ΔU is the change in internal energy of the system
  • Some part of heat supplied gets lost and remaining part is the work done on the surroundings. This remaining part is used up to increase or change the internal energy of the system.

ΔQ = ΔU + ΔW

Consider a system whose initial state is (P1, V1) and final state (P2, V2)

ΔU is the change in the energy of the system to change from initial state to final state.

  • Internal energy is a state variable which means it is path independent. It does not depend how state changes from initial to final.
  • But the work done and heat is path dependent. It depends on how the path changes from initial to final.
  • Consider a system whose initial state is defined as (P1,V1) and Final state is defined as (P2,V2).
  • The internal energy doesn’t depend on how the system has changed from initial state to final state. It only depends on how it has reached from initial state to final state.

Therefore:-

ΔQ – ΔW = ΔU    where

  • (ΔQ and ΔW are path dependent quantities whereas ΔU is path independent quantity)
  • This concludes ΔQ – ΔW is path independent quantity.
  • Case 1:- System undergoes a process such that ΔU = 0 which means internal energy is constant. From first law of thermodynamics

                            ΔQ = ΔU + ΔW putting ΔU = 0

  • This implies ΔQ = ΔW this means heat supplied by the surroundings is equal to the work done by the system on the surroundings. 
  • Case 2:- System is a gas in a cylinder with movable piston, by moving the piston we can change the volume of the gas.
  • If we move the piston downwards some work is done and it can be given as:-
  • Work done = ΔW
  • = Force x displacement
  • = P x Area x displacement
  • ΔW = PΔV (ΔV= Area x displacement) (Equation 1)
  • Therefore by first law of thermodynamics
  • ΔQ = ΔU + PΔV where ΔV= change in volume

(From Equation 1)

Class 11 Physics Chapter 12 Thermodynamics Notes and NCERT Solution
Thermodynamics Notes

Problem: An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing?

Answer:

Heat is supplied to the system at a rate of 100 W.

Heat supplied, Q = 100 J/s

The system performs at a rate of 75 J/s.

Work done, W = 75 J/s

From the first law of thermodynamics, we have:

Q = U + W

Where, U = Internal energy

U = Q – W

= 100 – 75

= 25 J/s

= 25 W

Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

Problem:  A cylinder filled with gas and fitted with a movable piston. In changing the state of a gas adiabatically from equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

 Answer:

The work done (W) on the system while the gas changes from state A to state B is 22.3 J.

This is an adiabatic process. Hence, change in heat is zero.

ΔQ = 0

ΔW = –22.3 J (Since the work is done on the system)

From the first law of thermodynamics, we have:

ΔQ = ΔU + ΔW

Where,

ΔU = Change in the internal energy of the gas

ΔU = ΔQ – ΔW = – (– 22.3 J)

ΔU = + 22.3 J

When the gas goes from state A to state B via a process, the net heat absorbed by the system is:

ΔQ = 9.35 cal = 9.35 × 4.19 = 39.1765 J

 Heat absorbed, ΔQ = ΔU + ΔQ

ΔW = ΔQ – ΔU

= 39.1765 – 22.3

= 16.8765 J

Therefore, 16.88 J of work is done by the system.  

Problem: Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:

(a) What is the final pressure of the gas in A and B?

(b) What is the change in internal energy of the gas?

(c) What is the change in the temperature of the gas?

(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?

Answer: 

(a) When the stopcock is suddenly opened, the volume available to the gas at 1 atmospheric pressure will become two times. Therefore, pressure will decrease to one-half, i.e., 0.5 atmosphere.

(b) There will be no change in the internal energy of the gas as no work is done on/by the gas.

(c) Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.

(d) No, because the process called free expansion is rapid and cannot be controlled. the intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.

Specific heat capacity

  • Specific heat is defined as the amount of heat required to raise the temperature of a body per unit mass.
  • It depends on:-
  • Nature of substance
  • Temperature
  • Denoted by ‘s’

 Mathematically:-

                        s= (ΔQ /mΔT)

  • where m= mass of the body
  • ΔQ = amount of heat absorbed or rejected by the substance
  • ΔT= temperature change
  • Unit – J kg–1 K–1
  • If we are heating up oil in a pan, more heat is needed when heating up one cup of oil compared to just one tablespoon of oil. If the mass s is more the amount of heat required is more to increase the temperature by one degree.

Specific heat capacity of water

 Calorie: – One calorie is defined to be the amount of heat required to raise the

temperature of 1g of water from 14.5 °C to 15.5 °C.

  • In SI units, the specific heat capacity of water is 4186 J kg–1 K–1e.

   4.186 J g–1 K–1.

  • The specific heat capacity depends on the process or the conditions under which heat capacity transfer takes place.
Class 11 Physics Chapter 12 Thermodynamics Notes and NCERT Solution
Class 11 Physics Chapter 12 Thermodynamics Notes and NCERT Solution

Problem:  A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g?

Answer:

Water is flowing at a rate of 3.0 litre/min.

The geyser heats the water, raising the temperature from 27°C to 77°C.

Initial temperature, T1 = 27°C

Final temperature, T2 = 77°C

∴Rise in temperature, ΔT = T2-T1

               = 77 – 27= 50°C

Heat of combustion = 4 × 104J/g°C

Specific heat of water, c = 4.2 J-1 g-1°C-1

Mass of flowing water, m = 3.0 litre/min = 3000 g/min

Total heat used, ΔQ = mc ΔT

= 3000 × 4.2 × 50

= 6.3 × 105 J/min

∴Rate of consumption = (6.3×105)/(4×104)= 15.75g/min

Molar Specific Heat Capacity

  • Heat capacity per mole of the substance is the defined as the amount of heat (in moles) absorbed or rejected (instead of mass m in kg) by the substance to change its temperature by one unit.

C = S/ μ= ΔQ / μ ΔT  

Where

  • μ= amount of substance in moles
  • C = molar specific heat capacity of the substance.
  • ΔQ = amount of heat absorbed or rejected by a substance.
  • ΔT = temperature change
  • Depends on :
  • nature of substance
  • Temperature
  • Conditions under which heat is supplied
  • SI Unit: J/mol/K

Examples: – All the cooking vessels are made by the material which has less specific heat and their bottom is polished so that they can be heated quickly by applying small amount of heat. Vessels made of copper, aluminium.

Molar Specific heat capacity at constant pressure (Cp)

  • If the gas is held under constant pressure during the heat transfer, then the corresponding molar specific heat capacity is called molar specific heat capacity at constant pressure (Cp).

Problem: – What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure?

(Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.)

Answer:-

Mass of nitrogen, m = 2.0 × 10–2 kg = 20 g

Rise in temperature, ΔT = 45°C

Molecular mass of N2, M = 28

Universal gas constant, R = 8.3 J mol–1 K–1

Number of moles, n =m/M

=2.0×10-2x103/28 = 0.714

Molar specific heat at constant pressure for nitrogen = Cp =7/2R

=7/2×8.3 = 29.05 J mol–1 K–1

The total amount of heat to be supplied is given by the relation:

ΔQ = nCP ΔT

= 0.714 × 29.05 × 45

= 933.38 J

Therefore, the amount of heat to be supplied is 933.38 J.

Molar Specific heat capacity at constant volume (Cv):-

  • If the volume of the gas is maintained during the heat transfer, then the corresponding molar specific heat capacity is called molar specific heat capacity at constant volume (Cv).

 To Prove: – Cp-Cv   = R for an ideal gas

                From First Law: – ΔQ = ΔU + ΔW.

  • Consider the case a gas is enclosed in a cylinder fitted with piston. Then the work done changes to
  • ΔQ = ΔU + PΔV
  • At constant volume ΔQ = ΔU (where ΔV=0)

Therefore

               Cv = (ΔQ/ ΔT) v = (ΔU/ ΔT) v = ΔU/ ΔT

  • At constant pressure:- Cp=( ΔQ/ ΔT)p  =  (ΔU + PΔV)/ ΔT

By solving and doing all calculations:

               C– Cv = R

     Hence proved.

 Specific Heat Ratio: –

  • It is denoted by γ.
  • γ = Cp/Cv
  • For Mono atomic gas:-
  • C=3/2R
  • Cp=5/2R
  • γ = 1.67
  • For Diatomic gas: –
  • C=5/2 R
  • Cp=7/2R
  • γ = 1.4
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