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*NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics*

**NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics****:**

Section Name | Topic Name |

12 | Thermodynamics |

12.1 | Introduction |

12.2 | Thermal equilibrium |

12.3 | Zeroth law of thermodynamics |

12.4 | Heat, internal energy and work |

12.5 | First law of thermodynamics |

12.6 | Specific heat capacity |

12.7 | Thermodynamic state variables and equation of state |

12.8 | Thermodynamic processes |

12.9 | Heat engines |

12.10 | Refrigerators and heat pumps |

12.11 | Second law of thermodynamics |

12.12 | Reversible and irreversible processes |

12.13 | Carnot engine |

**Introduction**

- Thermodynamics is that branch of physics which deals with concepts of heat and temperature and their relation to energy and work.
- We can also consider it as a macroscopic science which deals with bulk systems and tells us about system as a whole.
- In this chapter we will learn about the laws of thermodynamics which describes the system in terms of macroscopic variables, reversible and irreversible processes. Finally we will also learn on what principle heat engines, refrigerators and Carnot engine work.

Examples: – Refrigerator, steam engine

**Thermodynamics vs. Mechanics**

- In Thermodynamics we consider only the state of the object which means we will only consider macroscopic variables like pressure, volume and temperature.
- In mechanics we consider the motion, velocity and acceleration of the object.

For eg: –

- In mechanics if a bullet is fired from a gun we will consider the motion of bullet and its velocity, acceleration etc.

**Thermal Equilibrium**

- Two systems are said to be in thermal equilibrium if the temperatures of the two systems are equal.
- In mechanics if the net force on a system is zero, the system is in equilibrium.
- In Thermodynamics equilibrium means all the macroscopic variables (pressure, temperature and volume) don’t change with time. They are constant throughout.

For Example: –

- Consider two bodies at different temperatures one is at 30
^{0}C and another at 60^{0}C then the heat will flow from body at higher temperature to the body at lower temperature. - Heat will flow till both bodies acquire same temperature.
- This state when there is no heat flow between two bodies when they acquire the same temperature is known as thermal equilibrium.

**Types of Equilibrium**

** Thermal Equilibrium**: – Two systems are said to be in thermal equilibrium with each other if the temperatures of both the systems do not change with time.

** Chemical Equilibrium: – **Two systems are said to be in chemical equilibrium with each other if the composition of the system does not change over time.

** Mechanical Equilibrium: –** Two systems are said to be in mechanical equilibrium with each other if the pressure of the system doesn’t change with time.

A system is said to be in **Thermodynamic equilibrium** when all of its macroscopic variables are constant.

**System and Surroundings**

** System: – **System is defined as any part of universe enclosed by some boundary through which exchange of heat or energy takes place.

** Surroundings**: – Any part of the universe which is not a system.

- For example: – If we consider a hot coffee in a kettle then the kettle is the system and everything else is the surroundings.
- System and surroundings constitute Universe.

**Types of Systems**

In this system there is exchange of energy and matter. For Example: – Water boils in a pan without lid, a cup of coffee.__Open System: –__

**Types of Walls**

: – It is an insulating wall which doesn’t allow heat to flow from one system to another. This means temperature of both the systems won’t change with time.__Adiabatic wall__

- Consider 2 systems A and B as shown in the figure, which are separated by adiabatic wall. Let pressure and volume of A be(P
_{1},V_{1}) and B be(P_{2},V_{2})- Both are at the same temperature as there is no change in temperature with time.
- There is no heat flow between A and B as they are separated by an adiabatic wall.
- Both these systems are also separated from the surroundings by adiabatic wall which means there is no flow of heat between A and surroundings and also B and surroundings.

: – It is a conducting wall which allows the flow of heat between any 2 systems.__Diathermic Wall__

- Consider two systems A and B which are separated by a conducting wall. System A is at higher temperature T
_{1}, pressure P_{1}and volume V_{1}and System B is at lower temperature T_{2}, pressure P_{2}and volume V_{2}. - There is flow of heat from a system at a higher temperature to the system at a lower temperature till the systems reach thermal equilibrium.
- For Example: – A vessel made up of metals like copper or aluminium has diathermic walls.

**Zeroth Law of Thermodynamics**

- Zeroth law of thermodynamics states that when two systems are in thermal equilibrium through a third system separately then they are in thermal equilibrium with each other also.
- For eg: – Consider two systems A and B which are separated by an adiabatic wall. Heat flow happens between systems A and C, and between B and C, due to which all 3 systems attain thermal equilibrium.

**Thermodynamic state variables**

Thermodynamic state variables are the macroscopic quantities which determine the thermodynamic equilibrium state of a system.

- These macroscopic quantities are known as thermodynamics state variables.
- Since these macroscopic quantities describe the behaviour of thermodynamic system it is known as thermodynamic.
- As they determine the state of the system that is pressure, volume and temperature, at one particular time they are known as thermodynamic state variables.
- Pressures (P), Volume (V), Temperature (T), mass (m), Internal energy (U) are the thermodynamic state variables.
- These variables can tell us the position or the condition of any gas at that particular time.
- A system not in equilibrium cannot be described by state variables. It means the macroscopic variables are changing with time and they are not constant.

__Types of thermodynamic state variables:-__

: – They indicate the size of the system, which means extensive variables are those that depend on the mass of the system or the number of particles in the system. Example: volume, mass, internal energy. If we consider a system whose mass is greater than the size of that system is greater. All these depend on the size of the system.__Extensive variables__: –__Intensive variables__

- They don’t depend on size of the system.
- Pressure and Temperature are known as intensive variables.

**Equation of State**

- Equation of State depicts the relationship between the state variables (pressure, mass, volume, density).
- It describes about the state of the matter under given set of physical conditions.
- Consider an ideal gas the equation of state is

PV= μRT where

- P, V and T are state variables and μ = number of moles
- For a fixed amount of gas we can have two independent variables either volume and temperature or volume and pressure.
- All the variables cannot be independent at the same time.
- This is known
**equation of state for ideal gas**.

**Internal Energy**

- It is defined as the sum of kinetic energies and potential energies of the molecules constituting the system as a whole and not of individual molecule.
- It is macroscopic variable of the system.
- It is denoted by U.
- It is thermodynamic state variable.
- It is an extensive variable as it depends on the size of the system.
- It can be specified by values of pressure, volume and temperature at that particular time.
- It only depends on the state of the system at that particular time.

** Work**: – We can do some work on the system (in this case system is bottle with a balloon tied) as a result the internal energy of the system changes.

Internal energy increases when the system absorbs heat and some work is done on the system similarly internal energy decreases if we change the conditions.

__How Internal energy is different from Work and Heat__

- Heat and work are not state variables unlike internal energy.
- They are modes of energy transfer to system resulting in change in internal energy.

**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.

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 (P_{1}, V_{1}) and final state (P_{2}, V_{2})

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 (P
_{1},V_{1}) and Final state is defined as (P_{2},V_{2}). - 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)

**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^{–1}e.

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.

** 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, T_{1} = 27°C

Final temperature, T_{2} = 77°C

∴Rise in temperature, ΔT = T_{2}-T_{1}

= 77 – 27= 50°C

Heat of combustion = 4 × 10^{4}J/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 × 10^{5} J/min

∴Rate of consumption = (6.3×10^{5})/(4×10^{4})= 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 N_{2} = 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 N_{2}, M = 28

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

Number of moles, n =m/M

=2.0×10^{-2}x10^{3}/28 = 0.714

Molar specific heat at constant pressure for nitrogen = C_{p} =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 = nC_{P} 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 ****(C _{v})**:-

- 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 (C
_{v}).

** To Prove**: – C

_{p}-C

_{v }= 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

C_{v} = (ΔQ/ ΔT)_{ v} = (ΔU/ ΔT)_{ v} = ΔU/ ΔT

- At constant pressure:- C
_{p}=( ΔQ/ ΔT)_{p}= (ΔU + PΔV)/ ΔT

By solving and doing all calculations:

C_{p }– C_{v} = R

Hence proved.

** Specific Heat Ratio**: –

- It is denoted by γ.
- γ = C
_{p}/C_{v} - For Mono atomic gas:-
- C
_{v }=3/2R - C
_{p}=5/2R - γ = 1.67
- For Diatomic gas: –
- C
_{v }=5/2 R - C
_{p}=7/2R - γ = 1.4

**Thermodynamic processes – Quasi Static Process**

- Quasi static term means semi static .It is not purely moving.
- It is a hypothetical construct which means it is not in real.
- It is an infinitely slow process which means change from its original position is not at all significant.
- System changes its variables (P, T, and V) so slowly that it remains in equilibrium with its surroundings throughout.

- Consider a gas initially at Pressure (P) and Temperature (T) changes it to a new state whose Pressure is (P’) and Temperature (T’).
- If we change the surrounding pressure to P by very small amount then allow the system to reach that system.
- The characteristics for a system to be Quasi-static process
- Extremely very slow process.
- There should not be any accelerated motion. Not large temperature gradient. Temperature gradient means the

In a quasi-static process, the temperature of the surrounding reservoir and the external pressure differ only infinitesimally from the temperature and pressure of the system.

**Isothermal Processes**

- Isothermal: – Iso means same and thermal related to temperature. In Isothermal process the temperature remains constant throughout while all other variables change.
- Temperature is constant throughout.
- For an ideal gas
- PV = nRT where
- n=no. of moles (constant), R = universal gas constant, T =constant for isothermal process.

- This implies PV=constant
- Pressure and volume are inversely proportional to each other.
- Graphically if we plot pressure and volume

**Adiabatic Processes**

- Adiabatic is a process in which there is no heat flow takes place between the system and the surroundings.
- These processes are sudden.
- The walls of the container should be adiabatic
- For an adiabatic process of an ideal gas
- From Boyle’s law
- PV
^{γ}= constant

Where γ = C_{p}/C_{v} Specific heat ratio

Example: – Hot tea in Thermos flask. It will remain hot as there is no exchange of heat takes place because the walls of thermos is insulating.

**Problem**: – A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

**Answer**: The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.

Initial pressure inside the cylinder = P_{1}

Final pressure inside the cylinder = P_{2}

Initial volume inside the cylinder = V_{1}

Final volume inside the cylinder = V_{2}

Ratio of specific heats, γ = 1.4

For an adiabatic process, we have: P_{1}V_{1} ^{γ} = P_{2}V_{2} ^{γ}

The final volume is compressed to half of its initial volume.

V_{2}= V_{1}/2

P_{1}V_{1} ^{γ} = P_{2} (V_{1}/2)^{ γ}

=P_{2}/P_{1 } = V_{1} ^{γ}/ ( V_{1}/2)^{ γ} = (2) ^{γ }= (2)^{1.4} =2.639

Hence, the pressure increases by a factor of 2.639.

Graphically if we plot pressure and volume.

- If an ideal gas undergoes a change in its state adiabatically from (P
_{1}, V_{1}) to (P_{2}, V_{2}) - P
_{1}V_{1}^{ γ}= P_{2}V_{2}^{ γ} - The work done in an adiabatic change of an ideal gas from the

state (P_{1}, V_{1}, T_{1}) to the state (P_{2}, V_{2}, T_{2}).

W =∫ P V dV = P ∫V dV (Integrating between the limits V_{2} and V_{1})

For Adiabatic Process

- P V
^{γ}= constant This implies P= constant / V^{γ} - W = constant ∫dV/ V
^{γ} - constant [V
^{ γ-1}/- γ+1] - constant/1- γ [V
_{2}^{1- γ}– V_{1}^{1-γ}] - = constant/1- γ[1/ V
_{2}^{1- γ}– 1/ V_{1}^{1-γ}] - By solving Work done
**W= R/****(****γ-1)(T**where_{2}-T_{1}), - T
_{2}= final Temperature - T
_{1}=initial temperature - R=Universal gas constant
- γ = Specific heat ratio
- This is the work done during adiabatic change.
- Consider W= R/ (γ-1)(T
_{2}-T_{1})

Case 1: W>0 (when T1>T2)

Temperature of the gas decreases.

Case 2:- W< 0 (T1<T2)

Temperature of the gas increases.

**Isochoric Processes**

- Isochoric process means volume is constant while all other variables change.
- As volume is kept constant therefore no work is done on or by the gas.
- Heat absorbed by the gas is completely used to change its internal energy and its temperature.
- From First law of Thermodynamics
- ΔQ= Δ U + ΔW (ΔW =0)
- ΔQ= Δ U
- This means whatever heat is supplied to the system that is used up completely to change the internal energy and temperature of the system.

- C
_{v }= Δ U/ Δ T

Example: – If we heat a gas filled in a closed cylinder fitted with a piston. When we supply heat to this cylinder the piston won’t move as there is as there is no volume change. There will be no work is also being done. Whatever the heat is added it will be used to increase the internal energy of the system.

**Isobaric Processes**

- Iso means same and baric is related to pressure.
- In this process pressure is constant while all other variables change.
- Process in which pressure is constant.
- Work done is given as :
- W=P (V
_{2}-V_{1}) - =μ R (T
_{2}– T_{1})

- W=P (V
- From First law of thermodynamics
- ΔQ= Δ U + ΔW
- ΔQ= Δ U + μ R (T
_{2}– T_{1}) - We can see from the above equation that the heat absorbed goes partly to increase internal energy and partly to do work.

Example: –

- Consider a cylinder filled with gas fitted with piston. When we heat the cylinder the gas expands but the pressure remains the same because of piston.
- Boiling water in an open pot which is at atmospheric pressure. When the water boils it changes to steam and this steam expands and since it is not contained, it stays at atmospheric pressure. So the pressure remains constant but energy changes.

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F?

**Answer**:

Total work done by the gas from D to E to F = Area of ΔDEF

Area of ΔDEF = ½ x DE x EF

Where,

DF = Change in pressure = 600 N/m^{2}

– 300 N/m^{2}

= 300 N/m^{2}

FE = Change in volume

= 5.0 m^{3}

– 2.0 m^{3}

= 3.0 m^{3}

Area of ΔDEF =1/2x300x3 = 450 J

Therefore, the total work done by the gas from D to E to F is 450 J.

Consider a hot reservoir having a temperature T_{1} and a cold reservoir having a temperature T_{2}.

** Efficiency of a heat engine**:

- Efficiency indicates how much useful work we get as an output by the engine by using the amount of heat energy as input.
- It is denoted by η.
- Mathematically :
- η = W/Q
_{1}- where W= output and Q
_{1}= input - By calculating

- where W= output and Q
**η = 1- Q**_{2}/Q_{1}

Where

- Q1=heat input in 1 cycle
- Q2=work done in 1 cycle.
- η = 1 – Q
_{2}/Q_{1} - For 100% efficient η = 1 ( which means Q
_{2}/Q_{1}=0)- There is always some heat is lost to the surroundings

- Heat lost (Q
_{2}) = 0

There is no heat engine whose efficiency is 100%.There is always some of the losses associated with the heat engines.

**Problem**: A steam engine delivers 5.4×10^{8} J of work per minute and services 3.6 × 10^{9} J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

**Answer**:

Work done by the steam engine per minute, W = 5.4 × 10^{8} J

Heat supplied from the boiler, H = 3.6 × 10^{9} J

Efficiency of the engine =Output energy/Input energy

η = W/H = 5.4 x 10^{8}/ 3.6 × 10^{9} = 0.15

Hence, the percentage efficiency of the engine is 15 %.

Amount of heat wasted = 3.6 × 10^{9} – 5.4 × 10^{8}

= 30.6 × 10^{8} = 3.06 × 10^{9} J

Therefore, the amount of heat wasted per minute is 3.06 × 10^{9} J.

__Coefficient of Performance (COP) of the refrigerator__

- It is denoted by α.
- Mathematically given as :
**α = Q**_{2}/W- where Q
_{2}is the heat extracted from the cold reservoir - W is the work done on the system–the refrigerant

- α can be greater than 1
- Using Law of energy conservation :- Q
_{2}+W=Q_{1} - This implies W=Q2-Q1
- Therefore:-
- α =Q2/Q1-Q2
- W ≠ 0 then Q1-Q2 ≠0
- This implies α ≠infinite

**Second law of Thermodynamics**

There are 2 statements of second law of thermodynamics given by two scientists:

No process is possible whose result is the absorption of heat from a reservoir and the complete conversion of the heat into work.__Kelvin-Planck ∫Statement: –__: – No process is possible whose result is the transfer of heat from a colder object to a hotter object.__Clausius statement__

Explanation of Kelvin-Planck Statement: It is always impossible that the total amount of heat which is supplied to system will get converted to work, and there will always be loss of heat. Complete conversion of heat into work is not possible.

Explanation of Clausius statement: – Transfer of heat from colder body to hotter body won’t take place until some external work is done on the system.

**Reversible and Irreversible processes**

__Reversible Process__

- A thermodynamic process is reversible if the process can be turned back such that the system and surroundings return to their original states, with no other change anywhere else in the universe.

- This means in the Reversible processes if a process starts from initial state then it goes to final state and then it can reversed back from final state to initial state.
- Examples:- Isothermal expansion and compression, Electrolysis
- A process is reversible if :-

- It is quasi-static
- No dissipative forces (that is no loss of heat by friction etc.).
- Both initial and final states of the system are in thermodynamic equilibrium with each other.

__Irreversible Process__

- Irreversible processes are those that cannot be reversed.
- Two causes which give rise to irreversible processes
- Irreversible processes takes place at a very fast rate.
- Dissipative Effects.

- Examples:-Plastic deformation, Combustion, Diffusion, Falling of water from hill.

**Carnot engine**

- A Carnot engine is named after Carnot scientist.
- It is a reversible heat engine operating between two temperatures.
- It has a maximum efficiency which no other engine can have.

__Cycle of processes in a Carnot engine__

Basic Function of any heat engine is it will take heat Q_{1} from a hot reservoir at temperature T_{1} and give heat Q_{2 } to a cold reservoir at temperature T_{2}.

- As system is absorbing heat so it is isothermal expansion. Engine absorbs heat Q
_{1}at temperature T_{1}. - An adiabatic process takes place inside the engine because of which there is increase in the temperature of the engine from T
_{1}to T_{2}but no flow of heat. - As system is releasing heat so it is isothermal contraction. Engine releases heat Q
_{2}at temperature T_{2}. - An adiabatic process takes place again which changes the temperature of the system from T
_{2}to T_{1}. - One cycle of Carnot engine will have Isothermal expansion then adiabatic process, and then isothermal contraction followed by adiabatic process.
- This will keep on repeating.

The efficiency of Carnot engine is given by:-

η = 1 – T_{2}/T_{1}

The graph below depicts the Carnot cycle for a heat engine with an ideal gas as the working substance.

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