ThermodynamicsThermodynamics

Chapter 12Chapter 12

Thermodynamics VocabularyThermodynamics Vocabulary

bull Thermo (heat) dynamics (transfer)

bull Thermodynamic systems describe many many particles (molecules) which obey Newtonrsquos laws for dynamics but which would be difficult to analyze due to their numbers

bull We use macroscopic means for analysis of these systems of many particles - involving quantities such as pressure volume and temperature

Thermodynamics VocabularyThermodynamics Vocabulary

bull System ndash a definite quantity of matter enclosed by boundaries or surfaces Boundaries need not have definite shape or volume

bull Thermally Isolated System ndash system in which no heat is transferred into or out of the system

bull Heat Reservoir ndash a large separate system with unlimited heat capacity (any amount of heat can be withdrawn or added without appreciably changing the temperature)

State of a SystemState of a Systembull Kinematics equations describe the motion of an

object using variables which change but are related (displacement velocity acceleration) Motion can be described at any moment using these quantities

bull Thermodynamic systems can be described similarly using relationships between pressure volume and temperature Relationships between these variables are called Equations of State and describe the state of the system at any moment

p-V Diagramsp-V Diagrams

bull A pressure-volume diagram shows how pressure changes as a function of volume

bull Each coordinate (V p) specifies a state of the system Pressure and volume are given and temperature may be found indirectly

bull A process is any change in the state of a system Processes may be reversible or irreversible

p-V Diagramsp-V Diagrams

First Law of ThermodynamicsFirst Law of Thermodynamics

bull First Law describes how heat transferred to a system Q relates to internal energy of the system U and work done by the system W

bull Recall Q = heat added W = work done U = internal energy (see page 360)

bull First Law of Thermo Q = ΔU + W

First Law of ThermodynamicsFirst Law of Thermodynamics

bull Q = ΔU + W

bull If a system absorbs 2000J of heat and does 800J of work (expanding perhaps) then the internal energy will increase by 1200J

Example 1Example 1

bull A 65 kg worker shovels coal for 30 hours During this process he does work at a rate of 20 W and loses heat to the environment at a rate of 480 W Ignoring the loss of water by perspiration how much fat will the worker lose The energy value of fat is 93 kcalg

Keep Track of SignsKeep Track of Signs

bull Positive work is done when force and displacement are in the same direction (expanding gas)

bull Q is positive when heat is added negative when heat is lost

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Thermodynamics VocabularyThermodynamics Vocabulary

bull Thermo (heat) dynamics (transfer)

bull Thermodynamic systems describe many many particles (molecules) which obey Newtonrsquos laws for dynamics but which would be difficult to analyze due to their numbers

bull We use macroscopic means for analysis of these systems of many particles - involving quantities such as pressure volume and temperature

Thermodynamics VocabularyThermodynamics Vocabulary

bull System ndash a definite quantity of matter enclosed by boundaries or surfaces Boundaries need not have definite shape or volume

bull Thermally Isolated System ndash system in which no heat is transferred into or out of the system

bull Heat Reservoir ndash a large separate system with unlimited heat capacity (any amount of heat can be withdrawn or added without appreciably changing the temperature)

State of a SystemState of a Systembull Kinematics equations describe the motion of an

object using variables which change but are related (displacement velocity acceleration) Motion can be described at any moment using these quantities

bull Thermodynamic systems can be described similarly using relationships between pressure volume and temperature Relationships between these variables are called Equations of State and describe the state of the system at any moment

p-V Diagramsp-V Diagrams

bull A pressure-volume diagram shows how pressure changes as a function of volume

bull Each coordinate (V p) specifies a state of the system Pressure and volume are given and temperature may be found indirectly

bull A process is any change in the state of a system Processes may be reversible or irreversible

p-V Diagramsp-V Diagrams

First Law of ThermodynamicsFirst Law of Thermodynamics

bull First Law describes how heat transferred to a system Q relates to internal energy of the system U and work done by the system W

bull Recall Q = heat added W = work done U = internal energy (see page 360)

bull First Law of Thermo Q = ΔU + W

First Law of ThermodynamicsFirst Law of Thermodynamics

bull Q = ΔU + W

bull If a system absorbs 2000J of heat and does 800J of work (expanding perhaps) then the internal energy will increase by 1200J

Example 1Example 1

bull A 65 kg worker shovels coal for 30 hours During this process he does work at a rate of 20 W and loses heat to the environment at a rate of 480 W Ignoring the loss of water by perspiration how much fat will the worker lose The energy value of fat is 93 kcalg

Keep Track of SignsKeep Track of Signs

bull Positive work is done when force and displacement are in the same direction (expanding gas)

bull Q is positive when heat is added negative when heat is lost

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Thermodynamics VocabularyThermodynamics Vocabulary

bull System ndash a definite quantity of matter enclosed by boundaries or surfaces Boundaries need not have definite shape or volume

bull Thermally Isolated System ndash system in which no heat is transferred into or out of the system

bull Heat Reservoir ndash a large separate system with unlimited heat capacity (any amount of heat can be withdrawn or added without appreciably changing the temperature)

State of a SystemState of a Systembull Kinematics equations describe the motion of an

object using variables which change but are related (displacement velocity acceleration) Motion can be described at any moment using these quantities

bull Thermodynamic systems can be described similarly using relationships between pressure volume and temperature Relationships between these variables are called Equations of State and describe the state of the system at any moment

p-V Diagramsp-V Diagrams

bull A pressure-volume diagram shows how pressure changes as a function of volume

bull Each coordinate (V p) specifies a state of the system Pressure and volume are given and temperature may be found indirectly

bull A process is any change in the state of a system Processes may be reversible or irreversible

p-V Diagramsp-V Diagrams

First Law of ThermodynamicsFirst Law of Thermodynamics

bull First Law describes how heat transferred to a system Q relates to internal energy of the system U and work done by the system W

bull Recall Q = heat added W = work done U = internal energy (see page 360)

bull First Law of Thermo Q = ΔU + W

First Law of ThermodynamicsFirst Law of Thermodynamics

bull Q = ΔU + W

bull If a system absorbs 2000J of heat and does 800J of work (expanding perhaps) then the internal energy will increase by 1200J

Example 1Example 1

bull A 65 kg worker shovels coal for 30 hours During this process he does work at a rate of 20 W and loses heat to the environment at a rate of 480 W Ignoring the loss of water by perspiration how much fat will the worker lose The energy value of fat is 93 kcalg

Keep Track of SignsKeep Track of Signs

bull Positive work is done when force and displacement are in the same direction (expanding gas)

bull Q is positive when heat is added negative when heat is lost

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

State of a SystemState of a Systembull Kinematics equations describe the motion of an

object using variables which change but are related (displacement velocity acceleration) Motion can be described at any moment using these quantities

bull Thermodynamic systems can be described similarly using relationships between pressure volume and temperature Relationships between these variables are called Equations of State and describe the state of the system at any moment

p-V Diagramsp-V Diagrams

bull A pressure-volume diagram shows how pressure changes as a function of volume

bull Each coordinate (V p) specifies a state of the system Pressure and volume are given and temperature may be found indirectly

bull A process is any change in the state of a system Processes may be reversible or irreversible

p-V Diagramsp-V Diagrams

First Law of ThermodynamicsFirst Law of Thermodynamics

bull First Law describes how heat transferred to a system Q relates to internal energy of the system U and work done by the system W

bull Recall Q = heat added W = work done U = internal energy (see page 360)

bull First Law of Thermo Q = ΔU + W

First Law of ThermodynamicsFirst Law of Thermodynamics

bull Q = ΔU + W

bull If a system absorbs 2000J of heat and does 800J of work (expanding perhaps) then the internal energy will increase by 1200J

Example 1Example 1

bull A 65 kg worker shovels coal for 30 hours During this process he does work at a rate of 20 W and loses heat to the environment at a rate of 480 W Ignoring the loss of water by perspiration how much fat will the worker lose The energy value of fat is 93 kcalg

Keep Track of SignsKeep Track of Signs

bull Positive work is done when force and displacement are in the same direction (expanding gas)

bull Q is positive when heat is added negative when heat is lost

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

p-V Diagramsp-V Diagrams

bull A pressure-volume diagram shows how pressure changes as a function of volume

bull Each coordinate (V p) specifies a state of the system Pressure and volume are given and temperature may be found indirectly

bull A process is any change in the state of a system Processes may be reversible or irreversible

p-V Diagramsp-V Diagrams

First Law of ThermodynamicsFirst Law of Thermodynamics

bull First Law describes how heat transferred to a system Q relates to internal energy of the system U and work done by the system W

bull Recall Q = heat added W = work done U = internal energy (see page 360)

bull First Law of Thermo Q = ΔU + W

First Law of ThermodynamicsFirst Law of Thermodynamics

bull Q = ΔU + W

bull If a system absorbs 2000J of heat and does 800J of work (expanding perhaps) then the internal energy will increase by 1200J

Example 1Example 1

bull A 65 kg worker shovels coal for 30 hours During this process he does work at a rate of 20 W and loses heat to the environment at a rate of 480 W Ignoring the loss of water by perspiration how much fat will the worker lose The energy value of fat is 93 kcalg

Keep Track of SignsKeep Track of Signs

bull Positive work is done when force and displacement are in the same direction (expanding gas)

bull Q is positive when heat is added negative when heat is lost

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

p-V Diagramsp-V Diagrams

First Law of ThermodynamicsFirst Law of Thermodynamics

bull First Law describes how heat transferred to a system Q relates to internal energy of the system U and work done by the system W

bull Recall Q = heat added W = work done U = internal energy (see page 360)

bull First Law of Thermo Q = ΔU + W

First Law of ThermodynamicsFirst Law of Thermodynamics

bull Q = ΔU + W

bull If a system absorbs 2000J of heat and does 800J of work (expanding perhaps) then the internal energy will increase by 1200J

Example 1Example 1

bull A 65 kg worker shovels coal for 30 hours During this process he does work at a rate of 20 W and loses heat to the environment at a rate of 480 W Ignoring the loss of water by perspiration how much fat will the worker lose The energy value of fat is 93 kcalg

Keep Track of SignsKeep Track of Signs

bull Positive work is done when force and displacement are in the same direction (expanding gas)

bull Q is positive when heat is added negative when heat is lost

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

First Law of ThermodynamicsFirst Law of Thermodynamics

bull First Law describes how heat transferred to a system Q relates to internal energy of the system U and work done by the system W

bull Recall Q = heat added W = work done U = internal energy (see page 360)

bull First Law of Thermo Q = ΔU + W

First Law of ThermodynamicsFirst Law of Thermodynamics

bull Q = ΔU + W

bull If a system absorbs 2000J of heat and does 800J of work (expanding perhaps) then the internal energy will increase by 1200J

Example 1Example 1

bull A 65 kg worker shovels coal for 30 hours During this process he does work at a rate of 20 W and loses heat to the environment at a rate of 480 W Ignoring the loss of water by perspiration how much fat will the worker lose The energy value of fat is 93 kcalg

Keep Track of SignsKeep Track of Signs

bull Positive work is done when force and displacement are in the same direction (expanding gas)

bull Q is positive when heat is added negative when heat is lost

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

First Law of ThermodynamicsFirst Law of Thermodynamics

bull Q = ΔU + W

bull If a system absorbs 2000J of heat and does 800J of work (expanding perhaps) then the internal energy will increase by 1200J

Example 1Example 1

bull A 65 kg worker shovels coal for 30 hours During this process he does work at a rate of 20 W and loses heat to the environment at a rate of 480 W Ignoring the loss of water by perspiration how much fat will the worker lose The energy value of fat is 93 kcalg

Keep Track of SignsKeep Track of Signs

bull Positive work is done when force and displacement are in the same direction (expanding gas)

bull Q is positive when heat is added negative when heat is lost

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Example 1Example 1

bull A 65 kg worker shovels coal for 30 hours During this process he does work at a rate of 20 W and loses heat to the environment at a rate of 480 W Ignoring the loss of water by perspiration how much fat will the worker lose The energy value of fat is 93 kcalg

Keep Track of SignsKeep Track of Signs

bull Positive work is done when force and displacement are in the same direction (expanding gas)

bull Q is positive when heat is added negative when heat is lost

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Keep Track of SignsKeep Track of Signs

bull Positive work is done when force and displacement are in the same direction (expanding gas)

bull Q is positive when heat is added negative when heat is lost

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Work from a p-V DiagramWork from a p-V Diagram

bull Work = FΔxpA ΔxpΔV

bull On a pressure-volume diagram work done by a system is equal to the area under the p-V curve

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Isothermal ProcessIsothermal Processbull Isothermal means

constant temperaturebull Internal Energy

remains constant if temperature remains constant

bull Q = ΔU + W ΔU=0

bull Q = W

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Isothermal ProcessIsothermal ProcessFirst Law of Thermo

Q = ΔU + W

But for isothermal

Q = W

All added heat goes into

work done by expanding gas

Work = nRT ln(VfVi)

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Isobaric ProcessIsobaric Process

bull A constant pressure process is called isobaric

bull Work in isobaric process is

W = p(V2 ndash V1) = pΔV

bull So Q = ΔU + pΔV

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

ExampleExample

bull Two moles of a monatomic ideal gas initially at 0deg and 10 atm are expanded to twice their original volume using two different processes They are expanded isothermally and then starting in the same initial state they are expanded isobarically During which process does the gas do more work Find the work done in each case

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Isometric ProcessIsometric Processbull Isometric means

isovolumetric hellip it describes a process at constant volume

bull If volume doesnrsquot change no work is done

bull Q = ΔU

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

ExampleExample

bull Many lsquoemptyrsquo aerosol cans contain remnant propellant gases under approximately 1 atm of pressure and 20degC They display the warning ldquoDo not dispose of this can in an incineratorrdquo What is the change in internal energy of such a gas if 500J of heat is added to it raising the temperature to 2000degF What is the final pressure of the gas

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Adiabatic ProcessAdiabatic Process

bull Adiabatic means no heat is transferred into or out of the system

bull Q = 0 so first law of thermodynamics becomes

ΔU = -W

bull If internal energy increases work has been done

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

Adiabatic ProcessAdiabatic Process

bull adiabatic expansion

bull Pressure and volume are related along an adiabat by p1V1

γ = p2V2γ

bull Work done in an adiabatic process can be found using

Wadiabatic = (p1V1 ndash p2V2) (γ - 1)

γ = 167 for monotonic gas 140 for diatomic

SummarySummaryProcess Characteristic Result First Law

Isothermal T = constant ΔU = 0 Q = W

Isobaric p = constant W = pΔV Q = ΔU + pΔV

Isometric V = constant W = 0 Q = ΔU

Adiabatic Q = 0 ΔU = -W

ExampleExample

bull Blowing Air

bull A sample of helium expands to triple its initial volume adiabatically in one case and isothermally in another It starts from the same initial state in both cases The sample contains 200mol of helium initially at 20 ordmC and 10 atm

a) during which process does the gas do more work

b) Calculate the work done in each process

Second Law of Second Law of ThermodynamicsThermodynamics

bull Second Law provides a way to predict the direction in which thermal processes proceed

bull Heat flows spontaneously from a warmer body to a cooler body

bull Heat energy cannot be completely transformed into mechanical work

EntropyEntropy

bull Entropy is a measure of a systemrsquos ability to do work

bull Entropy is a measure of disorder

bull A system naturally moves towards greater disorder More order = lower entropy more disorder = higher entropy

bull ΔSgt0 for natural processes

EntropyEntropy

bull ΔS = Q T [Joules Kelvin] gives a change in entropy at a constant temperature T

bull ΔS = increase in entropy

bull Entropy increases if Q is positive temperature must be measured in Kelvin

ExamplesExamples

bull While doing physical exercise at 34degC an athlete loses 0400 kg of water per hour by evaporation of perspiration from his skin Estimate the change in entropy of the water as it vaporizes Latent heat of vaporization for perspiration is

242X105 Jkg

Heat EnginesHeat Engines

bull Heat engines convert heat energy into useful work

bull Examples are electric turbines (power plants) and internal combustion engines (automobiles)

bull lsquoHeat inrsquo generates work gas must be brought to original state via a compression which takes work

bull Wnet is the enclosed area

Heat EnginesHeat Enginesbull Net Work = Work done

by gas - Work done ongas

bull Wnet = Wexp ndash Wcom

bull For a cycle ΔU = 0 sohellip

bull Wnet = Wexp ndash Wcom

Wnet = Qh ndash Qc

EfficiencyEfficiency

bull Efficiency is the ratio of net work out to heat in

bull ε = Wnet Qin

bull ε = (Qh ndash Qc) Qh

bull ε = 1 ndash QcQh

EfficiencyEfficiency

bull No engine is perfectly efficient More heat (energy) is required (Qh) than useful work created

bull Why

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

ExampleExample

bull Blowing Air

bull A sample of helium expands to triple its initial volume adiabatically in one case and isothermally in another It starts from the same initial state in both cases The sample contains 200mol of helium initially at 20 ordmC and 10 atm

a) during which process does the gas do more work

b) Calculate the work done in each process

Second Law of Second Law of ThermodynamicsThermodynamics

bull Second Law provides a way to predict the direction in which thermal processes proceed

bull Heat flows spontaneously from a warmer body to a cooler body

bull Heat energy cannot be completely transformed into mechanical work

EntropyEntropy

bull Entropy is a measure of a systemrsquos ability to do work

bull Entropy is a measure of disorder

bull A system naturally moves towards greater disorder More order = lower entropy more disorder = higher entropy

bull ΔSgt0 for natural processes

EntropyEntropy

bull ΔS = Q T [Joules Kelvin] gives a change in entropy at a constant temperature T

bull ΔS = increase in entropy

bull Entropy increases if Q is positive temperature must be measured in Kelvin

ExamplesExamples

bull While doing physical exercise at 34degC an athlete loses 0400 kg of water per hour by evaporation of perspiration from his skin Estimate the change in entropy of the water as it vaporizes Latent heat of vaporization for perspiration is

242X105 Jkg

Heat EnginesHeat Engines

bull Heat engines convert heat energy into useful work

bull Examples are electric turbines (power plants) and internal combustion engines (automobiles)

bull lsquoHeat inrsquo generates work gas must be brought to original state via a compression which takes work

bull Wnet is the enclosed area

Heat EnginesHeat Enginesbull Net Work = Work done

by gas - Work done ongas

bull Wnet = Wexp ndash Wcom

bull For a cycle ΔU = 0 sohellip

bull Wnet = Wexp ndash Wcom

Wnet = Qh ndash Qc

EfficiencyEfficiency

bull Efficiency is the ratio of net work out to heat in

bull ε = Wnet Qin

bull ε = (Qh ndash Qc) Qh

bull ε = 1 ndash QcQh

EfficiencyEfficiency

bull No engine is perfectly efficient More heat (energy) is required (Qh) than useful work created

bull Why

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

Second Law of Second Law of ThermodynamicsThermodynamics

bull Second Law provides a way to predict the direction in which thermal processes proceed

bull Heat flows spontaneously from a warmer body to a cooler body

bull Heat energy cannot be completely transformed into mechanical work

EntropyEntropy

bull Entropy is a measure of a systemrsquos ability to do work

bull Entropy is a measure of disorder

bull A system naturally moves towards greater disorder More order = lower entropy more disorder = higher entropy

bull ΔSgt0 for natural processes

EntropyEntropy

bull ΔS = Q T [Joules Kelvin] gives a change in entropy at a constant temperature T

bull ΔS = increase in entropy

bull Entropy increases if Q is positive temperature must be measured in Kelvin

ExamplesExamples

bull While doing physical exercise at 34degC an athlete loses 0400 kg of water per hour by evaporation of perspiration from his skin Estimate the change in entropy of the water as it vaporizes Latent heat of vaporization for perspiration is

242X105 Jkg

Heat EnginesHeat Engines

bull Heat engines convert heat energy into useful work

bull Examples are electric turbines (power plants) and internal combustion engines (automobiles)

bull lsquoHeat inrsquo generates work gas must be brought to original state via a compression which takes work

bull Wnet is the enclosed area

Heat EnginesHeat Enginesbull Net Work = Work done

by gas - Work done ongas

bull Wnet = Wexp ndash Wcom

bull For a cycle ΔU = 0 sohellip

bull Wnet = Wexp ndash Wcom

Wnet = Qh ndash Qc

EfficiencyEfficiency

bull Efficiency is the ratio of net work out to heat in

bull ε = Wnet Qin

bull ε = (Qh ndash Qc) Qh

bull ε = 1 ndash QcQh

EfficiencyEfficiency

bull No engine is perfectly efficient More heat (energy) is required (Qh) than useful work created

bull Why

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

EntropyEntropy

bull Entropy is a measure of a systemrsquos ability to do work

bull Entropy is a measure of disorder

bull A system naturally moves towards greater disorder More order = lower entropy more disorder = higher entropy

bull ΔSgt0 for natural processes

EntropyEntropy

bull ΔS = Q T [Joules Kelvin] gives a change in entropy at a constant temperature T

bull ΔS = increase in entropy

bull Entropy increases if Q is positive temperature must be measured in Kelvin

ExamplesExamples

bull While doing physical exercise at 34degC an athlete loses 0400 kg of water per hour by evaporation of perspiration from his skin Estimate the change in entropy of the water as it vaporizes Latent heat of vaporization for perspiration is

242X105 Jkg

Heat EnginesHeat Engines

bull Heat engines convert heat energy into useful work

bull Examples are electric turbines (power plants) and internal combustion engines (automobiles)

bull lsquoHeat inrsquo generates work gas must be brought to original state via a compression which takes work

bull Wnet is the enclosed area

Heat EnginesHeat Enginesbull Net Work = Work done

by gas - Work done ongas

bull Wnet = Wexp ndash Wcom

bull For a cycle ΔU = 0 sohellip

bull Wnet = Wexp ndash Wcom

Wnet = Qh ndash Qc

EfficiencyEfficiency

bull Efficiency is the ratio of net work out to heat in

bull ε = Wnet Qin

bull ε = (Qh ndash Qc) Qh

bull ε = 1 ndash QcQh

EfficiencyEfficiency

bull No engine is perfectly efficient More heat (energy) is required (Qh) than useful work created

bull Why

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

EntropyEntropy

bull ΔS = Q T [Joules Kelvin] gives a change in entropy at a constant temperature T

bull ΔS = increase in entropy

bull Entropy increases if Q is positive temperature must be measured in Kelvin

ExamplesExamples

bull While doing physical exercise at 34degC an athlete loses 0400 kg of water per hour by evaporation of perspiration from his skin Estimate the change in entropy of the water as it vaporizes Latent heat of vaporization for perspiration is

242X105 Jkg

Heat EnginesHeat Engines

bull Heat engines convert heat energy into useful work

bull Examples are electric turbines (power plants) and internal combustion engines (automobiles)

bull lsquoHeat inrsquo generates work gas must be brought to original state via a compression which takes work

bull Wnet is the enclosed area

Heat EnginesHeat Enginesbull Net Work = Work done

by gas - Work done ongas

bull Wnet = Wexp ndash Wcom

bull For a cycle ΔU = 0 sohellip

bull Wnet = Wexp ndash Wcom

Wnet = Qh ndash Qc

EfficiencyEfficiency

bull Efficiency is the ratio of net work out to heat in

bull ε = Wnet Qin

bull ε = (Qh ndash Qc) Qh

bull ε = 1 ndash QcQh

EfficiencyEfficiency

bull No engine is perfectly efficient More heat (energy) is required (Qh) than useful work created

bull Why

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

ExamplesExamples

bull While doing physical exercise at 34degC an athlete loses 0400 kg of water per hour by evaporation of perspiration from his skin Estimate the change in entropy of the water as it vaporizes Latent heat of vaporization for perspiration is

242X105 Jkg

Heat EnginesHeat Engines

bull Heat engines convert heat energy into useful work

bull Examples are electric turbines (power plants) and internal combustion engines (automobiles)

bull lsquoHeat inrsquo generates work gas must be brought to original state via a compression which takes work

bull Wnet is the enclosed area

Heat EnginesHeat Enginesbull Net Work = Work done

by gas - Work done ongas

bull Wnet = Wexp ndash Wcom

bull For a cycle ΔU = 0 sohellip

bull Wnet = Wexp ndash Wcom

Wnet = Qh ndash Qc

EfficiencyEfficiency

bull Efficiency is the ratio of net work out to heat in

bull ε = Wnet Qin

bull ε = (Qh ndash Qc) Qh

bull ε = 1 ndash QcQh

EfficiencyEfficiency

bull No engine is perfectly efficient More heat (energy) is required (Qh) than useful work created

bull Why

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

Heat EnginesHeat Engines

bull Heat engines convert heat energy into useful work

bull Examples are electric turbines (power plants) and internal combustion engines (automobiles)

bull lsquoHeat inrsquo generates work gas must be brought to original state via a compression which takes work

bull Wnet is the enclosed area

Heat EnginesHeat Enginesbull Net Work = Work done

by gas - Work done ongas

bull Wnet = Wexp ndash Wcom

bull For a cycle ΔU = 0 sohellip

bull Wnet = Wexp ndash Wcom

Wnet = Qh ndash Qc

EfficiencyEfficiency

bull Efficiency is the ratio of net work out to heat in

bull ε = Wnet Qin

bull ε = (Qh ndash Qc) Qh

bull ε = 1 ndash QcQh

EfficiencyEfficiency

bull No engine is perfectly efficient More heat (energy) is required (Qh) than useful work created

bull Why

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

Heat EnginesHeat Enginesbull Net Work = Work done

by gas - Work done ongas

bull Wnet = Wexp ndash Wcom

bull For a cycle ΔU = 0 sohellip

bull Wnet = Wexp ndash Wcom

Wnet = Qh ndash Qc

EfficiencyEfficiency

bull Efficiency is the ratio of net work out to heat in

bull ε = Wnet Qin

bull ε = (Qh ndash Qc) Qh

bull ε = 1 ndash QcQh

EfficiencyEfficiency

bull No engine is perfectly efficient More heat (energy) is required (Qh) than useful work created

bull Why

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

EfficiencyEfficiency

bull Efficiency is the ratio of net work out to heat in

bull ε = Wnet Qin

bull ε = (Qh ndash Qc) Qh

bull ε = 1 ndash QcQh

EfficiencyEfficiency

bull No engine is perfectly efficient More heat (energy) is required (Qh) than useful work created

bull Why

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

EfficiencyEfficiency

bull No engine is perfectly efficient More heat (energy) is required (Qh) than useful work created

bull Why

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

ExampleExample

bull A leaf-blower engine absorbs 800J of heat energy from a high temperature reservoir (the ignited gas-air mixture) and exhausts 700J to a low temperature reservoir (the outside air) What is the efficiency

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

ExampleExamplebull You have 0100 mol of an

ideal monatomic gas that follows the cycle shown The pressure and temperature on the lower left is 100 atm and 20 degC Assume that the pressure doubles during the isometric pressure increase and the volume doubles during the isobaric expansion Find the efficiency

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

Otto Cycle ndash 4 Stroke EngineOtto Cycle ndash 4 Stroke Engine

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

Carnot CycleCarnot Cycle

bull A Carnot cycle is a cycle consisting of two isotherms and two adiabats

bull Sadi Carnot discovered that efficiency can be maximized by a cycle absorbing heat from a constant high temperature reservoir Th (isotherm) and exhausting it to a constant low temperature reservoir Tc (isotherm)

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh

Carnot CycleCarnot Cycle

bull Efficiency of a Carnot Cycle

bull ε = (Th ndash Tc)Th

= 1 ndash TcTh

= 1 ndash QcQh