• Chapter 20 The first law of thermodynamics

Department of Medicine General Physics, Week6 General Physics, Week6

Thermodynamics

Week 6 Thermodynamics

• Chapter 19 Temperature

• Chapter 20 The first law of thermodynamics

• Chapter 21 The kinetic theory of gases

• Chapter 21 The kinetic theory of gases

• Chapter 22 The second law of

thermodynamics

Chap. 19 Temperature

• Introduction

• Temperature and zeroth law of thermodynamics

• Thermometers and the Celsius

• Thermometers and the Celsius Temperature scale

• Thermal expansion of solids and liquids

• Macroscopic Description of an ideal gas

Introduction

• Temperature

• Heat

• Internal energy

19.1 Temperature and the Zeroth Law of Thermodynamics

19.1 Temperature and the Zeroth Law of Thermodynamics

Temperature: feeling of hot or cold of an object.

human sense do not represent correctly the real temperature Thermal contact: if energy can be exchanged between two

objects by temperature difference.

Thermal equilibrium: There is no energy transfer between two thermal contacting objects

Thermal equilibrium: There is no energy transfer between two thermal contacting objects

Zeroth

Zeroth law of law of thermodynamics thermodynamics

If objects A and B are separately in thermal equilibrium

with a third object C, then A and B are in themal equilibrium with

each other.

19.2 Thermometers and the Celsius Temperature Scale

19.2 Thermometers and the Celsius Temperature Scale

All thermometers are based on the principle that some physical property of a system changes as the system’s temperature changes.

Physical property changing with temperature (1) volume of liquid

(2) length of solid

(3) pressure of gas with constant volume (4) volume of gas with constant pressure

Alcohol or mercury bimetal

(4) volume of gas with constant pressure (5) electric resistance of conductor

(6) color of object

thermistor

Celsius temperature: define ice point of water as 0°C, boiling point of water as 100°C. equally divide by 100.

Fahrenheit temperature: define ice point of water as 32°C, boiling point of water as 212°C. equally divide by 180.

F T

T

=

+ 32 ° 5

9

19.3 The Constant-Volume Gas Thermometer and the Absolute Temperature Scale

19.3 The Constant-Volume Gas Thermometer and the Absolute Temperature Scale

Constant-Volume Gas Thermometer : variation of pressure of a fixed volume of gas with temperature.

The thermometer readings are virtually independent of the gas used when the pressure is low and temperature is high.

If the lines for various gases are

extended, the pressure is always zero when the temperature is –273.15

C -273.15°C: absolute zero

15 .

− 273

= T

T _C

Kelvin scale Kelvin scale

The kelvin is defined as 1/273.16 of the difference between absolute zero and the temperature of the triple point of water

TTriple riple point of point of water water

This is the combination of temperature and pressure where ice, water, and steam can all coexist. This point occurs at

0.01°C and

4.58mmHg.

F T

T = 9 + 32 ° F T

T

=

+ 32 ° 5

F

C

T T

T = ∆ = ∆

∆ 9

5

) 32

9 (

5 T F

T

=

− °

19.4 Thermal Expansion of Solids and Liquids 19.4 Thermal Expansion of Solids and Liquids

Thermal expansion:

the increase in the size of an object with an increase in its temperature. Thermal expansion is a consequence of the change in the average separation between the atoms in an object

If the expansion is small relative to the original

dimensions of the object, the change in any dimension is, to a good approximation, proportional to the first is, to a good approximation, proportional to the first power of the change in temperature

T Li L

∆

≡ ∆ / α

Average coefficient of linear expansion

T L

L =

∆

∆ α

^L

− ^L

= α ^L

^{( T}

− ^T

⁾

T V

V =

∆

∆ β

Average coefficient of volume expansion

] ) (

) (

3 3

1 [

) 1

(

) )(

)(

(

) )(

)(

(

3 2

3

T T

T V

T wh

T h h

T w w

T

h h

w w

V V

i i

∆ +

∆ +

∆ +

=

∆ +

=

∆ +

∆ +

∆ +

=

∆ +

∆ +

∆ +

=

∆ +

α α

α α

α α

α l

l l

l l

3

2

( )

) (

3

3 T T T

V V

i

∆ +

∆ +

∆

∆ =

α α

V

α

T V V

V T V

i i

∆

=

∆

→

∆

∆ =

) 3 (

3 α α

α

β = 3

Some Coefficients

Expansion of Railroad Track Example 19.2

A segment of steel railroad track has a length of 30.000m when the temperature is 0.0°C.

(A) What is its length when the temperature 40.0°C.

(B) Suppose the ends of the rail are rigidly clamped at 0.0°C so that expansion is prevented. What is the thermal stress set up in the rail if its temperature is raised to 40.0°C.

Solution (A)

C m

C T

L

L _i

) 0 . 40 )(

000 . 30 ](

) ( 6 10 11

[ × − ° ¹ °

=

∆

=

∆

−

α

m 013 .

= 0

m m

m

L _f = 30 .000 + 0.013 = 30 .013

(B)

Li

Y L A

F ∆

=

= stress Tensile

2 7

2 10

/ 10

7 . 8

000 . 30

013 . ) 0 /

10 20

(

m N

m m m

A N F

×

=

 

 



× 

=

◈

◈The The Unusual Behavior of Unusual Behavior of Water Water

Usually, liquid expands as the temperature is increase. The expansion coefficient is 10 times large that solid.

As the temperature increases from 0

C to 4

C, water

contracts. Above 4

C, water expands with increasing

temperature. The maximum

temperature. The maximum

density of water (1.000 g/cm

)

occurs at 4

C.

19.5 Macroscopic Description of an Ideal Gas 19.5 Macroscopic Description of an Ideal Gas

IIdeal gas deal gas Model:

1. The number of molecules in the gas is large, and the average separation between the molecules is large compared with their dimensions.

2. The molecules obey Newton’s laws of motion, but as a whole they move randomly.

3. The molecules interact only by short-range forces during elastic 3. The molecules interact only by short-range forces during elastic

collisions.

4. The molecules make elastic collisions with the walls.

5. The gas under consideration is a pure substance.

If the gases are at low pressures and dilute, this model adequately describes the behavior of real gases.

Mole: The amount of gas in a given volume

One mole of any substance is that amount of the substance that contains Avogadro’s number of constituent particles

Avogadro’s number N

= 6.022 x 10

The constituent particles can be atoms or molecules

M

n = m

In isolated gas system, In isolated gas system,

When a gas is kept at a constant temperature, its pressure is inversely proportional to its volume (Boyle’s law)

When a gas is kept at a constant pressure, its volume is directly proportional to its temperature (Charles law)

When the volume of the gas is kept constant, the pressure is directly proportional to the temperature (Guy-Lussac’s law)

2 2 1

1

V P V

P =

2 2 1

1

T V V = T

2 2 1

1

T

P

T P =

Molar mass

nRT PV =

The equation of state for an ideal gas

K mol

J

R = 8 . 314 / ⋅

K mol

atm L

R = 0 . 08206 ⋅ / ⋅

Modified equation of state for an ideal gas

N RT nRT N

PV

A

=

=

Modified equation of state for an ideal gas

T Nk PV =

K N J

k R

A

B

= = 1 . 38 × 10

/

Heating a Spray Can Example 19.4

Solution

nR

PV = T

f f f i

i i

T V P T

V P =

f f i

i

T P TP = kPa K kPa

P K T

P T _i

i f

f (202 ) 320

295

468  =



 



= 







= 

K T_i   295 



Suppose we include a volume change due to thermal expnasion of the steel can as the temperature increases. Does that alter our answer for the final pressure

significantly?

3 3

1

6( ) ](125 .00 )(173 ) 0.71

10 11 [ 3

3

cm C

cm C

T V T

V

V _{i i}

=

°

°

×

=

∆

=

∆

=

∆

−

−

α β

i f i i

f

f

P

V V T

P T  





 





 

 



= 

) 71

. 0 00

. 125 (

00 . 125

3 3

3

= + =

= cm cm

cm V

V

f i

kPa K kPa

P_f K 0.994 (202 ) 318 295

468 × × =



 



= 

∴

Chap. 20. The first law of thermodynamics

20.1 Heat and internal energy

20.2 Specific heat and Calorimetry 20.3 Latent heat

20.3 Latent heat

20.4 Work and heat in thermodynamic processes

20.5 The first law of thermodynamics 20.6 Some applications of the first law of thermodynamics

20.7 Energy transfer mechanism

20.1 Heat and Internal Energy 20.1 Heat and Internal Energy

Internal energy

all the energy of a system that is associated with its microscopic components.

These components are its atoms and molecules.

The system is viewed from a reference frame at rest with respect to the center of mass of the system.

Translational motion

Rotational motion

Vibrational motion

Potential energy

20.1 Heat and Internal Energy 20.1 Heat and Internal Energy

Heat: the transfer of energy across the boundary of a system due to a temperature difference

between the system and its surroundings

The term heat will also be used to represent the amount of energy transferred by this method

Both heat and work can change the internal energy of Both heat and work can change the internal energy of a system

The internal energy can be changed even when no energy is transferred by heat, but just by work

Units

Units of of Heat Heat

1 cal: the amount of energy transfer necessary to raise the temperature of 1 g of water from 14.5

C to 15.5

C.

1 Cal: the “Calorie” used for food is actually 1 kilocalorie 1 Btu: the amount of energy transfer necessary to raise the

temperature of 1 lb of water from 63

F to 64

F.

The

The Mechanical Equivalent of Mechanical Equivalent of Heat Heat

Joule established the equivalence between mechanical energy and internal energy.

J cal 4 . 186

1 =

20.2 Specific Heat and Calorimetry 20.2 Specific Heat and Calorimetry

The required heat depends on the material to increase a certain temperature for a given amount of material.

H

Heat capacity eat capacity C C :

the amount of energy needed to raise the temperature of that sample by 1

C

T C

Q = C ∆ T Q = ∆

SSpecific heat pecific heat c c : the heat capacity per unit mass

T m c Q

≡ ∆

T mc

Q = ∆

Calorimetry Calorimetry

One technique for measuring specific heat involves heating a material,

adding it to a sample of water, and recording the final temperature

hot

cold

Q

Q = −

calorimeter

hot

cold

Q

Q = −

) (

)

(

w

w

c T T m c T T

m − = − −

) (

) (

f x

x

w f

w w

x

m T T

T T

c c m

−

= −

Cooling a Hot Ingot Example 20.2

A 0.0500kg ingot of metal is heated to 200.0°C and then dropped into a calorimeter containing 0.400kg of water initially at 20.0°C. The final temperature of the mixed system is 22.4°C. Find the specific heat of the metal.

Solution

m

c

( T

− T

) = − m

c

( T

− T

)

) 0 . 200 4

. 22 )(

)(

0500 .

0 (

) 0 . 20 4

. 22 )(

/ 4186 )(

400 . 0 (

C C

c kg

C C

C kg J kg

x ° − °

−

=

°

−

°

°

⋅

C kg

J

c

= ⋅ °

∴ 453 /

Fun time for a Cowboy Example 20.3

A cowboy fires a silver bullet with a muzzle speed of 200m/s into a pine wall of a saloon. Assume all the internal energy generated by the impact remains with the bullet. What is the temperature change of the bullet?

Solution

∆ K + ∆ E

= 0 T mc Q

E = = ∆

∆

( ^{0 −}

^mv

) ^{+ mc ∆ T = 0} _J ^m _kg ^s _C ^C

c v mc

T mv

°

° =

= ⋅

=

=

∆

5 . ) 85

/ 234 ( 2

) / 200 (

2

2 2 2

2 1

20.3 Latent Heat 20.3 Latent Heat

PPhase change hase change:

a substance changes from one form to another LLatent heat atent heat:

If an amount of energy required to change the phase of a sample per mass

≡ Q m

L ≡ Q Q = ± mL

Latent heat of fusion L

: the phase change is from solid to liquid

Latent heat of vaporization L

: the phase change is from liquid

to gas

Sample Latent Heat Values

Heat energy required to convert a 1.00g cube of ice at -30.0°C to steam at 120.0°C.

Part A:

J

C C

kg J kg

T c m Q _{i i}

7 . 62

) 0 . 30 )(

/ 2090 )(

10 00 . 1

( ³

=

°

°

⋅

×

=

∆

=

−

Part B:

L m Q =

J

kg J kg

L m

Q _{i f}

333

) / 10 33 . 3 )(

10 00

. 1

( ^{3 5}

=

×

×

=

=

−

Part C: Q = m

c

∆ T = ( 1 . 00 × 10

kg )( 4 . 19 × 10

J / kg ⋅ ° C )( 100 . 0 ° C ) = 419 J Part D: Q = m

L

= ( 1 . 00 × 10

kg )( 2 . 26 × 10

J / kg ) = 2 . 26 × 10

J

Part E: Q = m

c

∆ T = ( 1 . 00 × 10

kg )( 2 . 01 × 10

J / kg ⋅ ° C )( 20 . 0 ° C ) = 40 . 2 J

Cooling the steam Example 20.4

What mass of steam initially at 130°C is needed to warm 200g of water in a 100g glass container from 20.0°C to 50.0°C?

Solution

( 1 ) Q

= − Q

) / 10 03 . 6 (

) 0 . 30 )(

/ 10 01 . 2 (

4

3 1

Kg J m

C C

kg J m

T c m Q

s

s s

s

×

−

=

°

−

°

⋅

×

=

∆

=

Steam of 130°C
steam of 100°C (2) Q_hot = −m_s(2.53×10⁶J /kg)

Water and glass of 20.0°C
50°C

s

) / 10 26 . 2

( ⁶

2 m J kg

Q =− _s ×

) / 10 09 . 2 (

) 0 . 50 )(

/ 10 19 . 4 (

5

3 3

Kg J m

C C

kg J m

T c m Q

s

s w

s

×

−

=

°

−

°

⋅

×

−

∆

=

] 10 09 . 2

10 26 . 2 10

03 . 6 [

5

6 4

3 2

1

× +

× +

×

−

=

+ +

=

s hot

m

Q Q

Q Q

steam
water of 100°C

Water of 100°C
water of 50°C

) 0 . 30 )(

837 )(

100 . 0 (

) 0 . 30 )(

10 19 . 4 )(

200 . 0

(

+

×

cold

= Q

J Q

2 . 77 10

)

3

( = ×

)]

/ 10 53

. 2 ( [

10 77

.

2 × ⁴ J = − −m_s × ⁶ J kg g

kg

m_s =1.09 ×10⁻² =10.9

20.4 Work and Heat in Thermodynamic Processes 20.4 Work and Heat in Thermodynamic Processes

State of a system can be described by pressure, temperature, volume, internal energy.

State variables describe the state of a system.

Quasi-statically: The compression is slow enough for all the system to remain essentially in thermal equilibrium

Transfer variables describe the change of state of a system.(heat and work)

PAdy Fdy

dy F

d

dW = F ⋅ r = − j ⋅ j = − = −

PdV dW = −

∫

−

=

i

v

v

PdV

W

Work done on a gas▶

The work done on a gas in a quasi-static process that takes the gas from an initial state to a final state is the negative of the area under the curve on the PV diagram, evaluated between the initial and final states.

W₂

W

)

1

P

( V

V

W = − −

) ( V V P

W = − −

W₁

W₂

W₃

W

= − P

( V

− V

)

∫

−

=

i

v

v

PdV

W

The work done does depend on the path taken.

Heat transfer also depends on the initial, final and path as work done..

Heat and work depends on the path, thus it can not be determined with initial and final states only.

20.5 The First Law of Thermodynamics 20.5 The First Law of Thermodynamics

a special case of the Law of Conservation of Energy. It takes into account changes in internal energy and energy transfers by heat and work.

W Q

E = +

∆ _int

Internal energy is a state variable as pressure, volume and temperature.

Internal energy is a state variable as pressure, volume and temperature.

dW dQ

dE

= +

An isolated system is one that does

not interact with its surroundings:

Q = W = 0 → ∆ E

= 0

A cyclic process is one that starts and ends in the same state:

W Q

E = → = −

∆

0

20.6 Some Applications of the First Law of Thermodynamics

20.6 Some Applications of the First Law of Thermodynamics

▷Adiabatic process: Q=0

W E =

∆

▷ Adiabatic free expansion: Q=0 & W=0

int

= 0

∆E

▷ Isobaric process: △P=0

▷ Isobaric process: △P=0

) ( V

V

P

W = − −

▷ Isovolumetric process: △V=0
△W=0

Q E =

∆

▷ Isothermal process: △T=0

int

= 0

∆E Q = − W

Isothermal

Isothermal Expansion of an Ideal Expansion of an Ideal Gas Gas

Expansion at fixed temperature,

V dV PdV nRT

W

i

f

i

V V

V

∫ ^{= −} ∫

−

=

nRT PV =

) ln

(ln

1

f i

V V V

V

V V

nRT

nV V nRT

nRT dV

W

i f

i

−

=

−

=

−

= ∫

 





 



= 

f i

V n V nRT

W 1

An isothermal expansion Example 20.5

A 1.0-mol sample of an ideal gas is kept at 0°C during an expansion from 3.0L to 10.0L.

(A) How much work is done on the gas during the expansion?

(B) How much energy transfer by heat occurs between the gas and its surroundings in this process?

(C) If the gas is returned to the original volume by means of an isobaric process, how much work is done on the gas?

Solution

(A) V

n V nRT W

f

1 i 



 



= 

J

L n L

K K

mol J

mol

f

103

7 . 2

0 . 10

0 . 1 3

) 273 )(

/ 31 . 8 )(

0 . 1 (

×

−

=



 



⋅ 

=





(B)

J W

Q

W Q

W Q E

3 int

10 7 . 2 0

×

=

−

= +

= +

=

∆ ^(C)

J

m m

m

K K

mol J

mol

V V V

V nRT V

P

W _{f i}

i i i

f

3

3 3 3

3

3 3

10 6 . 1

) 10

0 . 10 10

0 . 3 (

10 0 . 10

) 273 )(

/ 31 . 8 )(

0 . 1 (

) (

) (

×

=

×

−

×

×

= ⋅

−

−

=

−

−

=

−

−

−

20.7 Energy Transfer Mechanisms 20.7 Energy Transfer Mechanisms

▷

▷ Thermal Conduction Thermal Conduction

x A T t

Q

∆

∝ ∆

= ∆ P

dx kA dT

P =

dx

thermal conductivity

▷

▷ Convection Convection

Energy transfer by heated material movement

▷

▷ Radiation Radiation

Energy transfer in the form of

electromagnetic waves due to thermal vibrations of their molecules

Stefan’s law Stefan’s law

]

4

[

W σ AeT

P =

4 2

8

/

10 6696

.

5 ×

W m K σ =

e(emissivity): 0 ~ 1, same as absorptivity

e=1: ideal absorber or black body

Chap. 21 The Kinetic theory of Gases

21.1 Molecular model of an ideal gas 21.2 Molar specific heat of an ideal gas 21.3 Adiabatic processes for an ideal gas 21.3 Adiabatic processes for an ideal gas 21.4 The equipartition of energy

21.5 Distribution of molecular speed

21.1 Molecular Model of an Ideal Gas 21.1 Molecular Model of an Ideal Gas

Ideal gas Model Ideal gas Model:

1. The number of molecules in the gas is large, and the average separation between the molecules is large compared with their dimensions.

2. The molecules obey Newton’s laws of motion, but as a whole they move randomly.

3. The molecules interact only by short-range forces during elastic collisions.

4. The molecules make elastic collisions with the walls.

5. The gas under consideration is a pure substance.

If the gases are at low pressures and dilute, this model adequately describes the behavior of real gases.

Assume an ideal gas of N molecules is contained in a cube of edge length d and volume V.

xi xi

xi

xi

m v m v m v

p = −

− (

) = − 2

∆

Momentum change when a molecule of mass m

collide to wall.

p t

F ∆ = ∆

xi xi collision

onmolecule i

v m p t

F

0 ,

− 2

=

∆

=

∆

v

t 2 d

=

∆ F

∆ t = − 2 m

v

d v m d

v m t

v

F_i m ^{xi xi xi}

2 0 2

0 0

2 2

2 = − = −

− ∆

=

d v m d

v F m

F_{i onwall i} ^{xi xi}

2 0 2

0

,  =



 



−

−

=

−

=

By action-reaction law, the x component of force exerted onto wall by a molecule is

Total force on wall is

∑ ∑

=

=

=

= ^N

i

xi N

i

xi v

d m d

v F m

1 0 2 1

2 0

v v

N

i

∑

=

Since number of molecules is large 2 ^{2 x2}

N

xi

N v

∑ v = v

=

N

Since number of molecules is large

0 2

v

d N F = m

1

x i

xi

N v

∑ v =

=

2 2

2 2

zi yi xi

i v v v

v = + +

v

= v

+ v

+ v

= 3 v

 





 



= 

∴ d

v m F N

2 0

3

Pressure on the wall is

2 3 0

1 3

2 0 3

1

2 m v

V N d

v N m

d F A

P F 



 



= 

=

=

=

( )

V v N

V m

P N 



 



= 



 



= ₃²  ₂¹ ₀ ² ₃²

Pressure of an ideal gas is proportional to number of molecules per volume and average translational kinetic energy.

volume and average translational kinetic energy.

◈

◈ Molecular Molecular Interpretation of Interpretation of Temperature Temperature

(

)

1 3

2

N m v

PV =

(

)

1

3

2 m v

T k

B

=

Temperature is the measure of average kinetic energy of molecules.

T k v

m

2

1

=

T k v

m

2

1

=

2 2

2 2

2

v

v

v

3 v

v = + + =

T k v

m T

k v

m

2

1

= & =

Theorem of equipartition of energy

( ^{m v} ) ^{Nk T nRT}

N

K

=

=

=

Total translational kinetic energy of N molecules

M RT m

T v k

v _rms 3 ^B 3

0

2 = = =

Some Example v _rms Values

At a given temperature, lighter molecules move

faster, on the average, than heavier molecules

21.2 Molar Specific Heat of an Ideal Gas 21.2 Molar Specific Heat of an Ideal Gas

Heat depends on the path for the same temperature change.

pressure) constant

(

volume) constant

(

T nC

Q

T nC

Q

P V

∆

=

∆

=

C_V : molar specific heat at constant volume C_P : molar specific heat at constant pressure

Monoatomic ideal gas

nRT T

Nk K

E

=

=

=

Isovolumetric process, W=0

Q = ∆ E

− W = ∆ E

T nC

E =

∆

∆

dT dE C_V = n1 ^int molar specific heat at constant volume :

Monoatomic ideal gas

(

)

dt d n dT

dE

C_V 1n ^int 1 ₂³ ₂³

=

=

=

Isobaric process

T nC

Q =

∆

)

int Q W nC T ( P V

E = + = _P∆ + − ∆

∆

Internal energy is only depends on temperature

T nC

E =

∆

∆

V P T nR P

V V

P ∆ + ∆ = ∆ = ∆

T nR T

nC T

nC

∆ =

∆ − ∆

, nRT

Taking derivative of

PV =

R C

C

−

=

C

= R + C

=

R

67 . 3 1 5 2 / 3

2 /

5 = =

=

= R

R C

C

V

γ

Heating a cylinder of helium Example 21.2 Heating a cylinder of helium Example 21.2

A cylinder contains 3.00mol of helium gas at a temperature of 300 K.

(A) If the gas is heated at constant volume, how much energy must be transferred by heat to the gas for its temperature to increase to 500K?

(B) How much energy must be transferred by heat to the gas at constant pressure to raise the temperature to 500K?

Solution ^(A)

J

K K

K mol J

mol Q

3 1

10 50 . 7

) 300 500

)(

/ 5 . 12 )(

00 . 3 (

×

=

−

⋅

=

(B) Q₂ =nC_P∆T

J

K K

K mol J

mol Q

3 2

10 5 . 12

) 300 500

)(

/ 8 . 20 )(

00 . 3 (

×

=

−

⋅

=

Sample Values of Molar Specific

Heats

21.3 Adiabatic Processes for an Ideal Gas 21.3 Adiabatic Processes for an Ideal Gas

Since it is adiabatic process, Q=0

PdV dT

nC

dE

=

= −

C PdV VdP R

PdV + = −

nC dV dT P

V

−

= nRdT

VdP PdV + =

C

V

dV V

dV C

C C

P dP V

dV

V V

P

 = ( 1 − γ )



 



 −

−

= +

= 0 + V

dV P

dP γ

constant 1

1 nP + nV γ =

constant

γ

=

PV

^TV

^{= constant}

21.4 The Equipartition of Energy 21.4 The Equipartition of Energy

With complex molecules, other contributions to internal energy must be taken into account.

For dumbbell shape molecule, 3 translational, 2 rotational.

rotational.

( ^{k T} ) ^N ( ^{k T} ) ^{NK T nRT}

N

E

= 3

+ 2

=

=

( ^nRT ) ^R

dT d n dT

dE

C

n 1

1

=

=

=

R R

C

C

=

+ =

^{1 . 40}

5 7

2 5 2

7

= =

=

= R

R C

C

V

γ

Including vibrational energy

( ^{k T} ) ^N ( ^{k T} ) ^N ( ^{k T} ) ^nRT

N

E

= 3

+ 2

+ 2

=

( ^nRT ) ^R

dT d n dT

dE

C

n 1

1

=

=

=

But at high temperature it approaches to theoretical value.

◈

◈ A A Hint of Energy Hint of Energy Quantization Quantization

The energies of atoms and molecules are quantized.

The rotational and vibrational energy of a diatomic molecule is quantized The lowest allowed state is the ground state. The vibrational states are separated by larger energy gaps than are rotational states.

21.5 Distribution of Molecular Speeds 21.5 Distribution of Molecular Speeds

How many molecules has a specific energy?

Number density n_V(E) : number of molecules per unit volume having the energy between E ~ E+dE

T k E V

e

n E

n ( ) =

energy between E ~ E+dE

Boltzmann

◈

◈ Maxwell Maxwell--Boltzmann Boltzmann speed distribution speed distribution function function

T k v m B

v

B

e

T v k N m

N

2 / 3

0 ²

4 2 



 



=  π π

Number of molecules having speed between v ~ v+dv is N_V dv.

3k T k T

0

2 3 1.73

m T k m

T v k

v ^B

o B

rms = = =

0

60 . 8 1

m T k m

T

v k ^B

o B

avg = =

π

0

41 . 2 1

m T k m

T

v k ^B

o B

mp = =

mp avg

rms

v v

v > >

∴

Molecular speeds in hydrogen gas Example 21.6

A 0.500mol sample of hydrogen gas is at 300K.

(A) Find the average speed, the rms speed, and the most probable speed of the hydrogen molecules.

Solution (A)

s m

kg

K K

J m

T v_avg k^B

/ 10

78 . 1

) 10

67 . 1 ( 2

) 300 )(

/ 10

38 . 1 60 ( . 1 60

. 1

3

27 23

0

×

=

×

= ×

= ₋

−

s m

kg

K K

J m

T v

k

/ 10

93 . 1

) 10

67 . 1 ( 2

) 300 )(

/ 10

38 . 1 73 (

. 1 73

. 1

3

27 23

0

×

=

×

= ×

=

−

s m

kg

K K

J m

T v

k

/ 10

57 . 1

) 10

67 . 1 ( 2

) 300 )(

/ 10

38 . 1 41 (

. 1 41

. 1

3

27 23

0

×

=

×

= ×

=

(B) v e dv T

k N m

dv

N ^{m v k T}

B v

2 B

/ 2

2 / 3

0 0 ²

4 2 )

1

(  ⁻



 



=  π π

2 / 3

23

27

1 23

2 / 3 0

2 / 3 0

) 300 )(

/ 10

38 . 1 ( 2

) 10

67 . 1 ( 2

) 10

02 . 6 )(

500 . 0 ( 4

4 2 4 2

K K

J

kg

mol mol

T k nN m

T k N m

B A

B



 





×

×

×

=



 



= 



 





−

−

−

π π

π π π π

(B) Find the number of molecules with speeds between 400 m/s and 401 m/s.

3 3

14

23

/ 10

74 . 1

) 300 )(

/ 10

38 . 1 ( 2

m s

K K

J

×

=



 π × ⁻

0645 .

0

) 300 )(

/ 10

38 . 1 ( 2

) / 400 )(

10 67

. 1 ( 2

2 ²³

2 27

2 0

−

=

×

− ×

=

− ₋

−

K K

J

s m kg

T k

v m

B

molecules 10

61 . 2

) / 1 ( )

/ 400 )(

/ 10

74 . 1 (

19

0645 . 0 2 3

3 14

×

=

×

= s m m s e⁻ m s

dv N_v

Chap. 22 Heat engines,

Entropy, and the Second law of thermodynamics

22.1 Heat engines and the second law of thermodynamics

thermodynamics

22.2 Heat pumps and refrigerators

22.3 Reversible and irreversible processes 22.4 The Carnot engine

22.5 Gasoline and diesel engine 22.6 Entropy

22.7 Entropy changes in irreversible processes

22.8 Entropy on a microscopic scale

22.1 Heat Engines and the Second Law of Thermodynamics

22.1 Heat Engines and the Second Law of Thermodynamics

a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work

The working substance absorbs energy by heat from a high

temperature energy reservoir (Q ).

Heat engine:

Kelvin

Internal energy do not change.

temperature energy reservoir (Q_h).

Work is done by the engine

(W_eng). Energy is expelled as heat to a lower temperature reservoir (Q_c).

int

= 0

∆E

W

Q W

Q

E = + = −

∆

c h

eng

Q Q

W = −

Thermal efficiency e :

h c h

c h

h eng

Q Q Q

Q Q

Q

e W − = −

=

≡ 1 ( e < 1 )

Kelvin-Planck form of the second law of thermodynamics

It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the input of energy by heat from a reservoir and the performance of an equal amount of work.

The efficiency of an engine Example 22.1

An engine transfers 2.00×10³ J of energy from a hot reservoir during a cycle and transfers 1.50×10³ J as exhaust to a cold reservoir.

(A) Find the efficiency of the engine (B) How much work does this engine do in one cycle?

Solution ^(A)

(B)

% 0 . 25 . . , 250 . 10 0

00 . 2

10 50 . 1 1

1 ₃

3

e J i

J Q

e Q

h

c =

×

− ×

=

−

=

J

J J

Q Q

W_{eng h c}

2

3 3

10 0 . 5

10 50

. 1 10

00 . 2

×

=

×

−

×

=

−

=

22.2 Heat Pumps and Refrigerators 22.2 Heat Pumps and Refrigerators

Energy is extracted from the cold reservoir, Q_{C .} Energy is transferred to the hot reservoir, Q_{h .} Work must be done on the engine, W.

Clausius statement

It is impossible to construct a cyclical machine

whose sole effect is to transfer energy continuously by heat from one object to another object at a

higher temperature without the input of energy by work.

Coefficient of performance, COP) :

Heat pump can be used as a heater.

W COP ( cooling ) = Qc

W

COP = = Q^h

pump heat

on done work

rature high tempe

at nsferred energy tra

heating) (

Freezing water Example 22.2

A certain refrigerator has a COP of 5.00. When the refrigerator is running, its power input is 500W. A sample of water of mass 500 g and temperature 20.0°C is placed in input is 500W. A sample of water of mass 500 g and temperature 20.0°C is placed in the freezer compartment. How long does it take to freeze the water to ice at 0°C?

Assume all other parts of the refrigerator stay at the same temperature and there is no leakage of energy from the exterior, so the operation of the refrigerator results only in energy being extracted from the water.

Solution

J

kg J C

C kg J kg

L T c m mL

T mc

Q_{e f f}

5

5

10 08 . 2

] / 10 33 . 3 ) 0 . 20 )(

/ 4186 )[(

500 . 0 (

|

×

=

×

−

°

−

°

⋅

=

−

∆

=

−

∆

=

J J W

COP W Q

W

COP Q^{c c 4}

5

10 17 . 00 4

. 5

10 08 .

2 × ∴ = ×

=

=

→

=

W s J P

t W t

W 83.3

500 10 17 .

4 ⁴

× =

=

=

∆

∆ → P=

22.3 Reversible and Irreversible Processes 22.3 Reversible and Irreversible Processes

A reversiblereversible process is one in which every point along some path is an equilibrium state

And one for which the system can be returned to its initial state along the same path

An irreversibleirreversible process does not meet these requirements

All natural processes are known to be irreversible.

In real process, there is no equilibrium. But we can approximate as equilibrium, this process is called quasi-equilibrium or quasi-static process.

22.4 The Carnot Engine 22.4 The Carnot Engine

A heat engine operating in an ideal, reversible cycle (now called a Carnot cycle) between two reservoirs is the most efficient engine possible

This sets an upper limit on the efficiencies of all other engines

h c h

c h

c h

h eng

T T Q

Q Q

Q Q

Q

e W − = − = −

=

= 1 1

Ideal Carnot Q

W

Q’

Ideal Carnot

q

W

q