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14 THERMODYNAMICS 103

14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

14.2 Macroscopic Description of Matter . . . . . . . . . . . . . . . . . . . . . . . . . 104

14.2.1. State variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

14.2.2. Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

14.2.3. Phase changes, phase diagrams . . . . . . . . . . . . . . . . . . . . . . . 106

14.2.4. Ideal Gas Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

15 Heat, the First Law of Thermodynamics 113

15.1 Work and Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

15.1.1Work done on/by ideal-gas processes . . . . . . . . . . . . . . . . . . . . . 114

15.1.2 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

15.2 The First-Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 115

15.3 Thermal Properties of Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

15.3.1 Heat of transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

15.3.2 The specific heat of gases . . . . . . . . . . . . . . . . . . . . . . . . . . 117

15.3.3 More on adiabatic process . . . . . . . . . . . . . . . . . . . . . . . . . . 118

16 From Micro to Macro, Entropy of the 2nd Law of Thermodynamics 119

16.1 The Kinetic Theory of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

16.1.1 Maxwell speed Distribution . . . . . . . . . . . . . . . . . . . . . . . . . 119

16.1.2 Mean Free Path (MFP): the average distance between collision . . . . . . 121

16.1.3 Microscopic origin of PRESSURE . . . . . . . . . . . . . . . . . . . . . . 121

16.1.4 Microscopic View of TEMPERATURE . . . . . . . . . . . . . . . . . . . 121

16.2 Thermal energy and specific heat . . . . . . . . . . . . . . . . . . . . . . . . . . 121

16.3 Thermal interaction & Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

16.4 Irreversible Processes, Entropy and the 2nd Law of Thermodynamics . . . . . . . 124

17 Heat Engines & Refrigerators 125

17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

17.2 Heat to work and work to heat . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

17.3 Heat engines and refrigerators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

17.4 Ideal-gas engines and refrigerator . . . . . . . . . . . . . . . . . . . . . . . . . . 127

17.5 The Carnot Cycle and the limit of efficiency . . . . . . . . . . . . . . . . . . . . 129

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0 ( # .

( ( ' ( ( A 6 6 ( * ( .

' ( ( (( ( 6 ( (. * *6 6( (

F21Pi2

Pi1

Fe1

Fe2

F12m2

Before collision

m1

(. ( 8 (

( / / *6 ( &

( ( &

6 ( 6

(( 6

/ A A(

6 6

1 (&(

J

! #$

6 ( (( ( 6 ( (

J / K /

( ( (8 M

M/ J < K =M M/ J < K =M

. M/ ( ( 6 (

16( ( 5 ( 6

/ J / K M/ J / K < K =M

/ J / K M/ J / K < K =M

16( ( (( ( 6 ( (

J / K / J / K / K < K =

M K < K =

M

C J M J M M M

J

i

mN

m1

m3

m2

Fi3

Fi1

Fi2

FiN

Fei

m

/ @ $ ( ( (

6

5 J

B ( ( A A( 6

( 6 6 ( *

J

J

( A( 6 *( :/ ( ( 6

( (( ( 6 ( (

8 6 J , (

J ,

6 ( (( A( 6 ( ( 6 ( @ ( (( (

6 ( ( 8

! " ! %C

1' &

y1u

u2

v2

v1

m1

m1 m2

m2

After collision

Before collision

x

J I J

J I J

1'

B ** , * * 1 , = ** *) E* *

1 * , * , = , @ * 2 , 1 2* 4 * 8(

*4 * * 9 * 2 * * 2 * * 2* * 2

*4 ,) 2 * * 829 1 * *) * * *

* * 1 * 1* * 2 1 * , = , * * 1 * , =

+

> 1 * 2 * 0 /1 1 3 * 4 * 1

* , = , * , 4* 1 * 4

3 J I . 3 I 8&9

! #"

( (

6 ( 5( J

2 J 6 ,

1

< K"=

. . 8 ( ( ( ( 6 ( * ( ( *A . (8 (

K 2 J K

.( ( ( ( P 6 ( *A

/ 7( 6 ( 5( . * ( 8 *( 8( 8( <? <$=

((= /

2 J K

6

1

< K"=

!1

< K"=

B 6( ( ( * *A 8 8( C

2 J , 6 %!1

< K"=

:( (( @ 8( . 8 ( ( 6 6 ( ( .

( . ( A( 6 ( ( (

( (( ( *A 5( 8 ( ( 6( (

1

< K"=J

K

1

( 8 ? ( ( ( ( ( ( ( ( (( ( /

( T (( ( ( ( 6 ( *A 6 ( * ( ( ( (

( .( ( *A T .( ( ( 5( ( <*

( ( 6 6 ( ( ( ( (=

"

#

8 ( . 8 .( * 6 ( ( :. . (

6 ( ( ( ( *

3 ( # y

Nr

2r1rr3

Nm

1m3m

m2

x

( ( $ (

<= ( 6 ( (

<= 8( 6 ( (

<= ( 6 ( (

5(

( 6 J K K K K K K

J

. J

J

J

< K K K =

J

J

< K K K=

#

" # -,

C

J K K K J K K K <0=

( (( 6 A *

J K

. T (( ( 6 ( ( 6 ( (

T (( A( 6 (

m

mN

m2

mi

F1iFi1

Fi2F2i

FiN

FNi

Fext,i

1 J K K K < =

B <0=

J

K < K K K K =

K K < K K K K =

K K < K K K K =

K

K K < K K K K =

:( J (

J K K K J

( $ &( ( .( A( 6 ( 8 ( &

( 6 *(. 6 ( ( *8 6 ( ( (

( A 6 6

%A

CM

3N

2N

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x

y

G6 J '

J L K !L #L9 J L #L9

J J #

< ( 6 ( 6 '. *( (

( ((=

" # -

%A

0

0,y0(x )

explode at 0t

t0 ∆+ tm at2

m1 at t0 ∆+ t

x

y

projectilemotion

m0

v0

φ

( A ( (

( (. 6 M

6( ( A . 6

( < = B ( ( 6

( K M

(

6 ( A ( 6 ( ( 9( ( ( 6 (

J L9 J

J < !=L K < ! =L9

J < !=L K H< !=

!IL9

1( ( A (( ( /6 8 ( .

( <= <= ( ( A 88 ( 6 ( ;

( . 6. ( *

:. J K M

< K M= J L K L9 J K

< K M= J

J

!

L K

!

!

L9

" # -!

3 ( # # + ,

%A

1 6 6 ( / ( 6 6

( / 6 6 ( ( ( ((

1

x

y

x

Ω

y

x

y

Ω3

2Ω

1

2

2

R

R

R1R

U ( *( &A ( * &A

( 6 6

J

J

J

J

G ( ( 6 . 5

J

J ,

J

J

(

J

K

J

J,

< =

" # -#

3( 3 * ( ( 6 ( /

J .

3 J .

3

C

J.

3< =

J

K

%A

1 . *( ( &

yφ

φ

y

xR−R

d

E( *( &A ( ( * ( &A

( ( . ( 5

J !

6 ( (

J 3!

. 3 ( ( 6 ( .

J

J

3 !!

.3J

!!

.J

!

.

" # --

%A

y

At final statey

x

ground

x

ground

g

x

2R1 R

0

1 * .(

. .( . (

5 / * ( *( ( *

( *' 6( +( (

5 ?* (2 / 6 (

* ( *(

( &( A( 6

@ ( ( 6 ( (

6

J , K < =

!

J

!

16( ?*

J K

!

J J

!

3 *. # #

( 6 $ ( 8 ( 8(

/( ( 6 ( (

J

J

/ <0!=

( J

J

J

/ <0#=

" # -

* <0!= <0#=

J

J

( 5 ( (( ( ( ( / . ( ( * 5

( ( 6 $ ( *8 6 ( ( 8

8( (

8 6 ( ( ( ( ( * 6 ( (

J

J

/6

J

6 (( A( 6 @ (

J ,

T 8( 6 ( 6 ( 6 (O

%A

mcvvCEmM 1 6( 5

* / * 5

.( 6 (8 ( (

&( ( A( 6

( 8 .

, J K

( J K

, J K K

J

K

" # -$

3 $ # ,

−

Time t

t +∆tTime

vm

system

studiedbeing

system

studiedbeing

m∆+mv+∆vu

m∆

1( K M .

9( ( 6 (

( 8( 6 (

(8 ( ( %(

C.8 ( * (

(( ( 8(

8 (8 ( (

1( ( (( (

<= J

1( ( K M (( (

<K M= J < K M=< K M= K <M=

J KM K M K MM M

J KM K M< = K MM

C . 5

M J < K M= <= J M K M< = K MM

J

M M

J

K

< =

J

. J ( 8( 6 ( ( (8 ( (

:(

MM

M

J , MM , M ,

J J

J K J J J

" # -0

%A

M

momentum conserved

u

v

mass =

1 ( 6( .(

5 ( ( 6 4 *&

( 6 *(

momentum conserved

M

u

v1 '( 9( (

( 6

. 8( 6 9( *( (8 ( ( %(

B *( J ,

J

< =

J

<4=< = J 4< = 6 ( (

< = J

< = 6 ( '(

/ *( ( ( ( '( . ( 6 ( . 6 <

((= ( ( A( 6 ( ( ( ( '( *(

( ( ( 8( 6 (

" # -"

%A

rel

dMdt

M

+ve

g

v

'( ( * 9( ( (

6 .( 8( (8 ( ( '(

J

J

<=

J

%A

+ve

v

g

v

v∆+

∆M

∆+

M

M

M

6 .( 8 ' &

( .( ( 6

J

J

< =

J

<, = J ,

K

J

$

% &

x

z

y

Pr

/ * ( (( 6 * *( 5A A + *8 ( (

6 5A ( ( * / ( 6 ( *( ( A

6 ((

Prφ

y

x

s ( # J !

(

! J#

H( I

-

$ % & ,

x

tatP

t1atPφ

2

1φ

y

2

3' .( ( 8

8( 5 (( 8( 5

* 5

5 J! ! J

M!

M

5 J

M!

MJ!

H( I

E 8 (( ( 5 *

J5<= 5<=

JM5

M

J

M5

MJ5

H( I

1 8( 8(

z

w

x

y

P

Px

y

z

w

( C E.

4 " +.

E J5

J

5 J K J ((

1( J , 5 J 5 J . 5 ( ( 8(

5 J 5 K

$ % &

/6

5 J!

J 5 K ! J 5K

! K

1( J , ! J ! J . ! ( ( (

! J ! K 5 K

4 " , +. ,

y

r ∆s

r∆φ

x

( (8 M ( (( 8(

8 ( M!

6 M , M# J M# J M!

/ ( (( 8(

J

M#

MJ !

J 5

8 (( ( 8 *

J

J 5

J

B 8 ( . '. 6 ( ( .( &

(( ( ( ( (. ( ( <(( (=

/ ( ( ? (

J

J 5

C ( (( (

J K

$ % & !

aR

r

aT

a

y

x

: B 6 ( J , J 6OO

'

%

5 %6.

5(

y

r

F

P

x

1 6 . ( * ( ( /

(? *( ( ( ( 5

+ J

. J (

θ

rFT

F

P

x

y

+ J ( ( 6

#

' % -

x

r F

FT

y

P θ + J

( ( (

%A

τ

L

θ

mg

OPoint

r

+ J + J 1

.( ( ( .

θ

r

τ

OPoint

mg

+ J 1

.( ( ( (.

' %

5 " 2 ! "

z

Fsinθ

F

r

θ

m

y

x

E (

1 ( ( ( ( 6&A *

6 (

B 8 ( < ( ( (=

J

J J

( J

J

1 + J

+ J

: E*( 6 ( 6 ( A 6 ((

5 , J 6 ( ((

+ J ,

& :.( ! . 6 ((

( (

2rT

m1

T1

T2

m21r

2r

P

T1r

y

x

z

l

/. ' * &

( ( 6&A

' ( ( *

(( A 6&A

A( 6

' % $

/( 6 J K K

/( 6 J K

T1r

2rTh2 h11r

θ1

θ2

y

x

z

r2

/( (? ( ( *( 6&A

+ J <

= K <

=

J < = K < =

K< =

K < =

K< =

< "" "" =

J < = K < = K < =

:( ((

J

J <=

< = < = ( (

3( ( ( *(. * 7

J 7 S J 7

( 6 ( ( ? (

J

7 J 7

J <!=

E (&( 6 ? (

J <#=

E*((( ? <!= <#= ( ? <= . *(

J

' % 0

/( (?

+ J . ( A( 6 /? ( * ( 6

(

+ J

+

m1

m22r

1rF2TΣ

F1TΣ

τz

y

x

directionradial

radialdirection

( (( 6 .(

& J !

* ( (. &

( ( (( (

+ J <

= K <

=

J < =

K < =

( (*( ( (?

+ J <

= K <

=

J < K

=

. (( ( 6 ( ( 6 (

( *( ( 6 A

+ J ,

. , J K

6 $ &( *

x

1

m2

mNNr

2r1r

y

m ( (

( 6 ( 5

, J

' % "

/ (( (? ( ( (

+ J

J ,

: * ( ( .( ( * 8 5A (8 ( .(

( ( ( 6 ' *

%A

1 m3

m2

3

4

5

x

y

θ

30o

m

θ J !# '

J #! '

J '

<= B ( 6 ( 6 ( ( *( A (

(8

<*= 6 - : 6 ( . ( ( 6 ( (( *( (

A ( ( ( +( (

(2

1.

<= 5(

, J

J <!# '=<, = K <#! '=<# = K < '=<- =

J # '

E

, J <!# '=<# = K <#! '=<, = K < '=< = J " '

, J <!# '=<- = K <#! '=< = K < '=<, = J 0 '

Chapter 8 Rotational Dynamics 48

(b) θ = sin−1(3/5) ⇒ θ = 37

∴ τz = 4.5 N× 5 m× sin(30 + 37) = 20.7 Nm

But τz = Iαz

∴ αz = τz/I3 = 0.18 rad s−2 in clockwise direction

8.3 Parallel axis theorem

zaxis

M

C.M.

h

C.M.

Iz = ICM +Mh2

Iz = Moment of inertia rotating about z-axis,

ICM = Moment of inertia rotating about the axis

passing through C. M.,

z-axis is parallel to the C. M. axis and h is the

distance between the two parallel axes.

Proof

( , )

rn

xn yn

mnmass &coordinate

z & axisz’slab // to

xn yn( , )

( , )xCM yCM

zz’

h

z x’x

h

y’y

C.M.

For the Iz about the z-axis:

Iz =∑i

mir2i =

∑i

mi(x2i + y2i )

Let (xCM, yCM) be the x, y coordinates of the

C. M. measured from the x, y coordinate sys-

tem.xi = x′

i + xCM

yi = y′i + yCM

' % $,

, J

H< K = K < K =I

J

< K ! K K K ! K =

J

< K =

!

K!

K!

K < K =

J , K

: / A ( ( ( A (( ( ( ( 6

( ( ( A

5 " # ,

, J

( 6 , J

+( 5( (2

q

initial finalq

f(qi )

iq

∆qp=f(q)

p

q

"

"

<8=8 J

<8 =M8

' % $

%A

−L/2

x

xi

x

at the middleaxis through C.M.

+L/2

∆

G6 .( ( 1

(( ( . ( ( .(

( M

( ( (

M J 9M 9 J ( < ' =

, J

M J

#$

#$

J

#$

#$

9

( 9 J "1

, J

#H I

#$#$

1

J

#1

1

-

J

!1

z

L/2 L/2

C. M.

A (

, J , K

J

!1 K

1

!

J

#1

%A

x

z’z

ab

dx

1 6 ( ( (( *(

A ( ( (

(( ( ( ( ( .( .( A

( 6 ( (

J < =: : J (

( 6 ( ( ( ( *( @;

, J

! J

!:

( 6 ( 6 ( ( *( 6

, J , K J

!: K :

' % $!

( : J "<=

, J

!

K

J

!

K

/( ( 6 ( *( 6&A

, J

$

$

, J

$

$

! K

J

!<= K

#H I

$$

J

! K

!

J

!< K =

%A

dθ

z

dm

R

y

xR

θ

1 6 (( *( (

(

( ( (

J <=9 9 J ( <' =

C ( ( 6 ( 8 *

, J

J

9

J

!.

%

J

!.!.

J

%A

1 6 ' (( *( ( (

' % $#

y

R

ring with

x

drrradius and

thickness

z

( .( ('

J

.

6 ( <' =

H.< K = .I

J

.H.

K!.I

.!.

J!

( 6 ( 6 ( (( *( ( (

, J <= J!

/( ( 6 ( 6 ( ' *( ( (

, J, J

!

J

!

-J

!

5& /6.,. # + ,

B * ( ( ( ?* ( 6. ( ( * (5

= J ,

!=+ J , *( 6 6 (

' % $-

V( 6 . '. J ,

+ J , *( ( ( (

. ( (( (? *( ( 6 (2

irrP

Fi

r rPi−

Rigid body

x

z

y

O

P

6 ( ( (

+ J + K + K +

J < = K < = K K < =

J ,

:. . ( (? *( ( (

+& J < & = K < & = K K < & = J H< = K < = K K < =I

8 ( J ,

H& < K K K =I

J ,

6

J , < (( ?* (*=

+ J , *( 8

( (+ J , *( ( ( ?* ( * (*

%A

M

rR2

rM

Mgmg

R1 R2

4L

zx

y

R1r

m

L 1 ( * ?*

J ,

K K K J (

K K < K= J (

/' . (8

K < K= J , <-=

/' ( *( (

+ J K <= K

+ J 1! K

1

- K

1

! J ,

' % $

/6

! J , <=

<-= K <=

! J < K= K

! J K

#

!

J

! K

#

-

J

! J

! K

-

%A

3a

2a

R1

R2

M

ladder mass = m

Of

Mgmg

h

rough

frictionless wall

O

C.M.

a

/ *( ?*

J ,

J

J < K=

B (? *( ( A (

( ( (

K

J

J J K

' % $$

%A

θ

L

O

mgMg

x

y

T

α

θα

M

Fh

Fv

6 * 6

3( ( ( ( ( (

6

K J , <$=

J , <0=

/' ( (? *( ( A ( (

( (

1 < K =1 #

J ,

J< K"!=

< K =

B <$=

J <K= < K"!=

< K ='

B <0=

J< K"!=

< K ='

50 6.,. . .

%A

yT

mg

M

RT

m

B( .(

J

J

/? (

+ J J ,

. , J 6 ' (( *( ( (

J

!

! !

J

<"=

' % $0

( 6 ( ( ( .((

J

J

<"= *

! !

J

C

J

J

!

K !

J

J

!

K !

%A

T1

T1

T2

T2

R

mg

2mg

+ve

2m

m

B( .(

J <=

! J ! <,=

/( (?

+ J J <= <=

J <!=

( <= <,= ( <=

<! != <K= J

! <#=

E*((( <!= ( <#= . *(

! !<=<= J

!

J

! K #

J

"! K # J J

"! K #

' % $"

53 6.,. .

6

J ,

+ J , *( A ( ( 6 ( * *( 6&

(( ( ( 6 (

B ( ( . 6 ((

= 1A 6 (( (

*= (( A . ( (

%A

θ

C.M.

mass = M

θ

Rα

N

fMgsin

Mgcos

θa

.(

$ J

J '' <-=

/( (? (

+ J J , J

!

J

! <=

( <= ( <-=

! J ''

( J '' (

!'' J ''

'' J!

# J

'

J!

#

%A

oω

1 6 6 8

( 8( 5 ( . 6 &

@( 6 / Æ( 6 '( 6( *(.

( ( 6 - ( (

( 8 ( 6 *( 6( (

.(( *

' % $

<= +( ( 8( ( ( 2

<*= +( ( 8 6 2

E(

<= ( (8 ,

f

oω

aCM

N

α

Mg

J -$ J - <$=

(( .( ( + '. ((

9( ( J , 8( 6 ? ( @

J

J J

<0=

<$= <0= 8

- J

- J

<"=

, J

! J

J!

<=

3( 5 * ( 8( ( . 5 (( '. (

5 J 5 K J5 5

1( (

J 5 J5 "

<!,=

( <= ( <!,= . 8

!

J5 "

! J5 <!=

' % 0,

( <$= ( <!= . 8

!- J 5 !- J 5 <!!=

( <"= ( <!!= . 8

!-

-

J 5

J

#5 <!#=

<*= B <"=

J-

J5

#-

%A

M

oR

α

R

T

a

Mg

total mass=

/. (' (( (

. ( .( (

1 ( ( * 8 ( (

(

J

+ J J < = <!-=

+ J , J

! <!=

<!-= <!= 8

< = J

!

J !< =

<!$=

B

J <!0=

E*((( <!0= ( <!$= . *(

J !< =

J!

K !

' % 0

C

J

J!

K !

(

)

7 '1

1 ( 6 ( *( ( ( 8 *

l

vr

O

θ

z

x

y

7 J J / 7 J / 7

J

< /= J

/K /

J <= K /

7

J /

J J +

. ( (( 6 (

:( (( *( 7 + ( *8 ?( (

* 5 .( ( ( (

%A

mr

F=mg

θ

O

θ

b Px

y

1 ( 6 6 ( ( (

/? .( ( ( (

+ J J <.=

8

7 J

<.=

J

0!

( ) 0#

1 + J7

J

<= J

J

B ( 6 ( 6 ( 8 ( 6 7 7 7

/( ( 6 ( (

1 J

7 <* 5(=

1

J

7

J

+ <+ J + K + =

E(( .(( 6 ( (? :/ (*( ( 6 1

1

J

+

/ 6. 5 . ( *(. (

τ

Linear momentum Angular momentum

F// p

p

F

∆t

F// ∆tp//∆

p∆

//p∆

p∆p +∆ p

= F

= ∆t

//∆LL

//∆L//τ

+ L∆L

L

∆L

∆tτ ∆L

=

=

//τ

( ) 0-

1 (( *8(

C.M.

L

∆ L

∆ LL +

r

Mg

+ J < = <( .=

+ JM1

M

M1 . * ( .O

1 1 ( ( (

A ( A . 8 .

7 +. . +.

r

p

ω

z

Oy

x

r’

θ

θ

1 ( (

:. (' ( 6 ( (

7 J / ( ( 5

( ' ( (

< /S= ( 7

8( 5

(

G .( ( . ( 7 5

8( 2

( ) 0

r2

p1

p2

1

1 22

ω

Oy

x

z

= +

r’

r1

(. ?

9( * (

( 7 "" 5

7 ( 5 6 (

* ( *( ( ((

A

1 7 J ,5 < . 8 ( &

( / J 6 (= +(

( ( *(. 7 52

B 8 5 6 ( &

(

7 J 7 J /

J <=

J <5=

< J 5=

( J

7 J 5 J ,5

:(

/ 6&( 6 ( ? ( ,5 / ( (

*( 6 ( ( ( 6 *

6 ( * ( *( ( (( A 1 J ,5 B 1

5 *( ( ( (( 6&A

1 J ,5

6 + J +L& K +L' K +L

+ J1

/( ( 6 , (( + J , 5 . ( * O

( ) 0$

%A

mvR

mg

p= mv

rm m

OR

,

y

=

x

M

m

/( ( 6 ( (

1 J 7 K 7

1 J , 5

K

<(' ( (. ( * K8=

J

!5 K

/( A( (?

+ J + J

+ J1

J

!5 K

J

! K

. ( ( 6 ( ( ( 6 (

( J "

J

!

K

J!

K !

7 ( # +. *.

1 . '.

+ J1

6 ( (( A( (? ( ( ( @ ( #

J , /

( 8

( ) 00

%A

=f

wf

I

wi

Ii

, % ,

, 5 J ,5

1 8(

!=−LLw

sL w

stationaryturn table

1( ( . &

(

1 5 (( (&(*

(

1 J 1( 1 J 1 K <1(=

J 1 1(J 1

1 J !1(

7 $, # + ,)

Object is symmetric about the rotating axis

L∆//τ

Li

Lf + $$ 6 ( 6

M

M1 J + $$M

E *(

( ( ((

( ) 0"

θ

τ

Li

L∆Lf

Object is symmetric about the rotating axis

+ 6 ( 6

M

M1 J +M

E ((

( JM1

1 J+M

1

6 1 6 5A +M /( ( ,

8 (*( ((

%A

= 1 * / . ' ( (*(

= E ((

/ (

L

r

x

CM

Mg

z

Oy

θ

+L dL

dLτ= r Mgφd

θ

L

Lsin

y

x

z

O

( ( 1 ( ( (( A < 5=

%A( (? ( ( + J

8 ( (8 M M1 J +M J M

( ) 0

8

M1 J <1 =M!

M! JM

1 JM

1

:( (( ( A( (? ( ( 6 1 *( ( ( ( 6

1 < + 1 1 1=

/ 8( 1 < ( (( A= 8 . *( ( 8( A < &

=

18 6

5 JM!

MJ

1

*+, & -) !

-)

89 , #

S

θF

x

y

∆

( ( (8 6 M

A (( 6 ( (

(8 ( ( 6 M

+' * ( 6 (

* J

M J M

−ve work done

’

+ve work done

∆S

F

∆S

F

+' * ( (8 &

(8

. 5 *

J*

",

* +, ! & -) ! -) "

89 , , #

E ( ( ( 6 < =

x

i

x∆

xi+1

F(x )i

x0 xN

positivework done

F(x)

negativework done

F

x

8 ( . ( 6

( ( $ (( .( &

( M

( &&( (( ( 8 (&

8 A( (( (

< =

+' ( ( (8

M* < = J < =M

< = J*

/( .' 6 ( ( 6 (

* J

M* J

< =M J

< =

( ? ( ( (( 6 ( 5 .( (8 6 (8 < =

(8 6 (8 < =

* +, ! & -) ! -) "!

%A

F(x)

m

x

equilibrium positionx = 0

F = −kx

x

F = −kx

E 6

*) J

< =

J

J

!< ) =

%A

F(y)

F = −mg−ky

y (+ve)

y

equilibriumposition

y = 0

y (−ve)

−mg

m

( ( 6

J *) J

<=

J

< =

J <) =

!<) =

* +, ! & -) ! -) "#

x

∆r(t + t)

F(r(t))

r(t)r(t) ∆r

y

/9( 6 ( 8 *

<= J <=L&K <=L'

B ( ( 8 *

<= J <=L&K <=L'

( 8 6 <= ( < K M= ( ( (8 M

6 M , 6 A * ( ( ( (8 (( <<==

+' ( ( (8 .( ( M

M* J <<== M< ( (( 6 ( (*(OO=

* J

%A

mg

mTφ

φL

F

x

y

F

φ

1 * ( .( ( 1 ( 1 6 . .

@( ( 6( ( ( ! ( (

8 .( (( (( ( (( 6 * ( B (

.' * ( 6

* +, ! & -) ! -) "-

1.

6 (( 6 @ J , J ,

! J , ! J ,

J (!

S

x

y

φ∆φ

x

∆x+ x

∆

( ( M 6 ! !K M!

M* J M J M*( J M

(

J 1 ! J 1 ! !

M* J (! 1 ! ! J 1 ! !

C

* J

*

1 ! ! J 1< !=

89+

fvviFx

xi fxOx

( 6 (

( ( ( &( 6

( ( 6 (

J

J

J

<=

+' ( * ( 6

* J

J

< ? <==

J

* J

!< =

5 '( ; J

!

* J ; ; J M;

* +, ! & -) ! -) "

6 + (8 % M; % ,

6 + (8 M; ,

: ( 6 8 (8 ( ( *( 8 6 '(

( ( *( ( ( * J M; ( 6

89 9 +

θ

F

d

y

xP

dSφ

r

* 8 (

*( ( (( 6&A .( 6 (

(

+' * ( 6

* J < != J ! J +

. + ( 6&( 6 ( (? *( (

6 ( * ( 6 (

* J

+

6 ( (? ((

* J +

. J*

J +

J +5

m1

r1 v1

r2

v2m2

w ( 6 ( *

( 6 ( & 8 *

; J

!

J

! < 5= J

!

5

/( (( '( 6 ( *

; J

; J

!

5

J

!

5

* +, ! & -) ! -) "$

; J

!,5

, ( ( 6 ( 6 ( * *( ( (( A

& : +

B ( M; J , ; J ;

B ( M; , ; ;

B ( ( ( (. *9( (' (( 6(

%A

( . ( 5 *.

2m1

um

16(

K J <!=

K

J

<#=

B <!=

J

<-=

E*((( <-= ( <#= . (

K

J

K

K

!

J

< K=

!

K <

= J ,

E8 6 ( . 5

J

!

K

J

K

* +, ! & -) ! -) "0

0 ( #

(( 5 6 8(8 6 . ( 8

( 6 P ((

%A 6 8(8 6

= E

!= 8(( 6

#= * 6

%A 6 &8(8 6 & 6(

V( 1 5( 6 8(8 62

5(

1 8(8 6 6 (( 6 ( 8 ( P 6 (

6 ( .' * ( 6 ( ( 6 *( ( (

( *( ( . * ( *( (

:( ((

= <= 8 6 6 ( A( 6( !<= ((

!<= J <=

!= J ,

+' 6 ( < (( ( ( (= @

6

2

1B

A

x

y

2

A Bx

O

1

Chapter 11 Work, and Kinetic Energy and Potential Energy 88

∫path1

F · dr =∫path2

F · dr

∴ Travelling from point A to B, then back to A, the work done is:

WA→B→A = WA→B +WB→A =

∫path1

F · dr +(−∫path2

F · dr)

= 0

11.7 Potential Energy

Consider a particle moves in the influence of a conservative force, which is position de-

pendent, i. e. F (x). Now the particle displaces from xi to xf , potential difference ΔU is

defined:

ΔU = Uf − Ui = −W

where W is the work done by the force during the displacement xi to xf .

Or ΔU = U(xf )− U(xi) = −∫ xf

xi

F (x)dx

If for a particle reference point x0, the potential energy is defined as zero, i. e. U(x0)def= 0.

U(x) = −∫ x

x0

F (x)dx

In particular,

U(x)− U(0) = −∫ x

0

F (x)dx

∴ d

dx[U(x) − U(0)] = − d

dx

∫ x

0

F (x)dx

⇒ dU

dx= −F (x)

* +, ! & -) ! -) "

E

F = −kx

x = 0, U = 0equilibrium position

x

m

J /' ( ?* ( ( * J , ((

<<,= J ,

<< = <<,= J

< =

<< = J

< =

<< = J

!

/<

J

!<! = J J

B 6 8(

y

y = 0, U = 0

y

F = −mg

/' <<,= J ,

<<= <<,= J

<=

<<= J

<= <<= J

/<

J J

4 ( # * /+

M< J < < J * <=

position

i Ufvi vf

finalposition

initial

U

* +, ! & -) ! -) ,

( * J < = ( .' * ( 6 ( 9 6

B 8 (

* J

< = J

!< = J ; ; J M; <$=

E*((( <$= ( <= . 8

< < J ; ;

< K; J < K;

M< J M;

( ( . 8(8 6 A( 6 (

8

8( * (( (( (

8 ( . 8 6 (( < 8( 6

6 *= (( <*( 5A A= ( :. . ( (

(( *( ( 8 ( * ((

x

rCM

rn’rn

nm

CM

y * ( 6 (

/( % 6 ( *

; J

!

<0=

:( ((

J K J K

. J 8( 6 & .( ( ( ( %(; 6

J 8( 6 ( *; ( 6 .( ( ( ( %(; 6

J 8( 6 & .( ( ( ( *; ( 6

B <0= . *(

; J

!

< K = < K = J

!

< K ! K

=

( ( (

< = J

< = J

* +, ! & -) ! -)

1 J <

="

< = J J ,

1 ( ( (

!

J

!

< 5= J

!,5

. 5 ( 8( *( A ( ( ( 6

; J

! K

!,5

( ( /( ( 6 ( 6 ( ((

! ( (( ( .( (( *( ( A ( ( 6

( (( A ( 8

5

U(x)

E

U(x )f

3E4E

1E2E

U(x )g

K(xf )

K(x )g

xcxb xd xexa xf xg

0

x

( A 8(8 6 5 .( (( << =

< = J <

1( J + J ,

* +, ! & -) ! -) !

J + (* ?* & ( ( A (&

6

J (* ?* & ( A 6 (

( (

J ( ?* & ( A 6

<< = K

! J = . = ( 8 ((

%A

6 = J = . ( 8 5

= J ;< += K << += ( J +

= J ;< = K << = ( J

6 ( 6 ( ( = 7( ( . 8 7( *8 6.

= 6 = J = ( ( (( ( J

!= 6 = J = ( ( ( J

#= 6 = J = ( ( ( (. 8 C.8 6 ( 6 (

8 ( ( 8 ( ( ( ( 8

-= 6 = J = ( ( ( %

= 6 = = ( * .

6 << = '. ( * ( .' ( ( ( (

%A

6 ( J , < J ,= J < J ,= J , E << = J

= J

!H <,=I K

!H<,=I J

! J ((

* +, ! & -) ! -) #

1( (

<< = K

! J

!

! K

! J

! . J <= J <=

J

J

J

J

/ 8

( J J

J

J

J J

J

J

.

!

6 J K

.

!

(

J

.

!

J

.

!K

.

!J

J

6 J

.

!

(

J

.

!

J

.

!

K

.

!

J

J

. -)

6 A( 6 ( ( ( ( @ ( 8( 6 *

M; K M< J *

. * ( .' ( ( * ( A( 6

%A

U

gravWgrav

Wspring Wspring

K + UgravK + Ugrav

+Uspring

K

+ Kspring

W

Earth Earth Earth Earth

E( J

M; J

* K*

E( J K

E

M; K M< J

*

E( J K

%(

M; K M< J

*

E( J K

%( K E

M; K M< K

M< J ,

2 + #

M; K M< K M= J *

-

. -)

. = ( ( 6 ( (

( ( % ( .( ( ( 6 (

< ( ( ( *9( ((= ( % ( .( 6 *(.

(

= J ; K <

3' ( ( 6 *

J 6 *

:( (( ( ( (

dx

Fext

CMCMCM

( 6 ( 6 ( ( &

*

J

< ( ( 8 ?( * :

:( (( ( ( ( .'

( 5( T 6 ( ( (

( * =

J J

J

( 6 ( ( 8( 6 (

J

!

! J ; ;

< ;J

=

# J M; 6 ((

T ( 6 <= ?(

. # ( ( 6 ( ( 6

M; K M< K = J*

T 8( 6 <%= ?(

WWW/ ?( ( ( .'& ( 6 ( # ( (

6 ( *( ( ( ( 6 ( ( (( ( 6 (

Chapter 12 Conservation of Energy 96

12.2 Some examples of conservation of energy

1) A sliding block is stopped on a horizontal table with friction.

Center of mass (COM) energy equation: −fsCM = −12Mv2CM

Conservation of energy (COE) equation: Wf = −12Mv2CM +ΔEint,block

2) Pushing a stick on a horiozntal frictionless table.

S

FextS

CMCMCM

Center of mass (COM) energy equation:

FextsCM =1

2Mv2CM

Conservation of energy (COE) equation:

Fexts =1

2Mv2CM +

1

2Iω2

If Fext is acted on center of mass,

s = sCM

Fexts = FextsCM =1

2Mv2CM

3) Ball rolling down an inclined plane without slipping

θ

SCMf

Mg

Center of mass (COM) energy equation:

(Mg sin θ − f)sCM =1

2Mv2CM

Conservation of energy (COE) equation:

Mg sCM sin θ︸︷︷︸Mg acts on CM

=1

2Mv2CM +

1

2Iω2

Notice that the frictional force does no work in the COE eq. as the instantaneous point

of contact between the ball and the plane does not move.

Chapter 12 Conservation of Energy 97

Example

Two men are pushing each other. m2 is pushed away from m1 by straightening their arms

and the force between them is F .

(a) What is the speed of m2 just after losing contact?

(b) What is the change in internal energies for m1 and m2?

frictionless floor

m m 2

m2 is pushedto move forward

1

Answer:

(a) Consider m2 as one system, COM eq. is:

FsCM = ΔKCM =1

2m2v

2CM,m2

where sCM is the displacement of the center of mass of m2.

∴ vCM,m2 =

√2FsCM

m2

(b) For m2, COE equation is

ΔK +ΔEint,m2 = Wext

where

ΔK = ΔKCM = |FsCM|Wext = |Fs| .

Note that s is the total extension of m1’s hand (i.e. the displacement of m2’s hand

when a force F is acting on it, where s = sCM).

∴ ΔEint,m2 = |Fs| − |Fscm|

For m1, COE equation is

ΔEint,m1 = Wext = −|Fs| (F opposite to s)

/.

! # - ;

F21

r21

m1 m2 ( 6 (8 (

6 A * (

J >

L

1F12

r12

m2m ( 6 (8 (

6 A * (

J >

L

; / .#

6 . ( %( ( * (( 6

m

E

RE

M

, J 6 ( %(

, J 6 ( %(

8(( (

J>,

,

J>,

,

"

/.

/< # / "

Fc

mgo

T

mgo

T

Fcw

φ

φα

* ( ( ( 8( 6 ( ( 8(

J

K

3.

J K <=

! !

:( (( J # . # J 7(8 8(

B ( ( ( ?( ! J ,

J K <=

!

J # J 5, #<! J ,= J 5,

3( * ( *(. ( ( ( 8( A

G E 3. ( * ( ( *(. ( ( ( 6&A

J

<. !=

5J < K !=

/. ,,

5

J !K ( !

5 !

J (!K (

( J 5 !

5 !

; # . .#

/

1 6 ((( A( ( 6 ( 6 ( .

(( ( ( (

/ !

1 6 A( 6 ( ( (

& ; +

b

a

rb

F dr

M a

m

r

M< J < < J *

6 (

* J

J

>

J >

/. ,

I J>

3 % * ,M 3 * % ,

9( (( * ( .' * ( 7(( 3 * 3 (

* ( 3 (

L< I < < I * I >

9. . (' I < I <;< . <;< I ,

<;< <;< I >

<;< I >

I*

%

A ( 7 ( 3 5@ ( 3 3 ( %( 2 ( (

( 3 ( %(: 7(( 3 5 ( ( * 3 ( ( (7

( 5(

! J

>,

I ,

0 + #

3m1 r13

r23r12

m2

m

< I >

J>

J>

% > ( (' ( ( ( ( ( 5(

= I <

/. ,!

3 /+ #

ω

m

M

r

(( *( (

< J >

; J

! J

!<5= J

!5

6 ( 8(( 6 8 ( (( 6

>

J 5 >

J 5

; J

!

>

= J ; K < J

!

>

>

J >

!

Chapter 14 THERMODYNAMICS

14.1 Introduction

Thermodynamics is the science of energy & energy-conversion.

• Macroscopic description of matter

State variables, temperature and the zeroth law of thermodynamics, phase change,

ideal gas processes

• Heat, and the First law of thermodynamics

Heat as energy transfer, heat & work for ideal gas processes, 1st law of

thermodynamics, thermal property of matter

• From MICRO to MACRO, entropy, and the 2nd law of thermodynamics

Molecular properties of gases, thermal energy and specific heat, the concept of

entropy, 2nd law of thermodynamics

• Heat engine & refrigerator

Heat to work and vice versa, ideal gas engines, the Carnot cycle, limit of efficiency

(perfect vs. real engine)

Chapter 14 Thermodynamics 104

14.2 Macroscopic Description of Matter

Thermodynamics deals with MACROSCOPIC systems, rather than the “particles”. It is all

about energy and energy conversion, especially that of converting HEAT ENERGY into

MECHANICAL WORK. (So, the word thermo-dynamics)

14.2.1. State variables

The set of parameters used to characterize or describe the “state” of a macro-system.

e.g., mass, volume, pressure, temperature, thermal energy, entropy… (are not all independent,

however)

Change of state variable: f ix x x∆ = −

• A system is in THERMAL EQUILIBRIUM if the state variables stay constant with

time. Two or more systems are in thermal equilibrium with each other when their

respective variables are unchanged upon making thermal constant.

• If system A and B are each in thermal equilibrium with a third system, then A and B

are in thermal equilibrium with each other. (zeroth law)

• Mass density: MVρ = , Number density: N

V

Atomic/Molecular mass:

12

1

2

( ) 12( ) 1.0078 1( ) 32

m Cm Hm O

µ

µ µµ

=

= ≈≈

, µ --atomic mass unit

Moles and Molar Mass: 231 6.02 10mol ≈ × basic particles

23 16.02 10AN mol−= × --Avogadro’s number

The number of moles in a substance containing N basic particles is A

Nn N= .


• The number of atoms in a system of mass M (in kg) is found by MNm

= , m is the

atomic mass.

• The molar mass is the mass in grams of 1 mol. of substance. 12 12( )molgM C mol= ,

212( )mol

gM O mol≈ .

• For a system of mass M consisting of atoms/molecules with molar mass molM , the

number of moles of the atoms/molecules in system is molM

Mn = .

Example

The 12C atoms weigh 12 g by definition, so the mass of one 12C atom is

12 2612( ) 1.993 10A

gm C kgN−= = × .

On the other hand, we also defined that 12( ) 12m C µ= , so 12

27( )1 1.661 1012

m C kgµ −= = × .

For any other substance, one way to find its atomic mass is, e.g.

27 262( ) 32 32 1.661 10 5.315 10m O kg kgµ − −= = × × = ×

14.2.2. Temperature

A measure of system’s THERMAL ENERGY, the kinetic and potential energy of

atoms/molecules in a system as they vibrate and/or move around.

Two systems that are in thermal equilibrium have the same temperature. In other words, the

temperature of a system is a property that determines whether or not a system is in thermal

equilibrium with other systems!

• Temperatures scales

1. Kelvin scale ( K): 273.16trT K= , ( ) 0T K ≥


2. Celsius and Fahrenheit: 273.15cT T= − , 9 325F cT T= +

The temperature in Kelvin scale is adopted as fundamental in physics! It is sometimes

called the absolute temperature scale. At the absolute zero temperature ( 0T K= ),

0thE = .

• Measuring the temperature – thermometers

Use the properties of a substance that vary with temperature. For example the pressure

of a gas at constant volume, the electrical resistance of a wire, the length of a metal

strip, the color of a lamp filament, etc.

Let X be a parameter property that is linearly dependent on T , XT * α= . At the

triple point of water, 273.16 trK Xα= , from which, 273.16trXα = is found. Then

at any other temperature, * (273.16 )tr

XT KX

= .

The pressure in a constant-volume gas thermometer extrapolates to zero at

0 273oT C= − . This is the basis for the concept of absolute zero.

14.2.3. Phase changes, phase diagrams

A substance may change phase, e.g. from solid to liquid by heating. For example, water

solidify (freeze) at the freezing point, but vaporize (boil) at the boiling point.

At the freezing (melting) point, the solid phase (ice) and the liquid phase (water) are in Phase


equilibrium, meaning that any amount of solid can coexist with any amount of liquid.

Similarly, at the boiling (condensation) point, the liquid and vapor phases of water are in

phase equilibrium.

Note that only at those boiling and melting points that phase equilibrium can be maintained!

• A phase diagram is a diagram showing how the phases & phase changes of a

substance vary with both temperature and pressure.

Examples

The following figures show the phase diagrams of water & 2CO . Three phases of matter are

the solid, liquid and gas.


The right figure below shows the temperature as a function of time as water is transformed

from solid to liquid to gas.

14.2.4. Ideal Gas Processes

• Ideal Gas

The potential-energy diagram for the interaction of two atoms is shown in figure. Solid and

liquid are systems where the atomic separation is close to eqr . A gas is a system where the

average spacing of atoms is much greater than eqr , so atoms are usually not interacting.


• An idealized hard-sphere model of the interaction potential energy of two atoms

A gas of atoms obeying such an interacting potential is called Ideal Gas. The ideal gas model

can be good approximation of a real gas when its density is low and its temperature is well

above the condensation point.

• Molecular speed

Atoms in a gas are in random motion at 0T K> . The distribution of speed is

outlined as follows.

(a) The most probable speed 2

PkTvm

=

(b) The average speed 8av

kTvmπ

=

(c) The root-mean-square speed 3

rmskTvm

=


A histogram showing the distribution of speeds in a beam of 2N molecules at 20oT C= .

• The Ideal Gas Law and Ideal Gas Processes

The ideal gas law (thermal equilibrium): BpV nRT Nk T= =

Universal gas constant: 8.31 JR mol K= ⋅

Boltzmann's constant: 231.38 10BA

R Jk KN−= = ×

In a sealed container, we have pV nRT

= = constant, and the number density of atoms in a

gas is given by Tk

pVN

B

= .

An ideal gas process is the means by which the gas changes from one state to another. The

p-V diagram is a graph of PRESSURE against VOLUME. A point on the p-V diagram

represents a unique state of a (sealed) gas.


A quasi-static process is one that when the system changes state from, say 1 to 2, it is done so

slowly that the system remains (approximately) at thermal equilibrium. Thus, a quasi-static

process is reversible.

(a) Constant-Volume (ISOCHORIC) process: if VV =

(b) Constant-Pressure (ISOBARIC) process: if PP =

(c) Constant-Temperature (ISOTHERMAL) process: if TT =

(d) ADIABATIC (no heat transfer) process: 0Q =

Example

A gas at 2.0 atm pressure and a temperature of 200o C is first expanded isothermally until

its volume has doubled. It then undergoes an isobaric compression until its original volume is

restored. Find the final temperature and pressure.


Solutions:

For process 1 2→ , ttanconsTT == 12 . Hence, we have

1122 VPVP = , or. 1 12 1

2

1.02

V PP P atm

V= = =

For process 2 3→ , 3 2 1.0P P atm= = .

Since 2

2

3

3

TV

TV

= , we have

3 13 2 1 2

2 1

1 1 (200 273.16) 236.5 36.52 2 2

oV VT T T T K CV V

= = = = × + = = − .

Chapter 15 Heat, the First Law of Thermodynamics

15.1 Work and Heat

• Work is the energy transferred to or from a system due to force acting on it over a

distance.

• Heat is the energy that flows between a system and its environment due to a

temperature difference between them.

• Energy conservation says: sys mech th extE E E W Q∆ = ∆ + ∆ = + , (Note: not W∆ &

Q∆ !!) where thmechsys EEE += is the total energy of the system

UKEmech += is the mechanical energy associated with the motion of the system as a whole

(macroscopic E ), K is kinetic energy and U is potential energy.

micromicroth UKE += is the energy associated with the motion of atoms/molecules within the

system (microscopic E ). It is one form of the “internal” energy.

extW is the work done by external forces (environment). Q is the heat transferred to the

system from its environment. Work and heat are the energies transferred between systems and

the environment. They are NOT the state variables or state functions! Heat is transferred by


114

one of the following three mechanisms:

(a) Thermal conduction xTkAH∆∆

=

(b) Convection

(c) Radiation 4TI σ=

15.1.1 Work done on/by ideal-gas processes

The work done on a gas is defined by f

i

V

V

W pdV= − ∫ . It is the negative of the area under the curve

between iV and fV !

(a) Isochoric process ( constV = ): 0=W

(b) Isobaric process ( constp = ): VpW ∆−= , if VVV −=∆

(c) Isothermal process ( constT = , constpV = ):

)ln()ln()ln(i

fff

i

fii

i

fV

V VV

VpVV

VpVV

nRTdVV

nRTWf

i

−=−=−=−= ∫

(d) Adibatic process ( 0=Q , constpV =γ , γ : ratio of specific heats)

1 1 1 1( ) [( ) 1] ( )1 1 1

f f

i i

V Vi i i i i i i

i i i f f f i ifV V

p V p V p V VdVW dV p V V V P V PVVV V

γ γγ γ γ γ

γ γ γ γ γ− − −= − = − = − − = − = −

− − −∫ ∫


115

The above expression equals to vnC T∆ .

The work done during an ideal gas process depends on the path followed through the p-V

diagram! The work done during these two ideal-gas processes is not the same.

15.1.2 Heat

Heat is the energy transfer, it is process-specific.

One needs to distinguish heat from thermal energy and temperature.

• THERMAL ENERGY is a form of energy of the system.

• TEMPERATURE is a measure of “hotness” of the system. It is related to the thermal

energy per molecule. It is also a state variable.

• HEAT is the energy transferred between the system and its environment as they

interact. It is NOT a particular form of energy, nor a state variable.

15.2 The First-Law of Thermodynamics

It is about the conservation of energy of a thermodynamic system. A thermodynamic system

is one where the internal energy is the only type of energy the system may have. So, we have

( 0=∆ mechE )

QWE +=∆ int

If the change of internal energy is solely in the form of thermal energy, then the above


116

statement becomes QWEth +=∆ .

15.3 Thermal Properties of Matter

Here, we look at the consequences of thE∆ to a system, be the thermal energy change due to

work W or heat Q.

• Temperature change: TMcEth ∆=∆ , M is the mass, and c is specific heat. Specific

heat is the amount of energy that raises the temperature of 1 kg of a substance by 1 K.

It is material specific.

If QEth =∆ (i.e., 0=W ), then TMcQ ∆= .

Molar specific heat is the amount of energy that raises the temperature of 1 mol. of a

substance by 1 K. TnCQ ∆= , n is the number of moles of the substance and C is

molar specific heat. For most elemental solids, KmolJC ⋅25~ .

• Phase change, as characterized by a thermal energy change without changing the

temperature. (Solid↔Liquid↔Gas)

Q = ML, where M is the mass and L is the heat of transformation.

15.3.1 Heat of transformation

Heat of transformation is the amount of heat energy that causes 1 kg of a substance to


117

undergo a phase change. The heat of transformation for a phase change between a solid and a

liquid is called Heat of Fusion ( fL ).The heat of transformation for a phase change between a

liquid and a gas is called Heat of Vaporization ( vL ).

⎩⎨⎧

±

±=

v

f

MLML

Q .

15.3.2 The specific heat of gases

For a gas, one needs to distinguish between the molar specific heat at constant volume vC

and the molar specific heat at constant pressure pC , where

TnCQ v∆= (temperature change at constant volume “A”)

TnCQ p∆= (temperature change at constant pressure “B”)

The thermal energy of a gas is associated with temperature, so the change of thermal energy

thE∆ will be the same for any two processes that have the same T∆ . Similarly, any two

processes that change the thermal energy of the gas by thE∆ will cause the same

temperature change T∆ . Process A and B have the same T∆ and the same thE∆ , but they

require different amounts of heat.

For process “A”, TnCQQWE vAth ∆==+=∆ )(


118

For process “B”, TnCVpE pBth ∆+∆−=∆ )(

So, TnCVpTnC pv ∆+∆−=∆

Since nRTpV = , TnRnRTVppV ∆=∆=∆=∆ )()(

Thus, TnCTnRTnC pv ∆+∆−=∆

and RCC vp += , TnCE vth ∆=∆ .

Remarks:

1. The change in thermal energy when temperature changes by T∆ is the same for any

processes, i.e., TnCE vth ∆=∆ .

2. The heat required to bring about the temperature change depends on the process itself. It is

different for different processes. (Heat depends on path, just like the work does!)

15.3.3 More on adiabatic process

As TnCWQE vth ∆=+=∆ , so for an adiabatic process ( 0=Q ), TnCW v∆= .

As that dWdEth =

VdVnRTpdVdTnCv −=−= or

VdV

CR

TdT

V

−=

Note that 1−=−

= γv

vP

v CCC

CR , where

v

p

CC

=γ , the specific heat ratio (>1).

So ∫∫ −−=f

i

f

i

V

V

T

T VdV

TdT )1(γ

1)ln()ln( −= γ

f

i

i

f

VV

TT

or 1)( −= γ

f

i

i

f

VV

TT

For ideal gas, nRpVT = , so .constVpVp iiff == γγ .

Chapter 16 From Micro to Macro, Entropy of the 2nd Law of Thermodynamics 16.1 The Kinetic Theory of Gases A gas consists of a vast number of atoms/molecules ceaselessly colliding with each other and

the walls of their container. For an ideal gas,

(a) these atoms/molecules (referred to as “particles”) are in RANDOM motion and obey

Newton’s laws of motion;

(b) the total number of the particles is “large” and yet the volume occupied by these

particles is negligibly small comparing to the volume the gas occupies;

(c) no force acts on a molecule except during collision;

(d) all collisions are elastic and of negligible duration.

16.1.1 Maxwell speed Distribution

Particles in a gas move randomly with different speeds. The distribution of speeds is

described by the so-called Maxwell speed distribution:

kTmv

ev)kT

m(N)v(N 2223 2

24 −=

ππ

0

( )N N v dv∞

= ∫

Chapter 16 From Micro to Macro, Entropy of the 2nd Law of Thermodynamics

120

Note as 2

21 mvE = , one may derive the Distribution molecule energies, the so-called

Maxwell-Boltzman energy distribution:

kTE

eE)kT(

N)E(N−

= 21

23

12π

dE)E(NN ∫∞

=0

• The most probable speed ( 0=dvdN ):

mkTv p

2=

• The average speed mkTdv)v(vN

Nvavg π

810

== ∫∞

• The root-mean-square (RMS) speed mkT)v(v avgrms

32 ==

mkTdv)v(Nv

N)v( avg

310

22 == ∫∞


121

16.1.2 Mean Free Path (MFP): the average distance between collision

pkT

d)vN(d 22 2

12

1ππ

λ ==

16.1.3 Microscopic origin of PRESSURE

The pressure of a gas exerted on the walls of a container in due to the steady rain of a vast

number of atoms/molecules striking the walls.

• The force (averaged) exerted to the wall by a single atom upon collision is

coll

avgavg t

mvF

∆2

=

• The total force due to collision of all atoms/molecules

A)v(mVNA)v(m

VNFNF avgavgxavgcollnet

22

3===

• The pressure then is VNkTmv

VNp rms == 2

3

16.1.4 Microscopic View of TEMPERATURE

The average translational kinetic energy per molecule is

22

21

21

rmsavgavg mv)v(m ==ε

kTmkTv avgrms 2

33=∴= ε∵

So, temperature is simply a measure of translational kinetic energy per molecule!

16.2 Thermal energy and specific heat

Thermal energy th micro microE K U= +

• Monatomic Gases, 0microU =

3 32 2th micro avgE K N NkT nRTε= = = =

So, 3 12.52V

JC R mol K= = ⋅

For a monatomic gas, the energy is exclusively translational. As the translational motions are


122

INDEPENDENT on the three space-coordinate axes, the energy is stored in 3 independent

modes (degree of freedom).

The thermal energy of a system of particles is equally divided among all the possible energy

modes. For a system of N particles at temperature T , the energy stored in each mode (or

in each degree of freedom) is NkT21 or, in terms of moles, nRT

21 .

A monatomic gas has 3 degree of freedoms, so nRTEth 213×= .

Remarks: Solids: an atom in solid has 3 degrees of freedom associated with the vibrational

kinetic energy, another 3 modes associated with the stretch/compress of the bonds (potential

energy), so nRTnRTEth 3216 =×= , Kmol

JRC ⋅== 0.253 .

• Diatomic molecules

(a) 3 modes for translational kinetic energy

(b) 3 modes for rotational degrees of freedom

(c) 2 modes for stretching/compressing bonds

So, it would suggest nRTEth 218×= , and RC 4= which is inconsistent with

experiment. The reason is due to the “quantum effect”, which prevent 3 modes from

being active. So, for diatomic molecules

nRTEth 25

= , KmolJRC ⋅== 8.20

25


123

16.3 Thermal interaction & Heat

Atoms in system 1 have higher kinetic energy than those in system 2 ( 21 TT > ), upon collision

between a “fast” atom in 1 and a slow atom in 2, energy is transferred. So, heat is the energy

transferred via collisions between the more energetic (warmer) atoms on one side and the less

energetic (cooler) atoms on the other.

At thermal equilibrium, there is no net energy transfer, or atoms at both sides have the same

average translational kinetic energy.

1 2( ) ( )avg avgε ε=

32avg kTε =∵

1 2 fT T T∴ = = (at thermal equilibrium)

Two thermally interacting systems reach a common final temperature by exchanging energy

via collisions, until atoms on each side have, on average, equal translational kinetic energies!

avgth NE ε=∵

Therefore, at equilibrium, 212

2

1

1

NNE

NE

NE totff

+== , iitot EEE 21 +=

then 11

1 2f tot

NE EN N

=+

, totf ENN

NE21

22 +

= .

Hence, the heat flowed from “1” to “2” is 2 2 2f iQ E E E= ∆ = − .


124

16.4 Irreversible Processes, Entropy and the 2nd Law of

Thermodynamics An irreversible process is one that happens only in one direction.

e.g., heat is transferred from the warmer side to the cooler side, but not the other way.

Why? After all, heat transfer is by collision, a micro-process that is irreversible!

It lies with the “probability”. Reversible microscopic events lend to irreversible macroscopic

behavior because some macroscopic states are vastly more probable than others.

The equilibrium is actually the MOST probable state in which to be!

• Entropy is a variable that measures the “orderedness” of a system.

It is a measure of the probability that a macroscopic state will occur.

TQS =∆ (reversible, isothermal)

• The 2nd Law of Thermodynamics

The entropy of an isolated system never decreases. It either increases until the system

reaches equilibrium, or if the system is in equilibrium, stays the same. An isolated

system never spontaneously generates order out of randomness.

0≥∆S

Chapter 17 Heat Engines & Refrigerators

17.1 Introduction

• Heat engine is a device that uses a cyclical process to transform heat energy

into work.

• Refrigerator is a device that uses work to move heat from a cold object to a hot

object.

17.2 Heat to work and work to heat

• Energy-transfer-diagram

(1) Energy reservoir is an object or part of environment so large that its temperature

and thermal energy do not change when heat is transferred between the system

Chapter 17 Heat Engines & Refrigerators 126

and the reservoir.

(2) A reservoir at a higher temperature than the system is called Hot reservoir.

(3) A reservoir at a lower temperature than the system is called Cold reservoir.

The First Law of Thermodynamics: thCH EWQQQ ∆+=−=

• Work to heat is easy and straight forward, which can have 100% efficiency.

• Heat to work is difficult, the 2nd law makes the transfer efficiency <100%.

The reason lies on the fact that for a practical device that transform heat into work

must return to its initial state at the end of the process and be ready for continued use!

17.3 Heat engines and refrigerators

For any heat engine, the closed-cycle device periodically return to its initial

conditions, so for a full cycle. 0)( =∆ netthE

Therefore, . CHnet QQQW −==

The engine’s THERMAL EFFICIENCY H

C

H QQ

QW

−== 1η .

For a refrigerator, the close-cycle device uses external work to extract heat from a

cold reservoir and exhaust heat to a hot reservoir. Again, 0=∆ thE . So,

WQQ CH += .


The refrigerator Coefficient of Performance WQK C= .

The 2nd law suggests that there is no perfect refrigerator with ∞=K ! It also

suggests there is no perfect heat engine with 1=η .

For the 1st statement, 0=⇒∞= WK , then it implies the refrigerator spontaneously

draw heat from cold reservoir to hot reservoir!

For the 2nd statement, if 1=η , then WQH =1 , using this work, we draw heat from a

cold reservoir, so , which equivalently drawing heat from “cold” to

“hot” spontaneously.

WQQ CH += 22

17.4 Ideal-gas engines and refrigerator

We use a gas as the working substance, and the close-cycle is represented by a

closed-cycle trajectory in the p V− diagram.

The net work done for such a closed-cycle is simply the area inside the closed curve.


Summary of ideal gas processes

Process Gas Law Work W Heat Q Thermal Energy

Isochoric f

f

i

iT

pT

p = 0 TnCV∆ QEth =∆

Isobaric f

f

i

iT

VT

V = p V− ∆ pnC T∆ thE Q W∆ = +

Isothermal ffii VpVp = ln( )

ln( )

f

i

f

i

VnRT VVpV V

−

− Q W= − 0=∆ thE

Adiabatic 11 −− =

=γγ

γγ

ffii

ffii

VTVT

VpVp

( )1

f f i i

V

p V p V

nC Tγ

−−

∆ 0 thE W∆ =

Any f

ff

i

iiT

VpT

Vp = −(area under PV curve) TnCE Vth ∆=∆

Properties of monatomic and diatomic gases

Monatomic Diatomic

thE nRT23 nRT

25

VC R23 R

25

pC R25 R

27

γ 67.135= 4.1

57=

Note: For refrigerator, as the heat always transfers from hotter object to a colder

object, it has to use the ADIBATIC process to lower the temperature of the

gas to below and to increase it to . CT HT


with only their direction changed.

ficiency. Otherwise, it again violate

r can have a coefficient of performance larger than that of a

A perfectly reversible engine must use only two types of processes: (1) Frictionless mechanical interactions with no heat transfer,

17.5 The Carnot Cycle and the limit of efficiency

A perfectly reversible engine is one that can be operated as either a heat engine or a

refrigerator between the same two energy reservoirs and with the same energy transfer,

A perfectly reversible engine has MAXIMUM ef

the 2nd law.

Similarly, no refrigerato

perfectly reversible refrigerator.

0=Q (2) Thermal interactions in which heat is transferred in an isothermal process,

. 0E∆ = th

The engine that uses only there two types of processes is called CARNOT engine


Carnot engine has maximum thermal efficiency maxη , and it operated as a refrigerahas the maximum coefficient of performance maxK

• The Carnot cycle

tor

d find the

.

We now analyze the Carnot cycle an maxη .

The Carnot cycle is an ideal gas cycle that consists of two adiabatic ( ) and two

0=Q

isothermal ( 0E∆ = ) processes. th

H

CQQ

−= 1η

)ln(2

112 V

VnRTQQ CC ==

)ln(3

434 V

VnRTQQ HH ==

1 12 3C HT V T Vγ γ− −= , 1 1

1 4C HT V T Vγ γ− −= . For adiabatic process,

So, 1 4

2 3V V= and

V V

H

C

TT

−= 1maxη .

CCarnot

H C

TK

T T=

−. For refrigerator,

· 17.3 heat engines and refrigerators.126 17.4 ideal-gas engines and refrigerator .127 17.5 the...

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