· 17.3 heat engines and refrigerators.126 17.4 ideal-gas engines and refrigerator .127 17.5 the...
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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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Before collision
x
J I J
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B ** , * * 1 , = ** *) E* *
1 * , * , = , @ * 2 , 1 2* 4 * 8(
*4 * * 9 * 2 * * 2 * * 2* * 2
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t +∆tTime
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m∆+mv+∆vu
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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
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< =
J
. J ( 8( 6 ( ( (8 ( (
:(
MM
M
J , MM , M ,
J J
J K J J J
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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
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8 ( M!
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+ J
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θ
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P
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y
+ J ( ( 6
#
' % -
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r F
FT
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P θ + J
( ( (
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L
θ
mg
OPoint
r
+ J + J 1
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θ
r
τ
OPoint
mg
+ J 1
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' %
5 " 2 ! "
z
Fsinθ
F
r
θ
m
y
x
E (
1 ( ( ( ( 6&A *
6 (
B 8 ( < ( ( (=
J
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( J
J
1 + J
+ J
: E*( 6 ( 6 ( A 6 ((
5 , J 6 ( ((
+ J ,
& :.( ! . 6 ((
( (
2rT
m1
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T2
m21r
2r
P
T1r
y
x
z
l
/. ' * &
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' ( ( *
(( 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 <#=
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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
$$
J
! K
!
J
!< K =
%A
dθ
z
dm
R
y
xR
θ
1 6 (( *( (
(
( ( (
J <=9 9 J ( <' =
C ( ( 6 ( 8 *
, J
J
9
J
!.
%
J
!.!.
J
%A
1 6 ' (( *( ( (
' % $#
y
R
ring with
x
drrradius and
thickness
z
( .( ('
J
.
6 ( <' =
H.< K = .I
J
.H.
K!.I
.!.
J!
( 6 ( 6 ( (( *( ( (
, J <= J!
/( ( 6 ( 6 ( ' *( ( (
, J, J
!
J
!
-J
!
5& /6.,. # + ,
B * ( ( ( ?* ( 6. ( ( * (5
= J ,
!=+ J , *( 6 6 (
' % $-
V( 6 . '. J ,
+ J , *( ( ( (
. ( (( (? *( ( 6 (2
irrP
Fi
r rPi−
Rigid body
x
z
y
O
P
6 ( ( (
+ J + K + K +
J < = K < = K K < =
J ,
:. . ( (? *( ( (
+& J < & = K < & = K K < & = J H< = K < = K K < =I
8 ( J ,
H& < K K K =I
J ,
6
J , < (( ?* (*=
+ J , *( 8
( (+ J , *( ( ( ?* ( * (*
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zx
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m
L 1 ( * ?*
J ,
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K K < K= J (
/' . (8
K < K= J , <-=
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+ J K <= K
+ J 1! K
1
- K
1
! J ,
' % $
/6
! J , <=
<-= K <=
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! J K
#
!
J
! K
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-
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3a
2a
R1
R2
M
ladder mass = m
Of
Mgmg
h
rough
frictionless wall
O
C.M.
a
/ *( ?*
J ,
J
J < K=
B (? *( ( A (
( ( (
K
J
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' % $$
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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 !
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T2
T2
R
mg
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+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= .
Chapter 14 Thermodynamics 105
• 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 ≥
Chapter 14 Thermodynamics 106
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
Chapter 14 Thermodynamics 107
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.
Chapter 14 Thermodynamics 108
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.
Chapter 14 Thermodynamics 109
• 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
=
Chapter 14 Thermodynamics 110
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.
Chapter 14 Thermodynamics 111
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.
Chapter 14 Thermodynamics 112
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
Chapter 15 Heat, the First Law of Thermodynamics
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− − −= − = − = − − = − = −
− − −∫ ∫
Chapter 15 Heat, the First Law of Thermodynamics
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
Chapter 15 Heat, the First Law of Thermodynamics
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
Chapter 15 Heat, the First Law of Thermodynamics
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 ∆==+=∆ )(
Chapter 15 Heat, the First Law of Thermodynamics
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 == ∫∞
Chapter 16 From Micro to Macro, Entropy of the 2nd Law of Thermodynamics
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
Chapter 16 From Micro to Macro, Entropy of the 2nd Law of Thermodynamics
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
Chapter 16 From Micro to Macro, Entropy of the 2nd Law of Thermodynamics
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= ∆ = − .
Chapter 16 From Micro to Macro, Entropy of the 2nd Law of Thermodynamics
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 += .
Chapter 17 Heat Engines & Refrigerators 127
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.
Chapter 17 Heat Engines & Refrigerators 128
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
Chapter 17 Heat Engines & Refrigerators 129
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
Chapter 17 Heat Engines & Refrigerators 130
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,
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