Transcript Review Chapter 18,19 20

Review chapters 18,19,and 20
overflow  V ' Hg  V ' flask   HgV T  3 glassV T  (  Hg  3 glass )V T
 (1.82  104  3  4.0  106 )  250  (105  25)  3.4 cm 3
ice( 10o C )  ice(0o C )  water (0o C )  water(20o C )
Qgained (ice)  Qlost ( water )
micecice T  mice LF  mwcw T "
mice  2220  10  mice  3.33  105  4.0  4186  20
mice  0.94 kg
(100  Tx )
(Tx  0)
R  ks A
 kc A
L
L
k s (100  Tx )  kcTx
 Tx  11.5 C
o
Eint  0
internal energy decrease 
V  0 the volume remains constant   W  0
Q  0  an adiabatic process 
but Eint  Q  W  0  0  0  contradiction
W  nRT ln(
Vf
Vi
)
pV  nRT  n 
pV
i i
RT
Vi  30  103 m 3
Vf
Vf
pV
20
5
3
i i
W
RT ln( )  pV
ln(
)

10

(30

10
)

ln(
)  1.2 kJ
i i
RT
Vi
Vi
30
repalce the irreversable process by isothermal process :
V
nRT ln f
Vf
dQ Q
Vi
S  
 
 nR ln
T
T
T
Vi
5
for Oxygen : SOxygen  2  8.31  ln( )  26.75 J / K
1
5
for Nitrogen : S Nitrogen  2  8.31  ln( )  3.71 J / K
4
S system  SOxygen  S Nitrogen  26.75  3.71  30 J / K
ice(0o C )  water (0o C )
Q  mLF  0.400  3.33  105  133.2 kJ
Q
S  (isothermal )
T
133200

 488 J / K
(273)
W
TL

 1
QH
TH
W
300
 1
1500
700
W  857 J
QL
TL
K

W TH  TL
200 20  273

W
20  20
W  32 J / s
pV   constant ( adiabatic)
c p 25 R 5
  3 
( monoatomic)
cV 2 R 3


pV

p
V
i i
f f
(
Vf
Vi
Vf
) 
pi
pf
3
pi 1
pi 1
pi 1
5
 ( )  ( )  ( 1 )  3  1.9
Vi
pf
pf
3 pi
Eint  0 (cycle)
 Qnet  Wnet
Wab  0 (isochoric )
Wbc  pV  50000  (60  20)  103  2000 J
Wnet  Wab  Wbc  Wca  0  2000  Wca  Qnet  1000 J
Wca  1000 J