560
Temperature
P19.44
(a)
Initially the air in the bell satisfies
PV
nRT
i
0b
e
l
l
=
or
PA
n
R
T
i
0
250
.
m
af
=
(1)
When the bell is lowered, the air in the bell satisfies
Px
A
n
R
T
f
bell
m
.
−=
(2)
where
x
is the height the water rises in the bell. Also, the pressure in the bell, once it is
lowered, is equal to the sea water pressure at the depth of the water level in the bell.
PP
g
x
P
g
bell
m
m
=+
−≈+
00
82 3
82 3
ρρ
..
a
f
a
f
(3)
The approximation is good, as
x
<
.
m. Substituting (3) into (2) and substituting
nR
from
(1) into (2),
Pg
x
A
P
V
T
T
f
i
82 3
2 50
+−
=
ρ
m
bell
a
f
.
Using
P
0
5
11
0
1
3
1
0
==
×
atm
Pa
.
and
=×
1025 10
3
k
g
m
3
x
T
T
g
P
x
f
=−
+
F
H
G
I
K
J
L
N
M
M
O
Q
P
P
+
×
×
F
H
G
G
I
K
J
J
L
N
M
M
M
O
Q
P
P
P
=
−
−
1
1
82 3
1
277 15
1
980
823
1013 10
224
1
3
5
1
.
.
.
.
.
.
.
m
m
m
K
293.15 K
kg m
m s
m
Nm
m
32
2
a
f
a
f
ej
e
j
(b)
If the water in the bell is to be expelled, the air pressure in the bell must be raised to the
water pressure at the bottom of the bell. That is,
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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