15. An air bubble rises up from the bottom of a lake, 1000 m deep, to its surface. Find the change in pressur
experiences. Take density of lake water equal to 1.1 x 103 kg/m3
[Ans. 10.7 X 106 N/m2)
Answers
Volume of the air bubble, V
1
=1.0cm
3
=1.0×10
−6
m
3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T
1
=12
o
C=285K
Temperature at the surface of the lake, T
2
=35
o
C=308K
The pressure on the surface of the lake: P
2
=1atm=1×1.103×10
5
Pa
The pressure at the depth of 40 m: P
1
=1atm+dρg
Where,
ρ is the density of water =10
3
kg/m
3
g is the acceleration due to gravity =9.8m/s
2
∴P
1
=1.103×10
5
+40×10
3
×9.8=493300Pa
We have
T
1
P
1
V
1
=
T
2
P
2
V
2
Where, V
2
is the volume of the air bubble when it reaches the surface.
V
2
=
T
1
P
2
P
1
V
1
T
2
=
285×1.013×10
5
493300×1×10
−6
×308
=5.263×10
−6
m
3
or 5.263cm
3
Therefore, when the air bubble reaches the surface, its volume becomes 5.263cm
3
.
Answer:
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