A gas bubble at the bottom of a lake has a volume of 200cm3 at a temperature of 8°C. This bubble rises and when it reaches the surface where the temperature is 15°C and the pressure is 100kPa, its volume increases to 550cm3. Calculate the pressure on the bubble when it was at the bottom of the lake.
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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
1
P
1
=
T
2
2
P
2
Where, V
2
V
2
=
T
1
P
2
1
2
P
1
=
285×1.013×10
5
493300×1×10
−6
×308
= 5.263× 10
−6
m
3
5.263cm
3
Therefore, when the air bubble reaches the surface, its volume becomes 5.263cm
3
.
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