wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An air bubble of volume 1.0cm3 rises from the bottom of a lake 40 m deep at a temperature of 12C. To what volume does it grow when it reaches the surface which is at a temperature of 35C ?

Open in App
Solution

Volume of the air bubble, V1=1.0cm3=1.0×106m3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T1=12oC=285K
Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa
The pressure at the depth of 40 m: P1=1atm+dρg
Where,
ρ is the density of water =103kg/m3
g is the acceleration due to gravity =9.8m/s2
P1=1.103×105+40×103×9.8=493300Pa

We have P1V1T1=P2V2T2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=P1V1T2T1P2
=493300×1×106×308285×1.013×105
=5.263×106m3 or 5.263cm3

Therefore, when the air bubble reaches the surface, its volume becomes 5.263cm3.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
The Equation of State
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon