An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12°C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C.
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We have,
here,
Now,
Therefore, when the air bubble reaches the surface, its volume becomes
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Your Answer:-
Volume of the air bubble, V1=
Bubble rises to height, d=
Temperature at a depth of 40m T1 = 12°C=285K
Temperature at the surface of the lake, T2=35°C=308K
The pressure on the surface of the lake: P2= 1atm =1×1.03 ×10^5 m
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 T1P1V1=T2P2V2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=T1P2P1V1T2
=285×1.013×105493300×1×10−6×308
=5.263×10−6m3 or 5.263cm3
Therefore, when the air bubble reaches the surface, its volume becomes 5.263cm3.
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