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?
Solution 1
Volume of the air bubble, V1 = 1.0 cm3 = 1.0 × 10–6 m3
Bubble rises to height, d = 40 m
Temperature at a depth of 40 m, T1 = 12°C = 285 K
Temperature at the surface of the lake, T2 = 35°C = 308 K
The pressure on the surface of the lake:
P2 = 1 atm = 1 ×1.013 × 105 Pa
The pressure at the depth of 40 m:
P1 = 1 atm + dρg
Where,
ρ is the density of water = 103 kg/m3
g is the acceleration due to gravity = 9.8 m/s2
∴P1 = 1.013 × 105 + 40 × 103 × 9.8 = 493300 Pa
We have : `(P_1V_1)/T_1 = (P_2V_2)/T_2`
Where, V2 is the volume of the air bubble when it reaches the surface
`V_2 = (P_1V_1T_2)/(T_1P_2)`
`= ((493300)(1.0 xx 10^(-6))308)/(285 xx 1.013 xx 10^5)`
= 5.263 × 10–6 m3 or 5.263 cm3
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3
Solution 2
Volume of the bubble inside, `V_1 = 1.0 cm^3 = 1xx10^(-6) m^3`
Pressure on the bubble, `P_1` = Pressure of water + Atmospheric pressure
= `pgh + 1.01 xx 10^5 = 1000 xx 9.8 xx 40 + 1.01 xx 10^5`
`=3.92 xx 10^5 + 1.01 xx 10^5 = 4.93 xx 10^5` Pa
Temperature, `T_1 = 12 ^@C = 273 + 12 = 285 K`
Also, pressure outside the lake, `P_2 = 1.01 xx 10^5 n m^(-2)`
Temperature, `T_2 = 35 ^@C = 273 + 35 = 308` K, Volume `V_2` = ?
Now `(P_1V_1)/T_1 = (P_2V_2)/T_2`
`:. V_2 = (P_1V_1)/T_1. T_2/P_2 = (4.93xx10^5xx1xx10^(-6)xx 308)/(285 xx 1.01 xx 10^5) = 5.3 xx 10^(-6) m^(-3)`
