An air bubble of volume `1.0 cm^(3)` 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`. ? Given `1 atm = 1.01 xx 10^(5) Pa`.

Explanation

Volume of the air bubble, ` V1 = 1.0 cm^3 = 1.0 XX 10^(–6 )m^3`
Bubble rises to height, d = 40 m
Temperature at a depth of 40 m, `T_1 = 12 ^(@)C = 285 K`
Temperature at the surface of the lake, `T_2 = 35^(@)C = 308 K`
The pressure on the surface of the lake:
` P_2 = 1 atm = 1 xx 1.013 xx 105` Pa
The pressure at the depth of 40 m:
`P_1 = 1 atm + d rho g`
Where,
`rho" is the density of water" = 103 kg//m^3`
g is the acceleration due to gravity `= 9.8 m//s^2`
` ∴P1 = 1.013 xx 105 + 40 xx 103 xx 9.8 = 493300 Pa` `((P_1)(V_1))/(T_1)=((P_2)(V_2))/(T_(2))`n
`=((493300)(1.0xx10^-6)308)/(285 xx 1.013xx 10^5)`
` = 5.263 × 10^(–6) m_(3) "or" 5.263 cm^3 `
Therefore, when the air bubble reaches the surface, its volume becomes` 5.263 cm^3 `.