An air bubble of volume `1.0 xx 10^(-6)m^(3)` rises from the bottom of a lake 40 m deep at a temperature `12^(@)C`. To what volume does it grow when it reaches the surface, which is at a temperature of `35^(@)C` ?

Explanation

Given `V_(1)=1xx10^(-6)m^(3),P_(1)=P_(0)+hpg`
`i.e. hpg=40xx10^(3)xx9.8=3.92xx10^(5)Nm^(-2)`
`& " " P_(0)=1.013xx10^(5)Nm^(-2)`
`therefore " " P_(1)=(1.013+3.92)xx10^(5)Nm^(-2)`
`P_(1)=4.933xx10^(5)Nm^(-2)`
`T_(1)=273+12=285K`
`P_(2)=P_(0)=1.013xx10Nm^(-2)`
Applying `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)) "we get"`
`(4.933xx10^(5)xx10^(-6))/(285)=(1.013xx10^(5)xxV_(2))/(308)`
`therefore " " V_(2)=(4.933xx10^(-6)xx308)/(1.033xx285)=5.063xx10^(-6)m^(3).