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

At the bottom of the lake `V_(1)=1cm^(3)`
`P_(1)=P_(a)+40m`
`=10.3+40`
`=50.3m` of water
`T_(1)=273+12`
`=285K`
At the surface `P_(2)=P_(a)`
`=10.3`m of water
`T_(2)=273+35`
`=308K`
Using `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
or `V_(2)=(P_(1)V_(1)T_(2))/(P_(2)T_(1))`
`=(50.3xx1xx10^(-6)xx308)/(10.3xx285)`
`=5.275xx10^(6)m^(3)`