An air bubble starts rising from the bottom of a lake. Its diameter is ` 3.6 mm` at the bottom and `4 mm` at the surface. The depth of the lake is `250 cm` and the temperature at the surface is `40^@ C`. What is the temperature at the bottom of the lake? Given atmospheric pressure = `76 cm of Hg and g = 980 cm//s^2`.

Explanation

At the bottom of the lake, volume of the bubble
`V_(1) = (4)/(3)pi (0.18)^(3)cm^(3)`
Pressure on the bubble `P_(1) =` Atmospheric
pressure `+` Pressure due to a column of `250cm` of water `= (76 xx 13.6 + 250)980 "dyne"//cm^(2)`
At the surface of the lake, volume of the bubble `V_(2) = (4)/(3)pi(0.2)^(3)cm^(3)`
Pressure on the bubble , `P_(2)` = atm. pressure
`= (76 xx 13.6 xx 980)"dyne"//cm^(2)`
`T_(2) = 273 + 40^(0)C = 313 K`
Now `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`
`rArr ((76 xx 13.6 + 250)680xx ((4)/(3))pi(0.18)^(3))/(T_(1))`
`= ((76 xx 13.6 + 250)680xx ((4)/(3))pi(0.2)^(3))/(313)` or `T_(1) = 283.37 K , T_(1) = 283.37 - 273 = 10.37^(0)C`