An air bubble of volume `0.21 cm^3` at the bottom of a water tank, at `5^(@)C` and 3 atm pressure, rises to the surface where the temperature is `25^(@)C` and pressure is 1 atm. What will be the volume of the bubble when it reaches the surface?

Explanation

Given : `P_(1) = 3` atm , `T_(1) = (273 + 5) = 278 K , V_(1) = 0.21 cm^(3)` (at the bottom of the tank ) and `P_(2) = 1` atm , `T_(2) = (273 + 25) = 298 K` , Volume of bubble = `V_(2)` (at the surface of the tank)
Since no. of moles (n) of air contained in the bubble remains constant
`(P_(1) V_(1))/(T_(1)) = (P_(2) V_(2))/(T_(2)) implies V_(2) = (P_(1) V_(1) T_(2))/(P_(2) T_(1)) = (3 "atm" xx 0.21 cm^(3) xx 298 K)/(1 "atm" xx 278 K) = 0.67 cm^(-3)`