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)= 1atm, T_(2)= (273+25)= 298K`, Volume of bubble = `V_(2)` (at the surface of the tank)
Since no. 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)`.