The temperature at the bottom of a `40 cm` deep lake is `12^(@)C` and that at the surface is `35^(@)C`. An air bubble of volume `1.0 cm^(3)` rises from the bottom to the surface. Its volume becomes (atmospheric pressure `= 10 m` of water)

Options

A. `2.0 cm^(3)`
B. `3.2 cm^(3)`
C. `5.4 cm^`
D. `8.0 cm^(3)`

Answer

C

Explanation

`P_(1) = P_(atm) + P_(water) = 10 + 40 = 50 m` of water
`V_(1) = 1 c c, T_(1) = 12 + 273 = 285 K`
`P_(2) = P_(atm) = 10 m` of water
`V_(2) = ?, T_(2) = 35 + 273 = 308 K`
Since number of moles in the bubble remains same
`(P_(1)V_(1))/(RT_(1)) = (P_(2)V_(2))/(RT_(2))`
`(50 xx 1)/(285) = (10 V_(2))/(308) rArr V_(2) = 5.4 cm^(3)`