Answer
The bubble's volume is $3.10~cm^3$ when it reaches the surface.
Work Step by Step
We can find the gauge pressure at a depth of 20.0 meters:
$P_g = \rho~g~h = (1000~kg/m^3)(9.80~m/s^2)(20.0~m) = 1.96\times 10^5~Pa$
We can find the absolute pressure initially:
$P_1 = P_{atm}+P_g$
$P_1 = (1.01\times 10^5~Pa)+(1.96\times 10^5~Pa)$
$P_1 = 2.97 \times 10^5~Pa$
We can find an expression for the original volume:
$P_1~V_1 = nRT_1$
$V_1 = \frac{nRT_1}{P_1}$
We can find an expression for the final volume:
$P_2~V_2 = nRT_2$
$V_2 = \frac{nRT_2}{P_2}$
We can divide $V_2$ by $V_1$:
$\frac{V_2}{V_1} = \frac{\frac{nRT_2}{P_2}}{\frac{nRT_1}{P_1}}$
$V_2 = \frac{T_2~P_1~V_1}{T_1~P_2}$
$V_2 = \frac{(298~K)(2.97\times 10^5~Pa)(1.00~cm^3)}{(283~K)(1.01\times 10^5~Pa)}$
$V_2 = 3.10~cm^3$
The bubble's volume is $3.10~cm^3$ when it reaches the surface.
