## College Physics (4th Edition)

The bubble's volume is $3.10~cm^3$ when it reaches the surface.
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.