## College Physics (4th Edition)

The bubble's diameter is $2.10~mm$ when it reaches the surface.
We can find the gauge pressure at a depth of 80.0 meters: $P_g = \rho~g~h = (1000~kg/m^3)(9.80~m/s^2)(80.0~m) = 7.84\times 10^5~Pa$ We can find the absolute pressure initially: $P_1 = P_{atm}+P_g$ $P_1 = (1.01\times 10^5~Pa)+(7.84\times 10^5~Pa)$ $P_1 = 8.85 \times 10^5~Pa$ 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 $V_1$: $\frac{V_2}{V_1} = \frac{\frac{nRT_2}{P_2}}{\frac{nRT_1}{P_1}}$ $V_2 = \frac{T_2~P_1}{T_1~P_2}~V_1$ $V_2 = \frac{(291~K)(8.85\times 10^5~Pa)}{(277~K)(1.01\times 10^5~Pa)}~V_1$ $V_2 = 9.205~V_1$ We can find the radius $r_2$ of the bubble at the surface: $V_2 = 9.205~V_1$ $\frac{4}{3}\pi~r_2^3 = (9.205)~\frac{4}{3}~\pi~r_1^3$ $r_2^3 = (9.205)~r_1^3$ $r_2 = (9.205)^{1/3}~r_1$ $r_2 = (2.10)~(0.500~mm)$ $r_2 = 1.05~mm$ Since the diameter is twice the radius, the bubble's diameter is $2.10~mm$ when it reaches the surface.