Question

A spherical air bubble originating from a scuba diver at a depth of 18m has a diameter of 1cm. What will the bubble's diameter be when it reaches the surface? a) .7cm b) 1cm c) 1.4cm
d) 1.7cm e) 2.5cm Can someone please explain the solution?

Answer 1

\[r=d/2\]
\[ V_{sphere} = \frac{4}{3}\pi r^3\]

So I think we can treat the air bubble as an ideal gas at a constant temperature, so we can say

\[PV=nRT\]

where the pressure at the surface would just be the air pressure at sea level
\[P_{surface}=101325 kg/(ms^2)\]

and where the pressure under the water is given by the pressure at the surface plus the hydrostatic pressure
\[P=\rho g h\]
\[ \rho = \text{ density of water} = 1000kg/m^3\]
\[g = \text{gravitational constant} = 9.81m/s^2\]
\[h=18m\]



So if we divide

\[\frac{P_{1}V_1=nRT}{P_{2}V_2=nRT} \ \longrightarrow \frac{P_1V_1}{P_2V_2}=1\]

\[\frac{P_1V_1}{P_2V_2}=1\]
\[\frac{(\rho g h + P_s)(\frac{4}{3}\pi r_1^3)}{(P_s)(\frac{4}{3}\pi r_2^3)}=1\]
\[r_2^3=\frac{(\rho g h+P_s)(r_1^3)}{P_s} \]
\[r_2=r_1 \sqrt[3]{\frac{(\rho g h+P_s)}{P_s}}=(.5cm)\sqrt[3]{\frac{(1000kg/m^3)(9.81m/s^2)(18m)}{101325 kg/ms^2}+1}\]
\[r_2=.5cm(1.400)\]
\[d=2r_2 = 1.4cm\]

So the answer is C, 1.4cm

Answer 2

thank you for being so thorough