Question

A spherical air bubble originating from a scuba diver at a depth of 18.0 m has a diameter of 1.0 cm. What will the bubble’s diameter be when it reaches the surface? (Assume constant temperature.)a. 0.7 cmb. 1.0 cmc. 1.4 cmd. 1.7 cm

Answer 1

Assuming the bubble as an idel gas, we use the following

$PV=\text{nRT}$

Where the pressure at sea level is 101K Pa.

The pressure would be given by

$P=\rho gh$

Whew rho = 1000 kg/m^3, g = 9.8 m/s^2, and h = 18m.

$\begin{gathered} P=\frac{1000\operatorname{km}}{m^3}\cdot\frac{9.8m}{s^2}\cdot18m \\ P_{\text{under}}=176,400Pa+101,000Pa=277,400Pa \\ \end{gathered}$

The pressure under the water is 277,400 Pa.

Then, form a ratio between the initial condition of the ideal gas and the final conditions.

$\begin{gathered} \frac{P_iV_i}{P_fV_f}=\frac{n_iR_iT_i}{n_fR_fT_f}=\frac{r^3_i}{r^3_f} \\ \end{gathered}$

Observe that the radii are not simplified because there's a change. Solve for the final radius.

$\begin{gathered} r^3_f=\frac{P_{under}\cdot r_i3}{P} \\ r_f=\sqrt[3]{\frac{277,400Pa\cdot r^3_i}{101,000Pa}}=r_i\sqrt[5]{2.75}=1\operatorname{cm}\cdot1.4 \\ r_f=1.4\operatorname{cm} \end{gathered}$

Therefore, the answer is c. 1.4 cm.