70.5k views
1 vote
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

User Wberry
by
3.8k points

1 Answer

3 votes

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(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} (P_iV_i)/(P_fV_f)=(n_iR_iT_i)/(n_fR_fT_f)=(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=(P_(under)\cdot r_i3)/(P) \\ r_f=\sqrt[3]{(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.

User Vishnu S Babu
by
3.5k points