Chapter 13, Problem 101A

To determine

Volume of the bubble.

Answer to Problem 101A

Volume of bubble will be $6.2\times {10}^{-6}{\text{m}}^3$ .

Explanation of Solution

Given info:

Atmospheric pressure, $P_{atm}\text{= 1}\text{.01}\times {\text{10}}^5\text{Pa}$

Density of water, ${\rho }_{water}\text{= 1}\text{.00}\times {\text{10}}^3{\text{kg/m}}^3$

Acceleration due to gravity, $g\text{= 9}\text{.8}\text{N/kg}$

Height, $h\text{= 5}\text{m}$

Volume at $\text{5m,}{V}_2=4.2\times {10}^{-6}{\text{m}}^3$

Formula used:

The product of the initial volume and initial pressure is equal to the product of the volume and pressure after a change in one of them under constant temperature. Mathematical expression of this statement is,

  $P_1{V}_1=P_2{V}_2$

Where,

  $P_1=$ initial pressure

  $P_2=$ Final pressure

  $V_1=$ Initial Volume

  $V_2=$ Final Volume

Calculation:

By using the formula, $P_1{V}_1=P_2{V}_2$ , derive the formula of final volume.

  $V_2=\frac{P_1{V}_1}{P_2}$

  $V_2=\frac{\left(P_{atm}+{\rho }_{water}gh\right)V_1}{P_{atm}}$

  $\text{= }\frac{\left(\text{1}\text{.01}\times {\text{10}}^5\text{Pa}+\text{1}\text{.00}\times {\text{10}}^3{\text{kg/m}}^3\times \text{9}\text{.8}\text{N/kg}\times \text{5}\text{m}\right)4.2\times {10}^{-6}{\text{m}}^3}{\text{1}\text{.01}\times {\text{10}}^5\text{Pa}}$

  $\text{ = 6}\text{.2}\times {\text{10}}^{-6}{\text{m}}^3$

Conclusion:

Volume of bubble just before diver reaches the surface of the water will be $6.2\times {10}^{-6}{\text{m}}^3$ .