Chapter 13, Problem 79GP

To determine

To Find: The increase in the volume.

Answer to Problem 79GP

  $11\text{ L}$

Explanation of Solution

Given info:

Submerged volume, $V_{\text{sub}}=5.5\text{ L}$ .

Height, $h=10\text{ m}$ below thesurface.

Density of the water, $\rho =1000{\text{ kg/m}}^{\text{3}}$ .

Pressure, $P=1\text{ atm }=1.013\times {10}^5{\text{ N/m}}^{\text{2}}$ .

Formula used:

Use ideal gas law to find the surface volume.

  $P\times V=n\times R\times T$

Here, $P$ is the pressure of the gas,

  $V$ is the volume occupied by the gas,

  $n$ is the number of molecules,

  $R$ is the universal gas constant,

  $T$ is the Temperature of gas.

Calculation:

Here, n, R and T are constant parameters. So, the relationship of pressure and volume is given by the following formula,

  $P_{\text{surface}}\times V_{\text{surface}}=P_{\text{submerged}}\times V_{\text{submerged}}$

  $V_{\text{surface}}=\frac{P_{\text{submerged}}\times V_{\text{submerged}}}{P_{\text{surface}}}$ …………………………………………….. Equation (1)

Here, surface pressure and atmospheric pressure are same. Thus,

  $P_{\text{Submerged}}\text{ = }{P}_{\text{atm}}\text{ + }\left(\rho \times g\times h\right)$

Substitute the value of $P_{\text{Submerged}}$ in the equation (1)

  $\begin{array}{l}{V}_{\text{surface}}=\frac{V_{\text{submerged}}\times (P_{\text{atm}}+\rho \times g\times h)}{P_{\text{atm}}}\\ V_{\text{surface}}=\frac{5.5\times (1.013\times {10}^5+1000\times 9.8\times 10)}{1.013\times {10}^5}\\ V_{\text{surface}}=10.85\text{ L }\approx 11\text{ L}\end{array}$

Conclusion:

Thus, the expand in the volume of the lungs of a scuba diver when he quickly rose to the surface is $11\text{ L}$ and it is not advisable because the volume of lungs when it rose to surface is almost double compared to submerged volume.