Question

A diver named Jacques observes a bubble of air rising from the bottom of a lake (where the absolute pressure Is 3.50 atm) to the surface (where the pressure is 1.00 atm). The temperature at the bottom is 4.00 °C, and the temperature at the surface is 23.0 °C.

Part A 

What is the ratio of the volume of the bubble as it reaches the surface (V_s) to its volume at the bottom (V_b)? 

Part B 

If Jacques were to hold his breath the air in his lungs would be kept at a constant temperature. Would it be safe for Jacques to hold his breath while ascending from the bottom of the lake to the surface? 

Answer 1

Concepts and reason

The concept required to solve this problem is ideal gas law.

Initially, calculate the ratio of volume at surface ${V_{\rm{s}}}$ and volume at bottom ${V_{\rm{b}}}$ by using the ideal gas equation. Finally, use that ratio to find that holding breath is safe while ascending from bottom to surface or not.

Fundamentals

The Ideal gas law is the combination of Boyle's law, Charles' law, and Gay-Lussac's law. According to Ideal gas law, the ratio of product of pressure and volume with temperature is constant for an ideal gas. The equation of combined gas law can be written as,

$\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}$

Here, $P$ is pressure, $V$ is the volume, $T$ is the temperature, and 1 and 2 subscripts represent state 1 and state 2.

The relation used to convert units is as follows:

$T{\rm{ K}} = T^\circ {\rm{C}} + 273$

The ratio of volume at surface ${V_{\rm{s}}}$ to volume at bottom ${V_{\rm{b}}}$ is,

$\frac{{{V_{\rm{s}}}}}{{{V_{\rm{b}}}}} = \frac{{{P_{\rm{b}}}{T_{\rm{s}}}}}{{{T_{\rm{b}}}{P_{\rm{s}}}}}$

Here, s represents surface and b represents bottom.

Substitute $1.00{\rm{ atm}}$ for ${P_{\rm{s}}}$, $3.50{\rm{ atm}}$ for ${P_{\rm{b}}}$, $23.0^\circ {\rm{C}}$ for ${T_{\rm{s}}}$, and $4.00^\circ {\rm{C}}$ for ${T_{\rm{b}}}$ in the equation $\frac{{{V_{\rm{s}}}}}{{{V_{\rm{b}}}}} = \frac{{{P_{\rm{b}}}{T_{\rm{s}}}}}{{{T_{\rm{b}}}{P_{\rm{s}}}}}$ and calculate the ratio of volume at surface ${V_{\rm{s}}}$ and volume at bottom ${V_{\rm{b}}}$.

$\frac{{{V_{\rm{s}}}}}{{{V_{\rm{b}}}}} = \frac{{\left( {3.50{\rm{ atm}}} \right)\left( {23.0^\circ {\rm{C}}} \right)}}{{\left( {4.00^\circ {\rm{C}}} \right)\left( {1.00{\rm{ atm}}} \right)}}$

Convert the temperature from $^\circ {\rm{C}}$ to K.

$\begin{array}{c}\\\frac{{{V_{\rm{s}}}}}{{{V_{\rm{b}}}}} = \frac{{\left( {3.50{\rm{ atm}}} \right)\left( {23.0 + 273{\rm{ K}}} \right)}}{{\left( {4.00 + 273{\rm{ K}}} \right)\left( {1.00{\rm{ atm}}} \right)}}\\\\ = \frac{{3.50\left( {296{\rm{ K}}} \right)}}{{\left( {277{\rm{ K}}} \right)}}\\\\ = 3.74\\\end{array}$

The ratio of volume at surface to bottom that is 3.74. This means the volume occupied by air increases by 3.74 times on the surface as compared to bottom. If a person holds breath while rising from bottom to surface the air in his lungs will occupy 3.74 times more volume. The lungs are already full of air and expanding air can damage the lungs. Therefore, it is not safe to hold breath while ascending from the bottom of the lake to the surface.

Ans:

The ratio of volume at surface ${V_{\rm{s}}}$ to volume at bottom ${V_{\rm{b}}}$ is 3.74.

It is not safe to hold the breath while ascending from the bottom of the lake to the surface.

Answer 2

so

no it is not safe to hold breath while rising to the surface of water

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