Solved Problem on Gases

A fish at the bottom of a lake 15 m deep emits an air bubble of volume V₀, the temperature at the bottom of the lake being 5 °C and at the surface 17 °C. Calculate the volume of the bubble when it reaches the surface of the lake. Data: atmospheric pressure p₀ = 1.01×10⁵ Pa, water density μ = 1.0 g/cm³, and acceleration due to gravity g = 9.8 m/s².

Problem data:

Initial volume of bubble: V₀;
Lake depth: h = 15 m;
Temperature at the bottom of the lake: t₁ = 5 °C;
Temperature at the surface of the lake: t₂ = 17 °C;
Atmospheric pressure: p₀ = 1.01×10⁵ Pa;
Density of water: μ = 1.0 g/cm³;
Acceleration due to gravity: g = 9.8 m/s².

Problem diagram:

Figure 1

Solution

First, we must convert the density of water given in grams per cubic centimeter (g/cm³) to kilograms per cubic meter (kg/m³) used in the International System of Units (SI) and temperatures from degrees Celsius (°C) to Kelvins ( K).

\[ \begin{gather} \mu=1.0\;\frac{\cancel{\text{g}}}{\text{cm}^{3}}\times\frac{10^{-3}\;\text{kg}}{1\;\cancel{\text{g}}}\times\frac{(1\;\text{cm})^{3}}{(10^{-2}\;\text{m})^{3}}=1.0\;\frac{1}{\cancel{\text{cm}^{3}}}\;\text{kg}\times\frac{1\;\cancel{\text{cm}^{3}}}{10^{-6}\;\text{m}^{3}}=1.0\times 10^{-3}\times 10^{6}\;\frac{\text{kg}}{\;\text{m}^{3}}=1.0\times 10^{3}\;\frac{\text{kg}}{\;\text{m}^{3}}\\[10pt] T_{1}=t_{1C}+273=5+273=278\;\text{K}\\[10pt] T_{2}=t_{2C}+273=17+273=290\;\text{K} \end{gather} \]

At the bottom of the lake, the air bubble is under the pressure of the water column above it and the atmospheric pressure above the lake (Figure 2), using Stevin's Law

\[ \begin{gather} \bbox[#99CCFF,10px] {p_{1}=p_{0}+\mu gh} \tag{I} \end{gather} \]

the volume will be

\[ \begin{gather} V_{1} = V_{0} \tag{II} \end{gather} \]

Figure 2

When the bubble reaches the surface it is only under the action of atmospheric pressure (Figure 3)

\[ \begin{gather} p_{2} = p_{0} \tag{III} \end{gather} \]

Figure 3

Considering the air bubble to be a perfect gas, we use the General Gas Equation applied to situations at the bottom and surface of the lake.

\[ \begin{gather} \bbox[#99CCFF,10px] {\frac{p_{1}V_{1}}{T_{1}}=\frac{p_{2}V_{2}}{T_{2}}} \tag{IV} \end{gather} \]

substituting the values of (I), (II), and (III) into the expression (IV)

\[ \begin{gather} \frac{(p_{0}+\mu gh)V_{0}}{T_{1}}=\frac{p_{0}V_{2}}{T_{2}}\\[5pt] V_{2}=\frac{T_{2}}{T_{1}}\frac{(p_{0}+\mu gh)}{p_{0}}V_{0} \end{gather} \]

substituting the values

\[ \begin{gather} V_{2}=\frac{T_{2}}{T_{1}}\frac{(p_{0}+\mu gh)}{p_{0}}V_{0}\\[5pt] V_{2}=\frac{290}{278}\times \frac{(1.01\times 10^{5}+1.0\times 10^{3}\times 9.8\times 15)}{1.01\times 10^{5}}V_{0}\\[5pt] V_{2}=1.04\times \frac{(1.01\times 10^{5}+1.47\times 10^{5})}{1.01\times 10^{5}}V_{0}\\[5pt] V_{2}=1.04\times \frac{2.48\times 10^{5}}{1.01\times 10^{5}}V_{0} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {V_{2}=2.6V_{0}} \end{gather} \]

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