Bubble in a Lake
Task number: 1283
An air bubble with a radius of 5.0 mm rises from the bottom of a lake 20.7 m deep. The temperature at the bottom of the lake is 7 °C and the temperature at the surface is 27 °C. The atmospheric pressure is 100 kPa. How big will the bubble be when it reaches the surface?
Hint
When the bubble rises from the bottom of the lake, it is not only the temperature what changes. It is also the pressure of the gas inside the bubble. This pressure corresponds to the pressure of the water surrounding the bubble.
Determine the water pressure surrounding the bubble at the bottom and at the surface of the lake.
Numerical values
h = 20.7 m the depth of the lake r1 = 5.0 mm = 5.0·10-3 m the bubble radius at the bottom of the lake t1 = 7 °C => T1 = 280 K the water temperature at the bottom of the lake t2 = 27 °C => T1 = 300 K the water temperature at the surface of the lake pa = 100 kPa = 1.00·105 Pa the atmospheric pressure r2 = ? the bubble radius at the surface of the lake From The Handbook of Chemistry and Physics:
ρ = 1000 kg m-3 the density of water g = 9.81 m s-2 the acceleration of gravity Analysis
When calculating the radius of the bubble, we use the fact that the pressure of the gas inside the bubble must be the same as the water pressure in the bubble surroundings. Otherwise the pressure forces of the surroundings and of the gas inside the bubble at the air-water interface would be different and the bubble would either stretch or shrink.
The water pressure in the bubble surroundings is determined by the hydrostatic pressure (which is directly proportional to the depth, where the bubble is located) and the atmospheric pressure, which thanks to Pascal’s law contributes with the same value to the overall pressure all over the volume of the lake. The gas pressure inside the bubble can be determined by the state equation.
Note: In the bubble there is air along with the saturated water vapor. In The Handbook of Chemistry and Physics we can find that the value of the pressure of saturated vapor for a given temperature is a few kPa. So we can neglect this contribution of the saturated vapor pressure in the bubble to the atmospheric pressure.
Solution
First, we express the water pressure p1 at the bottom of the lake, which is a sum of the atmospheric and the hydrostatic pressure
\[ p_1\,=\,p_a\,+\,h\rho g \]Realize that the pressure p2 at the surface of the lake is equal to the atmospheric pressure
\[ p_2\,=\,p_a\,. \]We assume that the gas inside the bubble behaves like an ideal gas, and therefore satisfies the ideal gas equation, i.e. that
\[ \frac{p_1V_1}{T_1}\,=\,\frac{p_2V_2}{T_2}\, \]applies. The air pressure is the same as the pressure of the surrounding water, and the volume of the bubble can be determined by the volume formula of a sphere \(V\,=\,\frac{4}{3}\pi r^3\). Now we substitute these relations into the state equation.
\[ \frac{\left(p_a+h\rho g\right)\frac{4}{3}\pi r_1^3}{T_1}\,=\,\frac{p_a\frac{4}{3}\pi r_2^3}{T_2} \] \[ \frac{\left(p_a+h\rho g\right)r_1^3}{T_1}\,=\,\frac{p_ar_2^3}{T_2} \]and evaluate the unknown radius r2
\[ r_2^3\,=\,\frac{\left(p_a+h\rho g\right)r_1^3T_2}{p_aT_1} \] \[ r_2\,=\,\sqrt[3]{\frac{\left(p_a+h\rho g\right)r_1^3T_2}{p_aT_1}}\,. \]Now we substitute the given values
\[ r_2\,=\,5\cdot{10^{-3}}\cdot \sqrt[3]{\frac{\left(10^5\cdot{20{.}7}\cdot 1000\cdot{9{.}81}\right)\cdot 300}{10^5\cdot{280}}}\,\mathrm{m} \] \[ r_2\,\dot{=}\,7{.}4\cdot{10^{-3}}\,\mathrm{m}\,=\,7{.}4\,\mathrm{mm}\,. \]Note: It is obvious that if we substituted the value of r1 in millimeters, the resulting value would also be in millimeters
Answer
The radius of the bubble at the surface of the lake will be approximately 7.4 mm.
Comment on the other contributions to the pressure inside the bubble
The surface tension of water also contributes to the air pressure inside the bubble. The curved surface of liquid induces the capillary pressure given by:
\[ p_c\,=\,\frac{2\sigma}{r}\,\mathrm{,} \]where r is the radius of the curvature of the liquid surface and σ is the surface tension of the liquid, its value for liquid being σ = 73 mNm-1 = 0.073 Nm-1.
We determine the value of the capillary pressure for both radii of the bubble:
\[ p_\mathrm{c1}\,=\,\frac{2\cdot{0{.}073}}{5\cdot{ 10^{-3}}}\,\mathrm{Pa}\,\dot{=}\,29\,\mathrm{Pa} \] \[ p_\mathrm{c2}\,=\,\frac{2\cdot{0{.}073}}{7{.}4\cdot{ 10^{-3}}}\,\mathrm{Pa}\,\dot{=}\,20\,\mathrm{Pa}\,. \]We can see that the value of the capillary pressure pc is in comparison to the atmospheric pressure pa negligible
Furthermore, there is saturated water vapor in the bubble along with the air. The pressure of the saturated water vapor for both temperatures is listed in The Handbook of Chemistry and Physics:
for t1 = 7 °C the pressure is ps = 1.02 kPa
for t1 = 27 °C the pressure is ps = 3.60 kPa
This contribution is also negligible in comparison to the atmospheric pressure.
