Question

In a hot summer day, a spherical air bubble that has a volume of 1.20 cm3 is released at temperature 17.0 °C by a scuba diver 25.0 m below the surface of ocean. Calculate the radius of the spherical bubble when it reaches the surface at temperature 30 °C? Assume that the number of air molecules in the bubble remain the same (rhosalt water = 1.027 g/cm3 ).

Answer 1

Answer:

The radius of the bubble when it reaches the surface at 30 ºC is 1.015 centimeters.

Explanation:

Let suppose that air bubble behaves as ideal gas, whose equation of state is:

$P\cdot V = n\cdot R_{u}\cdot T$ (Eq. 1)

Where:

$P$ - Pressure of the bubble, measured in kilopascals.

$V$ - Volume of the bubble, measured in cubic meters.

$n$ - Molar amount of the bubble, measured in kilomoles.

$T$ - Temperature, measured in Kelvin.

$R_{u}$ - Ideal gas constant, measured in kilopascal-cubic meter per kilomole-Kelvin.

Then, we eliminate the molar amount and the ideal gas constant by constructing the following relationship:

$\frac{P_{A}\cdot V_{A}}{T_{A}} = \frac{P_{B}\cdot V_{B}}{T_{B}}$ (Eq. 2)

Where:

$P_{A}$, $P_{B}$ - Pressure of the bubble at bottom and surface, measured in kilopascals.

$V_{A}$, $V_{B}$ - Volume of the bubble at bottom and surface, measured in cubic meters.

$T_{A}$, $T_{B}$ - Temperature of the bubble at bottom and surface, measured in Kelvin.

The pressure experimented by the bubble at bottom and surface are, respectively:

$P_{A} = 101.325\,kPa+\left(1027\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (25\,m)\cdot \left(\frac{1}{1000}\,\frac{kPa}{Pa} \right)$

$P_{A} = 353.120\,kPa$

$P_{B} = 101.325\,kPa$

If we know that $P_{A} = 353.120\,kPa$, $P_{B} = 101.325\,kPa$, $V_{A} = 1.20\times 10^{-6}\,m^{3}$, $T_{A} = 290.15\,K$ and $T_{B} = 303.15\,K$, then the volume of the bubble at surface is:

$\frac{(353.120\,kPa)\cdot (1.20\times 10^{-6}\,m^{3})}{290.15\,K} = \frac{(101.325\,kPa)\cdot V_{B}}{303.15\,K}$

$1.460\times 10^{-6} = 0.334\cdot V_{B}$

$V_{B} = 4.372\times 10^{-6}\,m^{3}$

$V_{B} = 4.372\,cm^{3}$

And the volume of the air bubble is determined by this formula:

$V_{B} = \frac{4\pi\cdot R^{3}}{3}$ (Eq. 3)

Where $R$ is the radius of the air bubble, measured in centimeters.

If we know that $V_{B} = 4.372\,cm^{3}$, then the radius of the air bubble is:

$4.372 = \frac{4\pi\cdot R^{3}}{3}$

$R^{3} = 1.044$

$R \approx 1.015\,cm$

The radius of the bubble when it reaches the surface at 30 ºC is 1.015 centimeters.