Question

An air bubble originating from an under water diver has a radius of 5 mm at some depth h. When the bubble reaches the surface of the water, it has a radius of 6.6 mm. Assuming the temperature of the air in the bubble remains constant, determine the absolute pressure at this depth h. The acceleration of gravity is 9.8 m/s 2 . Answer in units of Pa.

Answer 1

Answer:

Absolute Pressure at depth h = $1.312\times 10^5\ Pa.$

Explanation:

Given:

• Radius of the bubble at depth h, $\rm r_i = 5\ mm = 5\times 10^{-3}\ m.$
• Radius of the bubble at the surface of the water, $\rm r_f = 6.6\ mm = 6.6\times 10^{-3}\ m.$
• Acceleration due to gravity, $\rm g = 9.8\ m/s^2.$

Assumptions:

• Atmospheric pressure, $\rm P_{atm} = 1.01\times 10^5\ Pa.$
• Density of water, $\rm \rho = 1000\ kg/m^3.$
• Pressure at depth h = $\rm P_i.$
• Pressure at the surface of water = $\rm P_f.$

According to Ideal Gas Law,

$\rm PV=nkT$

where,

• P = pressure.
• V = volume.
• n = number of molecules.
• k = Boltzmann constant.
• T = absolute temperature.

For the given case, the temperature of the air bubble is constant. The number of molecules of the bubble also does not change, therefore,

PV = constant.

$\rm P_iV_i=P_fV_f$

The pressure at the surface is equal to the atmospheric pressure, $\rm P_f=P_{atm}.$

The pressure at depth h is equal to sum of atmospheric pressure and the pressure of the water upto depth h, $\rm P_i=P_{atm}+\rho gh.$

Using these values,

$\rm (P_{atm} + \rho gh)V_i = P_{atm}V_f\\\dfrac{(P_{atm} + \rho gh)}{P_{atm}}=\dfrac{V_f}{V_i}=\dfrac{\dfrac 43 \pi r_f^3}{\dfrac 43 \pi r_i^3}=\dfrac{r_f^3}{r_i^3}\\\dfrac{(1.01\times 10^5)+1000\times 9.8\times h}{1.01\times 10^5}=\dfrac{(6.6\times 10^{-3})^3}{(5\times 10^{-3})^3}=2.299\\(1.01\times 10^5)+9800\times h=2.299\times 1.01\times 10^5=2.322\times 10^5\\9800\ h= 2.322\times 10^5-1.01\times 10^5=1.312\times 10^5\\h=\dfrac{1.312\times 10^5}{9800}=13.39\ m.$

The absolute temperature at that depth is given by

$\rm P=\rho gh=1000\times 9.8\times 13.39=1.312\times 10^5\ Pa.$