Question

An air bubble released by a deep-water diver, 115 m below the surface of a lake, has a volume of 1.60 cm3. The surface of the lake is at sea level, and the density of the lake water can be approximated as that of pure water. As the bubble rises to the surface, the temperature of the water and the number of air molecules in the bubble can each be approximated as constant. Find the volume (in cm3) of the bubble just before it pops at the surface of the lake.
___ cm3.

Answer 1

Answer:

The value is  $V_2 = 1.9396 *10^{-5} \ m^3$

Explanation:

From the question we are told that

   The depth at which the bubble is released is  $h = 115 \ m$

    The volume of the air bubble is  $V = 1.60 cm^3 = 1.60 *10^{-6} \ m^3$

Generally from the ideal gas law

     $PV = nRT$

Given that  n , R , T are constant we have that

    $PV = constant$

So    

    $P_1 V_1=P_2 V_2$

Here $P_1$ is the pressure of the bubble at the depth where it is released which i mathematically represented as

      $P_1 = P_a + P$

Here $P_a$  is the atmospheric pressure with value  $P_a = 101325 \ Pa$

and   $P$ is the pressure due to the depth which is mathematically represented as

       $P = \rho * g * h$

So

       $P = 1000 * 9.8*115$

=>    $P = 1127000\ Pa$

Here  $\rho$ is the density of pure water with value  $\rho = 1000 \ kg/m^3$

        $g = 9.8 \ m/s^2$

        $V_1$ is the volume of the bubble at the depth where it is released

        $P_2$ is the pressure of the bubble at the surface which is equivalent to the atmospheric temperature

        $V_2$ is the volume of the bubble at the surface

So

      $V_2 = \frac{ P_1 * V_1}{ P_2}$

=>    $V_2 = \frac{(Pa+ P) * V_1}{P_a}$

=>    $V_2 = \frac{101325 + 1127000 * (1.60 *10^{-6})}{ 101325 }$

=>    $V_2 = 1.9396 *10^{-5} \ m^3$