Question

At 30.0 m below the surface of the sea (density = 1 025 kg/m3), where the temperature is 5.00°C, a diver exhales an air bubble having a volume of 0.95 cm3. If the surface temperature of the sea is 20.0°C, what is the volume of the bubble just before it breaks the surface?

Answer 1

Answer:

The volume is $V_a = 1.510 *10^{-5} m^3$

Explanation:

From the question we are told that

     The depth below the see is  $d_1 = 30.0 \ m$

     The density of the sea is  $\rho_s = 1025 \ kg /m^3$

      The temperature at this level is $T_d = 5.00 ^oC = 278 \ K$

      The volume of the air bubble at this depth is  $V_d = 0.95 \ cm^3 = 0.95 *0^{-6}\ m$

     The temperature at the surface is  $T_a = 20^oC =293\ K$

Generally the pressure at the given depth is mathematically evaluated as  

        $P_d = P_o + \rho_s * g * d$

Where $P_o$ is the atmospheric pressure with a constant value

        $P_o = 1.013 *10^{5} \ Pa$

substituting values

       $P_d = 1.013 * 10^{5} * + (1025 * 9.8 * 30 )$

        $P_d = 4.02650 * 10^{5} \ Pa$

According to the combined gas law  

          $\frac{P_a * V_a }{T_a } = \frac{P_d * V_d }{T_d }$

=>      $V_a = \frac{4.026650 *10^{5} * 0.95 *10^{-6} * 293 }{278 * 1.013*10^{5} }$

=>     $V_a = 1.510 *10^{-5} m^3$