Answer:
The volume is [tex]V_a = 1.510 *10^{-5} m^3[/tex]
Explanation:
From the question we are told that
The depth below the see is [tex]d_1 = 30.0 \ m[/tex]
The density of the sea is [tex]\rho_s = 1025 \ kg /m^3[/tex]
The temperature at this level is [tex]T_d = 5.00 ^oC = 278 \ K[/tex]
The volume of the air bubble at this depth is [tex]V_d = 0.95 \ cm^3 = 0.95 *0^{-6}\ m[/tex]
The temperature at the surface is [tex]T_a = 20^oC =293\ K[/tex]
Generally the pressure at the given depth is mathematically evaluated as
[tex]P_d = P_o + \rho_s * g * d[/tex]
Where [tex]P_o[/tex] is the atmospheric pressure with a constant value
[tex]P_o = 1.013 *10^{5} \ Pa[/tex]
substituting values
[tex]P_d = 1.013 * 10^{5} * + (1025 * 9.8 * 30 )[/tex]
[tex]P_d = 4.02650 * 10^{5} \ Pa[/tex]
According to the combined gas law
[tex]\frac{P_a * V_a }{T_a } = \frac{P_d * V_d }{T_d }[/tex]
=> [tex]V_a = \frac{4.026650 *10^{5} * 0.95 *10^{-6} * 293 }{278 * 1.013*10^{5} }[/tex]
=> [tex]V_a = 1.510 *10^{-5} m^3[/tex]
