A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spherical air tank that measures 74.0 wide. The biologist estimates she will need of air for the dive. Calculate the pressure to which this volume of air must be compressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to significant digits.

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Complete Question

A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spherical air tank that measures 74.0 wide. The biologist estimates she will need 2600 L of air for the dive. Calculate the pressure to which this volume of air must be compressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to significant digits.

Answer:

The pressure required is [tex]P_2= 12.2 \ atm[/tex]

Explanation:

Generally the volume of a sphere is mathematically denoted as

             [tex]V_s = \frac{4}{3} * \pi r^3[/tex]

Substituting [tex]r = \frac{d}{2} = \frac{74}{2} = 37cm[/tex]

          [tex]V_s = \frac{4}{3} * 3.42 * (37)^2[/tex]

               [tex]V_s = 2.121746 *10^5 cm^3[/tex]

Converting to Liters

               [tex]V_s = \frac{2.121746 *10^5}{1000}[/tex]

                [tex]V_s= 212.1746L[/tex]

Assume that the pressure at which the air is given to the diver is 1 atm when the air was occupying a volume of 2600L

So

From Charles law

               [tex]P_1V_1 = P_2 V_s[/tex]

Substituting  [tex]V_1 =2600 L[/tex] ,   [tex]P_1 = 1 atm[/tex] , [tex]V_s =212.1746L[/tex] , and making [tex]P_2[/tex] the subject we have

                 [tex]P_2 = \frac{P_1 * V_1}{V_s}[/tex]

                     [tex]= \frac{1 * 2600}{212.1746}[/tex]

                    [tex]P_2= 12.2 atm[/tex]