At surface level, a diving bell has an air space of 4.25 m3. When used during an exploration, it is submerged 60.0 metres. At that point, what is the volume of the air space? Assume that the mean density of seawater is 1.025 g·cm−3, while the temperature is the same as on the surface.

volume of the airspace=0.605m^3

Explanation:

Patm = 10^5 N/m^2

Depth= 60 metres

Pressure at the depth = ?

Density = 1.025 g/cm3 = 1025 kg/m3

P = Patm + hσg

P = 10^5 + 60*1025*9.8

P = 702700 N/m^2

P = 7.027 atm

Since the temperature is constant, Boyle’s law holds

P1V1 = P2V2

P1 = 1 atm  

P1, initial pressure of the bell(atm pressure) = 1 atm

The initial volume of the airspace, V1 = 4.25 m^3

Final pressure at the depth = 7.027 atm

Applying the Boyle’s lay

1*4.25 = 7.027 * V2

V2 = 4.25/7.027

V2 = 0.605m^3

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