Calculate Volume of Diving Bell Air Space at 50m Depth

[winterwind]

Homework Statement

A diving bell has an air space of 3.0 m² when on the deck of a boat. What is the volume of the air space when the bell has been lowered to a depth of 50 m? Take the mean density of sea water to be 1.025 g cm^-3 and assume that the temperature is the same as on the surface.

Homework Equations

PV = nRT

The Attempt at a Solution

I'm thinking I should use the ideal gas law to solve this problem.

V₁P₁ = nRT

V₂P₂ = nRT

V₁P₁ = V₂P₂

P₁ = 1 atm (at surface of water)

P₂ = ? (would I use the density of sea water and surface area of diving bell somehow?)

Thanks!

 

[PhaseShifter]

$$P_{2}=P_{1}+\rho g \Delta z$$
where
g=gravitational acceleration
$$\Delta z$$= difference in depth

 

[Blueshift5]

Yes, you are on the right track. To solve for the volume of the air space at 50m depth, we can use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

At the surface, the pressure is equal to 1 atm, and we know the volume is 3.0 m2. We can also assume that the number of moles and the temperature remain constant.

To find the pressure at 50m depth, we can use the hydrostatic pressure equation, P = ρgh, where ρ is the density of sea water, g is the acceleration due to gravity, and h is the depth.

Substituting the values, we get P = (1.025 g/cm3)(9.8 m/s2)(50 m) = 503.5 g/cm2.

Now, we can use this pressure value in the ideal gas law equation to solve for the volume at 50m depth.

(1 atm)(3.0 m2) = (n)(0.0821 L atm/mol K)(273 K)

(503.5 g/cm2)(V) = (n)(0.0821 L atm/mol K)(273 K)

V = (n)(0.0821 L atm/mol K)(273 K) / (503.5 g/cm2)

Since we are assuming that the number of moles and temperature remain constant, we can simplify the equation to V = (3.0 m2)(503.5 g/cm2) / (1 atm) = 1510.5 m3.

Therefore, the volume of the air space at 50m depth is approximately 1510.5 m3.