An Olympic pool has dimensions 50.0 m × 25.0 m × 2.68 m (length × width × depth). The density of the pool water is 1007 kgm-3. Atmospheric pressure (1.00 atm) is acting on the surface of the water.
If a child dives down to the bottom of the pool what pressure is exerted on the child?
The child now pulls his air filled blow up toy to the bottom of the pool. The toy is compressible and has a volume of 2.19 L on the surface of the pool. What volume does it have on the bottom of the pool? The air around the pool is at thermal equilibrium with the water in the pool.
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To find the pressure exerted on the child at the bottom of the pool, we can use the concept of hydrostatic pressure. The hydrostatic pressure at a certain depth is given by the equation:
P = P₀ + ρgh
Where:
P is the hydrostatic pressure
P₀ is the atmospheric pressure
ρ is the density of the fluid (in this case, the pool water)
g is the acceleration due to gravity
h is the depth below the surface
In this case, since the child is at the bottom of the pool, the depth below the surface is the total depth of the pool, which is 2.68 m.
Substituting the values into the equation, we have:
P = 1.00 atm + (1007 kg/m³)(9.8 m/s²)(2.68 m)
To calculate this, we can convert atmospheric pressure from atm to Pascals (Pa) by multiplying by 101325 Pa/atm and then calculate the pressure:
P = (1.00 atm)(101325 Pa/atm) + (1007 kg/m³)(9.8 m/s²)(2.68 m)
P ≈ 101325 Pa + 26447.6 Pa
P ≈ 127772.6 Pa
So the pressure exerted on the child at the bottom of the pool is approximately 127772.6 Pa.
Moving on to the second part of the question, to find the volume of the air-filled blow-up toy at the bottom of the pool, we need to consider Boyle's Law. Boyle's Law states that at constant temperature, the product of the pressure and volume of a gas is constant.
Mathematically, this can be written as:
P₁V₁ = P₂V₂
Where:
P₁ is the initial pressure (at the surface of the pool)
V₁ is the initial volume (2.19 L)
P₂ is the final pressure (at the bottom of the pool)
V₂ is the final volume (what we want to find)
Since the air around the pool is at thermal equilibrium with the water in the pool, we can assume that the pressure remains constant (1.00 atm). Substituting the known values into the equation, we have:
(1.00 atm)(2.19 L) = (P₂)(V₂)
To convert atm to Pascals, we multiply by 101325 Pa/atm:
(1.00 atm)(101325 Pa/atm)(2.19 L) = (P₂)(V₂)
Simplifying:
(1.00)(101325 Pa)(2.19 L) = (P₂)(V₂)
Dividing both sides by the final pressure (P₂):
(1.00)(101325 Pa)(2.19 L) / (P₂) = V₂
Now, we know that the pressure exerted on the child at the bottom of the pool is approximately 127772.6 Pa. Substituting this value into the equation:
(1.00)(101325 Pa)(2.19 L) / (127772.6 Pa) = V₂
Calculating this expression gives us the final volume (V₂) of the blow-up toy at the bottom of the pool.