The greatest ocean depths on earth are found in the Marianas Trench near the Philippines, where the depth of the bottom of the trench is about 11.0 km. Calculate the pressure(atm)due to the ocean at a depth of 10.1 km, assuming sea water density is constant all the way down. (The validity of the assumption of constant density is examined in one of the integrated concept problems.)
To calculate the pressure due to the ocean at a depth of 10.1 km, we can use the hydrostatic pressure equation:
P = P₀ + ρgh
Where:
P = Pressure at the given depth
P₀ = Pressure at the surface of the ocean (which is approximately equal to atmospheric pressure, around 1 atm)
ρ = Density of sea water
g = Acceleration due to gravity
h = Depth
Given:
Depth, h = 10.1 km = 10,100 m
Density of sea water, ρ = Constant (we'll assume 1,025 kg/m³)
Acceleration due to gravity, g = 9.8 m/s² (approximate)
Let's substitute the values into the equation and calculate the pressure:
P = 1 atm + (1,025 kg/m³) × (9.8 m/s²) × (10,100 m)
P = 1 atm + 10,029,500 N/m²
Since 1 atm is equivalent to 101,325 N/m², we can convert the pressure to atm:
P = 1 + (10,029,500 N/m² ÷ 101,325 N/m²)
P = 1 + 99 atm
Therefore, the pressure due to the ocean at a depth of 10.1 km is approximately 100 atm.
To calculate the pressure due to the ocean at a depth of 10.1 km, we can use the equation for pressure at a given depth:
P = ρgh
where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height (or depth) of the fluid column.
First, we need to determine the density of sea water. The density of sea water can vary slightly depending on its temperature and salinity, but as given in the problem, we will assume it is constant.
The density of sea water is typically around 1,025 kg/m³.
Next, we need to determine the acceleration due to gravity at that location. The average value of the acceleration due to gravity on Earth is approximately 9.8 m/s².
Now, we can calculate the pressure:
P = (density)(gravity)(depth)
P = (1,025 kg/m³)(9.8 m/s²)(10,100 m)
P = 100,485,000 N/m²
To convert this pressure into atmospheres, we divide by the standard atmospheric pressure (1 atmosphere ≈ 101,325 N/m²):
P(atm) = (100,485,000 N/m²) / (101,325 N/m²)
P(atm) ≈ 992.7 atm
Therefore, the pressure due to the ocean at a depth of 10.1 km is approximately 992.7 atm.
Please note that this calculation assumes the seawater density is constant all the way down and does not take into account other factors such as variations in temperature and salinity, which can affect the density of seawater.
well you have to make some assumption about what that density is.
Fresh water is about1000 kg/m^3
so its weight is about 9.81 * 10^3 Newtons/m^3
but salt water is about 1027 kg/m^3
so 10074 N/m^3
so at the bottom the pressure is
10.1 * 10^3 * 10074 = 10.2*10^7 N/m^2 or Pascals
an atmosphere is 1.013*10^5 Pascals
so we have
(10.2/1.013)* 10^2 = 10.1*10^2 = 1010 atmospheres (plus 1 atm for the air on top.)