Kathleen Rausch

2021-12-19

The deepest point in the ocean is 11 km below sea level, deeper than MT. Everest is tall. What is the pressure in atmospheres at this depth?

vrangett

Density of sea water $p=1029k\frac{g}{{m}^{3}}$
Atmospheric pressure ${P}_{0}=1.01×{10}^{5}Pa$
Depth $h=11km=11×{10}^{3}m$
Pressure of water at the deepest point
$P=pgh=1029×9.81×11×{10}^{3}=1.11×{10}^{8}Pa$
$\frac{P}{{P}_{0}}=1099$
Therefore net pressure $=P+{P}_{0}=1099+1atm=1100atm$

zesponderyd

The following provides the pressure equation:
$P={P}_{0}+pgd=1.013×{10}^{5}+1030\left(9.8\right)\left(11000\right)=1.1×{10}^{8}=\frac{1.1×{10}^{8}}{1.013×{10}^{5}}=1096atm$

Don Sumner

Concepts and reason
The concepts required to solve the given questions is the pressure in the atmosphere at the depth.
Initially, calculate the depth form km to m. Later, write an expression for the pressure at the depth. Finally, calculate the pressure in the atmosphere at this depth.
Fundamentals
The following is the term for the atmosphere at a certain depth:
Here, is the initial atmospheric pressure, D is the density of water, g is the acceleration due to gravity, and h is the deepest point in the ocean.
The depth in km is equal to,

Then, change it to m as follows:

Solution:
The depth of the ocean is converted from km to m on multiplying by a factor of ${10}^{3}$
Substitute 1.00 atm for ${P}_{0}$ for D, $9.8m/{s}^{2}$ for g, and 11000 m for h in the equation $P={P}_{0}+{D}_{gh}$

The pressure in the atmosphere at the depth is equal to 1097.09 atm.
Explanation:
The density of the water is equal to . However, we must determine the depth of the ocean at that particular point. The ocean contains salt water so, we have to use density of the salt water.

xleb123

The pressure at the deepest point in the ocean can be calculated using the hydrostatic pressure formula. The formula is given by:
$P=\rho ·g·h$
where:
$P$ is the pressure,
$\rho$ is the density of seawater,
$g$ is the acceleration due to gravity, and
$h$ is the height of the water column.
At the deepest point in the ocean, $h$ is 11 km. We need to convert this depth to meters, so $h=11×1000$ m. The density of seawater, $\rho$, is approximately $1030\phantom{\rule{0.167em}{0ex}}{\text{kg/m}}^{3}$, and the acceleration due to gravity, $g$, is approximately $9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$.
Substituting the values into the formula, we have:
$P=\left(1030\phantom{\rule{0.167em}{0ex}}{\text{kg/m}}^{3}\right)·\left(9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}\right)·\left(11×1000\phantom{\rule{0.167em}{0ex}}\text{m}\right)$
Evaluating this expression gives:
$P=1.1334×{10}^{8}\phantom{\rule{0.167em}{0ex}}{\text{kg/(ms}}^{2}\text{)}$
Since 1 atmosphere is defined as the average atmospheric pressure at sea level, which is approximately $1.013×{10}^{5}\phantom{\rule{0.167em}{0ex}}{\text{kg/(ms}}^{2}\text{)}$, we can convert the pressure to atmospheres by dividing by this value:
$\frac{1.1334×{10}^{8}\phantom{\rule{0.167em}{0ex}}{\text{kg/(ms}}^{2}\text{)}}{1.013×{10}^{5}\phantom{\rule{0.167em}{0ex}}{\text{kg/(ms}}^{2}\text{)}}$
Simplifying this expression, we find that the pressure at the deepest point in the ocean is approximately $1,120$ atmospheres.

Andre BalkonE

11,169.98 atmospheres
Explanation:
In this case, we need to calculate the pressure at a depth of 11 km below sea level. We know that the density of seawater is approximately 1030 kg/m${}^{3}$, and the acceleration due to gravity is approximately 9.8 m/s${}^{2}$. To convert the depth from kilometers to meters, we multiply by 1000.
Therefore, the pressure can be calculated as follows:
$P=\left(1030\phantom{\rule{0.167em}{0ex}}{\text{kg/m}}^{3}\right)·\left(9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}\right)·\left(11,000\phantom{\rule{0.167em}{0ex}}\text{m}\right)$
Calculating this expression will give us the pressure at that depth.
Now, let's solve it step by step:
Step 1: Convert depth to meters:
$h=11\phantom{\rule{0.167em}{0ex}}\text{km}×1000=11000\phantom{\rule{0.167em}{0ex}}\text{m}$
Step 2: Calculate the pressure:
$P=\left(1030\phantom{\rule{0.167em}{0ex}}{\text{kg/m}}^{3}\right)·\left(9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}\right)·\left(11000\phantom{\rule{0.167em}{0ex}}\text{m}\right)$
Step 3: Simplify the expression:
$P=1,132,600,000\phantom{\rule{0.167em}{0ex}}\text{kg/(m}·{\text{s}}^{2}\right)$
Step 4: Convert the units to atmospheres (atm):
$P=1,132,600,000\phantom{\rule{0.167em}{0ex}}\text{kg/(m}·{\text{s}}^{2}\right)×\frac{1\phantom{\rule{0.167em}{0ex}}\text{atm}}{101325\phantom{\rule{0.167em}{0ex}}\text{kg/(m}·{\text{s}}^{2}\right)}$
Step 5: Perform the calculation:
$P=11,169.98\phantom{\rule{0.167em}{0ex}}\text{atm}$
Therefore, the pressure at a depth of 11 km below sea level is approximately 11,169.98 atmospheres.

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