At a depth of 500 m in an ocean what is the absolute pressure? Given that the density of seawater is `1.03xx10^3 kg m^(-3)` and `g=10m/s^2`

Options

A. 40 atm
B. 52 atm
C. 32 atm
D. 62 atm

Answer

To find the absolute pressure at a depth of 500 m in the ocean, we can use the following formula:

\[ P_{\text{absolute}} = P_{\text{atmospheric}} + P_{\text{depth}} \]

Where:
- \(P_{\text{absolute}}\) is the absolute pressure at depth.
- \(P_{\text{atmospheric}}\) is the atmospheric pressure at the surface (approximately \(1.01 \times 10^5 \, \text{Pa}\)).
- \(P_{\text{depth}} = \rho g h\) is the pressure due to the water column above the depth.

### Step 1: Calculate the pressure due to the water column
Given:
- Density of seawater, \(\rho = 1.03 \times 10^3 \, \text{kg/m}^3\)
- Gravitational acceleration, \(g = 10 \, \text{m/s}^2\)
- Depth, \(h = 500 \, \text{m}\)

Using the formula:
\[ P_{\text{depth}} = \rho g h \]
Substituting the values:
\[ P_{\text{depth}} = (1.03 \times 10^3 \, \text{kg/m}^3)(10 \, \text{m/s}^2)(500 \, \text{m}) \]
Calculating:
\[ P_{\text{depth}} = 1.03 \times 10^3 \times 10 \times 500 = 5.15 \times 10^6 \, \text{Pa} \]

### Step 2: Calculate the absolute pressure
Now, substitute \(P_{\text{depth}}\) and \(P_{\text{atmospheric}}\) into the absolute pressure formula:
\[ P_{\text{absolute}} = P_{\text{atmospheric}} + P_{\text{depth}} \]
Substituting the values:
\[ P_{\text{absolute}} = 1.01 \times 10^5 \, \text{Pa} + 5.15 \times 10^6 \, \text{Pa} \]
Calculating:
\[ P_{\text{absolute}} = 1.01 \times 10^5 + 5.15 \times 10^6 = 5.25 \times 10^6 \, \text{Pa} \]

### Step 3: Convert to atmospheres
To convert the pressure from Pascals to atmospheres, we use the conversion factor \(1 \, \text{atm} = 1.01 \times 10^5 \, \text{Pa}\):
\[ P_{\text{absolute}} = \frac{5.25 \times 10^6 \, \text{Pa}}{1.01 \times 10^5 \, \text{Pa/atm}} \approx 52 \, \text{atm} \]

### Final Answer
The absolute pressure at a depth of 500 m in the ocean is approximately \(52 \, \text{atm}\).

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