Question
Find the density of seawater at a depth where the pressure is 300 atm if the density at the surface is 1100 kg/m^3. Seawater has a bulk modulus of 2.3 × 109 N/m^2. Bulk modulus is defined to be: B ≡(ρ0)(∆P)/∆ρ. Answer in units of kg/m^3.
Answers
GPT-4o mini
To find the density of seawater at a depth where the pressure is 300 atm, we can use the relationship defined by the bulk modulus. Given the bulk modulus \( B \) and the pressure change \( \Delta P \), we can relate the change in density \( \Delta \rho \) to these quantities.
The bulk modulus \( B \) can be defined as:
\[
B = \frac{\rho_0 \Delta P}{\Delta \rho}
\]
where:
- \( \rho_0 \) is the density at the surface,
- \( \Delta P \) is the change in pressure,
- \( \Delta \rho \) is the change in density.
### Given values:
- Density at the surface, \( \rho_0 = 1100 \, \text{kg/m}^3 \)
- Bulk modulus, \( B = 2.3 \times 10^9 \, \text{N/m}^2 \)
- Pressure depth, \( P = 300 \, \text{atm} \)
### Convert pressure from atm to pascals:
1 atm is approximately equal to \( 101325 \, \text{N/m}^2 \).
So,
\[
P = 300 \, \text{atm} \times 101325 \, \text{N/m}^2/\text{atm} = 300 \times 101325 \approx 30397500 \, \text{N/m}^2
\]
### Calculate the change in pressure \( \Delta P \):
Assuming that the pressure at the surface (atmospheric pressure) is approximately \( 1 \, \text{atm} \) (or \( 101325 \, \text{N/m}^2 \)), the change in pressure \( \Delta P \) is:
\[
\Delta P = P - P_{\text{surface}} \approx 30397500 \, \text{N/m}^2 - 101325 \, \text{N/m}^2 \approx 30397500 - 101325 \approx 30397375 \, \text{N/m}^2
\]
### Calculate the change in density \( \Delta \rho \):
Rearranging the bulk modulus formula gives:
\[
\Delta \rho = \frac{\rho_0 \Delta P}{B}
\]
Substituting the known values:
\[
\Delta \rho = \frac{1100 \, \text{kg/m}^3 \times 30397375 \, \text{N/m}^2}{2.3 \times 10^9 \, \text{N/m}^2}
\]
Calculating \( \Delta \rho \):
\[
\Delta \rho = \frac{33437112500 \, \text{kg/m}^3 \cdot \text{N/m}^2}{2.3 \times 10^9 \, \text{N/m}^2} \approx \frac{33437112500}{2300000000} \approx 14.564 \, \text{kg/m}^3
\]
### Finding the density at depth:
The density at depth \( \rho \) is given by:
\[
\rho = \rho_0 + \Delta \rho \approx 1100 \, \text{kg/m}^3 + 14.564 \, \text{kg/m}^3 \approx 1114.564 \, \text{kg/m}^3
\]
So, rounding to three significant figures, the density of seawater at a depth where the pressure is 300 atm is:
\[
\boxed{1115 \, \text{kg/m}^3}
\]
The bulk modulus \( B \) can be defined as:
\[
B = \frac{\rho_0 \Delta P}{\Delta \rho}
\]
where:
- \( \rho_0 \) is the density at the surface,
- \( \Delta P \) is the change in pressure,
- \( \Delta \rho \) is the change in density.
### Given values:
- Density at the surface, \( \rho_0 = 1100 \, \text{kg/m}^3 \)
- Bulk modulus, \( B = 2.3 \times 10^9 \, \text{N/m}^2 \)
- Pressure depth, \( P = 300 \, \text{atm} \)
### Convert pressure from atm to pascals:
1 atm is approximately equal to \( 101325 \, \text{N/m}^2 \).
So,
\[
P = 300 \, \text{atm} \times 101325 \, \text{N/m}^2/\text{atm} = 300 \times 101325 \approx 30397500 \, \text{N/m}^2
\]
### Calculate the change in pressure \( \Delta P \):
Assuming that the pressure at the surface (atmospheric pressure) is approximately \( 1 \, \text{atm} \) (or \( 101325 \, \text{N/m}^2 \)), the change in pressure \( \Delta P \) is:
\[
\Delta P = P - P_{\text{surface}} \approx 30397500 \, \text{N/m}^2 - 101325 \, \text{N/m}^2 \approx 30397500 - 101325 \approx 30397375 \, \text{N/m}^2
\]
### Calculate the change in density \( \Delta \rho \):
Rearranging the bulk modulus formula gives:
\[
\Delta \rho = \frac{\rho_0 \Delta P}{B}
\]
Substituting the known values:
\[
\Delta \rho = \frac{1100 \, \text{kg/m}^3 \times 30397375 \, \text{N/m}^2}{2.3 \times 10^9 \, \text{N/m}^2}
\]
Calculating \( \Delta \rho \):
\[
\Delta \rho = \frac{33437112500 \, \text{kg/m}^3 \cdot \text{N/m}^2}{2.3 \times 10^9 \, \text{N/m}^2} \approx \frac{33437112500}{2300000000} \approx 14.564 \, \text{kg/m}^3
\]
### Finding the density at depth:
The density at depth \( \rho \) is given by:
\[
\rho = \rho_0 + \Delta \rho \approx 1100 \, \text{kg/m}^3 + 14.564 \, \text{kg/m}^3 \approx 1114.564 \, \text{kg/m}^3
\]
So, rounding to three significant figures, the density of seawater at a depth where the pressure is 300 atm is:
\[
\boxed{1115 \, \text{kg/m}^3}
\]