Find the density of seawater at a depth where the pressure is 740 atm if the density at the surface is \( 1100 \mathrm{~kg} / \mathrm{m}^{3} \). Seawater has a bulk modulus of \( 2.3 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2} \). Bulk modulus is defined to be \[ B \equiv \frac{\rho_{0} \Delta P}{\Delta \rho} \] Answer in units of \( \mathrm{kg} / \mathrm{m}^{3} \).

To find the density of seawater at a depth where the pressure is 740 atm, we can use the relationship given by the bulk modulus. First, we need to convert pressures into pascals and solve for the change in density. 1. **Convert pressure to pascals:** \[ 740 \, \text{atm} \times 101325 \, \text{Pa/atm} = 74970000 \, \text{Pa} \] 2. **Using the bulk modulus formula, we can rearrange it as follows:** \[ \Delta \rho = \frac{\rho_0 \Delta P}{B} \] Here, \(\Delta P\) is the pressure at depth, \(\rho_0\) is the initial density, and \(B\) is the bulk modulus. 3. **Calculate \(\Delta \rho\):** \[ \Delta \rho = \frac{1100 \, \text{kg/m}^3 \times 74970000 \, \text{Pa}}{2.3 \times 10^9 \, \text{N/m}^2} \] \[ \Delta \rho \approx \frac{82467000000 \, \text{kg/m}^3 \cdot \text{Pa}}{2.3 \times 10^9 \, \text{N/m}^2} \approx 35.9 \, \text{kg/m}^3 \] 4. **Calculate the new density at that depth:** \[ \text{New density} = \rho_0 + \Delta \rho \approx 1100 \, \text{kg/m}^3 + 35.9 \, \text{kg/m}^3 \approx 1135.9 \, \text{kg/m}^3 \] Therefore, the density of seawater at a depth where the pressure is 740 atm is approximately **1135.9 kg/m³**.