Question

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^–3?

Answer 1

Given

• Pressure at depth: \( P = 80.0 \,\text{atm} \)
• Density of water at surface: \( \rho_{1} = 1.03 \times 10^{3} \,\text{kg m}^{-3} \)
• Bulk modulus of water: \( B \approx 2.0 \times 10^{9} \,\text{N m}^{-2} \)
• \( 1 \,\text{atm} = 1.013 \times 10^{5} \,\text{Pa} \)

1. Pressure at depth in SI units

Water at the surface is at approximately 1 atm; at the depth, total pressure is 80 atm. The increase in pressure is effectively 79 atm, but since 1 atm is small compared with 80 atm, we take 80 atm for the compression estimate:

\( \Delta P \approx 80 \times 1.013 \times 10^{5} \approx 8.10 \times 10^{6} \,\text{Pa} \)

2. Use bulk modulus relation

Bulk modulus:

\( B = -\,\dfrac{\Delta P}{\Delta V / V} \Rightarrow \dfrac{\Delta V}{V} = -\,\dfrac{\Delta P}{B} \)

Fractional volume change:

\( \dfrac{\Delta V}{V} = -\,\dfrac{8.10 \times 10^{6}}{2.0 \times 10^{9}} \approx -4.05 \times 10^{-3} \)

3. Relate volume change to density change

For a fixed mass \( m \), \( \rho = \dfrac{m}{V} \). If \( V \to V + \Delta V \),

\( \dfrac{\rho_{2}}{\rho_{1}} = \dfrac{V}{V + \Delta V} \approx \dfrac{1}{1 + \Delta V/V} \)

With \( \Delta V/V \approx -4.05 \times 10^{-3} \),

\( \dfrac{\rho_{2}}{\rho_{1}} \approx \dfrac{1}{1 - 4.05 \times 10^{-3}} \approx 1 + 4.05 \times 10^{-3} \)

4. Density at depth

\( \rho_{2} \approx \rho_{1} (1 + 4.05 \times 10^{-3}) \approx 1.03 \times 10^{3} \times 1.00405 \)

\( \rho_{2} \approx 1.034 \times 10^{3} \,\text{kg m}^{-3} \)

Density of water at the depth where pressure is 80.0 atm: \( \rho_{2} \approx 1.034 \times 10^{3} \,\text{kg m}^{-3} \).