Question 10 of 13intermediate🔧 ApplyNumerical4 marks

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03×101.03 \times 10³ kg m⁻³?

Correct Answer

The density of water at a depth where pressure is 80.0 atm is approximately 1.034×101.034 \times 10³ kg m⁻³.

Exercise: EXERCISES | Q: 8.13
For More Understanding

Explanation

This numerical problem applies the concept of bulk modulus to find the density of water under high pressure. The bulk modulus relates the pressure applied to the fractional change in volume. Since mass remains constant during compression, density is inversely proportional to volume. Using the bulk modulus value for water from Table 8.3 in the textbook (B = 2.2×102.2 \times 10⁹ N m⁻²), we can calculate the new density. The pressure must be converted from atmospheres to Pascals using the conversion factor 1 atm = 1.013×101.013 \times 10⁵ Pa.

Solution Steps

  1. Step 1: Identify given values: Pressure p = 80.0 atm, Surface density ρ₀ = 1.03×101.03 \times 10³ kg m⁻³, Bulk modulus of water B = 2.2×102.2 \times 10⁹ N m⁻² (from Table 8.3).

  2. Step 2: Convert pressure to SI units: p = 80.0×1.01380.0 \times 1.013 × 10⁵ Pa = 8.104×108.104 \times 10⁶ Pa.

  3. Step 3: Apply bulk modulus formula: B = -p/(ΔV/V), which gives ΔV/V = -p/B = -8.104×108.104 \times 10⁶ / 2.2×102.2 \times 10⁹ = -3.68×103.68 \times 10⁻³.

  4. Step 4: Relate density to volume: Since mass is constant, ρ₀V₀ = ρV, so ρ = ρ₀(V₀/V) = ρ₀/(1 + ΔV/V₀) ≈\approx ρ₀(1 - ΔV/V₀).

  5. Step 5: Calculate final density: ρ = ρ₀/(1 - p/B) = 1.03×101.03 \times 10³ / (1 - 3.68×103.68 \times 10⁻³) = 1.03×101.03 \times 10³ / 0.9963 ≈\approx 1.034×101.034 \times 10³ kg m⁻³.

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