The density of water at a depth where pressure is 80.0 atm is, given that its density at the surface is 1.03 ×10³ kg m^–3 (Compressibility of water = 45.8 × 10^–11Pa^–1)

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1.214 x 10³ kg /m³

1.034 x 10³ kg /m³

1.404 x 10³ kg /m³

2.404 x 10³ kg /m³

answer is B.

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Detailed Solution

B = 1/K = 1/45.8 × 10^–11 = 2.1 x 10⁹
$ρ^{1} = \frac{ρ}{1 - \frac{p}{B}} = \frac{1.03 \times 10^{3}}{1 - \frac{80 \times 10^{5}}{2.1 \times 10^{9}}} = 1.034 x 10^{3} kg / m^{3}$

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