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

Density of water at the surface= 1.03 × 10³ Kg/m³
Compressibility of water (K) = 45.8 × 10^-11 m²/N
Pressure = 80 atm
=80×1.013 × 10^5 N/m²

now, volume of water at the surface (V) = m/d
Volume of water at given depth (V')= m/d'
change in volume ( ∆V) = V - V'
= m/d - m/d'
= m{ 1/d - 1/d' }
Now,
volumetric strain {∆V/V}
= m{1/d - 1/d' }/{m/d}
= (1 - d/d')

we know,
Compressibility (K)= 1/bulk modulus
K= 1/{∆P/(∆V/V)}
K= ∆V/∆P.V
45.8 × 10^-11 = ( 1 - d/d')/∆P
45.8 × 10^-11 = (1 -1.03×10³/d')×1/{80×1.013×10^5}
45.8×10^-11×80×1.013×10^5 = 1 - 1.03 × 10³/d'
3.712 × 10^-3 = 1 - 1.03×10³/d'
after solving we get ,
d' = 1.034 × 10³ kg/m³