Question

What is the density of water at a depth where pressure is 100.0 atm, given that its

density at the surface is 1.0 × 103 kg m–3? Given compressibility of water is 50 × 10-11

Pa-1, 1atm = 1.0 × 105 Pa.​

Answer 1

Answer:

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

Step-by-step explanation: