Question 9.13:
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 kgm-3?
Pressure at the depth, P = 80.0 atm = 80 × 1.01 × 105 Pa
The density of water at the surface, ρ1= 1.03 × 103 kg m-3
Let ρ2 be the density of water at the depth h.
Let ΔV be the change in volume.
Bulk modulus, B=PV1∆V⇒∆VV1=PB
Compressibility of water =1B=45.8×10-11 Pa-1
∴ ∆VV1=80×1.013×105×45.8×10-11=3.71×10-3
Change in volume is given by:
∆V=V1-V2=m(1ρ1-1ρ2)∴ Volumetric strain =∆VV1=m(1ρ1-1ρ2)×ρ1m∴ ∆VV1=1-ρ1ρ2
⇒1-ρ1ρ2=3.71×10-3⇒ρ2=1.03×1031-3.71×10-3=1.034×103Kgm-3
Therefore, the density of water at the given depth is 1.034×103 Kgm-3.
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