β = 1 / 180GPa.
Now we need to find the pressure at 3.9km under the ocean surface, which will be our Δp: Hydrostatic pressure p at depth h below the surface is given by the formula (see Source 3):
(6) p = p₀ + ρ∙g∙h,
where:
p₀ is the atmospheric pressure at water surface, usually p₀ = 1atm = 101325Pa
ρ is the density of seawater, given as 1050 kg/m^3
g is gravitational acceleration = 9.81m/s^2
h is the depth = 3.9e3 m
(Since p₀ << ρ∙g∙h, I'm going to ignore it)
Substituting these into (5) gives (7):
(7) V2 / V1 = (1 - β * Δp)
= (1 - ( 1 / 180GPa) * (ρ * g * h) )
= (1 - ρ * g * h / 180GPa )
= 1 - 1050 * 9.81 * 3.9e3 / 180e9
= 0.9997
Our least precise data value has only 2 significant digits (3.9km), so we need to round our result to 2 sig digs, which gives:
(8) V2 / V1 = 1.0
So the gold bar's contraction in volume due to undersea pressure is negligible (nugatory).
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