What is the approximate density of water at a depth where the pressure is $80.0 ext{ atm}$, given t

Question

What is the approximate density of water at a depth where the pressure is $80.0	ext{ atm}$, given that its density at the surface is $1.03 	imes 10^3	ext{ kg/m}^3$ and the bulk modulus of water is $2.2 	imes 10^9	ext{ Pa}$? ($1	ext{ atm} = 1.013 	imes 10^5	ext{ Pa}$)

Professor PC Thomas Classes · Answer

$\Delta P = 80 	imes 1.013 	imes 10^5 \approx 8.1 	imes 10^6	ext{ Pa}$. $\frac{\Delta V}{V} = \frac{\Delta P}{B} = \frac{8.1 	imes 10^6}{2.2 	imes 10^9} \approx 3.68 	imes 10^{-3}$. Since $ho \propto \frac{1}{V}$, density increases. $ho' \approx ho(1 + \frac{\Delta V}{V}) = 1.03(1 + 0.00368) 	imes 10^3 \approx 1.034 	imes 10^3	ext{ kg/m}^3$.