The density at a depth where the pressure is 320 atm is [tex]1064.7 kg/m^3[/tex]
Explanation:
The bulk modulus is defined as
[tex]B=\rho_0 \frac{\Delta p}{\Delta \rho}[/tex]
where:
[tex]\rho_0[/tex] is the density at the surface
[tex]\Delta p[/tex] is the change in pressure
[tex]\Delta \rho[/tex] is the change in density
In this problem, we have:
[tex]B=2.3\cdot 10^9 N/m^2[/tex] (bulk modulus of seawater)
[tex]\rho_0 = 1050 kg/m^3[/tex] (density at the surface)
[tex]\Delta p = 320 atm - 1 atm = 319 atm = 3.23\cdot 10^7 Pa[/tex] is the change in pressure (we subtract the pressure at the surface, which is 1 atm)
Therefore, the change in density is
[tex]\Delta \rho = \frac{\rho_0 \Delta p}{B}=\frac{(1050)(3.23\cdot 10^7)}{2.3\cdot 10^9}=14.7 kg/m^3[/tex]
So the density at a depth where the pressure is 320 atm is
[tex]\rho= \rho_0 + \Delta \rho = 1050 + 14.7 = 1064.7 kg/m^3[/tex]
Learn more about pressure in fluids:
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