Question

The deepest part of the ocean is about 11 km and has a pressure of about 1000 atm. If the bulk modulus of sea water is 2.2x10⁹ Paand the surface density of sea water is 1.03x10³kg/m², what is the density of sea water at the bottom of the ocean?

Answer 1

as we know,

if D is density and P is pressure,

bulk modulous=B=D*dP/dD

==>dP=B*dD/D

integrating, we get

P=B*ln(D)+c

where c is the constant of integration

at the surface, D=1.03*10^3 kg/m^3

and P= 1 atm=101325 Pa

101325=2.2*10^9*ln(1.03*10^3)+c

==>c-=-1.5262*10^10

hence P=B*ln(D)-1.5262*10^10

at the 11 km depth, pressure =1000 atm=101325*1000=1.01325*10^8 Pa

hence if density at this depth is D,

1.01325*10^8=2.2*10^9*ln(D)-1.5262*10^10

==>2.2*10^9*ln(D)=1.5363*10^10

==>ln(D)=6.9833

==>D=1078.5 kg/m^3

hence density of sea water at the bottom of the ocean is 1078.5 kg/m^3