What is the density of water at a depth where pressure is 80.0 atm. Given that its density at the surface is `1.03xx10^(3)kgm^(-3)`? Compressibility of water is `45.8xx10^(-11)Pa^(-1)`

Explanation

Given density at the surface `=1.03xx10^(3)kgm^(-3)`
Compressibility of waer `=45.8xx10^(-11)Pa^(-1)`
Bulk Modulus `=1/("compressibility")=1/(45.8xx10^(-11)Pa`
`K=2.183xx10^(9)Pa`
We know that mass is constant.
As volume changes, density of water changes so that,
`(DeltaV)/V=-(Delta rho)/(rho)`
but `DeltaP=80-1=79atm=79xx1.013xx10^(5)=80.027xx10^(5)Pa`
For compression bulk modulus `K=(-DeltaP)/((DeltaV)/V)`
i.e. `K=((DeltaP)/(Delta rho))rho`
Change in density `Delta rho=((Delta P)rho)/K=((80.027xx10^(5))(1.03xx10^(3)))/(2.183xx10^(9))`
`=37.758xx10^(-1)=3.78kgm^(-3)`
Hence density at the depth`=rho + Deltarho`
`=1.03xx10^(3)+0.00378xx10^(3)`
`=1.03378xx10^(3)`
`=rho ' =1.034xx10^(3)kgm^(-3)`