Question

The bulk modulus of lead is `4.6` GPa. By what fraction will the density of a piece of lead increase if it is lowered to the bottom of the Pacific Ocean where the pressure is `40` MPa?

Answer 1

We know that density `rho` of a material is given by the expression

`rho="mass"/"volume"=m/V`

When a piece of lead is lowered to the bottom of the Pacific Ocean, due to increased pressure its volume will change and mass will remain constant. Therefore using differentials with respect to respective variables we get

`Deltarho=-m/V^2DeltaV`
`=>Deltarho=-rho(DeltaV)/V` .....(1)

We also know that the bulk modulus `B` of the material is the ratio of the change in pressure to the fractional volume compression. It can be written as

`B=(-DeltaP)/((DeltaV)/V)` .....(2)

Using (2) to rewrite (1) in terms of bulk modulus we get

`Deltarho=rho(DeltaP)/B`

Fractional change in density of lead is

`(Deltarho)/rho=(DeltaP)/B`

Inserting given numbers

`(Deltarho)/rho=(40xx10^6)/(4.6xx10^9)`
`=>(Deltarho)/rho=0.0087`