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?
1 Answer
We know that density
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=(-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