What is the density of water at a depth where the pressure is 80.0 atm, given that its density at the surface is 1.03 × 10^{3} kg m^{–3}?

#### Solution 1

Let the given depth be *h*.

Pressure at the given depth,* p* = 80.0 atm = 80 × 1.01 × 10^{5} Pa

Density of water at the surface, *ρ*_{1 }= 1.03 × 10^{3} kg m^{–3}

Let *ρ*_{2} be the density of water at the depth *h*.

Let *V*_{1} be the volume of water of mass *m* at the surface.

Let *V*_{2} be the volume of water of mass *m* at the depth *h*.

Let Δ*V* be the change in volume.

ΔV = V_{1} - V_{2}

`=m(1/rho_1 - 1/rho_2)`

`:."Volumetric strain" = (triangle V)/V_1`

`= m(1/rho_1 - 1/rho_2) xx rho_1/m`

`:.triangleV/V_1 = 1 - rho_1/rho_2` ....(i)

Bulk modulus, `B = (rhoV_1)/(triangleV)`

`(triangleV)/V_1 = rho/B`

Compressibity of water = `1/B = 45.8 xx 10^(-11) Pa^(-1)`

`:.(triangleV)/V_1 = 80xx 1.013 xx 10^5 xx 45.8 xx 10^11 = 3.71 xx 10^(-3)` ...(ii)

For equation i and ii we get

`1 - rho_1/rho_2 = 3.71 xx 10^(-3)`

`rho_2 = (1.03 xx 10^3)/(1-(3.71xx10^(-3)))`

`=1.034 xx 10^3 kg m^(-3)`

Therefore, the density of water at the given depth (*h*) is 1.034 × 10^{3} kg m^{–3}.

#### Solution 2

Compressibility of water `k = 1/B = 45.8 xx 10^11 Pa^(-1)`

Change in pressure, `trianglep = 80 atm - 1 atm`

`= 79 atm = 79 xx 1.013 xx 10^5 Pa`

Density of water at the surface

`rho = 1.03 xx 10^3 kg m^(-3)`

As `B= (trianglep.V)/(triangleV) or (triangleV)/V = (trianglep)/B = trianglep xx 1/B = trianglep xx k`

or `triangleV/V = 79 xx 1.013 xx 10^5 xx 45.8 xx 10^(-11) = 3.665 xx 10^(-5)`

Now `(triangleV)/V = ((M/rho)-(M/(rho')))/(M/rho) = 1 - rho/rho'`

or `rho/(rho') = 1 - (triangleV)/V`

or `rho' = rho/(1-(triangleV/V))`

or `rho' = (1.03 xx 10^3)/(1-3.665 xx 10^(-3)) = (1.03 xx 10^3)/0.996`

`= 1.034 xx 10^3 "kg/m"^3`