### Q.13:- What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10^{3} kg m^{–3}?

**Answer:-**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.

Δ*V* = *V*_{1} – *V*_{2}

= *m* [ (1/*ρ*_{1}) – (1/*ρ*_{2}) ]

∴ Volumetric strain = Δ*V* / *V*_{1}

= *m* [ (1/*ρ*_{1}) – (1/*ρ*_{2}) ] × (*ρ*_{1} / *m*)

Δ*V* / *V*_{1} = 1 – (ρ_{1}/*ρ*_{2}) ……(i)

Bulk modulus, *B* = p*V*_{1} / Δ*V*

Δ*V* / *V*_{1} = *p* / *B*

Compressibility of water = (1/*B*) = 45.8 × 10^{-11} Pa^{-1}

∴ Δ*V* / *V*_{1} = 80 × 1.013 × 10^{5} × 45.8 × 10^{-11} = 3.71 × 10^{-3} ….(ii)

For equations (*i*) and (*ii*), we get:

1 – (ρ_{1}/*ρ*_{2}) = 3.71 × 10^{-3}

*ρ*_{2} = 1.03 × 10^{3} / [ 1 – (3.71 × 10^{-3}) ]

= 1.034 × 10^{3} kg m^{-3}

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