Question

Suppose that a liquid has an appreciable compresibility. Its density therefore varies with depth and pressure. The density at the surface is ρ₀

Show that the density varies with pressure according to ρ=ρ_oe^kp   (Where p is gauge pressure at any depth, and k is the compressibillity.)

Find p as a function of depth, y.

Answer 1

According to Compressiblity definition, we know that:

$k = \frac{1}{V}\frac{\Delta V}{\Delta P}\\ k \Delta P= \frac{\Delta V }{dV}$

Integrating both side we get:

$kP= \ln V\\ V=e^{kP}$

Again we know that:

$density=\frac{mass}{volume}\\ \rho =\frac{m}{V}$

Put the value of V we get:

$\rho =\frac{m}{e^{kP}}=\rho_0 e^{kP}$

Where $\rho_0$ is constant.

From Pascal's Law, we know that:

$\frac{dP}{dy} = \rho g\\ dP=\rho g dy$

Integrating both side we get:

$\int dP=\int \rho g dy\\ P=\rho g \int dy\\ P=\rho g y$

So p is a function of depth ,y.

Proved.