Answer to Question #223776 in Physics for Ekott Jnr

Question

The pressure of sea water increases by 1atm for each 10m increase in the depth. By what percentage is the density of water increased in the deepest ocean of about 12km
(K=5.0×10‐⁵atm)

Answer 1

The isothermal compressibility is given by

$$k=\frac{1}{\rho}\frac{\Delta\rho}{\Delta p}$$

Hence, the change in density of water

$$\frac{\Delta \rho}{\rho}=k\times\Delta p=5.0\times 10^{-5}\times 1200=0.06=6\%$$