A hydraulic press contains 0.25 m³ (250 L) of oil. Find the decrease in volume of the oil (in %) when it is subjected to a pressure increase $\mathrm{\Delta p}=1.6\times {10}^7$. The bulk modulus of the oil is $\mathrm{B}=5.0\times {10}^9$ Pa. 

$\begin{array}{c}\mathrm{B}=-\frac{\mathrm{\Delta p}}{\mathrm{\Delta V}/{\mathrm{V}}_0}\\ \mathrm{\Delta V}=-\frac{{\mathrm{V}}_0\mathrm{\Delta p}}{B}=-\frac{\left(0.25{\mathrm{m}}^3\right)\left(1.6\times {10}^7\mathrm{Pa}\right)}{5.0\times {10}^9\mathrm{Pa}}\\ \\ =-8.0\times {10}^{-4}{\mathrm{m}}^3=-0.80\mathrm{L}\end{array}$

Even through the pressure increase is very large, the fractional change in volume is very small.

$\begin{array}{l}\therefore \mathrm{\Delta V}/{\mathrm{V}}_0=\left(-8.0\times {10}^{-4}{\mathrm{m}}^3\right)/\left(0.25{\mathrm{m}}^3\right)=-0.0032\\ \text{ or }0.32\mathrm{\%}\end{array}$