Question

A hydraulic press contains $0.25\, m ^{3}(250\, L )$ of oil. Find the decrease in volume of the oil (in %) when it is subjected to a pressure increase $\Delta p=1.6 \times 10^{7} Pa$. The bulk modulus of the oil is $B=5.0 \times 10^{9} Pa$.

Answer 1

$B =-\frac{\Delta p}{\Delta V / V_{0}}=-\frac{\left(0.25\, m ^{3}\right)\left(1.6 \times 10^{7} Pa \right)}{5.0 \times 10^{9} Pa }$
$=-8.0 \times 10^{-4} m ^{3}=-0.80\, L$
Even through the pressure increase is very large, the fractional change in volume is very small.
$\therefore \Delta V / V_{0}=\left(-8.0 \times 10^{-4} m ^{3}\right) /\left(0.25\, m ^{3}\right)=-0.0032$
or $ = -0.32\%$