Answer
87.50 J
Work Step by Step
We find:
$P= 350.0\,kPa=350.0\times10^{3}\,Pa$
$\Delta V=250.0\,cm^{3}-500.0\,cm^{3}=-250.0\,cm^{3}$
$=-250.0\,cm^{3}\times\frac{10^{-6}m^{3}}{1\,cm^{3}}=-250.0\times10^{-6}\,m^{3}$
$w=-P\Delta V=-(350.0\times10^{3}\,Pa)(-250.0\times10^{-6}\,m^{3})$
$=87.50\,J$
