Answer
$P_2=3300\ kPa$
Work Step by Step
$\dot{W}=\dot{m}(\int vdP+\Delta ke + \Delta pe)$
$\dot{W}=\dot{m}v_1(P_2-P_1)$
Given $\dot{W}=16\ kJ/s,\ \dot{m}=5\ kg/s,\ v_1=0.001\ m^3/kg,\ P_1=100\ kPa$ and solving for the final pressure:
$P_2=3300\ kPa$
