Answer
$V_2=255\ m³$
$W_b=18000\ kJ$
Work Step by Step
For an ideal gas:
$Pv=RT$
Given $P_i=P_e=100\ kPa,\ T_i=T_e=35°C,\ R=0.287\ kJ/kg.K$:
$v_i=v_e=0.8840\ m³/kg$
For the mass flowrates:
$\dot{m}=\frac{A\mathcal{V}}{v}$
Given $A_i=1\ m²,\ \mathcal{V}_i=2\ m/s,\ A_e=0.5\ m²,\ \mathcal{V}_e=1\ m/s$:
$\dot{m}_i=2.262\ kg/s$
$\dot{m}_e=0.566\ kg/s$
For a time period of 2 min:
$\Delta m=\Delta t(m_i-m_e)$
$\Delta m=203.6\ kg$
$\Delta V=v\Delta m$
$\Delta V=180\ m³$
Hence given $V_1=75\ m³$:
$V_2=255\ m³$
For a constant pressure:
$W_b=P\Delta V$
$W_b=18000\ kJ$
