Answer
$\dot{V}_1=0.575\ m³/s$
Work Step by Step
From the energy balance:
$\dot{m}h_1=\dot{m}h_2+\dot{W}_s$
$\dot{W}_s=\dot{m}c_p(T_1-T_2)$
Given $\dot{W}_s=650\ kW,\ c_p=1.13\ kJ/kg,\ T_1=1200\ K,\ T_2=800\ K$:
$\dot{m}=1.438\ kg/s$
For ideal gases:
$\dot{m}=\dfrac{P_1\dot{V}_1}{RT_1}$
Since $P_1=900\ kPa,\ R=0.3\ kJkg.K$
$\dot{V}_1=0.575\ m³/s$
