Answer
$m=8.04\ kg$
Work Step by Step
Maximum work is produced with an isentropic process:
$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^\dfrac{k-1}{k}$
Given $T_1=257\ K,\ P_1=100\ kPa,\ P_2=400\ kPa,\ k=1.4$
$T_2=356.7\ K$
From the energy balance:
$-W=\Delta U=mc_v(T_2-T_1)$
With $c_v=0.718\ kJ/kg.K,\ W=1000\ kJ$:
$m=8.04\ kg$
