Answer
$m=6.94\ kg$
Work Step by Step
The minimum work input is achieved in as isentropic process:
$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^\dfrac{k-1}{k}$
Given\ kg $T_1=27°C,\ P_2=600\ kPa,\ P_1=100\ kPa,\ k=1.4$
$T_2=500.5\ K$
From the energy balance:
$-W=\Delta U=mc_v(T_2-T_1)$
Since $W=-1000\ kJ,\ c_v=0.718\ kJ/kg.K$:
$m=6.94\ kg$
